Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments

Savvas Learning Co Geometry Student Edition Chapter 1 Exercise 1.3 Measuring Segments Solutions Page 23  Exercise 1  Problem 1

Given: The point on \(\overrightarrow{D A}\) that is 2 units from D.

To name each of the following.By using geometry theory.

The ray moves in the direction of A therefore the only point 2 units away from D is point B, because we have to move 2 units toward A.

Solving the question, we found that it is B.

Page 23  Exercise 2  Problem 2

Given:

The coordinate of the midpoint of AG

Using properties of line.

Midpoint of \(\overrightarrow{A G}\)

= \(\frac{A+G}{2}\)

Midpoint of \(\overrightarrow{A G}\)

= \(\frac{3−3}{2}\)

Midpoint \(\overrightarrow{A G}\) = 0

The coordinate of the midpoint of \(\overrightarrow{A G}\) = 0 whcich is at D.

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Exercise 1.3 Measuring Segments Savvas Geometry Answers Page 23  Exercise 3  Problem 3

Solving the question, we found point Q and line ℓ.

Page 23  Exercise 4  Problem 4

Given: Two segments are congruent and saying that two segments have equal length.

To find When would you use each phrase.

Using theoretical method.

If two segments are equal in length, then they are congruent and if they are congruent then they are equal in length.

But the word equal is used to compare numbers; we can not say that two numbers are congruent, we say that two numbers are equal.

And the word congruent is used to compare shapes; we can not say that two triangles are equal, we say that two triangles are congruent.

The word equal is used to compare numbers. and the word congruent is used to compare shapes.

Chapter 1 Exercise 1.3 Measuring Segments Savvas Learning Co Geometry Explanation Page 24  Exercise 5  Problem 5

Given: You and your friend live 5 mi apart. He says that it is 5 mi from his house to your house and −5 mi from your house to his house.

To find What is the error in his argument.

Using the theoretical method.

The error is that the distance is never negative.

He applied the rule that the distance between two points on the number line is the absolute value of the difference between the two points, but he did not apply the absolute value which gives only positive numbers.

For example, if we calculate the distance on the number line between the point 7 and the point 2, it should be

∣7 − 2∣ = ∣2 − 7∣ = 5 ≠ −5

The error is that the distance is never negative.

Solutions For Measuring Segments Exercise 1.3 In Savvas Geometry Chapter 1 Student Edition Page 24   Exercise 6  Problem 6

Given:

To Find the length of BD

Using addition.

Length of  BD = 6 + 3

BD = 9 units

Length of BD = 9 units

Page 24  Exercise 7  Problem 7

Given:

To Find the length of AD

Using addition.

Length of AD = 3 + 8

AD = 11

Length of AD = 11

Exercise 1.3 Measuring Segments Savvas Learning Co Geometry Detailed Answers Page 24  Exercise 8  Problem 8

Given:

To Find the length CE

Using subtraction.

Length of CE = 7 − 1

CE = 6 units.

Length of CE = 6 units.

Page 24  Exercise 9  Problem 9

Given: RS =  15

ST = 9

To find RT

Using addition.

RT = RS + ST

RT = 15 + 9

RT = 24

Value of RT = 24.

Geometry Chapter 1 Measuring Segments Savvas Learning Co Explanation Guide Page 24  Exercise 10  Problem 10

Given:

To find value of y

Using RT = RS + ST

RT = RS + ST

15y−9 = 4y + 8 + 8y + 4

15y − 9 = 12y + 12

15y − 12y = 12 + 9

3y = 21

y = \(\frac{3}{21}\)

y = 7

Value of y = 7.

Page 24  Exercise 10  Problem 11

Given:

To find RS,ST, and RT

Using simple calculation

15y − 9 = 4y + 8 + 8y + 4

15y − 9 = 12y + 12

15y − 12y = 12 + 9

3y = 21

y =\(\frac{3}{21}\)

y = 7

ST = 4y + 8

ST = 4.7 + 8

ST = 28 + 8

ST = 36

RS = 8y + 4

RS = 8.7 + 4

RS = 60

RT = RS + ST

RT = 60 + 36

RT = 96

Value of ST = 36, RT = 96,RS = 60

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 24   Exercise 11  Problem 12

Given:

To Tell whether the segments \(\overline{M N}\) and \(\overline{P Q}\) are congruent.

Using properties of line.

\(\overline{M N}\) = |−3 − 3| = 6 units

\(\overline{P Q}\) = |6 − 12| = 6 units

Since distance between \(\overline{M N}\) and \(\overline{P Q}\) are equal so they are congruent.

\(\overline{M N}\) and \(\overline{P Q}\) are congurent.

Page 24   Exercise 12  Problem 13

Given:

Tell whether the segments \(\overline{L P}\) and \(\overline{M Q}\) are congruent.

Using properties of line.

From the above figure we have

\(\overline{L P}\) = ∣−8−6∣ = 14 units

\(\overline{M Q}\) =∣−3−12∣ = 15 units.

So, distance between these two lines are not equal.

So, \(\overline{L P}\) and \(\overline{M Q}\) are not congruent \(\overline{L P}\) and \(\overline{M Q}\) are not congurent.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 24  Exercise 13  Problem 14

Given: A is the midpoint of ​ \(\overline{X Y}\)
​

To find XA

Using the midpoint of the line segment method.

From the given figure, XA = 3x.

So, it is must to solve x, first.

It is also given that A is the midpoint of a line segment XY

So, XA = AY

Substitute the given expression

3x = 5x − 6

⇒  −2x = −6

⇒  x =\(\frac{−6}{-2}\)

⇒  x = 3

Value of XA is 3

Page 24  Exercise 13  Problem 15

Given: A is the midpoint of ​ \(\overline{X Y}\)
​

To find AY and XY.

Using the midpoint of line segment method

A is mid point of XY.

Therefore, value of XA  and  AY are equal.

⇒ 3x = 5 x− 6

⇒ 5x − 3x = 6

⇒ 2x = 6

⇒ x = 3

AY = 5x − 6

Substitute the value of x in equation for AY

AY = 5(3) − 6

​⇒  15 − 6

⇒   9
​
By segment addition postulate

XA + AY = XY

⇒  9 + 9 = 18

Value of AY is 9 and XY is 18.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 24  Exercise 14  Problem 16

Given:  PT = 5x + 3 ,TQ = 7x − 9

To find the value of PT

Using the midpoint of a line segment method

PT = 5x + 3

TQ = 7x − 9

Since T is the midpoint of a line segment PQ.

PT = TQ

5x + 3  =  7x − 9

Subtract 5x from both sides

3 = 2x − 9

Add 9 both sides

12 = 2x

x = \(\frac{12}{2}\)

6  =  x or x  =  6

Substitute value of x in PT = 5x + 3.

Then it results in ​PT = 5(6) +  3
​
⇒ 33

The value of PT n is 33

Page 24  Exercise 15  Problem 17

Given: ​PT = 4x − 6 , ​TQ = 3x + 4

​

To find the value of PT.

Using the midpoint of line segment method.

Given that

PT = 4x−6

So it is required to solve x first

PT = TQ

Substituting the expression

4x − 6 = 3x + 4

Solve for x

4x = 3x + 10

⇒ x = 10

Substituting value of x in PT

​⇒ 4(10) − 6

⇒ 40 − 6

⇒ 34

Value of PT is 34.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 24  Exercise 16  Problem 18

Given: X = −7, Y = −3,Z = 1 and W = 5

To find the length of the two-segmentUsing the formula distance between two points

Given the following coordinates:

X  = −7,Y = −3, Z = 1 , and W = 5

The length of the segments:

ZX = ∣1−(−7)∣

ZX = ∣8∣

ZX = 8, and

WY = ∣5−(−3)∣

WY = ∣8∣

WY = 8

Since, ZX = WY ,then

\(\overline{Z X}\)= \(\overline{W Y}\).

The length of the two segments are 8 and 8 respectively and also they are congruent

Page 24  Exercise 17  Problem 19

Given: The coordinate of A is 0, AR = 5, and AT = 7.

To find the possible coordinates of the midpoint of the \(\overline{A T}\).

Using the properties of geometry.

Let M be the midpoint of \(\overline{A T}\) , given that A = 0 so T may be located to the left or to the right of A as shown.

If T is to the left of A, then T = −7 so the midpoint is,

M =  \(\frac{A+T}{2}\)

=  \(\frac{0+(-7)}{2}\)

=  \(\frac{-7}{2}\)

M = − 3.5

If T is to the right of A, then T = 7 so the midpoint is

M =  \(\frac{A+T}{2}\)

M = \(\frac{0+(7)}{2}\)

M =  \(\frac{7}{2}\)

M =  3.5

The possible coordinates of the segment is M = −3.5 or M = 3.5

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 24   Exercise 18   Problem 20

Given: The coordinate of A is 0, and AR = 5 and AT = 7.

To find the possible coordinates of the midpoint of the RT

Using the properties of geometry.

Let M be the midpoint o \(\overline{R T}\)

We are given that A = 0 so R and T may be located to the left or to the right of A as shown

If both R and T is to the left of A ,the mid point is

M = \(\frac{R+T}{2}\)

M = \(\frac{-5+(-7)}{2}\)

M = \(\frac{-12}{2}\)

Therefore M = −6.

If both R and T is to the right of A ,the mid point is

M = \(\frac{R+T}{2}\)

M = \(\frac{5+(7)}{2}\)

M = \(\frac{12}{2}\)

Therefore M = 6.

If T is left of A and R is to the right of A ,the midpoint is

M = \(\frac{R+T}{2}\)

M = \(\frac{5+(-7)}{2}\)

M = \(\frac{-2}{2}\)

M = − 1

Therefore M = −1

If R is left of A and T is to the right of A , the midpoint is

M = \(\frac{R+T}{2}\)

M = \(\frac{-5+(7)}{2}\)

M = \(\frac{2}{2}\)

M = 1

Therefore M = 1

The possible coordinates of the segment is M = −6 or M = 6 or M = −1 or M = 1.

Page 25  Exercise 19  Problem 21

Given:  To sketch a segment without using a ruler.

To sketch a segment of 3 in.

Using geometrical method.

Sketch the segment with 3 in

The segment of 3 in is 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 25  Exercise 20  Problem 22

Given: To sketch a segment without using a ruler.

To sketch a segment of 6 in.

Using geometrical method.

The segment with 6 in is

Segment with 6 in is 

Page 25  Exercise 21  Problem 23

Given: To sketch a segment without using a ruler.

To sketch a segment of 10 cm.

Using geometrical method.

The segment with 10 cm is

Segment with 10 cm is

Page 26  Exercise 22  Problem 24

Given: To sketch a segment without using a ruler.

To sketch a segment of 65 mm.

Using geometrical method.

The segment with 65 mm is

Segment with 65 mm is

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.3 Measuring Segments Page 25  Exercise 23  Problem 25

Given: A map of Florida and assuming that the route 10 between Quincy and Jacksonville is straight.

If you drive an average speed of 55 mi/h ,how long it will take to get from Lake Oak to Jacksonville.

To find out how can you use mile markers to find distances between points.

o find how average speed, distance, and time all relate to each other.Using average speed formula.

From the figure we can see that: 

Live Oak is marked as 283

Jacksonville is marked as 357

Therefore the distance between the two places is 357 − 283 = 74

Average speed = distance traveled/time taken

Substitute average speed = 55 and distance = 74

To get 55 = ( 74 /Time taken)

Time taken =\(\frac{74}{55}\) ≈ 1.35 Hours.

The average speed, distance, and time all relate to each other asb Time taken =\(\frac{74}{55}\) ≈ 1.35 Hours.