Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise 5.1 Midsegments Of Triangles

Savvas Learning Co Geometry Student Edition Chapter 5 Exercise 5.1 Midsegments Of Triangles Solutions Page 288  Exercise 1 Problem 1

From the given figure we can see that \(\overline{N O}\) is parallel to \(\overline{J K} \)

The segment \(\overline{N O}\) is parallel to \(\overline{J K} \).

Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288  Exercise 2  Problem 2

Given: Length of the side of the triangle.

To Find – Length of NM.

Use the relationship between the length of a midsegment and the length of the side of a triangle.

We have,LK = 46

Using the relationship

Length of a midsegment = \(\frac{1}{2}\) (length of the third side)

⇒ NM  =  \(\frac{1}{2}\)(46)

⇒ NM = 23

The length of the midsegment NM is 23.

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Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288  Exercise 3  Problem 3

A line segment that connects two midpoints of the sides of a triangle is called a midsegment.

It is a line that is drawn from one side to another, parallel to the other side.

A midsegment connects two mid points of the side of a triangle.

Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288  Exercise 4  Problem 4

The two noncollinear segments in the coordinate plane have the same slope, which means that the segments are neither perpendicular nor intersecting.

Thus, the two segments are parallel to each other which is the reason that the slopes of both the noncollinear segments are the same.

The two noncollinear segments are parallel to each other in the coordinate plane.

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 288  Exercise 5  Problem 5

In the given figure, we know that NP = PT.

Which means that P is the midpoint of the side NT.

On the other hand, there is no information if OL is equal to LT, so we can say that L is not the midpoint of the side OT.

Thus, we can say that the conclusion given by the student is not correct and PL is not parallel to NO.

The error in the student’s reasoning is that L is not the midpoint of the side OT.

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 288  Exercise 6  Problem 6

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288  Exercise 7  Problem 7

Given: A segment of the triangle.

To Find – The segment that is parallel to the given segment.

Use the triangle midsegment theorem.

We have the given segment,BC of the triangle

Using the triangle midsegment theorem, BC ∥ GF

The segment that is parallel to the segment BC is the segment GF.

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288  Exercise 8  Problem 8

Given: A segment of the triangle.

To Find – The segment that is parallel to the given segment.

Use the triangle midsegment theorem.

We have the given segment,EF of the triangle ABC.

Using the triangle midsegment theorem, EF ∥ AB

The segment parallel to the segment EF is the segment AB.

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288  Exercise 9  Problem 9

Given: A segment of the triangle ABC.

To Find – The segment parallel to the given segment.

Use the triangle midsegment theorem.

We have the given segment,CA of the triangle ABC.

Using the triangle midsegment theorem, CA ∥ EG

The segment parallel to the segment CA is the segment EG.

Solutions For Midsegments Of Triangles Exercise 5.1 In Savvas Geometry Chapter 5 Student Edition Page 288  Exercise 10  Problem 10

Given: A segment of the triangle ABC.

To Find –  The segment parallel to the given segment.

Use the triangle midsegment theorem.

We have the segment,EG of the triangle FGE

Using the triangle midsegment theorem, EG ∥ CA

The segment parallel to the segment EG is the segment CA.

Solutions For Midsegments Of Triangles Exercise 5.1 In Savvas Geometry Chapter 5 Student Edition Page 288  Exercise 11  Problem 11

Given: Midpoints and lengths of the sides of the ΔTUV.

To Find – The length of HE.

Use the triangle midsegment theorem.

We have

⇒ UV = 80

⇒ TV = 100 and HD = 80 also,E,D and H are the midpoints of ΔTUV

Using the triangle midsegment theorem

⇒ HE = \(\frac{1}{2}\) VU

⇒ HE = \(\frac{1}{2}\) (80)

⇒ HE = 40

The length of HE is 40 units.

Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Detailed Answers Page 288  Exercise 12  Problem 12

Given: Midpoints and lengths of the sides of the ΔTUV.

To Find –  The length of ED.

Use the triangle midsegment theorem.

We have

UV = 80, TV = 100 and HD = 80 also E,D and H are the midpoints of the sides of the ΔTUV

Using the triangle midsegment theorem

⇒ ED = \(\frac{1}{2}\) TV

⇒ ED = \(\frac{1}{2}\) (100)

⇒ ED = 50

The length of ED is 50 units.

Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Detailed Answers Page 288  Exercise 13  Problem 13

Given: HD = 80

To Find  –  The length of TU

In triangle TUV by midpoint theorem.

⇒ ​HD = \(\frac{1}{2}\) TU

⇒ 80 = \(\frac{1}{2}\) TU

⇒ TU = 160

​The value of TU = 160

Geometry Chapter 5 Midsegments of Triangles Savvas Learning Co explanation guide Page 288  Exercise 14  Problem 14

Given: HD = 80

To Find – The length of TE

In triangle TUV by mid point theorem.

​HD = \(\frac{1}{2}\) TU

80 = \(\frac{1}{2}\) TU

⇒ TU = 160

⇒ TU = 2 TE

⇒ TE = \(\frac{TU}{2}\)

⇒ TE = \(\frac{160}{2}\)

⇒ TE = 80

The value of TE = 80

Geometry Chapter 5 Midsegments Of Triangles Savvas Learning Co Explanation Guide Page 288  Exercise 15  Problem 15

Given:  A diagram

To Find – The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

SO,  x = \(\frac{1}{2}\) × 26

⇒  x = 13
​
The value of x = 13

Geometry Chapter 5 Midsegments of Triangles Savvas Learning Co explanation guide Page 288  Exercise 16  Problem 16

Given: A diagram

To Find  – The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

SO,  5x = \(\frac{1}{2}\) × 45

⇒  x  = 4.5

The value of x = 4.5

Savvas Learning Co Geometry Student Edition Chapter 5 Page 288  Exercise 17  Problem 17

Given: A diagram

To Find – The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

So​, 6x = \(\frac{72}{2}\)

⇒  x  =  6
​
The value of x = 6

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289  Exercise 18  Problem 18

Given: A diagram

To Find – The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

So ​x − 4  =  \(\frac{1}{2}\) × 17

⇒  x  =  \(\frac{17}{2}\) + 4

⇒  x  =  12.5
​
The value of x = 12.5

Savvas Learning Co Geometry Student Edition Chapter 5 Page 289  Exercise 19  Problem 19

Given: A diagram

To Find –  The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

So​ x + 2 =   \(\frac{1}{2}\)  × 38

⇒ x = 19 − 2

⇒  x = 17

​The value of x = 17

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289  Exercise 20 Problem 20

Given: A diagram

To Find –  The value of x

In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

So, ​5x − 4 = \(\frac{1}{2}\) × 8

⇒  5x = 8

⇒  x = \(\frac{8}{5}\)

⇒  x = 1.6
​
The value of x = 1.6.

Savvas Learning Co Geometry Student Edition Chapter 5 Page 289  Exercise 21  Problem 21

Given: The number of strides is shown in the triangle and the average stride is 3.5 ft

To Find  – The length of the longest side of the triangle.

In the triangle, the longest side will be in which maximum strides are paddled so the maximum number of strides is 250

The length of the longest side ​= 3.5 × 250

⇒  875ft
​
The length of the longest side is 875ft

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289  Exercise 21  Problem 22

Given: The sides of the triangle.To find: Distance to be paddled across the lake.

Let ,a,b, c be sides of the triangle.

⇒ a = 80 + 80 = 160  (Say)

⇒ b = 250  (Say)

⇒ c = 150 + 150 = 300 (Say)

Distance to be paddled across the lake = The perimeter of the triangle × 3.5

=  (a + b + c) × 3.5

=  (160 + 250 + 300) × 3.5

=  710 × 3.5

=  2485 ft

The distance to be paddled across the lake is 2485ft.

Savvas Learning Co Geometry Student Edition Chapter 5 Page 289  Exercise 22  Problem 23

Given: The following triangular face of the Rock and Roll Hall of Fame in Cleveland, Ohio, that is isosceles is provided with congruent red segments and the length of the base is 229 feet 6 Inches or 229 ft 6 in.

Converting the given measurement of the base into inches 229 ft 6 in

=  (229 × 12) + 6 in

=  2748  +  6 in

=  2754 in

The red segments divide the legs into four congruent parts and the white segment divides each leg into two congruent parts. The white segment is a midsegment of the triangular face of the building, so its length is half the length of the base.

So, the length of the white segment

=  \(\frac{2754}{2}\)

=  1377 in

Converting to feet and inches

Length of the white segment

=   \(\frac{1377}{12}\) ft

=   114. 75 ft

=   114 ft (0.75 × 12) in

=  114 ft 9 in

The length of the white segment is 114 ft 9in.

Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289  Exercise 23  Problem 24

Given: A figure.

To find –  m ∠ QPR

S is the mid-point of QP

T is the mid-point of QR

Using the mid-point theorem, we get  ST ∥ PR

QP is a transversal and  ST ∥ PR

Therefore,∠QPR = ∠QST  (Corresponding angles)

∠QPR = 40°  (Since,∠QST  =  40°)

Using the mid-point theorem,m ∠QPR = 40°.

Savvas Learning Co Geometry Student Edition Chapter 5 Page 289  Exercise 24  Problem 25

Given: The coordinates of the vertices of a triangle  EFG.

To find – The coordinates of H & J.

Since, H is the mid-point of EG

Therefore, coordinates of

H = \(\left(\frac{1+3}{2}, \frac{2-2}{2}\right)\)

H = (2,0)

Since, J is the mid-point of FG

Therefore, coordinates of

J = \(\left(\frac{5+3}{2}, \frac{6-2}{2}\right)\)

J = (4,2)

The coordinates of H = (2,0) and of J = (4,2).

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289  Exercise 24  Problem 26

Given:  H,  J are the mid-points of the sides EG, FG respectively.

To prove – HJ ∥ EF

Since, H,J are the mid-points of the sides EG, FG respectively.

Therefore, using the mid-point theorem we get, HJ ∥ EF

Using the mid-point theorem we get, HJ ∥ EF

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289  Exercise 24  Problem 27

Given: H,J are the mid-points of the sides EG, FG respectively.

To prove –  HJ = \(\frac{1}{2}\) EF

Since, H,J  are the mid-points of the sides EG,FG, respectively.

Therefore, using the mid-point theorem we get

⇒ HJ = \(\frac{1}{2}\) EF

Using the mid-point theorem we get,HJ = \(\frac{1}{2}\) EF

Savvas Learning Co Geometry Student Edition Chapter 5 Page 289  Exercise 25  Problem 28

Given: X,Y are the mid-points of the sides UV, UW respectively.m∠UXY = 60

To find – m∠V

Since, X, Y are the mid-points of the sides UV, UW respectively.

Therefore, using the mid-point theorem we get, XY ∥ VW

Hence,∠V = ∠UXY (Corresponding angles)

⇒ m∠V = 60 (Since m∠UXY = 60)

According to the question,m∠V = 60°.

Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289  Exercise 26  Problem 29

Given: X,Y are the mid-points of the sides UV,UW respectively.

To find –  m∠UYX

Since ,X,Y are the mid-points of the sides UV, UW respectively.

Therefore, using the mid-point theorem we get, XY ∥ VW

Hence,∠UYX = ∠W (Corresponding angles)

⇒ m∠UYX = 45 (Since m∠W = 45)

According to the given condition, m∠UYX = 45°

Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise

Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise Solutions Page 281  Exercise 1  Problem 1

Given: Use a compass and straightedge

To find –  Construct the perpendicular bisector of a segment.

Open the compass more than half of the distance between A and B n, and scribe arcs of the same radius centered at A and B.

Call the two points where these two arcs meet C and D. Draw the line between C and D.

CD is the perpendicular bisector of the line segment AB.

We have

The perpendicular bisector of a segment:

Chapter 5 Relationships Within Triangles Exercise Savvas Geometry Answers Page 281  Exercise 2  Problem 2

Given: A(5,1),B(−3,3),C(1,−7).

To find –  Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using the midpoint formula and length using the distance formula.

We have

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We have,the​distance​between​(x₁,​y₁),​(x₂,​y₂) : \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

To find AB, use A(5,1)and B(−3,3)

⇒  \(\sqrt{(-3-5)^2+(3-1)^2}\)

= 2\(\sqrt{17}\)

To find BC, use B(−3,3)and C(1,−7)

⇒  \(\sqrt{(1-(-3))^2+(-7-3)^2}\)

= 2\(\sqrt{29}\)

To find AC, use A(5,1)and C(1,−7)

⇒  \(\sqrt{(1-5)^2+(-7-1)^2}\)

= 4\(\sqrt{5}\)

The coordinates of the midpoint of the sides of AB, BC, AC are(1,2),(−1,−2), (3,−3)  respectively.  The length of the sides AB, BC, AC  are 2\(\sqrt{17}\), 2\(\sqrt{29}\),4 \(\sqrt{5}\) respectively.

Chapter 5 Relationships Within Triangles Exercise Savvas Geometry Answers Page 281  Exercise 3  Problem 3

Given: A(−1,2), B(9,2), C(−1,8)

To find – Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using midpoint formula and length using distance formula.

We have

Midpoint​of​ (x₁,​y₁),​(x₂,​y₂):  \(\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}\right)\)

To find AB, use A(−1,2) and B(9,2)

\(\left(\frac{9-1}{2}, \frac{2+2}{2}\right)\) = (4,2)

To find BC, use B(9,2)and C(−1,8)

\(\left(\frac{-1+9}{2}, \frac{8+2}{2}\right)\) = (4,5)

To find AC, use A(−1,2)and C(−1,8)

\(\left(\frac{-1-1}{2}, \frac{8+2}{2}\right)\) = (−1,5)

We have,the ​distance​between (x₁,​y₁),​(x₂,​y₂): \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
​
To find AB , use A(−1,2) and B(9,2)

⇒  \(\sqrt{(9-(-1))^2+(2-2)^2}\)

= 10

To find BC, use B(9,2) and C(−1,8)

⇒  \(\sqrt{(-1-9)^2+(8-2)^2}\)

= 2\(\sqrt{34}\)

To find AC, use A(−1,2)and C(−1,8)

⇒  \(\sqrt{(-1-(-1))^2+(8-2)^2}\)

= 6

The coordinates of the midpoint of the sides of AB, BC, AC are (4,2),(4,5),(−1,5) respectively. The length of the sides AB, BC, AC are 10, 2\(\sqrt{34}\), 6 respectively.

Relationships Within Triangles Exercises Solutions Chapter 5 Savvas Geometry Page 281  Exercise 4  Problem 4

Given: A(−2,−3), B(2,−3), C(0,3).

To find – Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using the midpoint formula and length using the distance formula.

We have

Midpoint ​of (x₁,​y₁),​(x₂,​y₂): \(\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}\right)\)

The midpoint of AB using A(−2,−3) and B(2,−3) is

\(\left(\frac{2-2}{2}, \frac{-3-3}{2}\right)\) = (0,−3)

The midpoint of BC using B(2,−3) and C(0,3) is

\(\left(\frac{0+2}{2}, \frac{3-3}{2}\right)\) = (1, 0)

The midpoint of AC using A(−2,−3) and C(0,3) is

\(\left(\frac{0-2}{2}, \frac{3-3}{2}\right)\) = (−1, 0)

We have the​ distance​between​ (x₁,​y₁),​(x₂,​y₂): \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
​
To find AB, use A(−2,−3) and B(2,−3) is

⇒  \(\sqrt{(2-(-2))^2+(-3-(-3))^2}\)

= 4

To find BC, use B(2,−3)and C(0,3)

⇒  \(\sqrt{(0-2)^2+(3-(-3))^2}\)

=  2\(\sqrt{10}\)

To find AC, use A(−2,−3) and C(0,3)

⇒  \(\sqrt{(0-(-2))^2+(3-(-3))^2}\)

=  2\(\sqrt{10}\)

The coordinates of midpoint of the sides of AB, BC, AC are (0,−3),(1,0),(−1,0) respectively. The length of the sides AB, BC, AC are 4, 2\(\sqrt{10}\),  2\(\sqrt{10}\) respectively.

Relationships Within Triangles Exercises Solutions Chapter 5 Savvas Geometry Page 281  Exercise 5  Problem 5

Given:  It is not too late.

To find –  The negation of a statement.

The negation of a statement is the opposite of the given mathematical statement.

It is too late is the negation of It is not too late.

The negation of the given statement is it is too late.

Chapter 5 Relationships Within Triangles Savvas Learning Co Geometry Explanation Page 281  Exercise 6 Problem 6

Given:  m∠R > 60

To find –  The negation of a statement.

The negation of a statement is the opposite of the given mathematical statement.

he measure of the angle R is not greater than 60.

The measure of the ∠R  is not greater than 60.

Solutions For Relationships Within Triangles Exercises In Savvas Geometry Chapter 5 Student Edition Page 281  Exercise 7  Problem 7

Given:  A(9,6), B(8,12)

To find – The slope of the line passing through the given points.

Slope between two points:

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

We have

⇒ (x₁,y₁) = (9,6)

⇒ (x₂,y₂) = (8,-12)

⇒ m = \(\frac{12-6}{8-9}\)

⇒ m = -6

The slope of the line passing through the given points is m =−6.

Relationships Within Triangles Exercises Savvas Learning Co Geometry Detailed Answers  Page 281  Exercise 8  Problem 8

Given: The distance between your home and your school is the length of the shortest path connecting them.

To find – How might you define the distance between a point and a line in geometry?

Distance is a numerical measurement of how far apart objects or points are.

The distance from a point to a line is the shortest distance from a given point to any point on an infinite straight line.

It is the perpendicular distance of the point to the line, the length of the line segment which joins the point to the nearest point on the line.

The distance from a point to a line is the shortest distance from a given point to any point on an infinite straight line. It is the perpendicular distance of the point to the line, the length of the line segment which joins the point to the nearest point on the line.

Savvas Learning Co Geometry Student Edition Chapter 5 Page 281  Exercise 9  Problem 9

Given:  A midpoint of a segment.

To find – What do you think a midsegment of a triangle is?

The midpoint is the middle point of a line segment.

It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints.

A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.

A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.

Relationships Within Triangles Exercises Savvas Learning Co Geometry Detailed Answers Page 281  Exercise 10  Problem 10

Given: If two parties are happening at the same time, they are concurrent.

To find –  What would it mean for three lines to be concurrent?

Lines in a plane or higher-dimensional space are said to be concurrent if they intersect at a single point.

Three or more lines in a plane passing through the same point are concurrent lines.

When a third line also passes through the point of intersection made by the first two lines then these three lines are said to be concurrent lines.

The point of intersection of all these lines is called the ‘Point of Concurrency’.

Three or more lines in a plane passing through the same point are concurrent lines. When a third line also passes through the point of intersection made by the first two lines then these three lines are said to be concurrent lines. The point of intersection of all these lines is called the ‘Point of Concurrency’.

Savvas Learning Co Geometry Student Edition Chapter 4 Congruent Triangles Exercise 4.2 Triangles Congruence By SSS And SAS

Savvas Learning Co Geometry Student Edition Chapter 4 Exercise 4.2 Triangles Congruence by SSS and SAS solutions Page 230  Exercise 1  Problem 1

In the given situation, we need to name the angle that is included between the given sides.

Let us draw a figure of the triangle with given information

According to the above figure

1. The angle that is included between the sides, \(\overline{\mathrm{PE}} \text { and } \overline{\mathrm{EN}}\) is ∠PEN.

2. The angle that is included between the sides, \(\overline{N P} \text { and } \overline{P E}\) is ∠NPE.

1. The angle that is included between the sides \(\overline{\mathrm{PE}} \text { and } \overline{\mathrm{EN}}\) is ∠PEN.

2. The angle that is included between the sides,  \(\overline{N P} \text { and } \overline{P E}\) is ∠NPE.

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Exercise 4.2 Triangles Congruence By Side Side Side And Side Angle Side Savvas Geometry Answers
Page 230  Exercise 2  Problem 2

In the given situation, we need to name the sides between which given angles are included.

Let us draw a figure of the triangle with given information,

According to the above figure

1. The sides which includes  \(\angle H \text { are } \overline{A H} \text { and } \overline{\mathrm{HT}}\)

2.The sides which includes  \(\angle T \text { are } \overline{H T} \text { and } \overline{T A}\)

Exercise 4.2 Triangles Congruence By Side Side Side And Side Angle Side Savvas Geometry Answers
Page 230  Exercise 3  Problem 3

In the given situation, we need to name the postulate that will be used to prove the triangles congruent.

The diagram shows that  \(\overline{\mathrm{BC}} \cong \overline{E C}\) and also  \(\overline{A C} \cong \overline{D C}\).

As  \(\angle A C B \cong \angle D C E\)  by vertically opposite angle.

Therefore Δ ABC≅ Δ DEC by using the Side Angle Side postulate (SAS).

Side Angle Side postulate (SAS) postulate would be used to prove the given triangles congruent.

Triangles Congruence By Sss And Sas Solutions Chapter 4 Exercise 4.2 Savvas Geometry Page 230  Exercise 4  Problem 4

In the given situation, we need to find the postulate that would be used to prove the triangles congruent.

The diagram shows that \(\overline{A B} \cong \overline{C D}\) and \(\overline{A D} \cong \overline{B C}\)

Also by the Reflexive Property of Congruence. \(\overline{B D} \cong \overline{B D}\)

Therefore, the given triangles are congruent by using Side-Side-Side (SSS) theorem.

The given triangles are congruent by using Side-Side-Side (SSS) theorem.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 230  Exercise 5  Problem 5

In the given question, we need to find what are the similarities and differences between SAS and SSS postulates.

How are the SAS Postulate and SSS Postulate​ alike?

1. They both require three congruency statements.

2. They both use two sides and one angle.

3. They both use one side and two angles.

4. They both can be used to prove that triangles are congruent.

5. They both use three sides.

How are the SAS Postulate and SSS Postulate​ different?

1. They require different numbers of congruency statements.

2. They apply to different kinds of triangles​ (acute or​ obtuse).

3. They use different numbers of sides​ and/or angles.

4. SSS can be used to prove two triangles are​ congruent, while SAS can be used to prove that corresponding angles are congruent.

The SSS and SAS Postulates are different and alike as well.

Chapter 4 Exercise 4.2 Triangles Congruence By Side Side Side And Side Angle Side Savvas Learning Co Geometry Explanation Page 230  Exercise 6  Problem 6

In the given situation, we need to find that the able triangles are congruent by which theorem.

Let us re-draw the given triangles

The diagram shows that,  \(\overline{A B} \cong \overline{P Q}\)  and  \(\overline{\mathrm{BC}} \cong \overline{Q R}\)

Also, ∠BAC ≅ ∠QPR

Therefore, the given triangles can be proven right congruent by using the Side Angle Side postulate (SAS).

The given triangles can be proven congruent by using the Side Angle Side postulate (SAS).

Solutions For Triangles Congruence By Sss And Sas Exercise 4.2 In Savvas Geometry Chapter 4 Student Edition Page 230  Exercise 7  Problem 7

Given: The side length of the first triangle is 7 ft and the side length of the second triangle is 21 ft.

To Find –  If two triangles are congruent.

Proof: According to the question,

∵  Side length of Δ ABC = 7 ft and side length of Δ PQR = 21 ft

Since, side length if both the triangles are in ratio1:3

Therefore Δ ABC ≅ ΔPQR, by using Side-Side-Side (SSS) theorem.

Triangles with side 7 ft and  21 ft are congruent by SSS theorem.

Exercise 4.2 Triangles Congruence By Side Side Side And Side Angle Side Savvas Learning Co Geometry Detailed Answers Page 230  Exercise 8  Problem 8

Given: \(\overline{J K}\) ≅ \(\overline{L M}\),\(\overline{J M}\) ≅ \(\overline{K K}\)

Proof:

 \(\overline{J K}\) ≅ \(\overline{L M}\)  (Given)

\(\mathrm{J} \overline{\mathrm{M}} \cong \overline{K K}\)  (Given)

\(\overline{K M} \cong \overline{K M}\) (Reflexive property of congruence )

Therefore, ΔJKM≅ m ΔLMK, by SSS theorem.

ΔJKM and ΔLMK are congruent using the SSS theorem.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 230  Exercise 9  Problem 9

Given:    \(\overline{I E} \cong \overline{G H}\),   \(\overline{E F}\) ≅ \(\overline{H F}\)  and F is the midpoint of GI

The figure:

To Find – Prove EFI ≅ HFG

When we compare these two triangles, we get that they have equal corresponding sides as we can see from the figure.

⇒  \(\overline{I E} \cong \overline{G H}\)

⇒  ​\(\overline{E F} \cong \overline{H F}\)

⇒  \(\overline{F I} \cong \overline{F G}\)

We know, that F is the midpoint of \(\overline{G I}\)

The midpoint divides the side into two equal parts.

Based on the SSS congruence theorem, these two triangles are congruent.

Hence, proved.

By the SSS Congruence theorem, the given triangles are congurent.

Geometry Chapter 4 Triangles Congruence By Side Side Side And Side Angle Side Savvas Learning Co Explanation Guide Page 230  Exercise 10  Problem 10

Given: \(\overline{W Z}\) ≅ \(\overline{S D}\) ≅ \(\overline{D W}\)

The figure:

To find –  Prove \(\mathrm{WZD} \cong S D Z\)

When we compare these two triangles WZD and SDZ we get that they have equal corresponding sides, as we can see from the given figure

⇒  \(\overline{W Z}\) ≅ \(\overline{S D}\) ≅ \(\overline{D W}\)

⇒  \(\overline{Z D} \cong \overline{Z D}\)

⇒  \(\overline{Z D}\)

Is the common side for both triangles based on the SSS Congruence theorem , these two traiangles are congruent.

Hence , proved

\(\overline{Z D}\) common side for both triangles. Based on the SSS congruence theorem thes two triangles are congruent.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 11  Problem 11

Given: 

The figure

To Find – What other information, if any, do you need to prove the two triangles congruent by SAS? Explain.

When we compare these two triangles RTS and WVU we get that they have equal corresponding sides, as we see that the figure.

⇒   \(T R\) ≅   \(\overline{V W}\)

⇒  \(\angle R \cong \angle W\)

⇒  \(\overline{T S} \cong \overline{V U}\)

∠R are the included angles for sides \(\overline{T R}\), \(\overline{T S}\)

∠W are the included angles for sides  \(\overline{V W}\), and  \(\overline{V U}\)

Now, we have to find the included angles for the sides.

When we compare these two triangles we get that they have equal corresponding angles, as we can see from the figure.

⇒  \(\angle T \cong \angle V\)

∠T are the included angles for sides   \(\overline{T R}\), \(\overline{T S}\)

∠V are the included angles for sides \(\overline{V W}\), and \(\overline{V U}\)

The additional information that we need to prove that these two triangles are congruent using the SAS Congruence theorem is

⇒    ∠T ≅ ∠V

Now, we will find corresponding sides for the included angles R and W.

∠R are the included angles for sides \(\overline{T R}\), \(\overline{T S}\)

∠W are the included angles for sides \(\overline{V W}\), and  \(\overline{V U}\)

The additional information that we need to prove that these two triangles are congruent using the SAS Congruence theorem is

⇒  \(R S\)   ≅  \(\overline{W U}\).

The additional information we need is   ∠T ≅ ∠V,  \(R S\) ≅  \(\overline{W U}\)

The additional information we need is   ∠T ≅ ∠V,  RS ≅  \(\overline{W U}\)

Page 231  Exercise 12  Problem 12

Given: 

The figure

To Find – Would you use SSS or SAS to prove the triangles congruent? If there is not enough information to prove the triangles congruent by SSS or SAS, write not enough information. Explain your answer.

When we compare these two triangles PQT and SQR we get that they have equal corresponding sides, as we see that the figure

⇒  \(P T \cong \overline{S R}\)

⇒  \(Q T \cong \overline{Q R}\)

⇒ ∠PQT and ∠SQR are vertical angles.

So,∠PQT≅∠SQR

∠PQT is not the included for sides \(\overline{P T}\) and \(\overline{Q T}\)

∠SQR is not the included angles for sides \(\overline{S R}\) and \(\overline{Q R}\)

There is not enough information to prove that triangles PQT and SQR are congruent using the SSS and the SAS Congruence Theorem.

There is not enough information to prove that triangles PQT and SQR are congruent using the SSS and the SAS Congruence Theorem.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 13  Problem 13

Given:

The figure

To Find –  Would you use SSS or SAS to prove the triangles congruent? If there is not enough information to prove the triangles congruent by SSS or SAS, write not enough information.

Explain your answer.  When we compare these two triangles ABC and CDA we get that they have equal corresponding sides, as we see that the figure

⇒  \(A B\)  ≅ \(\overline{D C}\)

∠BAC ≅ ∠DCA

⇒  \(A C \cong \overline{A C}\)

∠AC are the common side for both triangles.

∠BAC are the included angles for sides \(\overline{A B} \text { and } \overline{A C}\)

∠DCA are the included angles for the sides \(\overline{D C} \text { and } \overline{A C}\)

Based on SAS Congruence Theorem, these two triangles ABC and CDA are congruent.

We would use the SAS Congruence theorem.

By the SAS Congruence Theorem, these two triangles are congruent.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 14  Problem 14

Given: Each triangle should have two 5inches sides and an angle of 40degrees

To find – Which postulate, SSS or SAS, are you likely to apply to the given situation?

SAS postulate says that two triangles are congruent if two sides of one triangle are congruent with the corresponding sides of the other triangle and the angle that overlaps those two sides are congruent with the corresponding angle of the other triangle.

These two triangles have two equal corresponding sides of 5 inches and 40 degrees angle.

Based on the SAS congruence theorem, these two triangles are congruent.

Diagram:

The SAS congruence theorem, prove two triangles are congruent. The diagram

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 15  Problem 15

Given:  \(B C \cong \overline{D A}\),  \(\angle \bar{C} B \bar{D} \cong \angle A D B\)

The figure

To find – Prove BCD ≅ DAB

When we compare these two triangles BCD and DAB we get that they have equal corresponding sides, as we see that the figure:

⇒  \(B C \cong \overline{D A}\)

⇒  \(\angle C B D \cong \angle A D B\)

⇒  \(B D \cong \overline{B D}\)

∠CBD are the included angles for side \(\overline{B C} \text { and } \overline{B D}\)

∠ADB are the included angles for sides \(\overline{D A} \text { and } \overline{B D}\)

Based on the SAS Congruence Theorem, these two triangles BCD and DAB are congruent.

The SAS Congruence Theorem probe triangle BCD and DAB congruent.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 16  Problem 16

Given: The pairs A(1,4), B(5,5), C(2,2), D(−5,1), E(−1,0), F(−4,3)​

To find – Use the Distance Formula to determine whether ABC and DEF are congruent. Justify your answer.

The distance formula:

AB = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)

Now, we will calculate the side length of triangles ABC based on this formula

Now calculate for AB, BC, AC

A(1,4) , B(5,4)

⇒ AB = \(\sqrt{(1-5)^2+(4-5)^2}\)

⇒ AB = \(\sqrt{16+7}\)

⇒ AB = \(\sqrt{17}\)

B (5,5), C(2,2)

⇒ BC = \(\sqrt{(5-2)^2+(5-2)^2}\)

⇒ BC = \(\sqrt{9+9}\)

⇒ BC = \(\sqrt{18}\)

A(1,40, C(2,2)

⇒ AC = \(\sqrt{(1-2)^2+(4-2)^2}\)

⇒ AC = \(\sqrt{1+4}\)

⇒ AC = \(\sqrt{5}\)

Now, we know the length of the sides of a triangle DEF.

Now, we will calculate the length of the sides of a triangle DEF.

Calculate for DE, EF, DF.

D (-5, 1), E(-1,0)

⇒ DE = \(\sqrt{(-5-(-1))^2+(1-0)^2}\)

⇒ DE = \(\sqrt(16+1)\)

⇒ DE = \(\sqrt{17}\)

E (-1,0), F (-4,30

⇒ EF = \(\sqrt{\left(-1-(-4)^2+(0-3)^2\right.}\)

⇒ EF = \(\sqrt{9+9}\)

⇒ EF = \(\sqrt{18}\)

D (-5,1), F (-4,3)

⇒ DF = \(\sqrt{\left(-5-(-4)^2+(1-3)^2\right.}\)

⇒ DF = \(\sqrt{1+4}\)

⇒ DF = \(\sqrt{5}\)

When we compare the length of the sides of these two triangles we get that:

⇒ ​AB = DE

⇒ BC = EF

⇒ AC = DF

Based on SSS Congruence Theorem, these two triangles are congruent.

The SSS Congruence Theorem proves that the two triangles are congruent.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 17  Problem 17

Given: The A(3,8), B(8,12), C(10,5), D(3,−1), E(7,−7), F(12,−2)

To find – Use the Distance Formula to determine whether ABC and DEF are congruent. Justify your answer.

Given,A(3,8), B(8,12), C(10,5)​,D(3,−1), E(7,−7)F(12,−2)
​
Now use the distance formula:

Formula: \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)

Now substitute for AB, BC, CA, DE, EF, FD

⇒ AB = \(\sqrt{(8-3)^2+(12-8)^2}\) = \(\sqrt{41}\)

⇒ BC =  \(\sqrt{(10-8)^2+(5-12)^2}\) = \(\sqrt{53}\)

⇒ CA = \(\sqrt{(10-3)^2+(5-8)^2}\)= \(\sqrt{58}\)

⇒ DE =  \(\sqrt{(7-3)^2+\left(-7-(-1)^2\right.}\)= \(\sqrt{52}\)

⇒ EF =  \(\sqrt{(12-7)^2+\left(-2-(-7)^2\right.}\)= \(\sqrt{50}\)

⇒ FD =  \(\sqrt{(12-3)^2+\left(-2-(-1)^2\right.}\)= \(\sqrt{82}\)

We compute the lengths of each triangle’s edges, using the formula of distance.

So, Triangle ABC ≅ DEF

Since the triangles have different lengths of edges, they are not congruent.

The triangle ABC and DEF are not congruent.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 18  Problem 18

Given: The pairs A(2,9), B(2,4), C(5,4), D(1,−3), E(1,2), F(−2,2)

To find – Use the Distance Formula to determine whether ABC and DEF are congruent. Justify your answer.

The distance formula:

AB = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)

Now, we will calculate the side length of triangles ABC based on this formula:

Now calculate for AB, BC, AC

A(2,9) , B(2,4)

⇒ AB =  \(\sqrt{(2-2)^2+(9-4)^2}=\sqrt{25}\) = 5

⇒ BC =  \(\sqrt{(2-5)^2+(4-4)^2}=\sqrt{9}\)= 3

⇒ AC =  \(\sqrt{(2-5)^2+(9-4)^2}=\sqrt{34}\)

Now , we caculated for DE, EF, DF

⇒ DE =  \(\sqrt{(1-1)^2+\left(-3-2)^2\right.}=\sqrt{25}\) = 5

⇒ EF =  \(\sqrt{(1-(-2))^2+\left(-2-2)^2\right.}=\sqrt{9}\)= 3

⇒ DF= \(\sqrt{(-1)-2)^2+\left(-3-2)^2\right.}=\sqrt{34}\).

We compare  the lengths of the sides of these triangles we get that

⇒ AB= DE

⇒ BC = EF

⇒ AC= DF

Based on SSS Congruence theorem.. these two triangles are congruent.

The  SSS Congruence theorem.. these two triangles are congruent. 

Savvas Learning Co Geometry Student Edition Chapter 4 Page 231  Exercise 19  Problem 19

Given: Congruent triangles.

To find – List three real-life uses of congruent triangles.

For each real-life use, describe why you think congruence is necessary.

Two triangles are said to be congruent if they are of the same size and same shape.

Two congruent triangles have the same area and perimeter.

When we consider the application of congruent triangles in real life, the first thing I remember is toys for children can be in the shape of triangles that are congruent.

Congruent triangles are used in the construction to reinforce the structure.

Jewelry can also consist of congruent triangles.

When we consider the application of congruent triangles in real life, toys for children can be in the shape of triangles that are congruent. Congruent triangles are used in the construction to reinforce the structure. Jewelry can also consist of congruent triangles.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 20  Problem 20

Given: Use a straightedge to draw any triangle JKL

To find –   ΔMNP ≅ ΔJKL using SSS and SAS theoremThe SSS theorem states that two triangles are congruent if three sides of one are equal respectively to three sides of the other The SAS theorem states that two triangles are equal if two sides and the angle between those two sides are equal.

To prove ΔMNP ≅ ΔJKL using SSS theorem

We’ll start by drawing a triangle JKL, then a line adjacent to it with the point M marked on it.

The length of the line JL will then be measured with the opening of the compass and transferred from point M to the line.

The point P is obtained by intersecting the arc with the line.

Then we’ll use a compass to transfer the distance JK from point M to point P, and then we’ll use the compass to transfer the distance LK from point P to point M.

The point N is formed by the intersection of these two arcs.

Finally, we shall connect these points to form the triangle MNP.

We have

To prove using SAS theorem.

We’ll start by drawing a triangle JKL , then a line adjacent to it with the point M marked on it.

Then, using the opening of the compass, we’ll measure the length of the line JK and transfer it from point M to the line.

The point N is found at the intersection of the arc and the line.

Then, from the point M , we’ll transfer the arc with the angle at the vertex J and its length, and then, from the point M , we’ll transfer the distance JL via the compass’s opening, and finally, from the point M , we’ll transmit the distance JL

The point P is formed by the intersection of these two arcs.

Finally, we’ll connect these points to form the MNP triangle.

We have

By SSS theorem ΔMNP ≅ ΔJKL

By SAS theorem ΔMNP≅ΔJKL

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 21  Problem 21

Given: Two triangle ΔYHD AND ΔKPT

​YH = KP

HD = PT
​
To find – Triangles are congruent to each other.

Equate the corresponding sides of the one triangle to the second triangle.

In the above figure

As Given

YH = KP

HD = PT
​
As We are missing the third information for congruency criteria to be applied

We need a side to apply SSS or an angle to apply SAS.

NO, We can not.

We need third information to apply congruency criteria which can be a side or an angle to the corresponding sides

Page 232  Exercise 22  Problem 22

Given: Two triangles ΔGJK and ΔGMK

GJ = GM

∠IGK = ∠MGK

To find: Triangles are congruent to each other.

Equate the corresponding sides of the one triangle to the second triangle.

In the above figure

As Given In ΔGJK and ΔGMK

​GJ = GM

∠IGK = ∠MGK

GK = GK
​
Using SAS congruency criterion ΔGJK ≅ ΔGMK

Using SAS congruency criterion ΔGJK ≅ ΔGMK

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 23  Problem 23

Given: Two triangles where AE and BD bisect each other.

To find –  Triangles are congruent to each otherEquate the corresponding sides of the one triangle to the second triangle.

In ΔACB and ΔECD

It is given that AE and BD biscet each other so

​AC = CE

BC = CD

Also, ∠ACB = ECD (Vertically opposite angle)

By using SAS congruency criterion

ΔACB ≅ ΔECD

Using SAS congruency criterion,ΔACB ≅ ΔECD

Page 232  Exercise 24  Problem 24

Given: Two triangles where AB⊥CM, AB⊥DB, CM = DB

To find –  Triangles are congruent to each other.

Equate the corresponding sides of the one triangle to the second triangle.

Mid-pointIn ΔAMC and ΔMBD

It is given that CM = AB
​
As ∠AMC = ∠MBD

= 90

Also as m is the mid -point AM = MB

By using the SAS congruency criterion ΔAMC ≅ ΔMBD

Using SAS congruency criterion, ΔAMC ≅ ΔMBD

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 25  Problem 25

Given: polygon ABCD whose four sides are congruent.

To find –  EFGH and ABCD and EFGH are congruent or not Using the definition of congruent polygons, ABCD and EFGH are congruent of the corresponding angles are also congruent.

A quadrilateral is not rigid because, for the same side lengths, the interior angles can be different

Using the definition of congruent polygons, ABCD and EFGH are congruent of the corresponding angles are also congruent A quadrilateral is not rigid because, for the same side lengths, the interior angles can be different

Page 232  Exercise 26  Problem 26

Given: Two triangles where HK = LG, HF = LJ, FG = JK

To find –  Triangles are congruent to each otherEquate the corresponding sides of the one triangle to the second triangle.

In ΔFGH and ΔJKL

It is given that

⇒ HK = LG,HF = LJ,FG = JK

⇒ As HK = LG

Adding both sides GK we get

⇒ HK + GK = LG + GK

⇒ HG = KL

By using SSS congruency criteria We get  ΔFGH ≅ ΔJKL

By using SSS congruency Criterion ΔFGH ≅ ΔJKL

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 27  Problem 27

Given: Two triangles where ∠N = ∠L,MN = OL,NO = LM

To find – Prove MN ∥ OL

Equate the corresponding sides of the one triangle to the second triangle.

In ΔONM and ΔMLO

It is given that

⇒ ∠N = ∠L,MN = OL,NO = LM

By using the SAS congruency criterion we can say that, ΔONM ≅ ΔMLO

Also Using CPCT in the congruent triangles

∠NOM = ∠LMO which is a pair of alternate interior angles.

So MN ∥ OL

MN ∥ OL as ∠NOM = ∠LMO is pair of alternate interior angles.

Page 232  Exercise 28  Problem 28

Given: Two Triangles ΔVWY and ΔVWZ where ∠VWY = ∠VWZ

To find  –  Conditions for the triangles to be congruent by SAS.

As given ∠VWY = ∠VWZ also, we have a common side VW = VW

So we need another side to apply the SAS congruency criterion which can be YW = ZW and no other option can satisfy it

The correct answer  is  which is YW = ZW to satisfy SAS congruency criterion

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232 Exercise 29 Problem 29

Given: Two angles of a triangle equal to 43 and 38.

To find – Measure of third angle

Let The third angle be ∠A.

Then we know that sum of all interior angles is 180

So, according to question

​⇒ 43 + 38 + ∠A = 180

⇒ 81 + ∠A = 180

⇒ ∠A = 180 − 81

⇒ ∠A = 99

​The right anser is ∠A = 99

The correct answer is which is equal to 99

Page 232  Exercise 30  Problem 30

Given: ABCD ≅ EFGH

To find –  Side 9s de that corresponds to BC

So, given here ABCD ≅ EFGH both are congruent. And the rule for writing congruency is to write the corresponding same angles and sides of shapes.

For eg; if in ΔABC and ΔDEF, ∠A = ∠E, AB = DE, and BC = DF. Then congruency will be written as ΔABC ≅ ΔEDF.

So, according to the given congruency here, we can figure out a few points.

Angles: ∠A = ∠E,∠B = ∠F,∠C = ∠G and ∠D = ∠H

Sides: AB = EF,BC = FG,CD = GH and DA = HE

Side corresponding to BC in congruency relation ABCD ≅ EFGH is FG.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 232  Exercise 3 1 Problem 31

Given: If x  = 3, then x² = 9

To find: Converse of statement and determine if true or false.

Here hypothesis is x = 3 and conclusion is  x²  = 9.

So, the converse of the given statement will be “If x² = 9, then x = 3″.

Simplifying the conclusion in the given statement by taking root both sides, we get x = ±3.

So, there are two solutions for the conclusion.

The statement is true, Converse is false.

Savvas Learning Co Geometry Student Edition Chapter 4 Congruent Triangles Exercise 4.1 Congruent Figures

Savvas Learning Co Geometry Student Edition Chapter 4 Exercise 4.1 Congruent Figures Solutions Page 221  Exercise 1  Problem 1

Given that,ΔBAT ≅ ΔFOR

⇒ \(\overline{T A} \cong \overline{R O}\)

⇒ ∠R ≅ ∠T

\(\overline{T A} \cong \overline{R O}\) ∠R ≅ ∠T.

Exercise 4.1 Congruent Figures Savvas Geometry Answers Page 221  Exercise 2   Problem 2

1. The relationship between ∠M and ∠T is that ∠M ≅ ∠T ,  2. If m ∠ A = 52 and m ∠ P = 36 , then m ∠ T = 92.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Congruent Figures Solutions Chapter 4 Exercise 4.1 Savvas Geometry Page 221  Exercise 3  Problem 3

In our daily life, the apartments have multiple buildings that are congruent, Vehicles of the same company and model are congruent, and classrooms of a school are congruent.

In our daily life, the apartments have multiple buildings that are congruent, Vehicles of the same company and model are congruent, and classrooms of a school are congruent.

Congruent Figures Solutions Chapter 4 Exercise 4.1 Savvas Geometry Page 222  Exercise 4  Problem 4

Given: ΔABC ≅ ΔABD

To find –  List the congruent corresponding parts.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

It is given that ΔABC and ΔABD ,so they will have equal corresponding sides and equal corresponding angles.

So, we conclude that vertex A,B,C corresponding to vertex A,B,D respectively.

Thus, we get

⇒   \(\overline{A B} \cong \overline{A B}\)

⇒   \(\overline{A C} \cong \overline{A D}\)

⇒   \(\overline{C B} \cong \overline{D B}\) And

⇒ ∠CAB ≅ ∠DAB

⇒ ∠ABC ≅ ∠ABD

⇒ ∠ACB ≅ ∠ADB

Thus, for ΔABC ≅ ΔABD we get

⇒   \(\overline{A B} \cong \overline{A B}\)

⇒   \(\overline{A C} \cong \overline{A D}\)

⇒  \(\overline{C B} \cong \overline{D B}\) 

And

⇒ ∠CAB ≅ ∠DAB

⇒ ∠ABC ≅ ∠ABD

⇒ ∠ACB ≅ ∠ADB

Chapter 4 Exercise 4.1 Congruent Figures Savvas Learning Co Geometry Explanation Page 222  Exercise 5  Problem 5

Given: ΔEFG ≅ ΔHIJ

To find – List the congruent corresponding parts.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

It is given that ΔEFG ≅ ΔHIJ, so they will have equal corresponding sides and equal corresponding angles.

So, we conclude that vertex E,F, G is congruent to vertex H,I ,J respectively

Therefore

⇒  \(\overline{E F} \cong \overline{H I}\)

⇒  \(\overline{F G} \cong \overline{I J}\)

⇒  \(\overline{G E} \cong \overline{J H}\) and

⇒ ∠GEF ≅ ∠JHI

⇒ ∠EFG ≅ ∠HIJ

⇒ ∠FGE ≅ ∠IJH

Thus, for ΔEFG ≅ ΔHIJ we get

⇒  \(\overline{E F} \cong \overline{H I}\)

⇒  \(\overline{F G} \cong \overline{I J}\)

⇒  \(\overline{G E} \cong \overline{J H}\)

And

⇒ ∠GEF ≅ ∠JHI

⇒ ∠EFG ≅ ∠HIJ

⇒ ∠FGE ≅ ∠IJH

Solutions For Congruent Figures Exercise 4.1 In Savvas Geometry Chapter 4 Student Edition Page 222   Exercise 6   Problem 6

Given: ΔLCM ≅ ΔBJK

To find –  Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

It is given that ΔLCM ≅ BJK so they will have equal corresponding sides and equal corresponding angles.

SO, we conclude that vertex L,C,M is congruent to vertex B ,J, K respectively.

Thus, \(\overline{K J} \cong \overline{M C}\)

Thus, \(\overline{K J} \cong \overline{M C}\)

Exercise 4.1 Congruent Figures Savvas Learning Co Geometry Detailed Answers Page 222   Exercise 7  Problem 7

Given: \(\overline{J B} \cong \overline{M C}\) ≅ ?

To find – Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

The complete congruence statement will be \(\overline{J B} \cong \overline{M L}\).

Geometry Chapter 4 Congruent Figures Savvas Learning Co Explanation Guide Page 222   Exercise 8   Problem 8

Given: ∠ L ≅?

To find –  Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ∠L ≅ ∠B

The congruence statement will be ∠L ≅ ∠B.

Geometry Chapter 4 Congruent Figures Savvas Learning Co Explanation Guide Page 222   Exercise 9  Problem 9

Given: ∠ K ≅?

To find – Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ∠K ≅ ∠C

The congruence statement will be ∠K ≅ ∠C.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222   Exercise 10  Problem 10

Given: ∠ M ≅?

To find –  Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ∠M ≅ ∠J

The congruence statement will be ∠M ≅ ∠J.

Page 222  Exercise 11  Problem 11

 Given: ΔCML ≅ ?

To find – Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ΔCML ≅ ΔKJB

The congruence statement will be ΔCML ≅ ΔKJB.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222   Exercise 12   Problem 12

Given:  ΔKBJ ≅ ?

To find – Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ΔKBJ ≅ ΔCLM

The congruence statement will be ΔKBJ ≅ ΔCLM

Page 222  Exercise 13  Problem 13

Given:  ΔMLC ≅ ?

To find –  Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ΔMLC ≅ ΔJBK

The congruence statement will be ΔMLC ≅ ΔJBK.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222   Exercise 14   Problem 14

Given: ΔJKB ≅ ?

To find – Complete the congruence statements.

Two triangles are congruent if they have equal corresponding sides and equal corresponding angles.

⇒ ΔJKB ≅ ΔMCL

The congruence statement will be ΔJKB ≅ ΔMCL

Page 222   Exercise 15  Problem 15

Given: POLY ≅ SIDE

To find – List the four pairs of congruent sides.

Two polygons are congruent if they have equal corresponding sides and equal corresponding angles.

⇒   \(\overline{P O} \cong \overline{S I}\)

⇒   \(\overline{O L} \cong \overline{I D}\)

⇒   \(\overline{L Y} \cong \overline{D E}\)

⇒   \(\overline{Y P} \cong \overline{E S}\)

The four pairs of congruent sides are:

⇒  \(\overline{P O} \cong \overline{S I}\)

⇒  \(\overline{O L} \cong \overline{I D}\)

⇒  \(\overline{L Y} \cong \overline{D E}\)

⇒  \(\overline{Y P} \cong \overline{E S}\)

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222  Exercise 16  Problem 16

Given: POLY ≅ SIDE

To find – List the four pairs of congruent angles.

Two polygons are congruent if they have equal corresponding sides and equal corresponding angles.

∠P ≅ ∠S

∠O ≅ ∠I

∠L ≅ ∠D

∠Y ≅ ∠E

The four pairs of congruent angles are:

∠P ≅ ∠S

∠O ≅ ∠I

∠L ≅ ∠D

∠Y ≅ ∠E

Page 222   Exercise 17   Problem 17

Given that: At an archeological site, the remains of two ancient step pyramids are congruent.

If ABCD ≅ EFGH

To find –  Find GH

These two shapes ABCD and FEGH are congruent to each other it means shape and size of ABCD and are same FEGH .

The size of GH = 45ft because GH ≅ CD

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222  Exercise 18  Problem 18

Given that: At an archeological site, the remains of two ancient step pyramids are congruent.

If ABCD ≅ EFGH

To find – Find EF

These two shapes ABCD and EFGH are congruent to each other it means shape and size of ABCD and EFGH are same.

The size of EF = 45 ft  because EF ≅ AB.

The size of EF = 45ft because EF ≅ AB for the problem

Page 222  Exercise 19  Problem 19

Given that: At an archeological site, the remains of two ancient step pyramids are congruent.

If ABCD ≅ EFGH

To find – Find BC

These two shapes ABCD and EFGH are congruent to each other it means shape and size of ABCD and EFGH are same.

The size of BC = 280ft because BC ≅ FG

The size of BC = 280ft because BC ≅ FG for the problem

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222  Exercise 20  Problem 20

Given that: At an archeological site, the remains of two ancient step pyramids are congruent.

If ABCD ≅ EFGH

To find –  Find m∠DCB

These two shapes ABCD and EFGH are congruent to each other it means shape and size of ABCD and EFGH are same.

The angle m ∠DCB = 128 because m ∠DCB ≅ m∠FGH

The angle m∠ DCB = 128 because m∠ DCB ≅ m∠FGH for the problem.

Page 222   Exercise 21  Problem 21

Given that: At an archeological site, the remains of two ancient step pyramids are congruent.

If ABCD ≅ EFGH

To find  – Find m ∠ EFG

These two shapes ABCD and EFGH are congruent to each other it means they have same shape and size.

The angle m ∠EFG = 128° because m ∠EFG ≅ m∠ABC

The angle m∠EFG = 128° because m ∠EFG ≅ m ∠ABC for the problem

Savvas Learning Co Geometry Student Edition Chapter 4 Page 222  Exercise 22  Problem 22

Given that: Δ SPQ and Δ TUV

To find – Can you conclude that the triangles are congruent.

The triangle SPQ,TUV are not congruent because shape and size of the triangles are not same.

The sides of triangle SPQ and TUV are not same.

The triangle SPQ, TUV are not congruent because shape and size of triangles are not same for the problem

Page 223  Exercise 23  Problem 23

Given that: Δ DEF ≅ Δ LMN

To find – Which of the following must be a correct congruence statement?

1. DE = LN

2. ∠N ≅ ∠F

3. FE ≅ NL

4. ∠M ≅ ∠F
​
Let the vertex E of triangle DEF corresponds to vertex N of triangle LMN, the vertex L of triangle LMN corresponds to vertex D of triangle DEF and the vertex M of triangle LMN corresponds to vertex F of triangle DEF.

If the triangle congruent than ED ≅ LN, EF ≅ NM, DF ≅ LM,∠M ≅ ∠F.

The option (4)  is correct rest are wrong.

The option (4) ∠M ≅ ∠F  is correct when ΔDEF ≅ ΔLMN for the problem

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise  24   Problem 24

Given that: Randall says he can use the information in the figure to prove  ΔBCD ≅ Δ DAB

To find –  Is he correct? Explain.

If the triangle ΔBCD ≅ ΔDAB  then BC ≅ BA, AD ≅ CD, ∠CDB ≅ ∠ADB, ∠CBD ≅ ABD

But Randall says  CD ≅ BA, AD ≅ BC, ∠CDB  ≅  ∠ABD, ∠CBD ≅ ∠ADB  which is incorrect.

The correct explanation of congruence of triangles ΔBCD ≅ ΔDAB is BC ≅ BA,AD ≅  CD, ∠CDB  ≅  ∠ADB,∠CBD ≅ ABD for Randall problem

And the diagram shown as

Page 223   Exercise 25   Problem 25

Given that: ΔABC ≅ ΔDEF ,m∠A = x + 10. m∠D = 2x

To find – Find the measures of the given angles or the lengths of the given sides.

If the two triangles are congruent than sides and angles are congruent.

The given m∠A = x + 10. m∠D = 2x and the triangles ΔABC ≅ ΔDEF than

m∠A ≅ m∠D it means

⇒ ​x + 10 = 2x

⇒ 10 = 2x – x

⇒ 10 = x

⇒ x = 10 and now

⇒ m∠A = x + 10

⇒ m∠A = 20°

⇒ m∠D = 2(10)

⇒ m∠D = 20°

The length x = 10 and angles m∠A = 20 ° and m∠D = 20° for the problem of congruency when ΔABC ≅ ΔDEF and m ∠A = x + 10,m∠D = 2x.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 26  Problem 26

Given that:  ΔABC ≅ ΔDEF , m∠B = 3y, m∠E = 6y − 12

To find – Find the measures of the given angles or the lengths of the given sides.

If the two triangles are congruent than sides and angles are congruent

The given m∠B = 3y, m∠E = 6y − 12 and the triangles ABC ≅ DEF than m ∠B ≅ m∠E  it means

⇒ ​3y = 6y − 12

⇒ 3y = 12

⇒ y = 4 and now

⇒ ​m∠B = 3(4)
​
⇒ m ∠B = 12°

⇒ m∠E = 6(4)−12

⇒ m ∠E = 12°

The length y = 4 and angles m ∠B = 12° and m ∠E = 12° the problem of congruency when ΔABC ≅ ΔDEF and m∠B = 3y,m∠E = 6y − 12

Page 223  Exercise 27  Problem 27

Given: Two triangle s ΔABC ≅ ΔDEF and  BC = 3z + 2, EF = z + 6

To find – Measures of the lengths of the given sides.

Since, the triangles are congruent, vertex B of ΔABC corresponds to E of ΔDEFTriangles corresponding sides are equal, i.e. BC = EF

Equating corresponding sides of two triangle.

⇒ BC = EF

⇒ 3z + 2 = z + 6

Bringing one side to one side of equation and constant terms to other side of triangle.

⇒ 3z − z = 6 − 2

⇒ 2z = 4

⇒ z = \(\frac{4}{2}\)

⇒ z = 2

Substituting value of z in lengths BC and EF

⇒ BC = 3 × 2 + 2

⇒ BC = 8 units

⇒ EF= 2 + 6

⇒ EF = 8 units

Measures of the lengths of the given sides are- BC = 8 units EF = 8 units

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 28  Problem 28

Given : Two triangles ΔABC ≅ ΔDEF and AC = 7a + 5, DF = 5a + 9

To find –  Measures of the lengths of the given sides.

Since, the triangles are congruent, vertex A of ΔABC corresponds to D of ΔDEF Triangles corresponding sides are equal, i.e. AC = DF

Equating corresponding sides of two triangle.

⇒ AC = DF

⇒ 7a + 5 = 5a + 9

Bringing one side to one side of equation and constant terms to other side of triangle.

⇒ 7a − 5a = 9 − 5

⇒ 2a = 4

⇒ a = \(\frac{4}{2}\)

⇒ a = 2

Substituting value of a in lengths AC and DF

⇒ AC = 7 × 2 + 5

⇒ AC = 19 units

⇒ DF = 5 × 2 + 9

⇒ DF = 19 units

Measures of the lengths of the given sides are- AC = 19 units, DF = 19 units

Page 223  Exercise 29  Problem 29

Given: Two triangles Δ ABC ≅ Δ DBE.

To find – Meaning for two triangles to be congruent.

Two triangles are said to be congruent, if there corresponding sides and angles are equal.

Meaning for two triangles to be congruent is – “Two triangles are said to be congruent, if there corresponding sides and angles are equal”.

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 29  Problem 30

Given: Two triangles ΔABC ≅ ΔDBE.

To find –  Which angle measures are already known.The angles known in ΔABC are m∠ CAB = (x + 5)°and m∠ABC = 51°

The angles known in ΔDBE are m∠

BED = 81°

Since, The given triangles are congruent, vertex C of Δ ABC corresponds to vertex E of

Δ DBE m∠DEB  =  m∠ACB = 81°

The angles known in ΔABC are m∠ CAB = (x + 5)°  and m ∠ACB = 51° and m∠ ACB = 81° , The angles known in ΔDBE are m ∠BED = 81°

Page 223  Exercise 29  Problem 31

Given: Two triangles ΔABC ≅ ΔDBE.

To find  – The value of x and the missing angle measure in a triangle.

Using Angle sum property of triangle,  Sum of angles of triangle is 180°

m∠BAC + m∠BCA+m∠

ABC = 180°

⇒ x + 5 + 81° + 51°

= 180°

⇒ x +137° =180°

⇒ x = 43°

Since, The given triangles are congruent, missing angles can be found out by equating the corresponding angles.

Vertex A of ΔABC corresponds to Vertex D of ΔDBE

∴ m ∠CAB = m ∠EDB = 43 + 5 = 48°

Vertex B is common in both the triangles.

⇒ m ∠ABC = m ∠DBE = 51°

Value of x is 43°

The given triangles ΔABC ≅ ΔDBE are congruent, missing angles can be found out by equating the corresponding angles.m ∠EDB = 48° m ∠DBE = 51°  m∠ACB = 81°

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 30  Problem 32

Given: Two triangles ΔABC ≅ ΔKLM.

To Find – The values of the variables.

Since, ΔABC≅ ΔKLM, vertex C of ΔABC corresponds to vertex M of ΔKLM.

Therefore, m ∠ACB = m ∠KML = 3x°

Using angle sum property in ΔABC

⇒ m∠ ACB + m ∠CBA + m ∠BAC = 1800

⇒ 3x + 90° + 45° = 180°

⇒ 3x = 180° − 135°

⇒ 3x = 45°

⇒ x = 15°

The value of the variable x is 15°

Page 223  Exercise 31  Problem 33

Given: Two triangles ΔACD ≅ ΔACB.

To Find –  The values of the variables.

Since A is the common vertex of two triangles.

The corresponding angles m ∠BAC = m ∠DAC 6x = 30° which on solving gives

⇒ 6x = 30°

⇒ \(\frac{300}{6}\)

⇒ x = 5°

The value of the variable x is 5°

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 32  Problem 34

Given: Figure of Two triangles ΔMLJ and ΔZRN.

To find – To Complete in two different ways that ΔJLM ≅ Δ NRZ.

Three corresponding angles of ΔJLM and ΔNRZ are equal.

From the figures of two triangles JLM and NRZ

⇒ ∠L = ∠R

⇒ ∠M = ∠Z

⇒ ∠J = ∠N

Three corresponding sides are also equal.

⇒ ML = ZR

⇒ LJ = RN

⇒ JM = NZ

Therefore, ΔJLM ≅ ΔNRZ, since corresponding angles and sides of two triangles are equal.

From the figures of two triangles, ΔJLM and ΔNRZ

Corresponding angles are equal

⇒ ∠L = ∠R

⇒ ∠M = ∠Z

⇒ ∠J = ∠N

Corresponding sides are equal

⇒ ML = ZR

⇒ LJ = RN

⇒ JM = NZ

Therefore, Δ JLM ≅ ΔNRZ

Savvas Learning Co Geometry Student Edition Chapter 4 Page 223  Exercise 33  Problem 35

Given: Terms “congruent sides and angles” and “triangles”.

To find – To write a congruence statement for two triangles and to list the congruent sides and angles.

Let two triangles are congruent ΔABC ≅ ΔXYZ

List of congruent sides of two triangles are

⇒ AB = XY

⇒ BC = YZ

⇒ AC = XZ

List of congruent angles of two triangles are

⇒ ∠A = ∠X

⇒ ∠B = ∠Y

⇒ ∠C = ∠Z

Congruence statement for two triangles is ΔABC ≅ ΔXYZ

List of congruent sides of two triangles are

⇒ AB = XY

⇒ BC = YZ

⇒ AC = XZ

List of congruent angles of two triangles are

⇒ ∠A = ∠X

⇒ ∠B = ∠Y

⇒ ∠C = ∠Z

Page 223   Exercise 34  Problem 36

Given: Two triangles ΔABD and ΔCDB, and AB⊥AD, BC⊥CD, AB ≅ CD, AD≅CB, AB ∥ CD

To find –  To prove ΔABD ≅ ΔCDB.

The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

In the two triangles given

AD side of ΔABD= BC side of ΔCDB

AB side of ΔABD = CD side of ΔCDB

∠A = ∠C (Since, both are right angles)

Using SAS rule, ΔABD and ΔCDB are congruent.

AD side of ΔABD= BC side of ΔCDB

AB side of ΔABD = CD side of ΔCDB

⇒ ∠A = ∠C

Using SAS rule, ΔABD ≅ ΔCDB

Savvas Learning Co Geometry Student Edition Chapter 4 Page 224  Exercise 35  Problem 37

Given: Two triangle ΔPRS and ΔQTS, PR ∥ TQ , PR ≅ TQ, PS ≅ QS

To find –  To prove ΔPRS ≅ ΔQTS

The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

In the two triangles given

PR ≅ TQPS ≅QS

Since, PR ∥ TQ and PQ is the transversal.

∠P = ∠Q, because they are alternate interior angles.

Using SAS rule, ΔPRS ≅ ΔQTS

In the two triangles ΔPRS and ΔQTS

⇒ PR ≅ TQ

⇒ PS ≅ QS

⇒ ∠P = ∠Q

Using SAS rule, ΔPRS ≅ ΔQTS

Page 224  Exercise 36  Problem 38

Given  The vertices of ΔGHJ are G(−2,−1), H(−2,3),J(1,3).

To find –  KL, LM, and KM.

Using distance formula to calculate distance between two points.

Distance = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)

Where (x₁,y₁) and  (x₂,y₂) are two points.

Since, ΔKLM ≅ ΔGHJ.

Side KL corresponds to side GH of ΔGHJ

KL = \(\sqrt{(-2+2)^2+(3+1)^2}\)

= \(\sqrt{16}\)

Side KL = 4 units.

Side LM corresponds to side HJ of ΔGHJ

LM = \(\sqrt{(1+2)^2+(3-3)^2}\)

= \(\sqrt{9}\)

Side LM = 3 units.

Side KM corresponds to side GJ of ΔGHJ

KM = \(\sqrt{(1+2)^2+(3+1)^2}\)

= \(\sqrt{9+16}\)

KM = 5 units.

Side KL = 4 units , LM = 3 units , KM = 5 units

Savvas Learning Co Geometry Student Edition Chapter 4 Page 224  Exercise 37  Problem 39

Given: The coordinates of L and M(3,−3),(6,−3)

To find – How many pairs of coordinates are possible for K, find one pair.

Given, L(3,−3) M(6,−3)
​
Now, let K as(x,y)

Centroid:

\(\frac{3-6+x}{3}\) = \(\frac{-3+(-3)+y}{3}\)

Now equate the x-coordinates and y-coordinates we get:

⇒ \(\frac{-3+x}{3}=\frac{1}{1}\)

= −3 + 1x = 3

x = 6

⇒ \(\frac{6+y}{3}=\frac{2}{1}\)

= 6 + 1y = 6

⇒ 1y = 6 − 6

⇒ y = 1
​
The one pair for, k is(6,1)

The pair for, k is(6,1)

Page 224  Exercise 38  Problem 40

Given: ΔHLN ≅ ΔGST and m∠H = 66, m∠S = 42

To find – To find the value of m∠T

Congruent triangles are triangles that have the same size and shape.

We conclude m∠H = m∠G = 66

And we know m∠S = 42

The sum of interior angles of a triangle is180°

In ΔGST

⇒ m ∠G + m ∠S + m ∠T = 180°

⇒ 66 + 42 + m ∠T = 180°

⇒ m∠T = 180° − 108°

⇒ m ∠T = 72°

The value of m ∠T = 72° 

Savvas Learning Co Geometry Student Edition Chapter 3 Parallel And Perpendicular Lines Exercise 3.3 Proving Lines Parallel

Savvas Learning Co Geometry Student Edition Chapter 3 Exercise 3.3 Proving Lines Parallel Solutions Page 160  Exercise 1  Problem 1

Given:

To find –  State the theorem to prove a ∥ b.

We can prove a ∥ b, by the converse of the alternate interior angles theorem.

The theorem states, if two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel.

If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel.

Exercise 3.3 Proving Lines Parallel Savvas Geometry Answers Page 160  Exercise 2  Problem 2

Given:

To find –  y Use the interior angle theorem.

The interior angle theorem state that, if two parallel lines and a transversal form the same side interior angles, then those angles are supplementary.

In the figure

By exterior angle theorem

65° + y = 180°

y = 180° − 65°

y = 115°

The required angle y°  is 115°.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Proving Lines Parallel Solutions Chapter 3 Exercise 3.3 Savvas Geometry Page 160   Exercise 3  Problem 3

Given: Alternate interior angle theorem and its converse.

To find –  Use of Alternate interior angle theorem and its converse.

When the two lines are given parallel and we have to prove alternate angles are congruent, then the Alternate interior angle theorem is used.

When Alternate angles are given congruent and we have to prove two lines are parallel, then the converse of alternate interior angle theorem is used.

When the two lines are given parallel and we have to prove alternate angles are congruent, then the Alternate interior angle theorem is used and when Alternate interior angles are given congruent and we have to prove two lines are parallel, then the converse of alternate interior angle theorem is used.

Proving Lines Parallel Solutions Chapter 3 Exercise 3.3 Savvas Geometry Page 160  Exercise 4  Problem 4

Given:  Flow proofs and two-column proofs.

To find – Similarities and differences of flow proofs and two-column proofs.

Flow proofs are diagrammatic representations of the proofs represented by arrows to reach the conclusion.

Two-column proofs are represented by two columns, one which consists of conclusions and the other consisting of reasons.

The similarity between Flow proofs and Two-column proofs is it is represented by arrows and reaches conclusion step-by-step.

The difference is Flow proof does not include reasons but column proof does include reasons.

The similarity between Flow proofs and Two-column proofs is it is represented by arrows and reach a conclusion step-by-step and the difference is Flow proof does not include reasons but column proof does include reasons.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 160  Exercise 5  Problem 5

Given:

To find – Parallel lines.

Use the converse of the corresponding angle theorem and find parallel lines.

In the figure

∠E ≅ ∠G

By converse of corresponding angle theorem

Lines BE ∥ CG are parallel and EG is transversal.

The required parallel lines are BE ∥ CG.

Chapter 3 Exercise 3.3 Proving Lines Parallel Savvas Learning Co Geometry Explanation Page 160  Exercise 6  Problem 6

Solutions For Proving Lines Parallel Exercise 3.3 In Savvas Geometry Chapter 3 Student Edition Page 160  Exercise 7  Problem 7

Given:

To find – Parallel lines. Use the converse of the corresponding angle theorem and find parallel lines.

In the figure

By converse of corresponding angle theorem

Lines CA ∥ HR are parallel and MR is transversal.

The required parallel lines are CA ∥ HR.

Exercise 3.3 Proving Lines Parallel Savvas Learning Co Geometry Detailed Answers Page 160  Exercise 8  Problem 8

Given:

To find –  Parallel lines.Use the converse of the corresponding angle theorem and find parallel lines.

In the figure

∠ JKR ≅ ∠LMT

By converse of corresponding angle theorem

Lines KR ∥ MT are parallel and JM is transversal.

The required parallel lines are KR ∥ MT.

Exercise 3.3 Proving Lines Parallel Savvas Learning Co Geometry Detailed Answers Page 161  Exercise 9  Problem 9

Given:

To find –  x.

Use the Alternate angle theorem.

In the figure

Lines l ∥ m.

By alternate interior angle theorem, 95° = (2x−5)°

⇒ 95 − 5 = 2x

⇒ 90 = 2x

⇒ \(\frac{90}{2}\)= x

⇒ 45 = x  or  x = 45

The required value of x is 45°.

Geometry Chapter 3 Proving Lines Parallel Savvas Learning Co Explanation Guide Page 161  Exercise 10  Problem 10

Given:

To find – x.

Use the vertical opposite and Alternate angle theorem.

In the figure

Lines l ∥ m

By vertically opposite angle

∠1 = 3x − 33  ……………………. (1)

By corresponding angle theorem

∠1 = 2x + 26 …………………………. (2)

From  equation (1) and (2)

⇒ 3x − 33 = 2x + 26

⇒ 3x − 2x = 26 + 33

⇒ x = 59

The required value of x is 59°.

Geometry Chapter 3 Proving Lines Parallel Savvas Learning Co Explanation Guide Page 161  Exercise 11  Problem 11

Given:

To find – x.

Use the Alternate exterior angle theorem.

In the figure

Lines l ∥ m.

By alternate exterior angle theorem

⇒ 105 = 3x − 18

⇒ 105 + 18 = 3x

⇒ 123 = 3x

⇒ \(\frac{123}{3}\) = x

⇒ 41 = x  or  x = 41

The required value of x is 41°.

Page 161  Exercise 12  Problem 12

Given: ∠2 is supplementary to ∠3.

To find – Parallel lines.

Use the converse of the interior angle theorem and find parallel lines.

In figure

∠2 is supplementary to ∠3

By converse of interior angle theorem

Line a ∥ b are parallel.

The required parallel lines are a ∥ b.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 13  Problem 13

Given: The given figure is

To find – We need to find if ∠1 ≅ ∠3 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠1 ≅ ∠3

If two lines and transversal form corresponding angles that are congruent, then the lines are parallel,∴a and b mare parallel, the line l is the transversal.

∠1 ≅ ∠3 If two lines and transversal form corresponding angles that are congruent, then the lines are parallel So and are parallel, the line is the transversal.

Page 161  Exercise 14  Problem 14

Given: The given figure is

To find – We need to find if ∠6 is supplementary to∠7 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠6 is supplementary to ∠7 if the Consecutive interior angles on the same side of the transversal are supplementary then the two lines are parallel.

∴ a ∥ b

∠6 is supplementary to ∠7 if the Consecutive interior angles on the same side of the transversal are supplementary then the two lines are parallel. so a ∥ b

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 15  Problem 15

Given: The given figure is

To find –  We need to find if ∠9 ≅ ∠12 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The information given ∠in 9 ≅ ∠12 does not prove that any lines are parallel.

∠9 and ∠12 are vertical angles.

We do not need parallel lines to have vertical angles.

The given information ∠9 ≅ ∠12 does not prove any lines are parallel or not. ∠9 and ∠12 are vertical angles. We do not need parallel lines to have vertical angles.

Page 161  Exercise 16  Problem 16

Given: The given figure is

To find – We need to find if m∠7 = 65, m∠9 = 115 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠7 = 65,m∠9 = 115 ∠7 and ∠9 are supplementary, yet they are not any special angle pair of lines.

So no lines are proved to be parallel.

m∠7 = 65,m∠9 = 115 ∠7 and ∠9 are supplementary, yet they are not any special angle pair of lines. So no lines are proved to be parallel.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 17  Problem 17

Given: The given figure is

To find – We need to find if ∠2≅∠10 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠2 ≅ ∠10 ∠2 and ∠10 are corresponding angles on the side of the transversal a so the line l is parallel to the line m

∠2 ≅ ∠10, ∠2, and ∠10 are corresponding angles on the side of the transversal so the line is parallel to the line

Page 161 Exercise 18 Problem 18

Given: The given figure is

To find –  We need to find if ∠1 ≅ ∠8 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠1 ≅ ∠8 ∠1 and ∠8 are alternate exterior angles on the side of the transversal.

Since they are congruent then a ∥ b by the converse of the alternate exterior angles theorem.

∠1 ≅ ∠8 by the converse of the alternate exterior angles theorem a ∥ b

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 19  Problem 19

Given: The given figure is

To find – We need to find if ∠8≅∠6 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠8 ≅ ∠6 ∠8 and ∠6 are the corresponding angles on the side of the transversall.

Since they are congruent, then a ∥ b by the converse of the corresponding angles postulate.

∠8 ≅ ∠6, ∠8 and ∠6 are the corresponding angles on the side of the transversal Since they are congruent, then by the converse of the corresponding angles postulate

Page 161  Exercise  20  Problem 20

Given: The given figure is

To find –  We need to find if ∠11 ≅ ∠7 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information ∠11 ≅ ∠7.

The congruency shown does not follow any theorem or postulate and, therefore, does not prove any lines parallel.

∠11 ≅ ∠7 The congruency shown does not follow any theorem or postulate and, therefore, does not prove any lines parallel.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 21  Problem 21

Given: The given figure is

To find  – We need to find if ∠5 ≅ ∠10 then proof which lines, if any, are parallel.

When any two parallel lines are intersected by another line called a transversal.

Any two lines are said to be parallel if the Corresponding angles so formed are equal.

Any two lines are said to be parallel if the Alternate interior angles so formed are equal.

Any two lines are said to be parallel if the Alternate exterior angles so formed are equal.

Any two lines are said to be parallel if the Consecutive interior angles on the same side of the transversal are supplementary.

The given information is ∠5 ≅ ∠10 ∠5 and ∠10 are alternate interior angles on the side of the transversal.

Since they are congruent,then l ∥ m by the converse of the alternate interior angles theorem.

∠5 and ∠10 are alternate interior angles on the side of the transversal Since they are congruent, then l ∥ m by the converse of the alternate interior angles theorem.

Page 161  Exercise 22  Problem 22

Given: The given figure is

To find – We need to find the value of x for which l ∥ m.

Parallel lines are straight lines that never meet each other no matter how long we extend them.

We know that l ∥ m from the given figure we know that

let ∠1 = 19x

⇒ ∠2 = 17x

⇒ ∠3 = 27x

Now name the angle

∠4 = ∠1 = 19x

The angle 1 and 4 are vertical.

∴ ∠4 + ∠2 = 180°

⇒ 19x + 17x = 180

⇒ 36x = 180

⇒ x = \(\frac{180}{36}\)

⇒x = 5

The value of x is 5

Savvas Learning Co Geometry Student Edition Chapter 3 Page 161  Exercise 23  Problem 23

Given: The given figure is

To find – We need to find the value of x for which l ∥ m

Parallel lines are straight lines that never meet each other no matter how long we extend them.

The vertical angles with the (5x + 40)° angle and the 2x° angle are the same side interior angles that must be supplementary so that l ∥ m by the converse of the same side interior angles theorem

So we find that

⇒ (5x + 40) + 2x = 180

⇒  7x + 40 = 180

⇒  7x = 140

⇒  x = 20

The value of x is 20

Page 161  Exercise 24  Problem 24

Given: The given figure is

To find – We need to prove the Converse of the Same-Side Interior Angles Theorem, the given value is m∠3 + m∠6 = 180, prove l ∥ m.

Parallel lines are straight lines that never meet each other no matter how long we extend them.

The given condition is m∠3 + m∠6 = 180 ∠3 and ∠6 are supplementary ∠6 and ∠7 are supplementary ∠3 ≅ ∠7

∴  From this condition we know that l ∥ m

∠3 and ∠6 are supplementary ∠6 and ∠7 are supplementary ∠3 ≅ ∠7

∴ From these all conditions we know that l ∥ m

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 25  Problem 25

Given: The given figure and condition is  m∠1 = 80−x, m ∠ 2 = 90 − 2x

To find –  We need to find m∠1 and m∠2

Parallel lines are straight lines that never meet each other no matter how long we extend them.

The given condition is  m∠1 = 80 − x, m∠2 = 90 − 2x

∴ 80 − x = 90 − 2x

⇒  −x = 10 − 2x

⇒  x = 10

∴ The angles are

m∠1 = 80 − 10

⇒  m∠1 = 70  and m ∠ 2 = 90 − 2(10)

⇒  m∠2 = 70

The values are x = 10, m ∠ 1 = 70,m ∠ 2 = 70

Page 162  Exercise 26   Problem 26

Given:  m∠1 = 60 − 2x, m∠2 = 70 − 4x

To Find –  Determine the value of x.

Given

⇒ ​m∠1 =  60 − 2x

⇒ m∠2 =  70 − 4x
​
Since corresponding angles are congruent, we have:

⇒ ​70 − 40 =  60 − 2x

⇒ 70 − 60 =  −2x + 4x

⇒ 10 = 2x

⇒ \(\frac{10}{5}\) = x

⇒ x = 5 or x = 5

Now substitute the value of x to find m:

⇒ ​m∠1 =  60 − 2(5)

⇒ m∠1 =  60 − 10

⇒ m∠1 =  50

⇒ m∠2 =  70 − 4(5)

⇒ m∠2 =  70 − 20

⇒ m∠2 =  50

The answer is x = 5,m ∠1 = 50 ,m ∠2 = 50

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162   Exercise 27   Problem 27

Given: m∠1 = 40 − 4x , m∠2 = 50 − 8x

To find –  Determine value of x and m∠1 and m∠2

Since r ∥ s therefore due to corresponding angles m∠1 = m∠2

Thus, 40 − 4x = 50 − 8x

⇒ 8x − 4x = 50 − 40

⇒ 4x = 10

⇒ x = \(\frac{10}{4}\)

Substituting value of x in m∠1 and m∠2

⇒ ​m∠1 = 40 − 4x = 40 − 4 × \(\frac{10}{4}\)

=  40 − 10

=  30
​
​m∠2 = 50 − 4x

=  50 − 8 × \(\frac{10}{4}\)

=  50 − 20

=  30
​
Thus, m∠1 = m∠2 = 30

Thus, value of x = \(\frac{10}{4}\) and m∠1 = m∠2 = 30

Page 162  Exercise 28  Problem 28

Given: m∠1 = 20 − 8x ,m∠2 = 30 − 16x

To find – Determine value of x and m∠1 and m∠2

It is given that r∥s

Thus ,due to corresponding angles m∠1 = m∠2

i.e. 20 − 8x  =  30 − 16x

⇒ 16x − 8x  =  30 − 20

⇒ 8x = 10

x = \(\frac{10}{8}\)

Substituting value of x in m∠1 and m∠2

​m∠1 = 20 − 8x

=  20 − 8 × \(\frac{10}{8}\)

=  20 − 10

=  10
​
​m∠1 = 30 − 16x

=  30 − 16 × \(\frac{10}{8}\)

=  30 − 20

=  10

Thus ,m∠1 = m∠2 = 10

Thus, value of x = \(\frac{10}{8}\) and m∠1 = m∠2 = 10.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 29  Problem 29

Given: m∠1 ≅ m∠2

To find –  Determine which lines will be parallel and why?

Considering the diagram

m∠1 = m∠2 ______ (Given)

m∠1 ≅ m∠9 ______ (We add the condition )

According to converse of corresponding angle postulate, the line j and line k are parallel i.e. j ∥ k

m∠1 ≅ m∠5______(We add the condition )

According to converse of corresponding angle postulate, the line l and linen are parallel i.e. l ∥ n

Thus, line j ∥ k and line l ∥ n

Page 162  Exercise 30  Problem 30

Given:  m∠8 = 110,m∠9 = 70

To find –  Determine which lines will be parallel and why?

Considering the diagram

m∠8 = 110, m∠9 = 70 _____ (Given)

m∠9 ≅ m∠3 ________ (We add the condition)

According to converse of alternate angle theorem line  k ∥ j

m∠3 + m∠9 = 110 + 70 = 180 so, angle 3 and 8 are supplementary according to converse of same side angle postulate line l and n are parallel i.e. l ∥ n

Thus, parallel lines are j ∥ k and l ∥ n

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 31  Problem 31

Given: ∠5 ≅ ∠11

To find – Determine which lines will be parallel

Considering the diagram

∠5 ≅ ∠11 ______ (Given)

∠11 ≅ ∠3 ______ (We add condition)

∠11 ≅ ∠3

⇒  j ∥ k

According to converse of corresponding angle postulate line j and k are parallel.

Also according to converse of alternate angle postulate line l and n are parallel

i.e.∠5 ≅ ∠3 ⇒ l ∥ n

Thus, line j ∥ k nand line l ∥ n

Page 162  Exercise 32  Problem 32

Given: ∠11 and ∠12 are supplementary.

To find –  Determine which lines will be parallel and why?

Considering the diagram

∠11, ∠12 are supplementary _______ (Given)

∠12 ≅ ∠4 ________ (Add condition )

According to converse of corresponding angle postulate line j and k are parallel.

i.e. ∠12 ≅ ∠4

⇒ j ∥ k

Thus, line j and k are parallel

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 33  Problem 33

Given: ∠1 ≅ ∠7

To find –  What postulate or theorem can be used to show that l ∥ n.

Since ∠1 and ∠7 are alternate exterior angle on the side of transversal k and since they are congruent then, l ∥ n by converse of exterior angle theorem.

Line l ∥ n by converse of exterior angle theorem.

Page 162  Exercise 34  Problem 34

Given:  l ∥ n, ∠12 ≅ ∠8

To find – Draw a flow proof to prove that j∥k

Considering the diagram

Filling the missing value,we cannot reason ∠12 ≅ ∠4 as the line involved are j and k which are yet to proven parallel.

The flow proof is shown as:

The flow proof that j ∥ k is as follows:

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 35  Problem 35

Given: m ∠P = 72, m∠L = 108, m∠A = 72, m∠N = 108

To find – Sides of the parallelogram that are parallel.

In parallelogram PLAN

m∠P = 72

m∠L = 108

m∠A = 72

m∠N = 108

Here, we observe that m∠P + m∠L = 72 + 108 = 108

I.e. angle P and angle L are supplementary because they are same side interior angle for the transversal PL and the line PN and LA

Hence by converse of same side interior angle theorem side PN ∥ LA

Also,m∠L + m∠A = 72 + 108 = 180

i.e. angle L and angle A are supplementary because they are same side interior angle for the transversal AL and line PL and AN

Hence by converse of same side interior angle theorem side PL ∥ AN

Thus, we got that in parallelogram PLAN PN ∥ LA and PL ∥ AN

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 36  Problem 36

Given: m∠P = 59, m∠L = 37,m∠A = 143, m∠N = 121

To find – Sides of the parallelogram that are parallel.

In parallelogram PLAN

m∠P = 59

m∠L = 37

m∠A = 143

m∠N = 121

Here,m∠P + m∠N = 59 + 121 = 180

Thus, angle P and angle N are supplementary because they are same side interior angle for the transversal PN and lines AN and PL.

Hence by converse of same side interior angle theorem side AL ∥ PL

Also, m∠P + m∠L = 59 + 37 = 96

Thus angle P and angle L are not supplementary i.e. PN ∦ LA

Therefore, AL ∥  PL

Thus, in parallelogram PLAN side AL ∥ PL

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 37  Problem 37

Given: m∠P = 56, m∠L = 124, m∠A = 124, m∠N = 56

To find – Sides of the parallelogram that are parallel.

Parallelogram PLAN can be sketched as:

∠P and ∠L are same side interior angle on transversal  PL and are supplementary i.e. (124 + 56 = 180)

So, by converse of same side interior angle theorem \(\overline{L A} \| \overline{P N}\) thus \(\overline{L A} \| \overline{P N}\)

Thus, in parallelogram PLAN \(\overline{L A} \| \overline{P N}\) thus \(\overline{L A} \| \overline{P N}\) 

Savvas Learning Co Geometry Student Edition Chapter 3 Page 162  Exercise 38  Problem 38

Given: Transversal t, line m ∥ n angle bisector a and b

To find – a ∥ b

Diagram :

Transversal t ,a ∥ b ∠1 + ∠2 ≅ ∠3 + ∠4 _______ (Corresponding angle postulate)

m∠1 + m∠2 = m∠3 + m∠4 ______ (Definition of congruence angle)

Angle bisector a and b ___________ (Given)

Therefore,m∠1 = m∠2, m∠3 = m∠4

m∠1 + m∠1 = m∠3 + m∠3 __________ (Substitution property)

2m∠1 = 2m∠3

m∠1 = m∠3

Therefore, a ∥ b ________ (Converse of corresponding angle postulate

Thus, a ∥ b

Savvas Learning Co Geometry Student Edition Chapter 3 Page 163  Exercise 39  Problem 39

The  corresponding angles between the parallel lines have equal measure and the measure of

m∠1 = 180 − 136 = 44 degree

m∠1 = 44 degree

The correct answer is 44 degrees

Savvas Learning Co Geometry Student Edition Chapter 3 Parallel and Perpendicular Lines Exercise 3.2 Properties Of Parallel Lines

Savvas Learning Co Geometry Student Edition Chapter 3 Exercise 3.2 Properties Of Parallel Lines Solutions Page 152  Exercise 1  Problem 1

Given that:  The given two parallel lines with angles

To find – Identify two pairs of supplementary angles. 

f,g are parallel lines with 8 angles. 

These are vertical, alternate, corresponding angles, supplementary angles.

The given two parallel lines with angles

Have two lines pairs of supplementary angles are (2,40)(1,3).

The two pairs of supplementary angles are (2,40), (1,3) for the parallel lines with angles

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Exercise 3.2 Properties Of Parallel Lines Savvas Geometry Answers Page 152  Exercise 2  Problem 2

The value of ∠m8 = 70 because of alternate angles. for the given parallel lines and ∠m1 = 70.

Properties Of Parallel Lines Solutions Chapter 3 Exercise 3.2 Savvas Geometry Page 152  Exercise 3  Problem 3

Given: Alternate Interior Angles Theorem and the Alternate Exterior Angles Theorem

To Find – The similarity and difference between the given theorems.

The alternate interior angles are nonadjacent interior angles that lie on the opposite side of the transversal.

The Alternate Interior Angles Theorem states that “If a transversal intersects two parallel lines, then the alternate interior angles are congruent.

“The alternate exterior angles are nonadjacent exterior angles that lie on the opposite side of the transversal.

The Alternate Exterior Angles Theorem states that “If a transversal intersects two parallel lines, then the alternate exterior angles are congruent.

“The two theorems are alike in the fact that both of them deal with a transversal intersecting two parallel lines and that the alternate angles are congruent.

The two theorems differ from each other by the fact that the alternate interior angles are nonadjacent angles that lie within the parallel lines and the alternate exterior angles are nonadjacent angles that lie outside the parallel lines.

The similarity between the Alternate Interior Angles Theorem and the Alternate Exterior Angles Theorem is that two parallel lines are cut by a transversal and the angles are congruent. The difference between the Alternate Interior Angles Theorem and the Alternate Exterior Angles Theorem is that the interior angles are between the parallel line whereas the exterior angles are not between the parallel lines.

Properties Of Parallel Lines Solutions Chapter 3 Exercise 3.2 Savvas Geometry Page 152  Exercise 4  Problem 4

Given that: The problem is

To find – You proved that ∠1 and ∠8, in the diagram below, are supplementary.

What is a good name for this pair of angles? Explain.

The lines a,b are parallel with each other and one line intersect both these parallel lines.

The angles ∠1,∠8 are supplementary angles because these are alternate same side exterior angles.

And the sum of two alternate same side angles is 180.

If the sum of two angles is 180 called supplementary angles.

The ∠1,∠8 are supplementary angles because these are alternate same side exterior angles. the good name of the pair of angle is alternate same side exterior angles. for the problem

Chapter 3 Exercise 3.2 Properties Of Parallel Lines Savvas Learning Co Geometry Explanation Page 153  Exercise 5  Problem 5

Given that: The given problem is 

To find – Identify all the numbered angles that are congruent to the given angle.

Justify your answers the given one angle is 65 degree and the congruent angles are ∠1,∠7,∠4.

The angle ∠1 is vertical opposite angle of 65 , the angle ∠7 is alternate angle with 65 and the angle ∠4 is correspondence angle with angle 65.

The congruent angles are ∠1,∠7,∠4  for the given problem is 

Chapter 3 Exercise 3.2 Properties Of Parallel Lines Savvas Learning Co Geometry Explanation Page 153  Exercise 6  Problem 6

Given that: The given problem is 

To find –  Identify all the numbered angles that are congruent to the given angle.

Justify your answers the given one angle is 51 degree and the angles ∠7,∠5,∠4
are congruent angles.

The ∠7 angle is vertical opposite angle of 51, the ∠4 angle is alternate angle with 51 and the angle ∠5 is correspondence angle with angle 51.

The congruent angles are ∠7,∠5,∠4 for the problem is

Solutions For Properties Of Parallel Lines Exercise 3.2 In Savvas Geometry Chapter 3 Student Edition Page 153  Exercise 7  Problem 7

Given that: The given problem is

To find – Identify all the numbered angles that are congruent to the given angle.

Justify your answers.The given one angle is 120 degree and the angles ∠3,∠1 are congruent angles.

The angle ∠3 is alternate angle with angle 120 and the angle ∠1 is correspondence angle with 120.

The congruent angles are ∠1,∠3 for the problem is

Solutions For Properties Of Parallel Lines Exercise 3.2 In Savvas Geometry Chapter 3 Student Edition Page 153  Exercise 8  Problem 8

Given: The angle 2

To Find – Write a two-column proof.

Given  a ∣∣ b

Two-column proof:

The two-column theorem:

Exercise 3.2 Properties Of Parallel Lines Savvas Learning Co Geometry Detailed Answers Page 153  Exercise 9  Problem 9

Given:  Figure in which two lines a∥b and q is transversal.

To find: m∠1 and m∠2 m∠1 is equal to 120°, because they are corresponding angles.

m∠1 and m∠2 are interior angles on same-side.

Therefore, they are supplementary angles.

⇒ m∠1 + m∠2 = 180°

⇒ 120° + m∠2 = 180°

⇒ m∠2 = 180° −120°

⇒ m∠2 = 60°

According to the given figure, m∠1 = 120° and m∠2 = 60°

Exercise 3.2 Properties Of Parallel Lines Savvas Learning Co Geometry Detailed Answers Page 153  Exercise 10  Problem 10

Given: Figure in which two lines AB∥CD and BC and AD are transversal.

To find –  m∠1 and m∠2 = 70° since they are alternate interior angles.

m∠1 and 80° are interior angles on same side.

Therefore, they are supplementary angles.

⇒ m∠1 + 80° = 180°

⇒ m∠2 = 180°− 80°

⇒ m∠2 = 100°

According to the given figure, m∠1 = 100° and m∠2 = 70°

Geometry Chapter 3 Properties Of Parallel Lines Savvas Learning Co Explanation Guide Page 153  Exercise 11  Problem 11

Given: A figure in which two lines are parallel.

To find – value of x and to find the measure of each labeled angle.

5x° and 4x°  are angles on same side.

Therefore, they are supplementary.

Value of x is 20° Measure of labeled angle 5x° = 100°,  4x° = 80°

Geometry Chapter 3 Properties Of Parallel Lines Savvas Learning Co Explanation Guide Page 154  Exercise 12  Problem 12

Given: Etchings on floors and walls in Rome suggest that the game required a grid of two intersecting pairs of parallel lines, similar to the modern game tick-tack-toe.

The measure of one of the angles formed by the intersecting lines is 90 degrees

To Find –  Find the measure of each of the other 15 angles. Justify your answers.

The diagram: Draw a diagram with two intersecting pairs of parallel lines where one angle is ninety degrees.

The 90 degrees angle forms a linear pair with ∠1 and ∠ 4.

Since linear pairs are supplementary, then the sum of their measure is 180degree.

It means both the angles are equal to 90 degrees.

The 90 degrees angle and ∠5 are also vertical angles so they are congruent by the vertical angles theorem.

Hence, ∠5 is equal to 90 degrees

Each of these four angles we found is corresponding angles to angles  2,3,6,7,8,9,12,13 in no particular order so by the corresponding angles theorem, these eight angles also measure 90 degrees.

Using the angles set 2,3,6,7 or8,9,12,1 they are corresponding angles with angles10,11,14,15 so they are also congruent.

So, these four angles are measures 90 degrees

Hence, all 15 angles are measures 90 degrees.

All 15 angles are measure 90 degrees, the diagram

Savvas Learning Co Geometry Student Edition Chapter Page 154  Exercise 13  Problem 13

Given: Camper pulls the rope taut between the two parallel trees.

The figure:

To Find – What is m∠1.

Given

Camper pulls the rope taut between the two parallel trees.

The given figure forms a triangle.

Given one angle: 63

We can see from the figure that the triangle is a right angle triangle so

The second angle is: 90 degree

Now, we know the sum of all three angles is 180 so

The third angle is:

⇒ 90 + 63 = 153

⇒ 180 − 153 = 27
​
Now the 1 point is perpendicular to 27degrees

And the sum of the perpendicular is180

So,∠1 180 − 27 = 153

The answer is 153 degrees

The answer is 153 degrees

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 13  Problem 14

Given: The diagram

To Find –  Are∠1 the given angle alternate interior angles, same-side interior angles, or corresponding angles?

Given , ∠1

If two parallel lines are transected by a third line, the angles which are inside the parallel lines and on alternate sides of the third line are called alternate interior angles.

 The angles which are inside the parallel lines and on the same side of the third line are called opposite interior angles.

In the given figure there is no alternate interior angle so, it is not an alternate angle.

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

The same-side interior angles sum up to 180 degrees.

Let check:

⇒ 153 + 63 = 216

So, it is not the same side interior angle.

Corresponding angles are the angles that are formed when two parallel lines are intersected by the transversal. These are formed in the matching corners or corresponding corners with the transversal.

Yes, it is a corresponding angle.

The given figure is a corresponding angle.

The answer is the corresponding angle.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 14  Problem 15

Given:  l ∥ m

To find – To prove ∠3 and ∠6 are supplementary.

Line l and m are Parallel ∠2 and ∠3 from a linear pair and linear pair are supplementary.

⇒ m∠2 + m∠3 = 180

⇒ ∠2 ≅ ∠6 are congurent

From congruence m∠2 = m∠6

By using substutition property

m∠6 + m∠3 = 180

∠3 and ∠6 are supplementary.

∠3 and ∠6 are supplementary proved.

Page 155  Exercise 15  Problem 16

Given:  a ∥ b, ∠1 ≅ ∠4

To find – To prove ∠2 ≅ ∠3

In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.

Line l and m are Parallel.

∠1 and∠2, ∠3 and ∠4 both are supplementary.

​⇒ m∠1 + m∠2 = 180

⇒ m∠3 + m∠4 = 180
​
Both have same value

⇒ m∠1 + m∠2 = m∠3 + m∠4

⇒ ∠1 ≅ ∠4 (given)

Which means

⇒ m∠1 = m∠4

By using substutition property

m∠4 + m∠2 = m∠3 + m∠4

Subtract

⇒ m∠2 = m∠3

By the defination of congruence

⇒ ∠2 ≅ ∠3

∠2 and ∠6 are congurent peoved.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 16  Problem 17

Given: The Diagram

To find – To error analysis the diagram contain.

(60−2x)° and (2x−60)° are corresponding angles formed by tranversal,so they must be congruent.

We write it as

​⇒ 60 − 2x = 2x − 60

⇒ −4x =−120

⇒ x = 30°
​
Subsittuting  x = 30°

We get

​⇒ 60 − 2(30) = 0°

⇒ 2(30) − 60 = 0°
​
We get 0° in both the expression which contradict the diagram and the tranversal must have form angle with parallel lines.

We get 0° in both the expression which contradict the diagram and the tranversal must have form angle with parallel lines.

Page 155  Exercise 17  Problem 18

Given: ∠1 and ∠2 are same-side interior angles formed by two parallel lines and ∠1 = 115.

To find –  To find∠2

The theorem for the “same side interior angle theorem” states.

If a transversal intersects two parallel lines, each pair of same-side interior angles are supplementary.

Because ∠1 and∠2 are same-side interior angles formed by two parallel lines.

By using same-side interior angle theorem

⇒ m∠1 + m∠2 = 180

⇒ m∠2 = 180 − m∠1

⇒ ​m∠2 = 180 − 115

⇒ m∠2 = 65
​
The valuse of m∠2 is 65°

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 18  Problem 19

Given: A rectangular swimmimg pool area 1500 ft 2 surround by the walkway.

To find –  Length of fencing to surround the walkway.

Length of fencing to surround the walkway is equal to the perimeterof the larger rectangle.

Let l and w are the length and width of the smaller rectangle respectively

Area

lw = \(\frac{1500}{l}\)

w = \(\frac{1500}{50}\)

Let L and W are the length and width of the larger Rectangle respectively

⇒ ​L = 50 + 2(4)

⇒ L = 59ft
​
​⇒ W = 20 + 3(4)

⇒ W = 38ft
​
Perimeter of the bigger reactangle

⇒ ​P =  2(L) + 2(W)

⇒ P =  2(59) + 2(38)

⇒ P =  192 ft

​Length of fencing to surround the walkway is 192 ft

Page 155  Exercise 19  Problem 20

Given: The sum of measure angle and two times of measure angle is complement.

To find – To find measured angle.

An angle and its complement add up to

This means we have  x + 2x = 90

⇒ 3x = 90

⇒ x = 30

The measure of the angle is 30°

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 20  Problem 21

Given: ∠1 and ∠2 are vertical angles,  m∠1 = 4x and m∠2 = 56

To find –  To find value of x

Because ∠1 and ∠2 are vertical angles

By vertical angle theorem

⇒ m∠1 = m∠2

Subsitutin property

⇒ 4x = 56

⇒ x = \(\frac{56}{4}\)

The value of x is 46°

Page 155 Exercise 21  Problem 22

Given: The statement “Skew lines are coplanar.”

To find: The given statement is always, sometimes, or never true.

Skew lines are two or more lines that do not intersect, are not parallel, and are not coplanar.

The only way that two non parallel can never be intersect is if they are not coplaner.

The given statement is never ture.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 155  Exercise 22  Problem 23

Given: The statement “Skew lines intersect.”

To find –  The given statementis always, sometimes, or never true.

Skew lines are two lines that do not intersect and are not parallel.

By the definaton we can conclude that skew line never intersect.

This statement is never be ture.

Page 155  Exercise 23  Problem 24

Given: The statement “Parallel planes intersect.”

To find –  The given statement is always, sometimes, or never true.

Parallel lines are coplanar straight lines that do not intersect at any point.

By the defination we can conclude that Parallel planes intersect.

This statement is never be true.

Savvas Learning Co Geometry Student Edition Chapter Page 155  Exercise 24  Problem 25

Given: The statement ” If a triangle is a right triangle, then it has a 90°”

To find – To Write the converse of the statement.

Converse of the satement is if triangle has 90° then it is a right angle triangle.

The converse of the given statement is if triangle has 90° then it is a right angle triangle.

Savvas Learning Co Geometry Student Edition Chapter 3 Parallel and Perpendicular Lines Exercise 3.1 Lines And Angles

Savvas Learning Co Geometry Student Edition Chapter 3 Exercise 3.1 Lines And Angles Solutions Page 143  Exercise 1  Problem 1

Given: A figure.

To Find –  Parallel segments in the given figure.A segment is a part of the line.

We have a figure

In the given figure, we can see that the parallel segments are

The parallel segments in the given figure are

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Exercise 3.1 Lines And Angles Savvas Geometry Answers Page 143  Exercise 2  Problem 2

Given:  A figure.

To Find – Skew segments in the given figure.

A segment is a part of the line.

We have a figure

In the given figure, we can see that skew segments are

Some possible pairs of skew segments in the given figure are

Exercise 3.1 Lines And Angles Savvas Geometry Answers Page 143  Exercise 3  Problem 3

Some possible parallel planes in the given figure are:

Plane EFGH ∥ Plane ABCD

Plane AEFB ∥ Plane DHGC

Plane AEHD ∥ Plane BFGC

Lines And Angles Solutions Chapter 3 Exercise 3.1 Savvas Geometry Page 143  Exercise 4  Problem 4

Given: A figure.

To Find – Alternate interior angles in the given figure.

An angle is a combination of two rays ( half-lines ) with a common endpoint.

We have a figure

In the given figure, we can see that the possible pairs of alternate interior angles are

⇒ ∠2 and ∠3

⇒  ∠8 and ∠6

The possible pairs of alternate interior angles in the given figure are: ∠2 and ∠3, ∠8 and ∠6

Lines And Angles Solutions Chapter 3 Exercise 3.1 Savvas Geometry Page 143  Exercise 5  Problem 5

Given: A figure.

To Find –  Same-side interior angles in the given figure.

An angle is a combination of two rays ( half-lines ) with a common endpoint.

We have a figure

In the given figure, we can see that the pairs of same-side interior angles are

⇒ ∠3 and ∠8

⇒ ∠6 and ∠2

The possible pairs of same-side interior angles are: ∠3 and ∠8, ∠6 and ∠2

Chapter 3 Exercise 3.1 Lines And Angles Savvas Learning Co Geometry Explanation Page 143  Exercise 6  Problem 6

Given: A figure.

To Find – Corresponding angles in the given figure.

An angle is a combination of two rays ( half-lines ) with a common endpoint.

We have a figure

In the given figure, we can see the possible pairs of the corresponding angles are

⇒ ∠1 and ∠3

⇒ ∠7 and ∠6

⇒ ∠8 and ∠5

⇒ ∠2 and ∠4

The possible pairs of the corresponding angles are: ∠1 and ∠3,∠7 and ∠6,∠8 and ∠5,∠2 and ∠4

Solutions For Lines And Angles Exercise 3.1 In Savvas Geometry Chapter 3 Student Edition Page 143  Exercise 7  Problem 7

Given: A figure.

To Find  – Alternate exterior angles in the given figure.

An angle is a combination of two rays ( half-lines ) with a common endpoint.

We have a figure

In the given figure, we can see that possible pairs of alternate exterior angles are

⇒ ∠1 and ∠4

⇒ ∠7 and ∠5

The possible pairs of the alternate exterior angles in the given figure are:∠1 and ∠4, ∠7 and ∠5

Solutions for Lines and Angles Exercise 3.1 in Savvas Geometry Chapter 3 Student Edition Page 143  Exercise 8  Problem 8

Parallel lines are the lines that do not intersect, and if we do not include the property of coplanarity, we can find the lines in different planes , and will be called skew lines.

Skew lines are the lines that do not intersect but are not in the same plane, thus parallel lines are coplanar and which do not meet.

Coplanar is included in the definition of parallel planes to differentiate from the definition of skew lines.

Exercise 3.1 Lines And Angles Savvas Learning Co Geometry Detailed Answers Page 143  Exercise 9  Problem 9

Alternate interior angles are formed by a transversal intersecting two lines.

The angles are located inside the two lines but on the opposite sides of the transversal.

Alternate interior angles are located inside the two parallel lines on the opposite sides of the transversal.

Exercise 3.1 Lines And Angles Savvas Learning Co Geometry Detailed Answers Page 143  Exercise 10  Problem 10

Given: A figure is given

To Find  – Who is correct between Juan and Carly?

As the question says lines appearing to be parallel are parallel.

In the figure, it can be seen clearly that AB ∥ HG

Since Carly is saying AB ∥ HG, that is correct.

But Juan is saying that AB and HG are skewed, so he is wrong.

Carly is correct because he is saying AB∥HG

Geometry Chapter 3 Lines And Angles Savvas Learning Co Explanation Guide Page 144  Exercise 11  Problem 11

Given: A figure is given

To Find – All lines that are parallel to AB

In the figure, a line that is parallel to AB is FG

Line parallel to AB is FG

Geometry Chapter 3 Lines And Angles Savvas Learning Co Explanation Guide Page 144  Exercise 12  Problem 12

Given: A figure is given

To Find  – All lines that are parallel to DH

In the figure, lines that are parallel to DH are GB, FA, JE, and CL

Lines parallel to DH are GB, FA, JE,  and CL

Savvas Learning Co Geometry Student Edition Chapter 3 Page 144  Exercise 13  Problem 13

Given: A figure is given

To Find –  All lines that are parallel to EJ

In the figure, lines that are parallel to EJ are FA, GB, DH, and CL.

Lines parallel to EJ FA, GB, DH, and CL

Page 144  Exercise 14  Problem 14

Given: A figure is given

To Find –  All lines that are parallel to the plane JF AE

Lines parallel to JF AE are GB, DH, and CL

Savvas Learning Co Geometry Student Edition Chapter 3 Page 144  Exercise 15  Problem 15

Given: A figure is given

To Find –  A plane parallel to LH

A plane parallel to LH is JFGDC

Page 144  Exercise 16  Problem 16

Given: A figure is given

To Find  – Alternate interior angles.

Alternate interior angles in the figure are 2 & 3

Savvas Learning Co Geometry Student Edition Chapter 3 Page 144  Exercise 17  Problem 17

Given: A figure is given

To Find –  Whether the angles labeled in the same color alternate interior angles, same-side interior angles, corresponding angles, or alternate exterior angles?

Angles 3 & 4 and 5 & 6 are corresponding angles and angles  1 & 2 are same side interior angles.

Page 144  Exercise 18  Problem 18

Given: A figure is given

To Find – Whether the angles labeled in the same color alternate interior angles, same-side interior angles, corresponding angles, or alternate exterior angles?

Only angles 5 & 6 are alternate interior angles.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 144  Exercise 19  Problem 19

Given: A figure is given

To Find  – Whether ∠1 & ∠2 are alternate interior angles, same-side interior angles, corresponding angles, or alternate exterior angles?

∠1 & ∠2 are corresponding angles.

Page 144  Exercise 20  Problem 20

Let, the lines p,q be cut by a transversal t.

Clearly,∠1 & ∠8 and ∠5 & ∠4  forms the pair of alternate exterior angles.

Two pairs of alternate exterior angles do two lines and a transversal form.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 144  Exercise 21  Problem 21

Given: \(\stackrel{\leftrightarrow}{E D} \| \overleftarrow{H} \hat{G}\)

To find –  The statement as true or false.

The lines are False they are making skew lines.

According to the figure given, we can say that \(\overleftrightarrow{E D}\) ∦ \(\overleftarrow{H} \hat{G}\)  the lines and planes that appear to be parallel are not parallel they are skew.

Page 145  Exercise 22  Problem 22

Given: Plane  AED∥ Plane FGH

To find – The statement as true or false.

The plane AED∥ plane FGH is true.

According to the figure given, we can say that plane AED∥ plane FGH. The lines and planes that appear to be parallel are parallel.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 145  Exercise 23  Problem 23

Given: Plane ABH ∥ Plane CDF

To find –  The statement as true or false.

The lines are False they intersect above \(\overrightarrow{C G}\)

According to the figure given, we can say that plane ABH ∥ plane CDF  the lines and planes that appear to be parallel are not parallel they intersect above  \(\overrightarrow{C G}\)

Page 145  Exercise 24  Problem 24

Given: \(\overrightarrow{A B}\) and \(\overrightarrow{H G}\) are skew line

To find –  The statement as true or false.

The lines are skew lines.

According to the figure given, we can say that \(\overrightarrow{A B}\) and \(\overrightarrow{H G}\) the lines and planes appear to be a skew line.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 145  Exercise 25  Problem 25

Given: \(\overrightarrow{A E}\) and \(\overrightarrow{B C}\) are skew line

To find – The statement as true or false.

The lines are not skew lines because they intersect at point A.

According to the figure given, we can say that \(\overrightarrow{A E}\) and \(\overrightarrow{B C}\) the lines and planes apperear to be is not a skew lines because they intersect at point A.

Page 145  Exercise 26  Problem 26

Given: A rectangular rug covers the floor in a living room.

One of the walls in the same living room is painted blue.

To find –  Are the rug and the blue wall parallel No, the rug and the blue wall are not parallel because they intersect.

The opposite wall can be parallel to the blue wall.

A rectangular rug covers the floor in a living room. One of the walls in the same living room is painted blue is not parallel because they intersect.

Savvas Learning Co Geometry Student Edition Chapter 3 Page 145  Exercise 27  Problem 27

Given: Two planes that do not intersect are parallel.

To find –  Determine each statement is always, sometimes, or never true.

Two planes that do not intersect are always parallel as a plane is a flat, two-dimensional surface that extends infinitely far.

A plane is the two-dimensional analog of a point, a line, and three-dimensional space.

The two planes that do not intersect are always parallel.

Page 145  Exercise 28  Problem 28

Given: A statement – Two lines that lie in parallel planes are parallel.

To find – Each statement is always, sometimes, or never true.

In order to be parallel, the two lines must be co-planer.

And as only some lines in two parallel planes are co-planer, the statement is sometimes true.

The given statement is sometimes true.

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.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

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.

Savvas Learning Co Geometry Student Edition Chapter 3 Parallel and Perpendicular Lines Exercise

Savvas Learning Co Geometry Student Edition Chapter 3 Parallel And Perpendicular Lines Exercise Answers Page 137  Exercise 1  Problem 1

Given: The following figure is provided.

To Find – All the linear pairs in the figure.

We observe the figure and pick out the adjacent angles that together form a straight angle.

The first pair of adjacent angles are ∠1 and ∠5.

The second pair of adjacent angles are ∠2 and ∠5.

The linear pairs in the figure are ∠1,∠5, and ∠2,∠5.

Page 137  Exercise 2  Problem 2

Given: The following figure is provided.

To Find – All the pairs of vertical angles.

We carefully observe the figure and pick out the angles that are opposite to each other in intersecting lines.

There is only one pair of vertical angles, which is ∠1 and ∠2.

The vertical angles in the figure are ∠1 and ∠2.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Chapter 3 Parallel And Perpendicular Lines Solutions Savvas Geometry Page 137  Exercise 3  Problem 3

Parallel And Perpendicular Lines Exercises Savvas Geometry Student Edition Solutions Page 137  Exercise 4  Problem 4

Given: The equation 3x + 11 = 7x − 5

To Find –  The solution of the given equation.

Using the properties of equality, the given equation can be solved.

Considering the given equation

3x + 11 = 7x − 5

Using the subtraction property of equality

3x + 11 − 11 = 7x − 5 − 11

⇒  3x = 7x − 16

Using the subtraction property of equality

3x − 7x = 7x − 16 − 7x

⇒ −4x = −16

Using the division property of equality

\(\frac{−4x}{−4}\) =\(\frac{−16}{−4}\)

⇒  x = 4

The solution of the equation 3x + 11 = 7x − 5 is x = 4.

Parallel And Perpendicular Lines Exercises Savvas Geometry Student Edition Solutions Page 137  Exercise 5  Problem 5

Given: The equation (x − 4) + 52 = 109

To Find – The solution of the given equation.

Using the properties of equality, the given equation can be solved.

Considering the given equation

⇒ (x − 4) + 52 = 109

Using the subtraction property of equality

⇒ (x − 4) + 52 − 52 = 109 − 52

⇒  x − 4 = 57

Using the addition property of equality

⇒ x − 4 + 4 = 57 + 4

⇒ x = 61

The solution of the equation (x − 4) + 52 = 109 is x = 61.

Chapter 3 Parallel And Perpendicular Lines Explanation Savvas Geometry Student Edition Page 137  Exercise 6  Problem 6

Given: The equation (2x + 5) + (3x − 10) = 70

To Find –  The solution of the given equation.

Using the properties of equality, the given equation can be solved.

Considering the given equation

⇒ (2x + 5) + (3x − 10) = 70

⇒  2x + 5 + 3x − 10 = 70

Combining like terms

⇒ 5x − 5 = 70

Using the Addition Property of Equality

⇒ 5x − 5 + 5 = 70 + 5

⇒  5x = 75

Using the Division Property of Equality

\(\frac{5x}{5}\) = \(\frac{75}{5}\)

⇒  x = 15

The solution of the equation (2x + 5) + (3x − 10) = 70 is x = 15.

Chapter 3 Parallel And Perpendicular Lines Explanation Savvas Geometry Student Edition Page 137  Exercise 7  Problem 7

Given: The given points are(−4,2) and (4,4).

To Find –  The distance between the two given points.

Use the formula of distance between two points.

We have the given points,(−4,2) and (4,4)

Distance between the two points

= \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

=  \(\sqrt{(4-(-4))^2+(4-2)^2}\)

=  \(\sqrt{(4+4)^2+(2)^2}\)

=  \(\sqrt{64+4}\)

=  \(\sqrt{68}\)

=  8.25

The distance between the two points is 8.25 units.

Solutions For Parallel And Perpendicular Lines Exercises Savvas Geometry Chapter 3 Page 137  Exercise 8  Problem 8

Given: The given points are(3,−1) and (7,−2).

To Find – The distance between the two given points.

Use the formula of the distance between two points.

We have the given points,(3,−1) and (7,−2)

Distance between the two points

=  \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

=  \(\sqrt{(7-3)^2+(-2-(-1))^2}\)

=  \(\sqrt{4^2+(-2+1)^2}\)

=  \(\sqrt{16+1}\)

=  \(\sqrt{17}\) ≈ 4.12

The distance between the given two points is 4.12 units.

Solutions For Parallel And Perpendicular Lines Exercises Savvas Geometry Chapter 3 Page 137  Exercise 9  Problem 9

Given: The core of an apple is in the interior of the apple.

The peel is on the exterior.

To Find –  The meaning of the terms interior and exterior in the context of geometric figures.

A figure is a shape drawn on a plane or a space that comprises of curves, points, and lines.

When the figure is drawn on the plane or space, the plane or space gets divided into three regions.

The first region is the boundary of the figure, which is the figure itself.

The second region is the space inside the figure which is called the interior of the figure.

The third region is the space outside the figure which is called the exterior of the figure.

A figure divides a plane or a space into three parts – the figure itself, the region inside the figure called the interior, and the region outside the figure called the exterior.

Parallel And Perpendicular Lines Exercises Savvas Geometry Student Edition Detailed Solutions Page 137  Exercise 10  Problem 10

Given:  A ship sailing from the United States to Europe makes a transatlantic voyage.

To Find – Meaning of the prefix trans- and what a transversal does.

The words “transatlantic voyage” refer to a voyage that involves crossing the Atlantic Ocean.

So, the prefix trans- means to cross.

A transversal is a special type of line in geometry.

A transversal is a word with the prefix trans- so it means that it is a line that crosses a system of lines.

The prefix trans- means cross and the term transversal refers to a line that crosses other lines.

Parallel And Perpendicular Lines Exercises Savvas Geometry Student Edition Detailed Solutions Page 137  Exercise 11  Problem 11

Given: People in many jobs use flowcharts to describe the logical steps of a particular process.

To Find – How a flow proof can be used in geometry.

A flow proof shows each statement that leads to the conclusion using a diagram.

The sequence of the proof is represented by arrows.

The diagram’s form isn’t crucial, but the arrows should clearly demonstrate how one statement leads to the next.

Each statement is accompanied by an explanation or reason written beneath or beside it.

A flow proof shows the individual steps of the proof and how each step is related to other steps.

Savvas Learning Co Geometry Student Edition Chapter 2 Reasoning And Proof Exercise

Page 79  Exercise 1  Problem 1

Given: An expression, 9x − 13

To find: The value of the expression at x = 7

Let, p(x) = 9x − 13

Now,p(7) = 9 × 7 − 13

⇒ p(7) = 63 − 13

⇒ p(7) = 50

The value of the given expression at x = 7 is 50.

Page 79  Exercise 2  Problem 2

Given: An expression, 90 − 3x

To find: The value of the expression at x = 31

Let,p(x) = 90 − 3x

Now, p(31) = 90 − 3 × 31

⇒ p(31) = 90 − 93

⇒ p(31) = −3

The value of the given expression at x = 31 is−3.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 3  Problem 3

Page 79  Exercise 4  Problem 4

Given: An algebraic equation, 2x − 17 = 4

To find: The value of x

We have

⇒ 2x − 17 = 4

⇒ 2x = 4 + 17

⇒ 2x = 21

⇒ x = \(\frac{21}{2}\)

The solution of the given algebraic equation is x = \(\frac{21}{2}\).

Page 79  Exercise 5  Problem 5

Given: An algebraic equation,(10x + 5) + (6x − 1) = 180

To find: The value of x.

We have

⇒ (10x + 5) + (6x − 1) = 180

⇒ 16x + 5 − 1 = 180

⇒ 16x + 4 = 180

⇒ 16x = 176

⇒ x = 11

The solution of the given algebraic equation is x = 11.

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 6  Problem 6

Given: An algebraic equation,14x = 2(5x + 14)

To find: The value of x.

We have

⇒ 14x = 2(5x + 14)

⇒ 14x = 10x + 28

⇒ 4x = 28

⇒ x = 7

The solution of the given algebraic equation is x = 7.

Page 79  Exercise 7  Problem 7

Given: An algebraic equation,2(x + 4) =  x + 13

To find: The value of x.

We have

⇒ 2(x + 4) =  x + 13

⇒ 2x + 8 = x + 13

⇒ x = 13 − 8

⇒ x = 5

The solution of the given algebraic equation is x = 5.

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 8  Problem 8

Given: An algebraic equation,7x + 5 = 5x + 17

To find: The value of x

We have

⇒ 7x + 5 = 5x + 17

⇒ 7x − 5x = 17 − 5

⇒ 2x = 12

⇒ x = 6

The solution of the given algebraic equation is x = 6.

Page 79  Exercise 9  Problem 9

Given: An algebraic equation,(x + 21) + (2x + 9) = 90

To find: The value of x

We have

⇒ (x + 21) + (2x + 9) = 90

⇒ x + 2x + 21 + 9 =90

⇒ 3x + 30 = 90

⇒ 3x = 60

⇒ x = 20

The solution of the given algebraic equation is x = 20.

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 10  Problem 10

Given: An algebraic equation,2(3x − 4) + 10 = 5(x + 4)

To find: The value of x

We have

⇒ 2(3x − 4) + 10 = 5(x + 4)

⇒ 6x − 8 + 10 = 5x + 20

⇒ 6x − 5x = 20 + 8 − 10

⇒ x = 18

The solution of the given algebraic equation is x = 18.

Page 79  Exercise 11  Problem 11

Given: m∠1 = 4y and m∠2 = 2y + 18

To find:  m∠1&m∠2.

We have ∠ACB = 90   (since∠ACB is a right angle)

⇒ m∠1 + m∠2 = 90

⇒ 4y + 2y + 18 = 90

⇒ 6y = 72

⇒ y = 12

Now,m∠1 = 4 × 12

⇒  m∠1 = 48°  and m∠2 = 2 × 12 + 18

⇒  m∠2 = 42°

The value of m∠1 = 48°  and that of m∠2 = 42°.

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 12  Problem 12

Given: A figure.

To find: Linear pairs.

According to the definition of linear pair, we can only find one linear pair in the given figure and i.e., ∠ADC & ∠BDC.

The pair of angles that form a linear pair is ∠ADC & ∠BDC.

Page 79  Exercise 13  Problem 13

Given: A figure.

To find: A pair of adjacent angles that are not supplementary.

Clearly, ∠1 & ∠2 form a pair of adjacent angles that are not supplementary.

A pair of adjacent angles that are not supplementary is ∠1 & ∠2

Page 79  Exercise 14  Problem 14

Given: m∠ADC + m∠BDC = 180.

To find: Straight angle form.

The straight angle form has a measure 180° and is the sum of  m∠ADC +  m∠BDC  is  ∠ADB.

The required straight angle is ∠ADB.

Savvas Learning Co Geometry Student Edition Chapter 2 Page 79  Exercise 15  Problem 15

Given: The conclusion of a novel answers questions raised by the story.

To find: How do you think the term conclusion applies in geometryHypotheses are the answer you think you’ll find.

Prediction is your specific belief about the scientific idea: If my hypothesis is true, then I predict we will discover this.

The conclusion is the answer that the experiment gives.

The part of a conditional statement after then.

For example, the conclusion of “If a line is horizontal then the line has slope 0 ” is “the line has a slope 0”.

The term conclusion applies in geometry is that it gives the final value of the question.

Page 79  Exercise 16  Problem 16

Given: A detective uses deductive reasoning to solve a case by gathering, combining, and analyzing clues.

To find: How might you use deductive reasoning in geometryDeductive geometry is the art of deriving new geometric facts from previously-known facts by using logical reasoning.

In elementary school, many geometric facts are introduced by folding, cutting, or measuring exercises, not by logical deduction. In geometry, a written logical argument is called proof.

Deductive reasoning in geometry is much like the situation described above, except it relates to geometric terms. For example, given that a certain quadrilateral is a rectangle, and that all rectangles have equal diagonals, what can you deduce about the diagonals of this specific rectangle. They are equal.