Carnegie Learning

Geometry Student Text 2nd Edition Chapter 3 Perimeter and Area

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.6 Solution Page 132 Problem 1 Answer

The total length of the fencing is given, that is 20 feet.

We have to describe the shape, location, and dimension of the garden that maximize the area.

For this we will use the formula of the perimeter of the square, that is 4a, where a is the length of the sides of the square.

Length of fencing=20feet.

Let, the side of a square garden be a feet.

∴Perimeter = length of fencing.

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⇒ 4a=20

⇒ a=5feet

∴ Area of garden=a²

=5²

=25 feet²

Therefore, the location of the garden is in the front of the house.

Hence, the required shape is square with an area 25 feet2

which is located in front of the house.

Carnegie Learning Geometry Chapter 3 Page 132 Problem 2 Answer

The total length of the fencing is given, that is30feet.

We have to describe the shape, location, and dimension of the garden that maximize the area.

For this we will use the formula of the perimeter of the square, that is 4a, where a is the length of the sides of the square.

Length of the fencing=30

Let, the side of a square garden be a feet.

∴ Perimeter= Length of fencing

⇒4a=30

⇒a=30/4

⇒a=7.5feet

∴ Area of garden=a²

=(7.5)²

=56.25feet²

Therefore, the location of the garden is in the front of the house.

Hence, the required shape is square with an area 56.25 feet² which is located in front of the house.

Page 132 Problem 3 Answer

The total length of the fencing is given, that is50feet.

We have to describe the shape, location, and dimension of the garden that maximize the area.

For this we will use the formula of the perimeter of the square, that is 4a, where a is the length of the sides of the square.

We have,

Length of the fencing=50 feet

Let, the side of the square garden be a feet.

∴Perimeter= Length of the fencing

⇒4a=50

⇒a=50/4

⇒a=12.5feet

∴Area of a garden=12.5×12.5

=156.25 feet²

Therefore, the location of the garden is in the front of the house.

Hence, the required shape is square with an area 156.25 feet² which is located in front of the house.

Carnegie Learning Geometry Chapter 3 Page 134 Problem 4 Answer

Each square on the grid represents a square of 1 foot long and 1 foot wide.

We have,

The length of the rug=7 feet

The breadth of the rug=3 feet.

Hence,ABCD is the required model of the rug.

Carnegie Learning Geometry Chapter 3 Page 134 Problem 5 Answer

The length and breadth of the rug is7 feet and 3 feet respectively.

We have to find the area of the rug.

For this, we will use the formula,

Area of a rectangle=Length×Breadth.

We have,

Length=7 feet

Breadth=3 feet

∴Area of rug=7×3

=21feet²

We can see from the grid paper that the number of squares bounded by rug are 21 and the area of 1 square is 1 feet, therefore the area of the rug is=21 feet2.

Hence, the area of the rug is 21 feet 2.

Solutions for Parallel And Perpendicular Lines Exercise 2.6 In Carnegie Learning Geometry Page 134 Problem 6 Answer

The length and breadth of the ring is 7 feet and 3 feet respectively.

We have to find the perimeter of the rug.

For this, we will use the formula,

Perimeter=2(length+breadth).

We have,

Length=7 feet

Breadth=3 feet

∴Perimeter=2(l+b)

=2(7+3)

=20 feet.

We can see from the grid that the boundary of the rug cover 20 square.

Therefore, the perimeter is20 feet.

Hence, the perimeter of the rug is20 feet.

Carnegie Learning Geometry Chapter 3 Page 135 Problem 7 Answer

A rectangle is a quadrilateral with four right angles

we need to draw six rectangles in this grided figure in the question

So, drawing six rectangles,

Hence, we have drawn six rectangles on the grid and named them A, B, C, D, E, and F respectively.

Carnegie Learning Geometry Chapter 3 Page 136 Problem 8 Answer

Here the question is: Can you determine the perimeter of a rectangle without drawing it if you know the rectangle’s length and width

Yes, I can determine the perimeter of a rectangle by given width and length without drawing the rectangle.

The perimeter of the rectangle is determined by:2(l+b)

where l= length of rectangle

b=width of rectangle

Yes, I can determine the perimeter of a rectangle by given width and length without drawing the rectangle.

The perimeter of the rectangle is determined by 2(l+b), where l and b are length and width of rectangle.

Carnegie Learning Geometry 2nd Edition Exercise 2.6 Solutions Page 136 Problem 9 Answer

The question is : Can you determine the area of a rectangle without drawing it if you know the rectangle’s length and width?

Yes, we can find the area of the rectangle without drawing it.

We can find the area of the rectangle by multiplying the given length and width of the rectangle.

Yes, we can find the area of the rectangle without drawing it.

We can find the area of the rectangle by multiplying the given length and width of the rectangle.

Carnegie Learning Geometry Chapter 3 Page 136 Problem 10 Answer

Given length of rectangle =e

Width of rectangle =w

Area=A

The formula for the area of the rectangle is written as

A=e×w

A=ew

​Formula for the area of the rectangle:A=ew

Page 136 Problem 11 Answer

We have to determine the area of the rectangle if its perimeter is given.

No, we can not find the area of the rectangle if its perimeter is known to us, because as we know for calculating the area we need the length and width of the rectangle,

If we have only perimeter available then we need either length or width to calculate the area of rectangle.

No, we can not find the area of the rectangle if its perimeter is known to us, because as we know for calculating the area we need the length and width of the rectangle,

If we have only perimeter available then we need either length or width to calculate the area of the rectangle.

Carnegie Learning Geometry Chapter 3 Page 136 Problem 12 Answer

We have to determine the perimeter of the rectangle if its area is given.

No, we can not find the perimeter of the rectangle if its area is known to us, because as we know for calculating the perimeter we need the length and width of the rectangle,

If we have only area available then we need either length or width to calculate the perimeter of rectangle.

No, we can not find the perimeter of the rectangle if its area is known to us, because as we know for calculating the perimeter we need the length and width of the rectangle,

If we have only area available then we need either length or width to calculate the perimeter of rectangle.

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.6 Carnegie Learning Geometry Page 137 Problem 13 Answer

Given the length of a rectangle is 11 meters and the width of the rectangle is 5 meters

We have to calculate the area and perimeter of this rectangle.

Use the formula of area and perimeter of the rectangle.

Double the given length and width then calculate the perimeter of the new rectangle using the formula.

Length =11 meters

width =5 meters

Area=11×5=55​

The area of the rectangle is 55 square meters

Perimeter of the rectangle is: P=2(11+5)

=2(16)

=32

Perimeter of the rectangle is 32 meters.

When we double the length and width

New length is 2×11=22 meters

New width is 2×5=10 meters

So the perimeter is P=2(22+10)

=2(32)​

=64

​Perimeter of new rectangle is 64 meters

The area of the rectangle is 55 square meters and Perimeter of the rectangle is 32 meters.Perimeter of new rectangle is 64 meters

Carnegie Learning Geometry Chapter 3 Page 137 Problem 14 Answer

Given that,  a rectangle that is 11 meters long and 5 meters wide.

We have to find the effect of doubling the length and width have on the perimeter.

Here, l=11m and w=5m

Then Perimeter of rectangle, P=2(l+w)

⇒  P=2(11+5)=32m

Now, If doubling the length and width of rectangle we get,

Perimeter of rectangle, P=2(2l+2w)

⇒      P=2(22+10)=64m

Therefore, The perimeter of original rectangle is 32m and the perimeter of rectangle after doubling the length and width is 64m.

After doubling the length and width of rectangle the perimeter becomes double.

Page 137 Problem 15 Answer

Given that, a rectangle that is 11 meters long and 5 meters wide.

We have to find the effect on area of rectangle after doubling the length and width of rectangle.

Here, l=11m and w=5m

Then the area of rectangle, A=l×w

=11×5

=55sq.m​

Now, If doubling the length and width of rectangle we get,

Area of rectangle, A=2l×2w

=22×10

=220sqm.

⇒Original area of rectangle is 55sqm and are after doubling the length and width of rectangle is 220sqm.

Therefore, If we double the length and width of rectangle the area becomes four times increases.

If we double the length and width of rectangle the area becomes four times increases.

Carnegie Learning Geometry Chapter 3 Page 137 Problem 16 Answer

given that, a rectangle that is 22 meters long and 10 meters wide.

We have to find the area of rectangle.

Here,l=22m and w=10m

Therefore, Area of rectangle is,

A=l×w

=22×10

=220sq.m

​Required area of given rectangle is, A=220sqm.

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.6 Page 137 Problem 17 Answer

We have to find What effect of doubling the length and width on the area of rectangle.

Given rectangle have length l

and width w

the its area is given by,

A=l×w     . . . .  . (1)

Now of we double the length and width of rectangle the we get, area of rectangle is,    A=2l×2w=4(l×w). . . .  .(2)

So from equation (1) and (2) we can say that,

If we double the length and width of rectangle the area becomes four times increases.

If we double the length and width of rectangle the area becomes four times increases.

Carnegie Learning Geometry Chapter 3 Page 139 Problem 18 Answer

Given that, One square rug is seven feet long and seven feet wide.

We have to draw a model of this rug on the grid shown having each square on the grid represents a square that is one foot long and one foot wide.

We can draw a 7 feet by 7 feet square rug as shown in figure below,

Required drawing of a model of this rug on the grid is

Carnegie Learning Geometry Chapter 3 Page 139 Problem 19 Answer

Given  rug is seven feet long and seven feet wide as shown in figure below,

We have to find the area of this rug.

Given rug have a side of length l=7feet

Therefore its area is given by,

Area of square shape rug is,A=l2

=72

=49sq.ft

​Required area of given rug is A=49sqft

Carnegie Learning Geometry Chapter 3 Page 140 Problem 20 Answer

Given that, Each square on the grid represents a square that is one foot long and one foot wide.

We have to draw the six different squares on the grid. Use the letters A through F to name each square.

The six different squares on the grid.

Using the letters A through F to name each square.

Required drawing is

Carnegie Learning Geometry Chapter 3 Page 141 Problem 21 Answer

Yes, we can determine the perimeter of a square without drawing the square if you know the length of one side of the square.

Because if we know the length of one side is l (say) then,

we have length of each side of square is same.

That means, perimeter of square is sum of length of all sides of square.

So, from that we get,

Perimeter of square=P=l+l+l+l=4l

We are able to find the perimeter of square without drawing the square if you know the length of one side of length l of the square by using formula,

Perimeter of square=P=4l.

Carnegie Learning Geometry Chapter 2 Exercise 2.6 Free Solutions Page 141 Problem 22 Answer

Perimeter is the total length of the boundary . To find perimeter we add all the side lengths  of the square .

We know that the side length of the square is s

There are 4 sides in a square. The perimeter of the square is s+s+s+s

P is the perimeter.

P=s+s+s+s=4s

P=4s

​Perimeter of square is P=4s

Carnegie Learning Geometry Chapter 3 Page 141 Problem 23 Answer

Let side be the side length of the square

We know that area of square is length times width

For a square side length is side

Area of the square =​ =side×side=side²

​Area of the square =side²

Page 141 Problem 24 Answer

Given : Let s be the side length of square and A is the area of the square

We know that area of the square is side times side that is equal to side²

Lets replace the formula with variables

Are of the square formula

A=side²

A=s²

Area of the square formula is A=s²

Carnegie Learning Geometry Chapter 3 Page 142 Problem 25 Answer

Given : the side length of the square is 5 centimeters.

Use the formula to find the area and perimeter of the square

Side length of the square s=5 cm

Area of the square =s²

=5²

=25cm²​

Perimeter of the square =4s

=4(5)

=20 cm​

Area of the square =25cm²

Perimeter of the square = 20 cm

Carnegie Learning Geometry Exercise 2.6 Student Solutions Page 142 Problem 26 Answer

Given : The area of the square is 169 square feet

Find out the side length and perimeter of the square

Area of the square =169

Area of the square formula =s²

s²=169

Take square root

s=√169

s=13

Side length 13 feet

Perimeter of square =4s

=4(13)

=52 feet​

Side length of the square =13 feet

Perimeter of square =52 feet

Carnegie Learning Geometry Chapter 3 Page 143 Problem 27 Answer

Side length of the square is 9 inches

Find the perimeter and area

Double the side length of the square and then calculate the perimeter

side length of the square = 9 inches

Area of the square =s²=9²

=81 square inches​

Perimeter P=4s

=4(9)

=36 inches

​Side length of the square is 9 inches

Double the side length =2×9

=18 inches

​Perimeter of square =4s

=4(18)

=72 inches

​Area of the square = 81 square inches

Perimeter = 36 inches

Perimeter of square after side length is doubled = 72 inches

Page 143 Problem 28 Answer

we got   part (a)

Area of the square = 81 square inches

Perimeter = 36 inches

​Perimeter of square after side length is doubled = 72 inches

Now we analyze the perimeter when side length is 9 inches and when side length is doubled

Perimeter of square after side length is doubled is 2 times the perimeter of square with side length 9 inches

72 = 2 ⋅36

72=72

​When side length is double then the perimeter is also doubled

Carnegie Learning Geometry Chapter 3 Page 143 Problem 29 Answer

Area of the square with side length is equal to 9 inches

Double the side length of the square and check the effect on the area of the square .

Lets find area of square for side length =9 inches

Area of the square = s²

=9²

=81 square inches

​Now we double the side length

2s=2(9)

=18 inches

​Area of the square after doubling side length =s²

=(18)²

=324

​Area of the square after doubling side length =x (area of square )

324=x(81)

x=324/81

x=4

Area of the square after doubling side length = 4 times area of the square

Doubling the side length of a square

Area of the square after doubling side length  is four times the area of original square

Page 143 Problem 30 Answer

side length equal to 10 meters we need to find the area of the square

Double the side length and find the area of new square .

Side length = 10 m

Area of the square =s²

=10²

=100​

Double the side length of the square

2s=2(10)

=20

Now side length of new square is 20 m

Area of the new square =202

=400 square meters

​Area of the new square is 400 square meters

Carnegie Learning Geometry Chapter 3 Page 143 Problem 31 Answer

Area of the square with side length 10 meters  is 100 square meter

From part (a) we got the area of new square after doubling the side length is 400 square meters

Now compare the areas

400 = 4 ⋅100

Area of new square = 4⋅ Area of old square

Doubling the length of a side of the square gives  four times the area of the old square

Parallel And Perpendicular Lines Exercise 2.6 Carnegie Learning 2nd Edition Answers Page 144 Problem 32 Answer

We are given the perimeter of a square

Find the length of side of square then construct the square.

We have to write a paragraph to explain the construction of this square.

Constructing a Square With a Compass

Draw a reference line AB of 6cm (say) using a ruler.

Draw the 90 degree angle at A with the aid of the compass.

Draw the 90 degree angle at B with the aid of the compass.

Using an open compass 6 cm wide.

Draw an arc with A as the centre; that cuts arms at a 90 degree angle.

And mark the intersection point as C.Using the ruler find length measure of AC.

The AC length should be 6 cm.Using the compass again and take 6 cm width of the compass.

With D as the middle, draw an arc that cuts arms at an angle of 90 degrees.

And mark the junction point as D.Using the ruler and the BD length measurement.

The BD length should be 6 cm and we’ll get: Join CD.Use the CD ruler and calculate its length.

CD should be 6 cm in length and we get the corresponding square ABCD.

The square is shown in the following figure.

Carnegie Learning Geometry Chapter 3 Page 144 Problem 33 Answer

We are given the perimeter of a square

We have to write a paragraph to explain the construction of this square.

Also, construct the square.

Constructing a Square With a Compass

Draw a reference line AB of 6cm (say) using a ruler.

Draw the 90 degree angle at A with the aid of the compass.

Draw the 90 degree angle at B with the aid of the compass.

Using an open compass 6 cm wide.

Draw an arc with A as the centre; that cuts arms at a 90 degree angle.

And mark the intersection point as C.

Using the ruler find length measure of AC.

The AC length should be 6 cm.

Using the compass again and take 6 cm width of the compass.

With D as the middle, draw an arc that cuts arms at an angle of 90 degrees. And mark the junction point as D.

Using the ruler and the BD length measurement.

The BD length should be 6 cm and we’ll get: Join CD.

Use the CD ruler and calculate its length.

CD should be 6 cm in length and we get the corresponding square ABCD.

Hence, the constructed square as show in the figure.

Carnegie Learning Geometry Chapter 3 Page 145 Problem 34 Answer

We are given the perimeter of a rectangle

We have to write a paragraph to explain the construction of rectangle that is not a square .

To construct a rectangle, we need to create a quadrilateral that only has right angles.

All rectangles have opposites sides that are parallel to each other and equal in length.

If we are given two line segments, AB and CD, we can construct a rectangle with both AB and CD sides of length.

Suppose CD is the shorter of the two.

First, we create a line AE perpendicular to AB at A.

Then, cut off a segment of AE equal in length to CD.

We will call this point F.

Now, we can create two new perpendicular lines.

The first will meet AF at a right angle at F.

The second will meet AB at a right angle at B.

Call the intersection of these lines G.

Now, ABGF is a rectangle with side lengths AB and CD

Hence, the constructed rectangle as show in the figure.

Carnegie Learning Geometry Chapter 3 Page 145 Problem 35 Answer

We are given the perimeter of a rectangle

We have to construct the rectangle that is not a square.

Hence, the constructed rectangle as show in the figure.

Geometry Student Text 2nd Edition Chapter 2 Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.5 Solution Page 113 Problem 1 Answer

To construct: An equilateral triangle using the side shown.

Given :

The triangle is

The equilateral triangle is

Page 113 Problem 2 Answer

To compare: The triangle that you constructed with the triangles that your classmates constructed.

What do you observe, why

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Given :

All the triangles are same, since the triangle is equilateral triangle.

All the sides and angles are same.

All the triangles are same, since the triangle is equilateral triangle.

Carnegie Learning Geometry Student Chapter 2 Page 114 Problem 3 Answer

To construct: An isosceles triangle using one of the congruent sides shown. Indicate the congruent sides.

The isosceles triangle is

The isosceles triangle is

Carnegie Learning Geometry Student Chapter 2 Page 114 Problem 4 Answer

To compare: The triangle that you constructed with the triangles that your classmates constructed. What do you observe, why

Given :

Solutions for Parallel And Perpendicular Lines Exercise 2.5 In Carnegie Learning Geometry Page 114 Problem 5 Answer

To draw: Three different scalene triangles.

A scalene triangle is a triangle in which all 3 sides have different lengths.

The triangles are

The triangles are

Carnegie Learning Geometry Student Chapter 2 Page 114 Problem 6 Answer

To use: Protractor to measure each angle of the triangle you constructed in Question 1. What do you observe

Given :

The angles are measured

All the angles are 60⁰

All the angles are 60⁰

Page 114 Problem 7 Answer

To explain: How are equilateral and equiangular triangles related

Both of them are same

If all angles are equal then all sides are also equal Angle is 60⁰

Hence equilateral and equiangular triangles are same

Equilateral and equiangular triangles are same

Carnegie Learning Geometry Student Chapter 2 Page 115 Problem 8 Answer

To draw : Three different acute triangles.

An acute

The acute triangles are

Carnegie Learning Geometry 2nd Edition Exercise 2.5 Solutions Page 115 Problem 9 Answer

To construct: Three different right triangles.

Given :

The triangles are

The right angled triangles are

Carnegie Learning Geometry Student Chapter 2 Page 115 Problem 10 Answer

To compare: The right triangles that you constructed with the right triangles your classmates constructed. What do you observe, Why

Given :

The triangles are different triangles.

Here only one side and one angle is only fixed.

The triangles drawn by different students are not same

Parallel and Perpendicular Lines solutions Chapter 2 Exercise 2.5 Carnegie Learning Geometry Page 115 Problem 11 Answer

To draw : Three different obtuse triangles.

An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 900

The triangles are

The obtuse triangles are

Carnegie Learning Geometry Student Chapter 2 Page 116 Problem 12 Answer

To construct: A square using the side shown.

The square is

The square is

Page 116 Problem 13 Answer

To compare: The squares that you constructed with the squares that your classmates constructed. What do you observe, Why

Given :

All the squares drawn by students will be same.

Since all the sides and angles are same the figures are same.

All the squares drawn by students will be same.

Carnegie Learning Geometry Student Chapter 2 Page 116 Problem 14 Answer

To construct: A rectangle using the two non-congruent sides shown.

The rectangle is

The rectangle is

Page 117 Problem 15 Answer

To construct A rhombus using the side shown.

The rhombus is

The rhombus is

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.5 Page 117 Problem 16 Answer

To compare: The rhombus that you constructed with the rhombi that your classmates constructed. What do you observe, Why

Given :

Here all the rhombi will be not be same

The angle of rhombi will be different

The sides will be equal

All the rhombi drawn by students will not be same.

Page 117 Problem 17 Answer

To construct: A parallelogram using the two non-congruent sides shown.

Given :

The parallelogram is

The parallelogram is

Carnegie Learning Geometry Student Chapter 2 Page 117 Problem 18 Answer

To compare: The parallelograms that you constructed with the parallelograms that your classmates constructed. What do you observe, Why

Given :

The parallelograms drawn by students will be same.

Here two sides are fixed and they are parallel.

Hence all the parallelograms will be same.

All the parallelograms drawn by students will be same.

Carnegie Learning Geometry Exercise 2.5 Student Solutions Page 118 Problem 19 Answer

To compare: The kite that you constructed with the kites that your classmates constructed. What do you observe, Why

Given :

All the kites constructed by students will not be same.

All the sides will be matching

But angles won’t be matching

All the kites constructed by students will not be same.

Carnegie Learning Geometry Student Chapter 2 Page 118 Problem 20 Answer

To construct: A trapezoid using the starter line.

Given :

The trapezoid is

The trapezoid is

Page 118 Problem 21 Answer

To compare: The trapezoid that you constructed with the trapezoids that your classmates constructed. What do you observe, Why

Given :

All the trapezoids drawn by students will not be same.

Here only starter line is provided.

The angles and sides drawn by students will be different

All the trapezoids drawn by students will not be same.

Carnegie Learning Geometry Student Chapter 2 Page 119 Problem 22 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All equilateral triangles are equiangular triangles. ”

The statement is correct.

All the angle in equilateral triangle is 60⁰

In equiangular triangle also all the angles are 60⁰

All the sides are also same.

All equilateral triangles are equiangular triangles.

Hence the statement cannot be false,

The statement is true

Parallel And Perpendicular Lines Exercise 2.5 Carnegie Learning 2nd Edition Answers Page 119 Problem 23 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” An isosceles triangle can be an obtuse, acute, or right triangle. ”

The statement is false

An isosceles triangle is always acute.

In isosceles triangle two angles are same

Hence two angles are less than 900

Thereby all the angles are less than 90⁰

Counterexample is

Only acute triangle satisfy this

The statement is false

Carnegie Learning Geometry Student Chapter 2 Page 119 Problem 24 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” A scalene triangle can be an obtuse, acute, or right triangle”

The statement is true

A scalene triangle means all the lengths are different

In this case the triangle can be an obtuse, acute, or right triangle

A scalene triangle can be an obtuse, acute, or right triangle.

The statement cannot be false

The statement is true, a scalene triangle can be an obtuse, acute, or right triangle.

Page 119 Problem 25 Answer

To decide : Whether each statement about triangles or quadrilaterals is true or false.

Given : ” A right triangle can be an obtuse triangle.”

The statement is false

If one angle is right angle the sum of other two angle is 900

Hence the triangle cannot be obtuse.

The counter example is

Here one angle is 90⁰,all other angles are less than 90⁰

The statement is false a right triangle can’t be an obtuse triangle.

Carnegie Learning Geometry Student Chapter 2 Page 119 Problem 26 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All squares are rectangles .”

The statement is correct.

All the squares are rectangles.

Square is a rectangle with equal length and breadth having all angles as right angles.

All squares are rectangles

Hence the statement cannot be false,

The statement is true all squares are rectangles.

Page 120 Problem 27 Answer

To decide : Whether each statement about triangles or quadrilaterals is true or false.

Given : ” All rectangles are squares.”

The statement is false

All rectangles are not squares.

in rectangle length and breadth may be different

In square all lengths are same

Counter example is

The given figure is a rectangle but not a square

The statement is false

Carnegie Learning Geometry Student Chapter 2 Page 120 Problem 28 Answer

To decide : Whether each statement about triangles or quadrilaterals is true or false.

Given : ” All squares are rhombi.”

The statement is correct

In a rhombus all sides are equal angles are not same

In square all sides and angles are same.

All squares are rhombi.

Hence the statement cannot be false,

The statement is true all squares are rhombi.

Page 120 Problem 29 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All rhombi are squares.”

The statement is false

In a rhombus all sides are equal angles are not same

In square all sides and angles are same.

Counter example is

The figure is rhombus not a square

The statement is false

Carnegie Learning Geometry Student Chapter 2 Page 120 Problem 30 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All squares are parallelograms.”

The statement is correct.

In a parallelograms opposite sides are parallel

In a square opposite sides are parallel

All squares are parallelograms.

Hence the statement cannot be false

The statement is true all squares are parallelograms

Page 120 Problem 31 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All rectangles are parallelograms. ”

The statement is correct.

In a parallelograms opposite sides are parallel

In a rectangle opposite sides are parallel

All rectangles are parallelograms.

Hence the statement cannot be false

The statement is true all rectangles are parallelograms.

Carnegie Learning Geometry Student Chapter 2 Page 120 Problem 32 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given: ” All rhombi are parallelograms.”

The statement is correct.

In a parallelograms opposite sides are parallel

In a rhombi opposite sides are parallel

All rhombi are parallelograms.

Hence the statement cannot be false

The statement is true

Page 120 Problem 33 Answer

To decide: Whether each statement about triangles or quadrilaterals is true or false.

Given : ” All trapezoids are parallelograms. ”

The statement is false

In a parallelograms opposite sides are parallel

In a trapezoid one side is parallel

Counter example is

The figure is trapezoid not parallelogram

The statement is false all trapezoids are not parallelograms

Carnegie Learning Geometry Student Chapter 2 Page 120 Problem 34 Answer

To find : What you use inductive or deductive reasoning to determine if each statement was true or false.

We used an existing theory in finding true or false

Hence the reasoning used is deductive reasoning

The reasoning used is deductive reasoning

Geometry Student Text 2nd Edition Chapter 2Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.5 Solution Page 106 Problem 1 Answer

We are given a triangle ΔRAD

we have to explain how many line segments, angles, and vertices are needed to form a triangle.

A line does not have any endpoints.

A ray has only one fixed point.

A line segment is a part of a line that has a fixed length.

Perpendicular lines have a 90∘ angle between them.

Parallel lines do not meet.

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We need three sides, three angles to draw the triangle, without which the triangle is not possible.

A triangle has three sides, three angles, and three vertices.

The sum of all internal angles of a triangle is always equal to180°.

This is called the angle sum property of a triangle.

The sum of the length of any two sides of a triangle is greater than the length of the third side.

Carnegie Learning Geometry Chapter 2 Page 106 Problem 2 Answer

We are given a triangle

We have to label the vertices of the triangles and then use symbols to name each triangle.

We have labeled all the three triangles.

ΔLMN is an scalene triangle in which all sides are unequal.

ΔACB is a obtuse triangle in which ∠ACB is an obtuse angle.

ΔPQR is a right triangle in which ∠PQR is right angle.

Hence, ΔLMN is a scalene triangle, ΔACB is obtuse triangle, ΔPQR is right triangle.

Page 106 Problem 3 Answer

We are given a triangle ΔRAD.

We have to shade the interior of ΔRAD.

We need three sides, three angles to draw the triangle, without which the triangle is not possible.

The shaded area lies inside the triangle.

Hence, the shaded area is shown in the following tringle.

Solutions for Parallel and Perpendicular Lines Exercise 2.5 In Carnegie Learning Geometry Page 106 Problem 4 Answer

A triangle has three line segments and three angles.

We have to name the three sides and the three angles of ΔRAD.

A triangle has three sides, three angles, and three vertices.

The sum of all internal angles of a triangle is always equal to180∘.

This is called the angle sum property of a triangle.

The sum of the length of any two sides of a triangle is greater than the length of the third side.

Three sides of ΔRAD: RA,AD, RD

Three angles ofΔRAD:∠RAD,∠ADR,∠ARD

Carnegie Learning Geometry Chapter 2 Page 106 Problem 5 Answer

We are given

We have to construct a triangle with the three sides shown.

Steps to draw triangle

1. Draw a straight line.
2. Span the segment with your compass.
3. Trace a quarter-circle arc.
4. Switch the compass around.
5. Draw a second arc
6. Mark the point where the two arcs cross
7. Finish the triangle.

The required figure is,

Hence, the triangle figure is,

Page 106 Problem 6 Answer

We have to compare the triangle that you constructed with the triangles that your classmates constructed.

The triangle that we constructed,

The triangle that our classmate constructed.

We can see the shape of both triangles is the same but the position of the line is at different places.

Both triangles are congruent to each other.

Both are the same because the length of the sides is the same.

Hence, both triangles are the same because the length of the sides is the same.

Carnegie Learning Geometry Chapter 2 Page 107 Problem 7 Answer

To draw : Quadrilateral ABCD.

The drawing is done using MS paint

The quadrilateral is

The drawing is

ABCD is a quadrilateral.

Carnegie Learning Geometry 2nd Edition Exercise 2.5 Solutions Page 107 Problem 8 Answer

To explain : How many angles, sides, and vertices are needed to form a quadrilateral

A quadrilateral has 4 sides, 4 angles and 4 vertices.A quadrilateral can be regular or irregular.

The sum of all the interior angles of a quadrilateral is 360°.

Number of Angles=4

Sides=4

Vertices=4

Carnegie Learning Geometry Chapter 2 Page 107 Problem 9 Answer

To name: Two pairs of consecutive sides.

Given :

The pairs of consecutive sides are

AB,BC

CD,BC

The pairs of consecutive sides are

AB,BC

BC,CD

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.5 Carnegie Learning Geometry Page 107 Problem 10 Answer

To name : Two pairs of consecutive angles .

Given :

The pairs of  consecutive angles are

∠ABC,∠BCD

∠ADC,∠BCD

The pairs of  consecutive angles are

∠ABC,∠BCD

∠ADC,∠BCD

Carnegie Learning Geometry Chapter 2 Page 107 Problem 11 Answer

To name: Two pairs of opposite sides.

Given :

Two pairs of opposite sides are

AD,BC

AB,CD

Two pairs of opposite sides are

AD,BC

AB,CD

Page 107 Problem 12 Answer

To name: Two pairs of opposite angles .

Given :

Two pairs of opposite angles are

∠B,∠D

∠A,∠C

Two pairs of opposite angles are

∠B,∠D

∠A,∠C

Carnegie Learning Geometry Chapter 2 Page 108 Problem 13 Answer

To construct: A quadrilateral with the four sides shown. Label and name the quadrilateral.

Given :

The quadrilateral is

The quadrilateral is

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.5 Page 108 Problem 14 Answer

To compare: The quadrilateral that you constructed with the quadrilaterals that your classmates constructed.

What do you observe, why

Given :

Here only sides is given, angles are not given.

A student can draw this many ways with different angles

Hence the quadrilaterals drawn by student will not be same

The quadrilaterals drawn by student will not be same

Carnegie Learning Geometry Chapter 2 Page 108 Problem 15 Answer

To draw : And name the diagonals in each figure.

Given :

The drawing with diagonals is

In first figure diagonals are AC,BD

In second figure diagonals are EG,FH

The drawing with diagonals is

Page 109 Problem 16 Answer

To explain: What is the difference between the diagonals of the quadrilaterals

Given :

In first figure both diagonals are inside quadrilateral

In second figure one diagonal is inside and other is outside quadrilateral

In first figure both diagonals are inside quadrilateral, in second figure one diagonal is inside and other is outside quadrilateral

Carnegie Learning Geometry Chapter 2 Page 109 Problem 17 Answer

To classify : Each quadrilateral as convex or concave.

If the quadrilateral is concave, draw a line segment that connects two points in the interior such that the line segment is not completely in the interior of the figure.

Given :

Convex

b) Concave

c) Convex

d) Concave

Quadrilaterals are classified

Carnegie Learning Geometry Chapter 2 Exercise 2.5 Free Solutions Page 109 Problem 18 Answer

To sketch A convex pentagon, a concave pentagon, and a regular pentagon.

Convex pentagon

Concave pentagon

Regular pentagon

Convex pentagon

Concave pentagon

Regular pentagon

Carnegie Learning Geometry Chapter 2 Page 110 Problem 19 Answer

To construct A pentagon with the five sides shown.

Given

The pentagon is

Page 110 Problem 20 Answer

To compare : The pentagon that you constructed with those that your classmates constructed, What do you observe, why

Some pentagons are similar some are different.

Given :

With above sided students can many polygons

Student 1

Student 2

We can see both pentagons are not same.

This differs from student to student

The pentagons are not same for all the students.

The order and angle of joining the lines are different in different students. Hence the pentagons are different.

Carnegie Learning Geometry Chapter 2 Page 110 Problem 21 Answer

To name : Two pairs of consecutive sides and two pairs of consecutive angles in pentagon ABCDE.

The pentagons is

Consecutive sides

AE,AB

DE,EA

Consecutive angle

∠DEA,∠EAB

∠ABC∠EAB

Consecutive sides AE,AB and DE,EA

Consecutive angles∠DEA,∠EAB and ∠ABC,∠EAB

Page 111 Problem 22 Answer

To sketch A convex heptagon, a concave heptagon, and a regular heptagon.

Concave heptagon

Convex heptagon

Regular heptagon

Concave heptagon

Convex heptagon

Regular heptagon

Carnegie Learning Geometry Chapter 2 Page 111 Problem 23 Answer

To sketch A convex octagon, a concave octagon, and a regular octagon.

Convex octagon

Concave octagon

Regular octagon

Convex octagon

Concave octagon

Regular octagon

Parallel and Perpendicular Lines Exercise 2.5 Carnegie Learning 2nd Edition Answers Page 111 Problem 24 Answer

To sketch : A convex nonagon, a concave nonagon, and a regular nonagon.

Convex nonagon

Concave nonagon

Regular nonagon

Convex nonagon

Concave nonagon

Regular nonagon

Carnegie Learning Geometry Chapter 2 Page 112 Problem 25 Answer

To classify : Each polygon as regular or irregular, and then classify each polygon as convex or concave.

Explain your reasoning.

Given :

Here all sides are same but all angles are not same

Hence irregular polygon

Here all angles are not less than 1800

Hence concave polygon

Hence the given figure is concave irregular polygon

Page 112 Problem 26 Answer

To classify : Each polygon as regular or irregular, and then classify each polygon as convex or concave. Explain your reasoning.

Given :

Here all angles are same but all sides are not same

Hence irregular polygon

Here all angles are less than900

Hence convex polygon

Hence the given figure is Convex irregular polygon

Carnegie Learning Geometry Chapter 2 Page 112 Problem 27 Answer

To classify : Each polygon as regular or irregular, and then classify each polygon as convex or concave. Explain your reasoning.

Given :

Here all sides and all angles are same

Hence regular polygon

Here all angles are less than1800

Hence convex polygon

Convex regular polygon

Page 112 Problem 28 Answer

To classify: Each polygon as regular or irregular, and then classify each polygon as convex or concave. Explain your reasoning.

Given :

The figure is not a polygon, this figure cannot be formed using lines.

The figure is not a polygon, this figure cannot be formed using lines.

Geometry Student Text 2nd Edition Chapter 2 Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.4 Solution Page 98 Problem 1 Answer

The statement given is that if two parallel lines are intersected by a transversal, then the alternate interior angles are congruent.

The hypothesis (p) of the statement is that’ If two parallel lines are intersected by a transversal’.

The hypothesis (p) of the statement is that’ If two parallel lines are intersected by a transversal’.

Page 98 Problem 2 Answer

The statement given is that if two parallel lines are intersected by a transversal, then the alternate interior angles are congruent.

The conclusion (q) of the statement is that’ then the alternate interior angles are congruent’.

The conclusion (q) of the given statement is : ‘then the alternate interior angles are congruent’.

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

Page 98 Problem 3 Answer

The statement given is that if two parallel lines are intersected by a transversal, then the alternate interior angles are congruent.

The Alternate Interior Angle Converse Conjecture of the statement is that ‘If two lines intersected by a transversal form alternate interior angles, then the lines are parallel.’

The Alternate Interior Angle Converse Conjecture of the given statement is that ‘If two lines intersected by a transversal form alternate interior angles, then the lines are parallel.’

Page 98 Problem 4 Answer

The statement given is that if two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.

The hypothesis (p) of the statement is that’ If two parallel lines are intersected by a transversal’.

Hence, the hypothesis (p) of the statement is that’ If two parallel lines are intersected by a transversal’.

Page 98 Problem 5 Answer

The statement given is that if two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.

The conclusion (q) of the statement is that ‘ the alternate exterior angles are congruent ‘.

Hence, the conclusion (q) of the statement is that ‘ the alternate exterior angles are congruent ‘.

Solutions For Parallel And Perpendicular Lines Exercise 2.4 In Carnegie Learning Geometry Page 98 Problem 6 Answer

The statement is given:

Page 98 Problem 7 Answer

The statement is given:

Same-Side Interior Angle Theorem: if two parallel lines are intersected by a transversal, then the same-side interior angles are supplementary.

The hypothesis (p) of the statement is that ‘ if two parallel lines are intersected by a transversal.’

Hence, the hypothesis (p) of the statement is that ‘ if two parallel lines are intersected by a transversal.’

Page 98 Problem 8 Answer

The statement given :

Same-Side Interior Angle Theorem: if two parallel lines are intersected by a transversal, then the same-side interior angles are supplementary.

The conclusion (q) of the statement is that ‘ the same-side interior angles are supplementary.’

Hence, the conclusion (q) of the statement is that ‘ the same-side interior angles are supplementary.’

Page 98 Problem 9 Answer

For a statement: if p then q  the converse of the statement is if q then p

p is the hypothesis and q is the conclusion

Now we analyze the given statement

Same-Side Interior Angle Theorem: If two parallel lines are intersected by a transversal, then the same-side interior angles are supplementary.

The hypothesis  p is ‘two parallel lines are intersected by a transversal ‘

Conclusion (q) is ‘same side interior angles are supplementary ‘

The converse is If a transversal intersects two lines such that same side interior angles are supplementary  then  the two lines are parallel

The hypothesis p is ‘two parallel lines are intersected by a transversal ‘

Conclusion (q) is ‘same side interior angles are supplementary ‘

Same-Side Interior Angle Converse Conjecture:  If a transversal intersects two lines such that same-side interior angles are supplementary  then  the two lines are parallel

Carnegie Learning Geometry 2nd Edition Exercise 2.4 Solutions Page 98 Problem 10 Answer

Given: Same-Side Exterior Angle Theorem: If two parallel lines are intersected by a  transversal, then the same-side exterior angles are supplementary.

For a statement : if p then q  the converse of the statement is if  q then p

p is the hypothesis and q is the conclusion

Now we analyze the given statement

If two parallel lines are intersected by a  transversal, then the same-side exterior angles are supplementary.

The hypothesis p is ‘two parallel lines are intersected by a transversal ‘

Hypothesis p:   ‘two parallel lines are intersected by a transversal ‘

Page 98 Problem 11 Answer

For a statement: if p then q  the converse of the statement is if  q then p

p is the hypothesis and q is the conclusion

Now we analyze the given statement

Same-Side Interior Angle Theorem: If two parallel lines are intersected by a transversal, then the same-side exterior angles are supplementary.

Conclusion (q) is ‘same side exterior angles are supplementary ‘

Conclusion q:  ‘same side exterior angles are supplementary ‘

Page 98 Problem 12 Answer

Given: Same-Side Exterior Angle Theorem: If two parallel lines are intersected by a  transversal, then the same-side exterior angles are supplementary.

For a statement: if p then q  the converse of the statement is if  q then p

p is the hypothesis and q is the conclusion

Now we analyze the given statement

If two parallel lines are intersected by a  transversal, then the same-side exterior angles are supplementary.

The hypothesis p is ‘two parallel lines are intersected by a transversal ‘

Conclusion (q) is ‘the same-side exterior angles are supplementary ‘

The converse is:  If a transversal intersects two lines such that same side exterior angles are supplementary  then  the two lines are parallel

Same-Side Exterior Angle Converse Conjecture:

If a transversal intersects two lines such that same side exterior angles are supplementary  then  the two lines are parallel

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.4 Carnegie Learning Geometry Page 99 Problem 13 Answer

Given:  line r and transversal s the Corresponding Angle Converse Postulate  says If two lines intersected by a transversal

We need two lines that is intersected by a transversal. Here we have only line ‘r’ that is intersected by transversal ‘s’

(1)Construct another line t

(2) The line  t forms congruent corresponding angles

(3) We can say that the lines r and t are parallel

Steps are :

(1)Construct a another line  t

(2) The line  t forms congruent corresponding angles

(3) We can say that the lines  r and t are parallel

Page 99 Problem 14 Answer

Given:  line r and transversal s from first part of the question

the Corresponding Angle Converse Postulate  says If two lines intersected by a transversal

We need two lines that is intersected by a transversal. Here we have only line r that is intersected by transversal s

Construct a another line t . The line t forms congruent corresponding angles. We can say that the lines r and t are parallel

From the we can say that , line s is the transversal  line s is the transversal line

Page 99 Problem 15 Answer

Given :  line r and transversal s from first part of the question

the Corresponding Angle Converse Postulate  says If two lines intersected by a transversal

We need two lines that is intersected by a transversal.

Here we have only line r that is intersected by transversal s

Construct a another line t . The line t forms congruent corresponding angles . We can say that the lines r and t are parallel

From the we can say that , line  s is the transversal

Parallel lines are line r and t

Parallel lines are line r and t

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.4 Page 101 Problem 16 Answer

The Alternate exterior Angle Converse Conjecture states: “If two lines intersected by a transversal form congruent alternate exterior angles, then the lines are parallel.”

To prove this, we need to consider that two lines w  and x form congruent alternate exterior angles

Given: ∠2≅∠7 ( alternate exterior angles 2  and 7  are congruent )

Prove: w∥x   ( line w is parallel to line x)

Given:  ∠2≅∠7 alternate exterior angles are congruent

Prove: w∥x  lines w  and x  are parallel.

Page 101 Problem 17 Answer

If two lines intersected by a transversal form congruent alternate exterior angles, then the lines are parallel.

Given that congruent alternate exterior angles​∠1=∠8

∠2=∠7

We have to show that the Alternate Exterior Angle Converse Conjecture.

We have

Also, alternate exterior angles are congruent. So, ∠1=∠8

∠2=∠7

​But angles∠5 and ∠8 are vertically opposite angles, that is∠5=∠8 and also∠1=∠4 Vertically opposite angles are equal.

We get∠4=∠5 these are alternate interior angles.

When alternate angles are equal, then the lines are parallel.

Hence, the statement  “If two lines intersected by a transversal form congruent alternate exterior angles, then the lines are parallel.” is proved.

Carnegie Learning Geometry Chapter 2 Exercise 2.4 Free Solutions Page 102 Problem 18 Answer

Given that If two lines intersect by a transversal form supplementary same-side interior angles, then the lines are parallel.

Supplementary same-side interior angles that is∠4+∠6=180∘ and ∠3+∠5=180∘

We have to prove that statements for the Same Side Interior Angle Converse Conjecture.

We have

Also, given that∠4+∠6=180∘ −−−−−−−−(1) and ∠3+∠5=180∘

If a ray stands on a line, then the angles so formed is linear pair are parallel, then∠2+∠4=180∘−−−−−−−−(2)

From(1) and (2)

We get,​∠4+∠6=∠2+∠4

∠6=∠2​

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal. Then, the lines are parallel.

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal. Then, the lines are parallel.

Hence, the statement “If two lines intersected by a transversal form supplementary same-side interior angles, then the lines are parallel.” is proved.

Page 102 Problem 19 Answer

Given that If two lines intersected by a transversal form supplementary same-side interior angles, then the lines are parallel.

Also∠4+∠6=180∘ and ∠3+∠5=180∘

We have to prove the Same-Side Interior Angle Converse Conjecture.

We have

Also, given that∠4+∠6=180∘ −−−−−−−−(1) and ∠3+∠5=180∘

If a ray stands on a line, then the angles so formed is linear pair ∠2+∠4=180∘−−−−−(2)

From(1) and (2), we get​∠4+∠6=∠2+∠4

∠6=∠2​

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Hence, the statement “If two lines intersected by a transversal form supplementary same-side interior angles, then the lines are parallel.” is proved.

Carnegie Learning Geometry Exercise 2.4 Student Solutions Page 103 Problem 20 Answer

Given that If two lines intersected by a transversal form supplementary same-side exterior angles, then the lines are parallel.

Also∠1+∠7=180∘ and ∠2+∠8=180∘

We have to prove statements for the Same-Side Exterior Angle Converse Conjecture.

We have∠1+∠7=180∘−−−−−(1) and ∠2+∠8=180∘−−−−−(2)

The angles​∠6, ∠7 and ∠2,∠3 are vertically opposite angles. So they, are equal

∠6=∠7 and ∠2=∠3​

We have​∠4+∠6=180∘ −−−−−−−−(3) and ∠3+∠5=180∘

If a ray stands on a line, then the angles so formed is linear pair ∠2+∠4=180∘−−−−−(3)

From (3) and (4), we get

∠4+∠6=∠2+∠4

∠6=∠2​

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Hence, the statement “If two lines intersected by a transversal form supplementary same-side exterior angles, then the lines are parallel.” is proved.

Parallel and Perpendicular Lines Exercise 2.4 Carnegie Learning 2nd Edition Answers Page 103 Problem 21 Answer

We are given If two lines intersected by a transversal form supplementary same-side exterior angles, then the lines are parallel.

Given that∠1+∠7=180∘ and ∠2+∠8=180∘

We have to prove the Same-Side Exterior Angle Converse Conjecture.

We have​∠4+∠6=180∘ −−−−−−−−(1) and ∠3+∠5=180∘

If a ray stands on a line, then the angles so formed is linear pair ∠2+∠4=180∘−−−−−(2)

From (1) and (2), we get

∠4+∠6=∠2+∠4

∠6=∠2

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Thus, the lines are intersected by a transversal such that a pair of corresponding angles are equal.

Then, the lines are parallel.

Hence, the statement “If two lines intersected by a transversal form supplementary same-side exterior angles, then the lines are parallel.” is proved.

Geometry Student Text 2nd Edition Chapter 2 Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.3 Solution Page 91 Problem 1 Answer

Given: The statement is∠3≅∠6.

The Corresponding Angles Theorem states that the corresponding angles are congruent if the set of parallel lines are cut by a transversal line.

The angles that have the same measure are said to be the congruent angles.

A linear pair can be defined as the two adjacent angles formed by two intersecting lines that add up to 180°.

Prove the given statement for the Alternate Interior Angle Conjecture.

From the figure, observe that w and x are two parallel line which is intersected by a transversal line z.

Sincew∣∣x, conclude that the angle1 and 5 are congruent angles.

So, by congruent angle theorem,∠1≅∠5.

By the definition of congruent theorem, the measures of both angles are equal.

So,m∠1=m∠5…(a)

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

The angle 1 and 3 form a linear pair, the angles1 and 3 are supplementary.

Thus,m∠1+m∠3=180∘…(b)

The angles 5 and 6 are supplementary since the angle 5 and the angle 6 form a linear pair.

So,m∠5+m∠6=180∘…(c)

Substituting∠1 form∠5 into the equation(c).

m∠1+m∠6=180∘…(d)

Substitute the value of180∘=m∠1+m∠6 into (b).

m∠1+m∠3=m∠1+m∠6

Cancel∠1 on both sides.

m∠3=m∠6

Hence,∠3≅∠6 proved.

Statement∠3≅∠6 is proved for the Alternate Interior Angle Conjecture.

Page 92 Problem 2 Answer

Given:

To  Complete the flow chart proof of the Alternate Interior Angle Conjecture by writing the reason for each statement in the boxes provided.

Given

As per the definition of the Alternate interior angle

∠1=∠4

∠2=∠3

∠5=∠8

∠6=∠7

Also as per the transitive property

∠1=∠5

∠2=∠6

∠3=∠7

∠4=∠8

Because vertical angles are congruent

Hence,  the flow chart proof of the Alternate Interior Angle Conjecture have been completed.

Solutions for Parallel and Perpendicular Lines Exercise 2.3 In Carnegie Learning Geometry Page 92 Problem 3 Answer

Page 93 Problem 4 Answer

The objective of the problem is to draw labelled diagram for alternate exterior angle thoerem.

According to the alternate exterior angle theorem, If two parallel lines are intersected by a transversal, then alternate exterior angles are congruent.

The required labeled diagram for the given theorem is as shown below:

Here, two parallel lines L and M intersected by a transversal line P then,

​∠1≅∠8

∠2≅∠7

​Hence, the required diagram with a label for the alternate exterior angle conjecture is shown below:

Page 93 Problem 5 Answer

The objective of the problem is to write the given and prove statements for the alternate exterior angle conjecture.

Given:

According to the alternate exterior angle theorem, If two parallel lines are intersected by a transversal, then alternate exterior angles are congruent.

As it is seen in the given diagram, two parallel lines are given and one transversal line is given.

Therefore,

Given: L∣∣M

Prove: ∠1≅∠8

Hence, the two required statements forgiven and prove is,

Given: L∣∣M

Prove: ∠1≅∠8

Carnegie Learning Geometry 2nd Edition Exercise 2.3 Solutions Page 93 Problem 6 Answer

The objective of the problem is to prove the alternate exterior angle theorem.

According to the alternate exterior angle thoerem, If two parallel lines are intersected by a transversal, then alternate exterior angles are congruent.

The required diagram for the solution:

As two lines L∣∣M, by corresponding angles postulate,

∠1≅∠5

Also by the vertical angles theorem,

∠5≅∠8

Then, by the transitive property of congruence,

∠1≅∠8

Hence, alternate exterior angle theorem, i.e., ∠1≅∠8 is proved.

Page 94 Problem 7 Answer

The objective of the problem is to draw a labeled diagram that illustrates the same side interior angle conjecture.

According to the same side interior angle conjecture, if two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.

The required labeled diagram for the given theorem is as shown below:

Here, two parallel lines are L and M intersected by the transversal line P then,

∠4+∠6=180∘

∠3+∠5=180∘

​Hence, the required diagram with a label for the same side interior angle conjecture is shown below:

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.3 Carnegie Learning Geometry Page 94 Problem 8 Answer

The objective of the problem is to prove the same side interior angle conjecture.

When two parallel lines are cut by a transversal line, then a total of 8 angles are formed.

Given: Two parallel lines L and M are cut by the third line called transversal line P.

Prove: ∠3+∠5=180∘

Proof: Consider the below figure where two parallel lines L and M are cut by the transversal line P:

Here, ∠3 and ∠1 form a linear pair and same way ∠2 and ∠4 are also linear pairs.

According to the supplement postulate, the above two pairs are also supplementary,

​∠1+∠3=180∘

∠2+∠4=180∘

By applying corresponding angles,

​∠1≅∠5

∠2≅∠6

Therefore,

​∠5+∠3=180∘

∠6+∠4=180∘

Hence, it is proved that if two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.3 Page 94 Problem 9 Answer

Given: The Same-Side Interior Angle Conjecture states: “If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.” To Prove the Same-Side Interior Angle Conjecture.

Two column table proof

Hence,  the Same-Side Interior Angle Conjecture have been proved

Page 95 Problem 10 Answer

The objective of the problem is to draw and label a diagram that illustrates alternate exterior angle conjecture.

According to the alternate Exterior angle conjecture, if two parallel lines are intersected by a transversal, then the alternate exterior angles on the same side of the transversal are supllymentary.

The required labeled diagram for the given theorem is as shown below:

Here, two parallel lines are L and M intersected by the trasversal line P

The two pairs of alternate exterior angles are:

∠1 + ∠7=180∘

∠2 + ∠8=180∘

Hence, the required labeled diagram for alternate exterior angle conjecture is as shown below:

Carnegie Learning Geometry Chapter 2 Exercise 2.3 Free Solutions Page 95 Problem 11 Answer

The objective of the problem is to prove the same side exterior angle conjecture.

When two parallel lines are cut by a transversal line, then total of 8 angles are formed.

Given: two parallel lines L and M are cut by the third line called transversal line P.

Prove: ∠1 +∠7=180∘

Proof: Here, two parallel lines L and M are cut by the transversal line P which is shown below figure:

As two lines L and M are parallel, corresponding angles are congruent, i.e., ∠1 ≅∠5

Also, ∠5 and ∠7 form a straight line, i.e., ∠5 +∠7=180∘

By substitution, ∠1 +∠7=180∘

Hence, it is proved that if two parallel lines are intersected by a transversal, then exterior angles on the same side of transversal are supplymentary.

Page 95 Problem 12 Answer

Given: The diagram shows us a pair of parallel lines intersected by a transversal.

There are two pairs of exterior angles on the same side of the transversal, one pair being ∠2 and ∠8 and another pair being ∠1 and ∠7.

We need to prove that these exterior angles on the same side of the transversal are supplementary, that is m∠2+m∠8=180˚ and m∠1 + m∠7=180˚.

Proof: From the diagram, we see that m∠2+m∠4=180˚ (supplementary angles)

Again, m∠4=m∠8 (corresponding angles)

So, putting the value in the first equation, we have

m∠2+m∠4=180˚

⇒m∠2+m∠8=180˚

Hence, ∠2 and ∠8 are supplementary.

Similarly,​m∠1+m∠3=180˚

⇒m∠1+m∠7=180˚ (m∠3=m∠7 since they are corresponding angles)

​Hence, m∠1 and m∠7 are supplementary angles.

Therefore, exterior angles on the same side of the transversal are supplementary.

Hence proved, that if two parallel lines are intersected by a transversal, then exterior angles on the same side of the transversal are supplementary.

Carnegie Learning Geometry Exercise 2.3 Student Solutions Page 96 Problem 13 Answer

corresponding angles are congruent: The Corresponding Angle Postulate states that if two parallel lines are intersected by a transversal, then corresponding angles are congruent.

We used deductive reasoning to prove the postulate.

alternate interior angles are congruent: The Alternate Interior Angle Theorem states that if two parallel lines are intersected by a transversal, then alternate interior angles are congruent.

We used deductive reasoning to prove the postulate.

alternate exterior angles are congruent: The Alternate Exterior Angle Theorem states that if two parallel lines are intersected by a transversal, then alternate exterior angles are congruent.

We used deductive reasoning to prove the postulate.

same-side interior angles are supplementary: Same-Side Interior Angle Theorem states that if two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.

We used deductive reasoning to prove the postulate.

same-side exterior angles are supplementary: If two parallel lines are intersected by a transversal, then exterior angles on the same side of the transversal are supplementary.

We used deductive reasoning to prove the postulate.

corresponding angles are congruent: We used deductive reasoning to prove the postulate.

alternate interior angles are congruent: We used deductive reasoning to prove the postulate.

alternate exterior angles are congruent: We used deductive reasoning to prove the postulate.

same-side interior angles are supplementary: We used deductive reasoning to prove the postulate.

same-side exterior angles are supplementary: We used deductive reasoning to prove the postulate.

Geometry Student Text 2nd Edition Chapter 2 Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.2 Solution Page 87 Problem 1 Answer

Given:

To identify: The corresponding angles.

The pair of corresponding angles are

∠1=∠3

∠2=∠4

∠5=∠7

∠6=∠8

The pair of corresponding angles are

∠1=∠3

∠2=∠4

∠5=∠7

∠6=∠8

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

Page 88 Problem 2 Answer

Page 88 Problem 3 Answer

Given:

To write: The conjecture about the alternate exterior angles.

The pair of alternate exterior angles are congruent.

∠8=∠6

As they are corresponding angles.

∠1=∠6

As they are vertically opposite angles.

It can be said that:∠1=∠8

The pair of alternate exterior angles are congruent.

Solutions For Parallel And Perpendicular Lines Exercise 2.2 In Carnegie Learning Geometry Page 88 Problem 4 Answer

Given:

To write: A conjecture for same side angles and prove it.

The same side interior angles are supplementary.

The proof is ∠1+∠2=180∘

∠3+∠4=180∘

They are linear pair angles.

As the corresponding angles are equal:

∠1=∠3

∠2=∠4

Substituting the values we can say that:

∠3+∠2=180∘

∠2+∠3=180∘

Adding the equations: 2(∠2+∠3)=360∘

∠2+∠3=180∘

The same side interior angles are supplementary.

Page 88 Problem 5 Answer

Given:

To write: A conjecture about the dame side exterior angles.

The same side exterior angles are supplementary.

Proof: ∠1+∠2=180∘

∠3+∠4=180∘

They are linear pair angles.

Adding the equations: ∠1+∠2+∠3+∠4=360∘

∠2+∠3=180∘

As they are the same side interior angles.

∠1+∠4=180∘

The same side exterior angles are supplementary.

Carnegie Learning Geometry 2nd Edition Exercise 2.2 Solutions Page 88 Problem 6 Answer

To find: Whether the conjectures are inductive or deductive.

The conjectures are inductive as we have the data and we wrote the conjectures.

The conjectures are inductive as we have the data and we wrote the conjectures.

Page 89 Problem 7 Answer

Given: Two parallel lines and transversal, m∠1=38∘

To find: All unknown angles.

m∠1=m∠6=38∘

As they are vertically opposite angles.

38∘+∠2=180∘

∠2=142∘

∠2=∠5=142∘

As they are vertically opposite angles.

∠7=∠5=142∘

As they are corresponding angles

∠7=∠4=142∘

As they are vertically opposite angles.

m∠1=m∠3=38∘

As they are corresponding angles

m∠8=m∠3=38∘

As they are vertically opposite angles.

∠1 = 38∘

∠2 = 142∘

∠3 = 38∘

∠4 = 142∘

∠5 = 142∘

∠6 = 38∘

∠7 = 142∘

∠8 = 38∘

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.2 Carnegie Learning Geometry Page 89 Problem 8 Answer

Given: ∠1=67∘

To find: The unknown angles.

∠1=∠6=67∘

As they are vertically opposite angles.

As the corresponding angles are equal: ∠1=∠3=67∘

∠6=∠8=67∘

Linear pair angles: ∠1+∠2=180∘

∠2=113∘

Vertically opposite angles: ∠2=∠5=113∘

As the corresponding angles are equal: ∠2=∠4=113∘

Vertically opposite angles: ∠7=∠4=113∘

∠1=67∘

∠2=113∘

∠3=67∘

∠4=113∘

∠5=113∘

∠6=67∘

∠7=113∘

∠8=67∘

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.2 Page 90 Problem 9 Answer

Explain the Alternate Interior Angle Conjecture.

The Alternate Interior Angle Conjecture states that the alternate interior angles are congruent if the set of parallel lines are cut by a transversal line.

The Alternate Interior Angle Conjecture states that the alternate interior angles are congruent if the set of parallel lines are cut by a transversal line.

Page 90 Problem 10 Answer

Explain the Alternate Exterior Angle Conjecture.

The Alternate Exterior Angle Conjecture states that the alternate exterior angles are congruent when the set of parallel lines are intersected by a transversal line.

The Alternate Exterior Angle Conjecture states that the alternate exterior angles are congruent when the set of parallel lines are intersected by a transversal line.

Carnegie Learning Geometry Chapter 2 Exercise 2.2 Free Solutions Page 90 Problem 11 Answer

Explain the Same-Side Interior Angle Conjecture.

The Same-Side Interior Angle Conjecture states that the same side interior angles are supplementary if the set of parallel lines are cut by a transversal line.

The Same-Side Interior Angle Conjecture states that the same side interior angles are supplementary if the set of parallel lines are cut by a transversal line.

Carnegie Learning Geometry Exercise 2.2 Student Solutions Page 90 Problem 12 Answer

Explain the Same-Side Exterior Angle Conjecture.

The Same-Side Exterior Angle Conjecture states that the same side exterior angles are supplementary if the set of parallel lines are cut by a transversal line.

The Same-Side Exterior Angle Conjecture states that the same side exterior angles are supplementary if the set of parallel lines are cut by a transversal line.

Geometry Student Text 2nd Edition Chapter 2 Parallel and Perpendicular Lines

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.1 Solution Page 78 Problem 1 Answer

Given: You need to make a large letter “V” out of the poster board to complete a school project.

You are given two pieces of rectangular poster board, each measuring 1″ x 12″To Make your letter “V” similar to the one shown.

we will proceed step by step as per the question.

Take two pieces of poster board

Draw a rough sketch of V on the poster board

Then cut side GDLH and EFLD

Now take the protractor and measure the ∠GDE=40o

Also with the help of scale measure the segment HL=FL=12 inches

Hence, by following the above steps we can make a large letter “V” out of poster board to complete a school project.

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

Page 78 Problem 2 Answer

We have given the following diagram of letter V:-

We have to find that how many interior angles are located in V.

The given diagram is:-

We know that when two lines are joined then there exists an angle between them.

Here we can see that the line HG is joined with lines HL and GD.

So there are two angles ∠LHG and ∠HGD.

Also, the line EF joined with lines ED and FL.

So there are two angles ∠DEF and ∠EFL

Further, the line HL joined with FL, and line GD joined with ED.

So there are two more angles ∠GDE and ∠ELH

Hence there exist total of six angles interior the letter V.

The are total six interior angles in the following letter V.

Carnegie Learning Geometry Student Text 2nd Edition Chapter 2 Exercise 2.1 Solution Page 78 Problem 3 Answer

Page 79 Problem 4 Answer

A figure is given,

we have to identify which case is shown in figure.

Here, two lines coincide with each other.

that means two lines intersect at an infinite number of points.

so, two intersecting lines will be coplanar on the same plane.

Hence, the given figure  identifies case 2: two coplanar lines intersect at an infinite number of points.

Page 79 Problem 5 Answer

A figure is given.

we have to identify which case is shown in figure.

when two lines lie on the same plane, two lines are coplanar.

But, each line lies on own plane.

so, two lines will not coplanar.

Hence, the given figure identifies case 4: two lines are not coplanar.

Page 79 Problem 6 Answer

A figure is given,

we have to identify which case is shown in the figure.

we know that two intersecting lines are always coplanar and these lines lie on the same plane.

so, here two lines are intersecting at a single point and they lie on the same plane.

that is the two intersecting lines are coplanar on the same plane.

Hence, the given figure identifies case 1: two coplanar lines intersect at a single point.

Page 79 Problem 7 Answer

A figure is given,

we have to identify which case is shown in figure.

we know that two lines are parallel if they are coplanar on the same plane and do not intersect.

so, two coplanar lines do not intersect.

Hence, the given figure identifies case 3: two coplanar lines do not intersect.

Page 80 Problem 8 Answer

Given, two lines intersect at a single point.

Explain: Are the lines are always coplanar?

Two intersecting lines must be always coplanar,

because each line exists in many plane, but the two intersect means they share at least one plane.

so , the two lines will not always share all planes.

Though, they can be coplanar on the same plane

Hence, the two intersecting lines are always coplanar on the same plane.

Solutions for Parallel and Perpendicular Lines Exercise 2.1 in Carnegie Learning Geometry Page 80 Problem 9 Answer

Given, two lines intersect at an infinite number of points.

Explain: Are the lines always coplanar :

since, two lines coincide with each other and intersect at an infinite number of points.

so, each line exists in many plane but two lines intersect means that means they share at least on plane.

so, the two lines will not share all planes.

though, they can be coplanar on the same plane.

Hence, two intersecting lines are always coplanar on the same plane.

Page 80 Problem 10 Answer

Given, Two lines are parallel.

Describe the distance between a point on one line and the other line.

A figure is drawn,

since , two lines are parallel and do not intersect.

that means two lines are coplanar in the same plane.

and the length of segment is drawn from the point perpendicular to the line.

so , the distance between two parallel lines in the plane is the minimum distance between a point on one line and other line.

thus, it equals the perpendicular distance from any point on one line to other line.

Hence, it equals the perpendicular distance from any point on one line to the other line.

Page 80 Problem 11 Answer

To explain:  why the skew lines can not intersect.

Any two intersecting lines must lie in the same plane, are called as co planar lines.

we know that The skew lines are non-co planar lines.

By the definition of the skew lines, any two lines in the different planes, which are not parallel and do not intersect each other.

Therefore, The skew lines cannot intersect.

Hence, The skew lines cannot intersect.

Carnegie Learning Geometry 2nd Edition Exercise 2.1 solutions Page 81 Problem 12 Answer

Given: Two parallel lines l1,l2 and the linely intersects the parallel lines.

To find: The lines that each transversal intersects.

A transversal is a line that intersects two or more lines at distinct points.

Given: Two parallel lines l1, and l2 intersected by the linely.

By the definition of the transversal line: A transversal is a line that intersects two or more lines at distinct points.

Since, The linely intersect the two parallel lines.

Therefore, The linely is the transversal line.

The line ly is the transversal line.

Page 81 Problem 13 Answer

Given: Two non parallel lines l1,l2 and the line l3 intersects the non- parallel lines.

To find: The lines that each transversal intersects.

Given:Two non parallel lines l1,l2 and the line l3 intersects the non- parallel lines.

By the definition of the transversal line:A transversal is a line that intersects two or more lines at distinct points.

Since, The linel3 intersect the two non- parallel lines.

Therefore, The linel3 is the non-transversal line.

The linel3 is the transversal line.

The lines l1,l2 are two non parallel lines.

Page 82 Problem 14 Answer

Given: Two non-parallel lines m,n cut by the transversal line p.

To find: all interior angles.

Given: Two non- parallel linesm,n cut by the transversal line p.

To find: all interior angles.

Interior angles are a pair of angles that forms a straight lines.

In the above diagram

∠1,∠2 and ∠5,∠6 are the pair interior angles formed by the linem and p.

∠3,∠4 and ∠7,∠8 are the pair interior angles formed by the linen and p.

Hence,The interior angles formed by two lines are​∠1,∠2 and ∠5,∠6

∠3,∠4 and ∠7,∠8.​

Page 82 Problem 15 Answer

Given: Two non- parallel linesm,n

cut by the transversal line p.

To find: other interior and exterior angles.

Four of the eight angles are interior angles. The other four are exterior angles.

Angle 2 is an interior angle. Angle 1 is an exterior angle.

Interior angles are the angles that lie in the area enclosed between two parallel lines.

Exterior angles are the angles that lie in the area outside the parallel lines.

In the above diagram ∠2,∠3,∠6,∠7 are interior angles.

∠1,∠5,∠4,∠8 are exterior angles.

Hence, The interior angles are∠2,∠3,∠6,∠7

The exterior angles are∠1,∠4,∠5,∠8

Page 82 Problem 16 Answer

Given: Two non- parallel lines m,n cut by the transversal line P.

To find: all other pairs of same-side interior angles.

Given: One pair of same-side interior angles is∠2 and ∠3.

To find: all other pairs of same-side interior angles.

Same side interior angles are two angles that are on the interior of (between) the two lines and on the same side of the transversal.

In the above diagram, ∠2,∠3,∠6 and ∠7 are same side interior angles.

Hence,The same side interior angles formed by two lines are∠2,∠3 and ∠6,∠7.

Parallel And Perpendicular Lines Solutions Chapter 2 Exercise 2.1 Carnegie Learning Geometry Page 83 Problem 17 Answer

Given:Two non- parallel lines m,n cut by the transversal line  P.

To find: all other pairs of alternate exterior angles.

Given: One pair of alternate exterior angles is∠2 and ∠3.

To find: all other pairs of alternate exterior angles.

Alternate exterior angles are the angles formed on the opposite sides of the transversal.

In the above diagram, ∠2,∠3 and ∠6,∠7 are alternate exterior angles.

Hence,The alternate exterior angles formed by two lines are∠2,∠3 and ∠6,∠7.

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 2 Exercise 2.1 Page 83 Problem 18 Answer

Given:Two non- parallel lines m,n cut by the transversal line P.

To find: All other pairs of same-side exterior angles.

Given: One pair of same-side exterior angles is∠1 and∠4.

To find: All other pairs of same-side exterior angles.

Same side exterior angles are two angles that are on the exterior of the two lines and on the same side of the transversal.

In the above diagram, ∠1,∠4 and ∠5,∠8 are same side exterior angles.

Hence,The same side exterior angles formed by two lines are∠1,∠4 and ∠5,∠8.

Page 84 Problem 19 Answer

Given: Draw two non-parallel lines cut by a transversal, number each angle, and then use a protractor to measure each angle.

To Draw two non-parallel lines cut by a transversal and to measure each angle.

Two non-parallel lines can be drawn as

Place the hole of the protractor at the vertex of ∠1 with the flat part of the protractor lined with the horizontal side of the angle and then read the measure of the angle among the lower numbers of the protractor.

∠1=98°

Similarly, measure all the angles

∠2=82°

∠3=98°

∠4=82°

∠5=92°

∠6=88°

∠7=92°

∠8=88°

Hence, by following the above steps we can draw two non-parallel lines cut by a transversal and then use a protractor to measure each angle.

Carnegie Learning Geometry Chapter 2 Exercise 2.1 Free Solutions Page 84 Problem 20 Answer

To draw: Two parallel lines and transversal and measure the angles by the protractor.

All the angles are 90∘.

All the angles are 90∘.

Page 84 Problem 21 Answer

Given: The two parallel and non-parallel lines and their transversals.

For parallel lines, the alternate interior angles are equal but not in the case of the non-parallel lines.

For parallel lines, the alternate interior angles are equal but not in the case of the non-parallel lines.

Page 84 Problem 22 Answer

Given: The two parallel and non-parallel lines and their transversals.

For parallel lines, the alternate exterior angles are equal but not in the case of the non-parallel lines.

For parallel lines, the alternate exterior angles are equal but not in the case of the non-parallel lines.

Carnegie Learning Geometry Exercise 2.1 Student Solutions Page 84 Problem 23 Answer

Given: The two parallel and non-parallel lines and their transversals.

For parallel lines, the corresponding angles are equal but not in the case of the non-parallel lines.

For parallel lines, the corresponding angles are equal but not in the case of the non-parallel lines.

Page 84 Problem 24 Answer

Given: The two parallel and non-parallel lines and their transversals.

The same side interior angles are supplementary for parallel lines but not in the case of the non-parallel lines.

The same side interior angles are supplementary for parallel lines but not in the case of the non-parallel lines.

Page 84 Problem 25 Answer

Given: The two parallel and non-parallel lines and their transversals.

The same side exterior angles are supplementary for the parallel lines but not for the non-parallel lines.

The same side exterior angles are supplementary for the parallel lines but not for the non-parallel lines.

Page 85 Problem 26 Answer

Given:

To specify inductive or deductive reasoning to summarize the conclusion

As per the question we have used deductive reasoning to summarize the conclusion.

Hence, deductive reasoning was used to summarize the conclusion.

Page 85 Problem 27 Answer

Given:

To specify  inductive or deductive reasoning to summarize the conclusion

As per the question we have used deductive reasoning to summarize the conclusion.

Hence, deductive reasoning was used to summarize the conclusion.

Parallel And Perpendicular Lines Exercise 2.1 Carnegie Learning 2nd Edition Answers Page 85 Problem 28 Answer

Given:

To Compare the measures of the angles everyone used and your chart to the charts of the rest of your class.

After comparing the measures of the angles everyone used and your chart to the charts of the rest of your class we conclude the following table

Hence, After comparing the measures of the angles everyone used and your chart to the charts of the rest of your class we conclude the following table

Page 85 Problem 29 Answer

Given:

To  state the relationships in the chart as conjectures or theorems

The relationship in the char can be stated as conjectures and theorems as from the table

We can say that for alternate interior angles when two parallel lines are cut by a transverse

they are congruent, which can be specified by theorem.

Whereas when two non-parallel lines are cut by a transversal they are non-congruent, which can be specified by conjectures.

Similarly, all other angles can be stated as conjectures and theorems.

There are three instances that would be enough evidence to be considered proof of the relationship.

In the case of alternate interior angles when two parallel lines are cut by a transverse

they are congruent, which can be specified by theorem.

Whereas when two non-parallel lines are cut by a transversal they are non-congruent, which can be specified by conjectures.

Similar is the case for alternate interior angles and corresponding angles.

Hence, we can state the relationships in the chart as conjectures or theorems.

There are three instances that would be enough evidence to be considered proof of the relationship.

Geometry Student Text 2nd Edition Chapter 1 Tools of Geometry

Carnegie Learning Geometry Student Text 2nd Edition Chapter 1 Exercise 1.6 Solution Page 55 Problem 1 Answer

Question 1.

How can the addition property of equality be applied to angle measures?

Answer:

The addition property of equality

The addition property of equality can be applied to angle measures as follows:

If m∠1=m∠2, then m∠1+m∠3=m∠2+m∠3.

Therefore, the addition property of equality to angles is m∠1+m∠3=m∠2+m∠3.

Page 55 Problem 2 Answer

Question 2.

How can the addition property of equality be applied to segment measures?

Answer:

The addition property of equality can be applied to segment measures as follows:

If mABˉ=mCDˉ, then mABˉ+mEFˉ=mCDˉ+mEFˉ.

The addition property of equality to segments is mABˉ+mEFˉ=mCDˉ+mEFˉ.

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

Carnegie Learning Geometry Student Chapter 1 Page 56 Problem 3 Answer

Question 3.

How can the subtraction property of equality be applied to angle measures?

Answer:

The subtraction property of equality can be applied to angles as follows:

If m∠1=m∠2, then m∠1−m∠3=m∠2−m∠3.

Therefore, the Subtraction Property of Equality to angles is m∠1−m∠3=m∠2−m∠3.

Page 56 Problem 4 Answer

Question 4.

How can the subtraction property of equality be applied to segment measures?

Answer:

Page 56 Problem 5 Answer

Question 5.

How can the reflexive property be applied to angle measures?

Answer:

The reflexive property can be applied to angle measures as m∠1=m∠1.

The reflexive property to angles is m∠1=m∠1.

Carnegie Learning Geometry Student Chapter 1 Page 56 Problem 6 Answer

Question 6.

How can the reflexive property be applied to segment measures?

Answer:

The reflexive property can be applied to segment measures as mABˉ=mABˉ.

Therefore, the reflexive property to segments is mABˉ=mABˉ.

Solutions for Tools of Geometry Exercise 1.6 in Carnegie Learning Geometry Page 57 Problem 7 Answer

Question 7.

How can the substitution property be applied to angle measures?

Answer:

The substitution property can be applied to angle measures as follows:

If m∠1=56∘,m∠2=56∘, then m∠1=m∠2.

Therefore, the substitution property to angles is m∠1=m∠2.

Carnegie Learning Geometry Student Chapter 1 Page 57 Problem 8 Answer

Question 8.

How can the substitution property be applied to segment measures?

Answer:

The substitution property can be applied to segment measures as follows:

If mABˉ=4 mm and mCDˉ

=4 mm then mABˉ=mCDˉ.

The substitution property to segments is mABˉ=mCDˉ.

Page 57 Problem 9 Answer

Question 9.

How can the transitive property be applied to angle measures?

Answer:

The transitive property can be applied to angle measures as follows:

If m∠1=m∠2 and m∠2=m∠3 then m∠1=m∠3.

Therefore, the transitive property to angle measures is, “If m∠1=m∠2 and m∠2=m∠3 then m∠1=m∠3”.

Carnegie Learning Geometry Student Chapter 1 Page 57 Problem 10 Answer

Question 10.

How can the transitive property be applied to congruent segment measures?

Answer:

The transitive property can be applied to congruent segment measures as follows:

If mABˉ=mCDˉ and mCDˉ=mEFˉ then mABˉ=mEFˉ.

The transitive property to congruent segment measures is, “If mABˉ=mCDˉ and mCDˉ=mEFˉ

then mABˉ=mEFˉ”.

Carnegie Learning Geometry 2nd Edition Exercise 1.6 Solutions Page 58 Problem 11 Answer

Question 11.

Given four collinear points A, B, C, and D such that B lies between A and C, C lies between B and D, and \(\overline{A B} \cong \overline{C D}\). Draw the line segment showing the collinear points and their relationships.

Answer:

Given

Given four collinear points A, B, C, and D such that B lies between A and C, C lies between B and D, and \(\overline{A B} \cong \overline{C D}\).

Draw four collinear points A,B,C,D such thatB is lies betweenA and C,C is lies betweenB and D,

And ABˉ≅CDˉ

Since pointsA,B,C,D is collinear and ABˉ≅CDˉ

⇒ mABˉ

=mCDˉ

So required line is

A line with four collinear points A,B,C,D, such that B is between A and C,C is lie between B and D,and ABˉ≅CD ˉ is

Carnegie Learning Geometry Student Chapter 1 Page 58 Problem 12 Answer

Question 12.

Line segment \(\overline{A B}\) is congruent to line segment \(\overline{B D}\)

To Prove: Line segment \(\overline{A C}\) is congruent to line segment \(\overline{B D}\)

Answer:

We haveABˉ≅CDˉ then  ACˉ≅BDˉ

So the hypothesis is LineABˉ is congruent to LineCDˉ and  the conclusion is Line AC is congruent to Line BDˉ.

Given: Line ABˉ is congruent to Line CDˉ.

Prove: Line ACˉ is congruent to Line BDˉ.

Carnegie Learning Geometry Student Chapter 1 Page 58 Problem 13 Answer

Question 13.

Given: Line segment \(\overline{A B}\) is congruent to line segment \(\overline{C D}\).

Prove: Line segment \(\overline{A C}\) is congruent to line segment \(\overline{B B}\).

Answer:

We have Given: Line ABˉ is congruent to Line CDˉ.

Prove: LineACˉ is congruent to Line BDˉ.

SinceABˉ≅CDˉ

We have LineABˉ is equal to CDˉ

In segment measure form we get

⇒mABˉ=mCDˉ −−−−−−−(1)

By Reflexive PropertyBCˉ≅BCˉ.

In segment measure form we get

⇒mBCˉ

=mBCˉ −−−−−−−(2)

By Addition Property of Equality add equation (1) and (2) we get

⇒mABˉ+mBCˉ=mCDˉ+mBCˉ

⇒mABˉ+mBCˉ=mBCˉ+mCDˉ −−−−−−−(3)

By Transitive Property ABˉ+BCˉ≅ACˉ and BCˉ+CDˉ≅BDˉ

In segment measure form we get

⇒m(ABˉ+BCˉ)=mACˉ and m(BCˉ+CDˉ)=mBDˉ

⇒mABˉ+mBCˉ=mACˉ and mBCˉ+mCDˉ=mBDˉ −−−−−−−(4)

By Substitution Property of Equality Using (4) in (3) we get

⇒mACˉ=mBDˉ

⇒ACˉ≅BDˉ

Hence the result is proved.

So flow chart becomes:

The flow chart for result IfABˉ≅CDˉ then ACˉ≅BDˉ is:

Tools Of Geometry Solutions Chapter 1 Exercise 1.6 Carnegie Learning Geometry Page 59 Problem 14 Answer

Question 14.

Given the conditional statement: If \(\overline{A B} \cong \overline{C D}\), then \(\overline{A C} \cong \overline{B D}\)

Write a two-column proof for the given conditional statement

Answer:

Given the conditional statement is IfABˉ≅CDˉ then ACˉ≅BDˉ

And flow chart is

A two-column proof becomes

A two-column proof of the conditional statement: IfABˉ≅CDˉ then ACˉ≅BDˉis

Carnegie Learning Geometry Student Chapter 1 Page 59 Problem 15 Answer

Question 15.

Given the conditional statement: If \(\overline{A B} \cong \overline{C D}\) and \overline{A C} \cong \overline{B D}

Answer:

Given the conditional statement is IfABˉ≅CDˉ then ACˉ≅BDˉ and A two-column proof is

So A paragraph proof is

Given ABˉ≅CDˉ and we have to prove AC ≅BDˉ

IfABˉ≅CDˉ then by segment measure form we get

⇒mABˉ

=mCDˉ −−−−−−−(1)

We know that by reflexive property BCˉ≅BCˉ

⇒mBCˉ

=mBCˉ −−−−−−−(2)

Now by addition property of equality add equation (1) and (2) we get

⇒mABˉ+mBCˉ

=mCDˉ+mBCˉ

⇒mABˉ+mBCˉ

=mBCˉ+mCD −−−−−−−(3)

We know that by transitive property⇒ABˉ+BCˉ=ACˉ and BCˉ+CDˉ=BDˉ

⇒m(ABˉ+BCˉ)=mACˉ and m(BCˉ+CDˉ)=mBDˉ

⇒mABˉ+mBCˉ=mACˉ and mBCˉ+mCDˉ=mBDˉ −−−−−−−(4)

Now by substitution property of equality using (4) in (3) we get

⇒mACˉ

=mBDˉ

⇒ACˉ≅BDˉ​

Hence the result is proved.

A paragraph proof of the conditional statement If ABˉ≅CDˉ then ACˉ≅BDˉ is given in explanation part.

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 1 Exercise 1.6 Page 60 Problem 16 Answer

Question 16.

Given : ACD and ∠BCD are right angles.

Prove: ∠ACD≅∠BCD

Here is the completed flow chart with the statements for each reason:

Answer:

We have to Complete the flow chart of the Right Angle Congruence Theorem by writing the statement for each reason in the boxes provided.

Given:∠ACD and ∠BCD are right angles.

Prove:∠ACD≅∠BCD

In first row write the statement of given.

Given:∠ACD is right angles, ∠BCD is right angles.

In second row write the results of right angles.

i.e.​If ∠ACD is right angles, ∠BCD is right angles.

⇒∠ACD=90°,∠BCD=90°

​In third row write the result of transitive property of equality

If ∠ACD=90°,∠BCD=90°

⇒∠ACD≅∠BCD

​So we have which is required flow chart of the Right Angle Congruence Theorem.

The flow chart of the Right Angle Congruence Theorem:

Carnegie Learning Geometry Student Chapter 1 Page 61 Problem 17 Answer

Question 17.

Prove that ∠1≅∠3.

Answer:

Consider the diagram:

We have to write the “Given” statement such that “Prove:∠1≅∠3”.

For this by The Congruent Supplement Theorem we get the “Given” Statements

∠1 and ∠2 are supplement angles i.e.∠1+∠2=180°

∠3 and ∠4 are supplement angles i.e.∠3+∠4=180°

∠2=∠4 which is required solution.

By The Congruent Supplement Theorem

Given: ∠1 and ∠2 are supplement angles i.e.∠1+∠2=180°

Given: ∠3 and ∠4 are supplement angles i.e.∠3+∠4=180°

Given: ∠2=∠4  is required “Given” for Prove:∠1≅∠3

Carnegie Learning Geometry Chapter 1 Exercise 1.6 Free Solutions Page 61 Problem 18 Answer

Question 18.

Proof of the Congruent Supplement Theorem:

1. ∠1 + ∠2 = 180
2. ∠3 + ∠4 = 180
3. ∠2 = ∠4

Answer:

We have to complete a flow chart proof of the Congruent Supplement Theorem by drawing arrows to connect the steps in a logical sequence.

In first row write given parts portion.In second row write Definitions downwards to given parts.

In third row write substitution property downwards-center of definition of supplementary angles.

In forth row write subtraction property of equality downwards-center of supplementary angles and definition of congruent angles.

In fifth row write definition of congruent angles downwards to subtraction property of equality

So we have which is required flow chart.

A flow chart proof of the Congruent Supplement Theorem:

Carnegie Learning Geometry Student Chapter 1 Page 62 Problem 19 Answer

Question 19.

Congruent Supplement Theorem: If two angles are supplementary to the same angle (or to congruent angles), then those two angles are congruent.

To prove that ∠1≅∠3

Answer:

We have to create a two-column proof of the Congruent Supplement Theorem.

In first row write given parts portion.

In second row write Definitions downwards to given parts.

In third row write substitution property downwards-center of definition of supplementary angles.

In forth row write subtraction property of equality downwards-center of supplementary angles and definition of congruent angles.

In fifth row write definition of congruent angles downwards to subtraction property of equality

And flow chart is

A two-column proof becomes

A two-column proof of the Congruent Supplement Theorem is:

Carnegie Learning Geometry Student Chapter 1 Page 62 Problem 20 Answer

Question 20.

Given:

• ∠1 and ∠2 are complementary angles.
• ∠3 and ∠4 are complementary angles.
• ∠2 ≅ ∠4

To Prove: ∠1 ≅ ∠3

Answer:

The angles ∠1 and ∠2 are complimentary. Also, the angles ∠3 and∠4 are complimentary angles.

It is also given that it conclude that ∠1 and ∠3 are congruent

Carnegie Learning Geometry Exercise 1.6 Student Solutions Page 62 Problem 21 Answer

Question 21.

• ∠1 and ∠2 are complementary.
• ∠3 and ∠4 are complementary.
• ∠2 ≅ ∠4

To Prove: ∠1 ≅ ∠3

Answer:

To use : The diagram to write the “Given” and “Prove” statements for the Congruent Complement Theorem.

Diagram for the given Theorem is given below:

Now from the given Theorem the Given and concluding statement is as-

Given : ∠1 and∠2 are complimentary.

Given :∠3 and∠4 are complimentary.

Given :∠1+∠2=90° and∠3+∠4=90°

Prove :∠1≅∠3

Given :∠1 and∠2 are complimentary.

Given :∠3 and∠4 are complimentary

Given :∠1+∠2=90 and∠3+∠4=90

Prove :∠1≅∠3

Carnegie Learning Geometry Student Chapter 1 Page 63 Problem 22 Answer

Question 22.

To create a flow chart proof of the Congruent Complement Theorem:

Given:

• ∠1 and ∠2 are complementary.
• ∠3 and ∠4 are complementary.
• ∠2 ≅ ∠4

To Prove: ∠1 ≅ ∠3

Answer:

To create: A flow chart proof of the Congruent Complement Theorem.

We have

The flowchart is

The flow chart is

Carnegie Learning Geometry Student Chapter 1 Page 64 Problem 23 Answer

Question 23.

write a formal proof to demonstrate that vertical angles are congruent.

Answer:

We know that,

∠1+∠2=180 and∠2+∠3=180   ( Linear Pair )

Therefore, ​∠1+∠2=∠2+∠3 ⇒∠1=∠3​

Similarly, ∠2=∠4

These are vertical angles hence it is proved that vertical angles are congruent.

Therefore, vertical angles are congruent.

Tools Of Geometry Exercise 1.6 Carnegie Learning 2nd Edition Answers Page 64 Problem 24 Answer

Question 24.

Using the diagram and the given statements, write a proof to demonstrate that ∠1 and ∠3 are congruent and that ∠2 and ∠4 are congruent.

Answer:

The diagram for the vertical angle theorem is given below-

From the given statements and the the definition of vertical angle theorem we can state the Prove statements as-

Prove :∠1and∠3 are congruent

Prove :∠2 and∠4 are congruent

Therefore, the Prove statements are as-

Prove :∠1 and∠3 are congruent

Prove :∠2 and∠4 are congruent

Carnegie Learning Geometry Student Chapter 1 Page 64 Problem 25 Answer

Question 25.

To Shows that ∠1 and ∠3 are congruent, in accordance with the vertical angle theorem.

Answer:

First Prove:∠1 and∠3 are congruent

Page 65 Problem 26 Answer

Question 26.

To Prove that ∠2 and ∠4 are congruent.

Answer:

For the vertical angle theorem

Given :∠2 and∠3 are linear pair

Given :∠3 and∠4 are linear pair

Prove :∠2 and∠4 are congruent

We have to Create a two-column proof of the “Prove” statement

Carnegie Learning Geometry Student Chapter 1 Page 67 Problem 27 Answer

Question 27.

To prove that ∠5 ≅ ∠7

Answer:

A two-column proof uses a table to present a logical argument and assigns each column to do one job to take a reader from premise to conclusion.

Paragraph proof is a logical argument written as a paragraph, giving evidence and details to arrive at a conclusion.

So, everything needs to fit in an appropriate place.

Two-Column proof is easiest to understand as it makes the reader see the statement and conclusion side by side.

While as Paragraph proof makes the proof wordier and harder to follow thus it is hardest to understand.

Therefore, Two column proof is easiest to understand while as Paragraph proof is hardest to understand.

Page 67 Problem 28 Answer

Question 28.

Given the forms of mathematical proofs, which type of proof is easiest to write and which is hardest to write? Explain your reasoning.

Answer:

Given: Which form of proof is easiest to write? Hardest to write

To explain Which form of proof is easiest to write? Hardest to write.

A two-column proof is a easiest proof in which the steps are written in the left column and the corresponding reasons in the right column.

Each step is numbered and the same number is used for the corresponding reason.

A flow chart proof is a hardest proof in which the steps and corresponding reasons are written in boxes.

Arrows connect the boxes and indicate how each step and reason is generated from one or more other steps and reasons.

Hence, A two-column proof is a easiest proof and a flow chart proof is a hardest proof to write.

Carnegie Learning Geometry Student Chapter 1 Page 67 Problem 29 Answer

Question 29.

Which form of proof generally has the fewest steps? Which form of proof has the most steps? Explain your reasoning.

Answer:

Given: Which form of proof has the fewest steps? The most stepsTo Explain Which form of proof has the fewest steps? The most steps

A two-column proof is a proof which has the fewest steps, as in this the steps are written in the left column and the corresponding reasons in the right column.

Each step is numbered and same number is used for the corresponding reason.

A flow chart proof is a proof which has the most steps and corresponding reasons are written in boxes.

Arrows connect the boxes and indicates how each step and reason is generated from one or more other steps and reasons.

Hence, A two-column proof is a proof which has the fewest steps and a flow chart proof is a proof which has the most steps.

Carnegie Learning Geometry Student Chapter 1 Page 67 Problem 30 Answer

Question 30.

Which form of proof do you prefer? Explain why you prefer this form of proof.

Answer:

Given:  Which form of proof do you prefer To explain  Which form of proof do you prefer

I use to prefer a two-column proof because in it steps are written in the left column and the corresponding reasons in the right column.

Hence, I use to prefer a two-column proof because in it steps are written in the left column and the corresponding reasons in the right column.

Geometry Student Text 2nd Edition Chapter 1 Tools of Geometry

Carnegie Learning Geometry Student Text 2nd Edition Chapter 1 Exercise 1.5 Solution Page 46 Problem 1 Answer

Question 1.

What is the hypothesis p in the given statement?

Answer:

Given : If the measure of an angle is 32∘, then the angle is acute.

To find : What is the hypothesis p

The form is if p then q

Hence p= The measure of an angle is 32⁰

p is the measure of an angle is32⁰

Page 46 Problem 2 Answer

Question 2.

What is the conclusion q in the given statement?

Answer:

Given : If the measure of an angle is 32∘, then the angle is acute.

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

To find : What is the conclusion q

The form is if p then q

Henceq= the angle is acute.

Q is the angle is acute.

Page 46 Problem 3 Answer

Question 3.

What does the phrase “If p is true” mean in terms of the conditional statement?

Answer:

“If p is true” mean in terms of the conditional statement as follows

Given : If the measure of an angle is 32∘, then the angle is acute.

To find : What does the phrase “If p is true” mean in terms of the conditional statement

pis measure of an angle is320

If p is true can be written as if measure of angle 320 is true

Measure of angle 320 is true

Page 46 Problem 4 Answer

Question 4.

What does the phrase “If q is true” mean in terms of the conditional statement?

Answer:

Given  phrase “If q is true”

Given : If the measure of an angle is 32∘, then the angle is acute.

To find : What does the phrase “If q is true” mean in terms of the conditional statement

Q is  the angle is acute.

If q is true is, the angle is acute is true

The angle is acute is true

Solutions For Tools Of Geometry Exercise 1.5 In Carnegie Learning Geometry Page 46 Problem 5 Answer

Question 5.

If the measure of an angle is 32°, then the angle is acute so the truth value of the conditional statement is true.

Answer:

Page 46 Problem 6 Answer

Question 6.

If p is true and q is false, then the truth value of a conditional statement is false. If the measure of an angle is 32°, then the angle is acute. What does the phrase “If p is true” mean in terms of the conditional statement?

Answer:

Given :  If p is true and q is false, then the truth value of a conditional statement is false.

If the measure of an angle is 32°, then the angle is acute.

To explain : What does the phrase “If p is true” mean in terms of the conditional statement

P is measure of an angle is320

If p is true can be written as if measure of angle 320 is true Measure of angle 320 is true

Page 46 Problem 7 Answer

Question 7.

If p is true and q is false, then the truth value of a conditional statement is false. If the measure of an angle is 32°, then the angle is acute. What does the phrase “If q is false” mean in terms of the conditional statement?

Answer:

Given :  If p is true and q is false, then the truth value of a conditional statement is false.

If the measure of an angle is 32°, then the angle is acute.

To explain : What does the phrase “If q is false” mean in terms of the conditional statement

Q is the angle is acute.

Q is false means

Angle is not acute

Angle is not acute

Page 46 Problem 8 Answer

Question 8.

If p is true and q is false, then the truth value of a conditional statement is false. If the measure of an angle is 32°, then the angle is acute. Why is the truth value of the conditional statement false?

Answer:

Given :  If p is true and q is false, then the truth value of a conditional statement is false.

If the measure of an angle is 32°, then the angle is acute.

To explain : Why the truth value of the conditional statement is false.

Given that if p is true and q is false, then the truth value of a conditional statement is false.

If the angle is not acute then the angle cannot be 320

Why the truth value of the conditional statement is false is explained

Page 47 Problem 9 Answer

Question 9.

What does the phrase “If q is true” mean in terms of the conditional statement?

Answer:

Given that “If p is false and q is true, then the truth value of a conditional statement is true.”

We need to explain what does the phrase “If p is false” mean in terms of the conditional statement.

Since p is the hypothesis of the conditional statement and given thatp

is false, therefore the hypothesis of the conditional statement is false.

For example,

If the measure of an angle is32∘, then the angle is acute.

Here the statement p is “The measure of an angle is 32∘.”

If p is false, then the measure of an angle cannot be 32∘.

The phrase “If p is false” means the hypothesis of the conditional statement is false.

Page 47 Problem 10 Answer

Question 10.

What is the truth value of the conditional statement if p is false and q is true? Explain why this is the case using an example.

Answer:

Given that “If p is false and q is true, then the truth value of a conditional statement is true.”

We need to explain what does the phrase “If q is true” mean in terms of the conditional statement.

Since q is the conclusion of the conditional statement and given that q is true , therefore the conclusion of the conditional statement is true.

For example,If the measure of an angle is32∘then the angle is acute.

Here the statement q is “The angle is acute.”

If q is true, then the angle is acute.

The phrase “If q is true” means the conclusion of the conditional statement is true.

Carnegie Learning Geometry 2nd Edition Exercise 1.5 Solutions Page 47 Problem 11 Answer

Question 11.

Explain why the truth value of the conditional statement is true if p is false and q is true.

Answer:

Given: a conditional statement

To find:  We have explain why the truth value of the conditional statement is true.

a conditional statement being true requires it to be true under all possible circumstances

We try to explain this through one example

Consider A number is even then it is an Integer

In the above statement

p= A number is even

q= then it is an integer

Consider an number 3 we know that 3 is not an even number

Hence the statement p is false  but we also know that 3 is an integer Hence the statement q

is true  then we just  have found a counterexample.

3 would be a counterexample proving that not all even numbers are integers. But that does not fit . since 3 not even but odd.

A true counterexample would have to be an even number which is not an integer, which is clearly impossible.

From this we conclude that A is necessary condition for B but not sufficient.

Hence If p is false and q is true, then p→q  is  true.

Hence if p is false and q is true then p→q is true because A is a necessary condition for B

but not sufficient condition.

Page 48 Problem 12 Answer

Question 12.

A conditional statement of the form “If p, then q”. What is the conclusion q in terms of the conditional statement?

Answer:

A conditional statement is of the form ”  If p, then q “, where p is hypothesis  and q is conclusion of the hypothesis.

Here,The statement of condition or hypothesis  is  p:m ABˉ

=6 inches and m BCˉ=6 inches.

In other words given line segments ABˉ  and BCˉ are of equal length.

For the given problem ,the hypothesis is  p:m ABˉ=6 inches and m BCˉ=6 inches.

Page 48 Problem 13 Answer

Question 13.

Given the conditional statement “If \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches, then \(\overline{A B} \cong \overline{B C}\),” identify the hypothesis p and explain why the conclusion q is valid.

Answer:

A conditional statement is of the form ”  If p, then q “, where p is hypothesis  and q is conclusion of the hypothesis.

Here, The conclusion is q: ABˉ ≅ BCˉ

In other words given line segments ABˉ and BCˉ are congruent.

For the given problem,the conclusion is q: ABˉ ≅ BCˉ

Page 48 Problem 14 Answer

Question 14.

The conditional statement “If \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches, then \(\overline{A B} \cong \overline{B C}\)” with p: \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches and q: \(\overline{A B} \cong \overline{B C}\);

1. If p is true (i.e., \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches), why is q (i.e., \(\overline{A B} \cong \overline{B C}\)) Also true?
2. Explain why the truth value of the conditional statement p→q is true when both p and q are true.

Answer:

Let   p:m ABˉ=6 inches and m BCˉ=6 inches  and q:ABˉ≅BCˉ.

Given that p is true. Hence m ABˉ=m BCˉ=6 inches.

Hence they are of equal length.

So the above line segments ABˉ andBC ˉ are said to be congruent lines.

Hence q:ABˉ≅BCˉ is true.

So the truth value of p→q is true.

The truth value of the conditional statement if both  p and q are true is true.

Page 48 Problem 15 Answer

Question 15.

p: \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches, and q: \(\overline{A B} \cong \overline{B C}\). What is the truth value of the conditional statement p → q when p is true and q is false?

Answer:

Let  p:m ABˉ=6 inches and m BCˉ=6 inches and q:ABˉ≅BCˉ.

Given that p is true.

Hence m ABˉ=mBCˉ=6 inches.

So the length of ABˉ andBCˉ are same.

So they are congruent lines.

But given that q:ABˉ≅BCˉ is false.

So the truth valuep→q of is false.

For the given problem, the truth value of the conditional statement if p is true and q is false is false.

Tools Of Geometry Solutions Chapter 1 Exercise 1.5 Carnegie Learning Geometry Page 48 Problem 16 Answer

Question 16.

p: \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches, and q: \(\overline{A B} \cong \overline{B C}\).

What is the truth value of the conditional statement p → q when p is false and q is true?

Answer:

Let p:m ABˉ=6 inches and m BCˉ=6 inchesand q:ABˉ≅BCˉ.

Give that p is false. Hence mABˉ≠6 inches and mBCˉ≠6 inches.

Let us assume mABˉ=7 inches. mBCˉ=7 inches.

Hence ABˉ and BCˉ are congruent lines.

Implies that q:ABˉ≅BCˉ is true.

So the truth value of p→q is true.

For the given problem, the truth value of the conditional statement if p is false and q is true is true.

Page 48 Problem 17 Answer

Question 17.

p: \(m \overline{A B}=6\) inches and \(m \overline{B C}=6\) inches, and q: \(\overline{A B} \cong \overline{B C}\).

What is the truth value of the conditional statement p q when p is false and q is true?

1. Determine the true value of p.
2. Determine the true value of q.
3. Use the values of p and q to determine the true value of the conditional statement p → q.

Answer:

Let p:mABˉ=6 inches and mBCˉ=6 inchesand q:ABˉ≅BCˉ.

Given that p is false. mABˉ≠6 inches and mBCˉ≠6 inches.

Let us assume mABˉ=7 inches. mBCˉ=5 inches.

Hence ABˉ and BCˉ are not congruent lines.

So it implies that q is false, which is the correct conclusion.

Hence the truth value p→q of is true.

For the given problem, the truth value of the conditional statement, if both p and q are false, is true.

Page 49 Problem 18 Answer

Question 18.

Given that BD bisects ∠ABC, prove that ∠ABD is congruent to ∠CBD using the properties of angle bisectors.

Answer:

Consider the given conditional statement –

If BD bisects ∠ABC then∠ABD≅∠CBD

We have to draw the diagram for the given conditional statement as well write the hypothesis as given and conclude as Prove statement

Diagram for the given conditional statement is given below:

Now from the given conditional statement the Given and concluding statement is as

Given : BD bisects∠ABC

Prove :∠ABD≅∠CBD

Given :BD bisects∠ABC

Prove :∠ABD≅∠CBD

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 1 Exercise 1.5 Page 49 Problem 19 Answer

Question 19.

Consider the given conditional statement – AM ≅ MB, if M is the midpoint of AB.

1. Draw the diagram for the given conditional statement.
2. Using the properties of congruency, prove that AM is congruent to MB.

Answer:

Consider the given conditional statement -AM≅MB, ifM is the mid-point of AB.

We have to draw the diagram for the given conditional statement as well write the hypothesis as given and conclude as Prove statement by using the properties of congreuncy.

Diagram for the given conditional statement is given below:

Now from the given conditional statement the Given and concluding statement is as

Given : M is the mid-point of AB.

To prove : AM≅MB

Given : M is the mid-point of AB.

Prove : AM≅MB

Page 50 Problem 20 Answer

Question 20.

If AB ⊥ CD at point C, then ∠ACD is a right angle and ∠BCD is a right angle.

1. Draw the diagram for the given conditional statement.
2. Using the definition of perpendicular lines, prove that ∠ACD and ∠BCD are right angles.

Answer:

Consider the given conditional statement –

If AB⊥CD at pointC, then∠ACD is a right angle and∠BCD is a right angle.

We have to draw the diagram for the given conditional statement as well write the hypothesis as given and conclude as Prove statement

Diagram for the given conditional statement is given below:

Now from the given conditional statement the Given and Concluding are as-

Given : AB⊥CD

To prove : ∠ACD and∠BCD are right angles.

Given :AB⊥CD

Prove :∠ACD and∠BCD are right angles.

Carnegie Learning Geometry Chapter 1 Exercise 1.5 Free Solutions Page 50 Problem 21 Answer

Question 21.

m∠DEG + m∠GEF = 180°, if ∠DEG and ∠GEF are a linear pair.

1. Draw the diagram for the given conditional statement.
2. Using the definition of a linear pair, prove that m∠DEG + m∠GEF = 180°.

Answer:

Consider the given conditional statement -m∠DEG+m∠GEF=180°,if∠DEF and∠GEF are liner pair.

We have to draw the diagram for the given conditional statement as well write the hypothesis as given and conclude as Prove statement.

Diagram for the given conditional statement is given below:

Now from the given conditional statement the Given and concluding statement is as

Given : ∠DEG and∠GEF are linear pair.

To prove: m∠DEG+m∠GEF=180°

Given :∠DEG and∠GEF are linear pair.

Prove :m∠DEG+m∠GEF=180

Page 50 Problem 22 Answer

Question 22.

Consider the “W is the perpendicular bisector of PR if WX ⊥ PR and WX bisects PR.”

1. Draw the diagram for the given conditional statement.
2. Write the hypothesis as given and conclude as a proof statement.

Answer:

Consider the given conditional statement -W is the perpendicular bisector of PR, if WX⊥PR

And WX bisects  PR

We have to draw the diagram for the given conditional statement as well write the hypothesis as given and conclude as Prove statement

Diagram for the given conditional statement is given below:

Now from the given conditional statement the Given and concluding statement is

Given : WX⊥PR

Prove : WX bisects PR

Given : WX⊥PR

Prove : WX bisects PR

Carnegie Learning Geometry Exercise 1.5 Student Solutions Page 51 Problem 23 Answer

Question 23.

Consider the “If ∠ABD and ∠DBC are complementary, then BA ⊥ BC.”

1. Draw the diagram for the given conditional statement.
2. Write the hypothesis as given and conclude as a proof statement.
3. Provide a proof to show that BA ⊥ BC.

Answer:

Given: If ∠ABD and ∠DBC are complementary then BA⊥BC.

To prove that BA⊥BC

From the given data ∠ABD and ∠DBC are complementary

So from the definition of the complimentary angle we cans say that, two angles are called complementary if their measures add to 90 degrees.

Therefore from the figure, we can say that BA⊥BC

Hence, from above we can say that BA⊥BC

Page 53 Problem 24 Answer

Question 24.

Given two angles ∠A and ∠B, which form a linear pair:

1. Explain why ∠A and ∠B must add up to 180 degrees.
2. Draw the diagram to illustrate the linear pair.
3. Provide a mathematical proof to show that ∠A and ∠B add up to 180 degrees.

Answer:

Two angles are said to be linear if they are adjacent angles formed by two intersecting lines.

The measure of a straight angle is 180∘, Therefore, a linear pair of angles must add to180∘ .

The diagram of a linear pair is as follows;

From the sketch and linear pair postulate ∠A and ∠B are linear pairs.

Page 53 Problem 25 Answer

Question 25.

The two angles, 130°, and 50°:

1. Explain why these two angles are considered supplementary.
2. Draw the diagram to illustrate these supplementary angles.
3. Provide a mathematical proof to show that these angles add up to 180 degrees.

Answer:

If two angles sum up to 180 degrees, they are considered to be supplementary angles.

When supplementary angles are added together, they produce a straight angle (180 degrees).

Here, 130∘and 50∘are supplementary angles as their sum gives 180∘.

Definition From the defination of supplementary angles 130∘+50∘=180∘.

Page 53 Problem 26 Answer

Question 26.

The collinear points D, E, and F with point E between points D and F, draw the situation and explain how DE + EF = DF.

Answer:

Given: collinear points D, E, and F with point E between points D and F.

We have to draw the above situation.

Here, E is in between D and F.

In the figure above E is between D and F such that DE+EF =DF.

Page 53 Problem 27 Answer

Question 27.

Using the segment addition postulate, if B is a point on \(\overline{A C}\) such that AB + BC = AC, then A, B, and C are collinear points. Draw the diagram of the given situation and explain the segment addition postulate.

Answer:

According to the segment addition postulate if B is a point ACˉ such that AB+BC = AC then, A,B, and C are collinear points.

The diagram of the situation above is;

From the sketch above and the segment postulate AB+BC=AC.

Page 53 Problem 28 Answer

Question 28.

Using the segment addition postulate, verify the lengths of segments when a point B is on line segment \(\overline{A C}\) such that AB + BC = AC. Given that AC = 8m and B is 3m from A, find the length of BC and explain the verification process.

Answer:

Given

AC = 8m and B is 3m from A

According to the segment addition postulate if B is a point on AC such that AB+BC = AC then, A, B, and C are collinear points.

Let AC = 8m and B be 3 m  from A. Such that AB = 3 m. Then BC = 5 m , because 3 m +5 m= 8 m .

Since, AB = 3 m, BC = 5m then by collinear postulate AC=8 m.

3 m +5 m= 8 m.

Tools of Geometry Exercise 1.5 Carnegie Learning 2nd Edition answers Page 53 Problem 29 Answer

Question 29.

Given ∠DEF with EG in the interior, verify that ∠DEG + ∠GEF = ∠DEF.

Answer:

Let draw ∠DEF with EGin the interior, as shown below;

Clearly, line EG is drawn such that ∠DEG+∠GEF =∠DEF.

Page 53 Problem 30 Answer

Question 30.

Given ∠DEF with line segment EG in the interior, show that ∠DEG + ∠GEF = ∠DEF.

Answer:

Let us draw the line EG in ∠DEF as shown below;

Clearly, line EG  is drawn such that ∠DEG+∠GEF=∠DEF.

Geometry Student Text 2nd Edition Chapter 1 Tools of Geometry

Carnegie Learning Geometry Student Text 2nd Edition Chapter 1 Exercise 1.4 Solution Page 39 Problem 1 Answer

Question 1.

How did Emma reach the conclusion that raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power?

Answer:

Given: Emma is watching her big sister do homework. She notices the following:- nine cubed is equal to nine times nine times nine

10 to the fourth power is equal to four factors of 10 multiplied together

To specify How did Emma reach this conclusion

Emma notices

4²=4×4

9³=9×9×9

10⁴=10×10×10×10

​So by raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power.

Hence,  Emma reach this conclusion by raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power.

Read and learn More Carnegie Learning Geometry Student Text 2nd Edition Solutions

Carnegie Learning Geometry Chapter 1 Page 39 Problem 2 Answer

Question 2.

How did Ricky reach the conclusion that seven to the fourth power means multiplying seven by itself four times?

Answer:

It is given that 7 to 4^th power is to be calculated.

Seven to fourth power means that the number seven has to be multiplied to itself four times which is mathematically expressed as: 7×7×7×7.

Ricky reached the conclusion using the definition of nth power of a number x.

Ricky reached the conclusion using the definition of nth power of a number x.

Carnegie Learning Geometry Chapter 1 Page 40 Problem 3 Answer

Question 3.

How did Emma and Ricky reach their conclusions about raising numbers to a power, and are their observations correct according to the rule of exponents?

Answer:

Carnegie Learning Geometry Chapter 1 Page 40 Problem 4 Answer

Question 4.

How much total salary did Aaron receive if he worked for 4 hours at a rate of  8.25 dollars per hour?

Answer:

It is given that Salary per hour =8.25 dollars, Total number of hours worked =4 hours

Substitute Salary per hour =8.25 dollars, Total number of hours worked =4 hours into the formula:

Salary per hour=Total salary received

Total number of hours worked.

so that 8.25=Total salary received/4

Hence Total salary recived=8.25×4=33 dollars

If Aaron worked for four hours, the total salary received is 33 dollars

Carnegie Learning Geometry Chapter 1 Page 40 Problem 5 Answer

Question 5.

How can the formula Salary per hour = \(\frac{\text { Total salary received }}{\text { Total number of hours worked }}\) be used to determine Aaron’s total salary if he worked for 4 hours at an average rate of 8.25 dollars per hour?

Answer:

The average salary per hour and the number of hours worked are given to be 8.25 dollars and 4 hours respectively.

To answer the task 6(a), the formula Salary per hour=Total salary received

Total number of hours worked. is used to make a conclusion.

To answer task 6(a), the formula Salary per hour=Total salary received

Total number of hours worked. is used to make conclusions.

Solutions For Tools Of Geometry Exercise 1.4 In Carnegie Learning Geometry Page 40 Problem 6 Answer

Question 6.

What is the general term used for the process that starts with a broad, general idea and then verifies it for specific cases?

Answer:

“Thinking down from” starts with a very broad and general idea and then idea is verified for specific cases.

Generally “the act of thinking down” would mean that a general theory or idea is known and it has to be tested or verified to more specific cases with certain restrictions or conditions.

The general term for such a phrase should be “DEDUCTIVE REASONING”.

The general term for the phrase “the act of thinking down” should be “DEDUCTIVE REASONING”.

Carnegie Learning Geometry Chapter 1 Page 40 Problem 7 Answer

Question 7.

What is the general term used for the process that starts with specific statements and then tests the idea for general cases?

Answer:

“Thinking toward or up to” starts with a very specific statement and then idea is tested for general cases.

Generally “the act of thinking toward or up to” would mean that some idea holds true for some specific cases with certain restrictions and it has to be tested or verified for a

a general case with no restrictions.

The general term for such a phrase should be “INDUCTIVE REASONING”.

The general term for the phrase “the act of thinking toward or up to” should be “INDUCTIVE REASONING”.

Carnegie Learning Geometry 2nd Edition Exercise 1.4 solutions Page 41 Problem 8 Answer

Question 8.

What type of reasoning did Emma use when she observed that raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power?

Answer:

In the first question, Emma observes the following:

4²=4×4

– nine cubed is equal to nine times nine times nine.

– 10 to the fourth power is equal to four factors of 10 multiplied together.

In the first question, Emma observed three things and made an observation based on those three things.

The observation made was “raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power”.

This is an example of Inductive reasoning.

Emma used INDUCTIVE REASONING.

Carnegie Learning Geometry Chapter 1 Page 41 Problem 9 Answer

Question 9.

What type of reasoning did Emma use when she concluded that raising a number to a power is the same as multiplying the number by itself as many times as indicated by the power?

Answer:

In the first problem, Emma make a conclusion that raising a number to  power is same as multiplying the number by itself as many times as indicated by power.

Thus, Emma used deductive reasioning.

Emma used deductive reasioning.

Page 41 Problem 10 Answer

Question 10.

What is the specific information provided in the given problem?

Answer:

Specific information in a given problem is that “his neighbor Matilda smokes”.

Specific information in a given problem is that “his neighbor Matilda smokes”.

Carnegie Learning Geometry Chapter 1 Page 41 Problem 11 Answer

Question 11.

What is the general information provided in the given problem?

Answer:

General information in this problem is that tobacco greatly increases the risk of cancer.

General information in this problem is that tobacco greatly increases the risk of cancer.

Page 41 Problem 12 Answer

Question 12.

What conclusion can be drawn from the given specific and general information?

Answer:

The conclusion of this problem is that Matilda has a high risk of cancer.

The conclusion of this problem is that Matilda has a high risk of cancer.

Tools of Geometry Solutions Chapter 1 Exercise 1.4 Carnegie Learning Geometry Page 41 Problem 13 Answer

Question 13.

How does your friend use reasoning to make the conclusion that Matilda has a high risk of cancer?

Answer:

Friend uses deductive reasoning to make the conclusion.

Reason: As the conclusion is based on general information.

Friend uses deductive reasoning to make the conclusion as he uses general information.

Carnegie Learning Geometry Chapter 1 Page 41 Problem 14 Answer

Question 14.

Is your friend’s conclusion that “Matilda has a high risk of cancer” correct? Explain your reasoning.

Answer:

Yes, my friend’s conclusion is correct.

As general information tells that tobacco greatly increases the risk of cancer and his neighbor Matlida smokes.

So, the conclusion based on general and specific information is “Matilda has a high risk of cancer”.

Yes, the conclusion is correct as it is based on general information.

Step-By-Step Solutions For Carnegie Learning Geometry Chapter 1 Exercise 1.4 Page 42 Problem 15 Answer

Question 15.

Is your friend’s conclusion correct, and what information supports this conclusion?

Answer:

Specific information in a given problem is that it rained each of the five days she was on a London trip.

Specific information in a given problem is that it rained each of the five days she was on a London trip.

Carnegie Learning Geometry Chapter 1 Page 42 Problem 16 Answer

Question 16.

What is the general information provided in this problem?

Answer:

General information in this problem is that “It rains every day in England”.

General information in this problem is that “It rains every day in England”.

Page 42 Problem 17 Answer

Question 17.

What conclusion can be drawn from the general information that “It rains every day in England”?

Answer:

The conclusion of this problem is that “It rains every day in England!”.

The conclusion of this problem is that “It rains every day in England!”.

Carnegie Learning Geometry Chapter 1 Page 42 Problem 18 Answer

Question 18.

If Molly uses a specific example to make a conclusion, what type of reasoning is she using?

Answer:

Molly uses a specific example to make a conclusion. So, Molly uses inductive reasoning to make the conclusion.

Molly uses a specific example to conclude. So, Molly uses inductive reasoning to conclude.

Page 42 Problem 19 Answer

Question 19.

Molly returned from a trip to London and tells you, “It rains every day in England!” She explains that it rained each of the five days she was there.

To find out: Is Molly’s conclusion correct? Explain.

Answer:

Given: – Molly returns from a trip to London and tells you, “It rains every day in England!” She explains that it rained each of the five days she was there.

To find out: – Is Molly’s conclusion correct? Explain.

The process used: – The conclusion of Molly’s is that “It rains every day in England!”. It is based on Inductive reasoning

Inductive reasoning based on example, so maybe Molly’s is not correct.

Hence, Molly’s conclusion based on Inductive reasoning so may be conclusion is not correct.

Carnegie Learning Geometry Chapter 1 Page 42 Problem 20 Answer

Question 20.

Detailed notes in history class and math class.

To find out: What conclusion did your classmate make? Why?

Answer:

Given: -Detailed notes in history class and math class.

To find out: – What conclusion did your classmate make? Why?

Process used: – Classmate use Inductive reasoning by taking example of math and history notebook.

so my classmate think that i also completed my Biology notebook.

Hence, classmate make conclusion that i also have complete notebook of Biology by inductive reasoning because he take example of my history and math notebook.

Page 42 Problem 21 Answer

Question 21.

Detailed notes in history class and math class

To find out: What type of reasoning did your classmate use? Explain.

Answer:

Given: -Detailed notes in history class and math class

To find out: -What type of reasoning did your classmate use? Explain.

Process used: -Classmate use Inductive reasoning by taking example of math and history notebook.

so my classmate think that i also completed my Biology notebook.

Hence, classmate use inductive reasoning because he taking the example of my history and math notebook.

Carnegie Learning Geometry Chapter 1 Page 42 Problem 22 Answer

Question 22.

Detailed notes in history class and math class

To find out: What conclusion did the biology teacher make? Why?

Answer:

Given: -Detailed notes in history class and math class

To find out: –  What conclusion did the biology teacher make? Why?

Process used: – Because when biology teacher asks him(classmate) if he knows someone in class who always takes detailed notes.

He(classmate) gives my name to the teacher.

Then biology teacher suggests he borrow your biology notes because he concludes that they will be detailed on inductive reasoning.

Classmate give example of me that is the reason Biology teacher also think that i have detailed notebook of biology.

Hence, Biology teacher make conclusion that I also have detailed notebook of biology by inductive reasoning

Carnegie Learning Geometry Chapter 1 Exercise 1.4 Free Solutions Page 42 Problem 23 Answer

Question 23.

Detailed notes in history class and math class

To find out: What type of reasoning did the biology teacher use? Explain.

Answer:

Given: -Detailed notes in history class and math class

To find out: – What type of reasoning did the biology teacher use? Explain.

Process used: – Biology teacher use inductive reasoning by taking example that i have detailed notebook of biology because i have detailed notebook of math and history.

Because when he ask about detailed notebook of biology in the classroom then he gave my example.

Hence, Biology teacher use inductive reasoning because my classmate give my name as i have detailed notebook of biology.

Carnegie Learning Geometry Chapter 1 Page 42 Problem 24 Answer

Question 24.

Detailed notes in history class and math class.

To find out: Will your classmate’s conclusion always be true? Will the biology teacher’s conclusion always be true? Explain.

Answer:

Given: -Detailed notes in history class and math class.

To find out: – Will your classmate’s conclusion always be true? Will the biology teacher’s conclusion always be true? Explain.

Process used: – Inductive reasoning based on the example not on the rule.

My classmate take example of my detailed notebook of history and math, and my biology teacher believe on me because my classmate give my name as i have detailed notebook of biology.

Therefore, both my classmate and my teacher conclusion not always be true.

Hence, both my classmate and my teacher conclusion not always be true because they uses inductive reasoning.

Page 43 Problem 25 Answer

Question 25.

Sequence 4, 15, 26, 37

a₁ = 4, a₂ = 15, a₃ = 26, a₄ = 37

To find out: What is the next number in the sequence (a₅)? How did you calculate the next number?

Answer:

Given: – 4, 15, 26, 37

a₁=4, a₂=15, a₃=26, a₄=37

To find out: – What is the next number in the sequence (a₅)? How did you calculate the next number?

Formula used: -nth term,an=a+(n−1)d and d=a₂−a₁

a₂−a₁=15−4=11

a₃−a₂=26−15=11

Hence, the series in A.P.

d=a₂−a₁=15−4=11

If n=5

a₅=4+(5−1)11

a₅=4+44

a₅=48

Hence, by using an =a+(n−1)d

we calculate the next number a₅=48

Carnegie Learning Geometry Chapter 1 Page 43 Problem 26 Answer

Question 26.

Sequence -4, 15, 26, 37

To find out: What types of reasoning did you use and in what order to make the conclusion?

Answer:

Given: -4, 15, 26, 37

To find out: – What types of reasoning did you use and in what order to make the conclusion?

Process used: – By using d=a₂−a₁=a₃−a₂=a₄−a₃

we find Constant quantity in common difference,

d=15−4=26−15=37−26

d=11

Hence, the series is in Arithmetic progression

Carnegie Learning Geometry Exercise 1.4 Student Solutions Page 43 Problem 27 Answer

Question 27.

Explain the differences between inductive and deductive reasoning. Provide an example of each type of reasoning and outline the guidelines for ensuring logical and valid arguments in both cases.

Answer:

The differences between inductive and deductive reasoning:

Induction: A process of reasoning (arguing) which infers a general conclusion based on individual cases, examples, specific bits of evidence, and other specific types of premises.

Example: In Chicago last month, a nine-year-old boy died of an asthma attack while waiting for emergency aid.

After their ambulance was pelted by rocks in an earlier incident, city paramedics wouldn’t risk entering the Dearborn Homes Project (where the boy lived) without a police escort.

Thus, based on this example, one could inductively reason that the nine-year-old boy died as a result of having to wait for emergency treatment.

Guidelines for logical and valid induction:

1. When a body of evidence is being evaluated, the conclusion about that evidence that is the simplest but still covers all the facts is the best conclusion.
2. The evidence needs to be well-known and understood.
3. The evidence needs to be sufficient. When generalizing from a sample to an entire population, make sure the sample is large enough to show a real pattern.
4. The evidence needs to be representative. It should be typical of the entire population being generalized.

Deduction: A process of reasoning that starts with a general truth, applies that truth to a specific case (resulting in a second piece of evidence), and from those two pieces of evidence (premises), draws a specific conclusion about the specific case.

Example: Free access to public education is a key factor in the success of industrialized nations like the United States.

(major premise) India is working to become a successful, industrialized nation. (specific case)

Therefore, India should provide free access to public education for its citizens. (conclusion) Thus, deduction is an argument in which the conclusion is said to follow necessarily from the premise.

Guidelines for logical and valid deduction:

1. All premises must be true.
2. All expressions used in the premises must be clearly and consistently defined.
3. The first idea of the major premise must reappear in some form as the second idea in the specific case.
4. No valid deductive argument can have two negative premises.
5. No new idea can be introduced in the conclusion.

Hence, the Induction and Deduction are explain with help of example.

Carnegie Learning Geometry Chapter 1 Page 44 Problem 28 Answer

Question 28.

What are the reasons why a conclusion may be false? Explain using an example.

Answer:

To find out: – Reasons why a conclusion may be false?

Process used: -Derek tells his little brother that it will not rain for the next thirty days because he “knows everything.”

It is an assumed information based on inductive reasoning.

Inductive reasoning based on examples, so may be it false.

Hence, Either the assumed information is false or the argument is not valid, conclusion may false of his little brother because it is based on example there is no rule for his little brother assumption.

Tools of Geometry Exercise 1.4 Carnegie Learning 2nd Edition answers Page 44 Problem 29 Answer

Question 29.

The given conclusion is: Two lines are not parallel so the lines must intersect. Justify whether this conclusion is true or false.

Answer:

The given conclusion is: Two lines are not parallel so the lines must intersect.

The main objective is to justify why the conclusion is false.

The conclusion is true  as if two lines are not parallel then, they can be coincident as well as intersecting.

So, the conclusion is true because if the two lines are not parallel it means at any point they are going to meet or intersect.

Hence, the conclusion true and valid .

Carnegie Learning Geometry Chapter 1 Page 44 Problem 30 Answer

Question 30.

Provide an example of a conclusion that is false because the assumed information is false.

Answer:

To write : An example of a conclusion that is false because the assumed information is false.

Example is

“I have a very strong feeling that my ticket is the winning lottery ticket, so I’m quite confident I will win a lot of money tonight.”

Here argument is strong

But the information is false

The statement is

“I have a very strong feeling that my ticket is the winning lottery ticket, so I’m quite confident I will win a lot of money tonight.”

Carnegie Learning Geometry Chapter 1 Page 44 Problem 31 Answer

Question 31.

Provide an example of a conclusion that is false because the argument is not valid.

Answer:

To write : An example of a conclusion that is false because the argument is not valid.

The statement is

” Two lines are perpendicular so the lines must intersect. ”

There are lines which are perpendicular without intersection.

Hence the conclusion is wrong.

The statement is ” Two lines are perpendicular so the lines must intersect. “