# Chapter 7 practice test geometry

Need instant help while preparing the BIM Geometry  Chapter 7 topics? Then, here is the perfect guide for you all ie., Big Ideas Math Geometry Answers Chapter 7 Quadrilaterals and Other Polygons. Make use of this easy and helpful study resource at times of preparation and boost up your confidence to attempt the exam. Learn & Prepare by taking the help of BIM Book Geometry Answer Key and clarify all your doubts on the Ch 7 concepts. Big Ideas Math Geometry Chapter 7 Solution Key covers all 7.1 to 7.6 exercise questions, Chapter Tests, Review Tests, Assessments, Cumulative Practice, etc. which enhance your subject knowledge.

### Big Ideas Math Book Geometry Answer Key Chapter 7 Quadrilaterals and Other Polygons

Learning the concepts of Ch 7 Quadrilaterals and Other Polygons with the help of our provided BIM Geometry Answers is the best way to answer all the questions asked in the annual exams. The fact that you can observe in this Big Ideas Math Book Geometry Solution Key is covering all exercise questions of Chapter 7 Quadrilaterals and Other Polygons. So, just click on the quick links mentioned below and score better marks in any final exams or competitive examinations.

### Quadrilaterals and Other Polygons Maintaining Mathematical Proficiency

Solve the equation by interpreting the expression in parentheses as a single quantity.

Question 1.
4(7 – x) = 16
The given equation is:
4 (7 – x) = 16
7 – x = $$\frac{16}{4}$$
7 – x = 4
7 – 4 = x
x = 3
Hence, from the above,
We can conclude that the value of x is: 3

Question 2.
7(1 – x) + 2 = – 19
The given equation is:
7 (1 – x) + 2 = -19
7 (1 – x) = -19 – 2
1 – x = –$$\frac{21}{7}$$
1 – x = -3
1 + 3 = x
x = 4
Hence, from the above,
We can conclude that the value of x is: 4

Question 3.
3(x – 5) + 8(x – 5) = 22
The given equation is:
3 (x – 5) + 8 (x – 5) = 22
(x – 5) (3 + 8) = 22
x – 5 = $$\frac{22}{11}$$
x – 5 = 2
x = 2 + 5
x = 7
Hence, from the above,
We can conclude that the value of x is: 7

Determine which lines are parallel and which are perpendicular.

Question 4.

The given figure is:

From the given figure,
The coordinates of line a are: (-2, 2), (4, -2)
The coordinates of line b are: (-3, -2), (0, -4)
The coordinates of line ‘c’ are: (-3, 0), (3, -3)
The coordinates of line ‘d’ are: (1, 0), (3, 4)
Compare the given coordinates with (x1, y1), (x2, y2)
Now,
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
So,
The slope of line a = $$\frac{-2 – 2}{4 + 2 }$$
= $$\frac{-4}{6}$$
= –$$\frac{2}{3}$$
The slope of line b = $$\frac{-4 + 2}{0 + 3}$$
= $$\frac{-6}{3}$$
= -2
The slope of line c = $$\frac{-3 – 0}{3 + 3}$$
= $$\frac{-3}{6}$$
= –$$\frac{1}{2}$$
The slope of line d = $$\frac{4 – 0}{3 – 1}$$
= $$\frac{4}{2}$$
= 2
Hence, from the above,
We can conclude that line c and line d are perpendicular lines

Question 5.

The given figure is:

From the given figure,
The coordinates of line a are: (3, 1), (0, -3)
The coordinates of line b are: (0, 1), (-3, -3)
The coordinates of line ‘c’ are: (2, 1), (-2, 4)
The coordinates of line ‘d’ are: (4, -4), (-4, 2)
Compare the given coordinates with (x1, y1), (x2, y2)
Now,
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
So,
The slope of line a = $$\frac{-3 – 1}{0 – 3}$$
= $$\frac{-4}{-3}$$
= $$\frac{4}{3}$$
The slope of line b = $$\frac{-3 – 1}{-3 – 0}$$
= $$\frac{-4}{-3}$$
= $$\frac{4}{3}$$
The slope of line c = $$\frac{2 + 4}{-4 – 4}$$
= $$\frac{6}{-8}$$
= –$$\frac{3}{4}$$
The slope of line d = $$\frac{4 + 2}{-4 – 4}$$
= $$\frac{6}{-8}$$
= –$$\frac{3}{4}$$
Hence, from the above,
We can conclude that
line a and line b are parallel lines
line c and line d are parallel lines
line b and line c and line a and line c are perpendicular lines

Question 6.

The given figure is:

From the given figure,
The coordinates of line a are: (4, -4), (-2, -2)
The coordinates of line b are: (-3, -2), (-2, 2)
The coordinates of line ‘c’ are: (3, 1), (2, -3)
The coordinates of line ‘d’ are: (0, 3), (-4, 4)
Compare the given coordinates with (x1, y1), (x2, y2)
Now,
We know that,
The slope of the line (m) = $$\frac{y2 – y1}{x2 – x1}$$
So,
The slope of line a = $$\frac{-2 + 4}{-2 – 4}$$
= $$\frac{2}{-6}$$
= –$$\frac{1}{3}$$
The slope of line b = $$\frac{2 + 2}{-2 + 3}$$
= $$\frac{4}{1}$$
= 4
The slope of line c = $$\frac{-3 – 1}{2 – 3}$$
= $$\frac{-4}{-1}$$
= 4
The slope of line d = $$\frac{4 – 3}{-4 – 0}$$
= $$\frac{1}{-4}$$
= –$$\frac{1}{4}$$
Hence, from the above,
We can conclude that
line b and line c are parallel lines
line b and line d and line c and line d are perpendicular lines

Question 7.
ABSTRACT REASONING
Explain why interpreting an expression as a single quantity does not contradict the order of operations.
We know that,
In the order of operations, “Parenthesis” occupies the top position according to the BODMAS rule
So,
The interpreting of an expression as a single quantity or as different quantities don’t change the result
Hence, from the above,
We can conclude that the interpreting of an expression as a single quantity does not contradict the order of operations

### Quadrilaterals and Other Polygons Mathematical Practices

Monitoring Progress

Use the Venn diagram below to decide whether each statement is true or false. Explain your reasoning.

Question 1.
Some trapezoids are kites.
The given statement is:
Some trapezoids are kites
From the given Venn diagram,
We can observe that there is no relation between trapezoids and kites
Hence, from the above,
We can conclude that the given statement is false

Question 2.
No kites are parallelograms.
The given statement is:
No kites are parallelograms
From the given Venn diagram,
We can observe that there is no relation between kites and parallelograms
Hence, from the above,
We can conclude that the given statement is true

Question 3.
All parallelograms are rectangles.
The given statement is:
All parallelograms are rectangles
From the given Venn diagram,
We can observe that rectangles are a part of parallelograms but not all parallelograms are rectangles because parallelograms contain rhombuses, squares, and rectangles
Hence, from the above,
We can conclude that the given statement is false

Question 4.
The given statement is:
From the given Venn diagram,
We can observe that squares are a small part of quadrilaterals and quadrilaterals contain other than squares
Hence, from the above,
We can conclude that the given statement is true

Question 5.
Example 1 lists three true statements based on the Venn diagram above. Write six more true statements based on the Venn diagram.
The given Venn diagram is:

From the above Venn diagram,
The six more true statements based on the Venn diagram are:
a. Some parallelograms are rhombuses
b. Some parallelograms are squares
c. Some parallelograms are rectangles

Question 6.
A cyclic quadrilateral is a quadrilateral that can be circumscribed by a circle so that the circle touches each vertex. Redraw the Venn diagram so that it includes cyclic quadrilaterals.

it is given that a cyclic quadrilateral is a quadrilateral that can be circumscribed by a circle so that the circle touches each vertex.
Hence,
The completed Venn diagram that includes cyclic quadrilaterals is:

### 7.1 Angles of Polygons

Exploration 1

The Sum of the Angle Measures of a Polygon

Work with a partner. Use dynamic geometry software.

a. Draw a quadrilateral and a pentagon. Find the sum of the measures of the interior angles of each polygon.
Sample

The representation of the quadrilateral and the pentagon are:

From the above figures,
The angle measures of the quadrilateral are 90°, 90°, 90°, 90°, and 90°
So,
The sum of the angle measures of a quadrilateral = 90° + 90° + 90° + 90°
= 360°
Now,
From the above figure,
The angle measures of a pentagon are: 108°, 108°, 108°, 108°, and 108°
So,
The sum of the angle measures of a pentagon = 108° + 108° + 108° + 108° + 108°
= 540°

b. Draw other polygons and find the sums of the measures of their interior angles. Record your results in the table below.

We know that,
The sum of the angle measures of a polygon = 180° (n – 2)
Where
n is the number of sides
Hence,
The completed result of the sums of the internal measures of their internal angles is:

c. Plot the data from your table in a coordinate plane.
The table from part (b) is:

Hence,
The representation of the data in the table in the coordinate plane is:

d. Write a function that fits the data. Explain what the function represents.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data.
From part (c),
When we observe the coordinate plane,
The function that fits the data is:
y = 180 (x – 2)
Where
y is the sum of the measures of the internal angles
x is the number of sides

Exploration 2

The measure of one Angle in a Regular Polygon

Work with a partner.

a. Use the function you found in Exploration 1 to write a new function that gives the measure of one interior angle in a regular polygon with n sides.
From Exploration 1,
From part (d),
The function that fits the sum of the angle measures of the internal angles of n sides is:
y = 180° (n – 2) ——-(1)
Now,
To find the one interior angle in a regular polygon with n sides,
Divide eq. (1) by n
Hence,
The measure of an interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$

b. Use the function in part (a) to find the measure of one interior angle of a regular pentagon. Use dynamic geometry software are to check your result by constructing a regular pentagon and finding the measure of one of its interior angles.
From part (a),
We know that,
The measure of an interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
So,
The measure of an interior angle of a regular pentagon = $$\frac{180° (5 – 2)}{5}$$
= $$\frac{180° (3)}{5}$$
= 36° × 3
= 108°
Hence, from the above,
We can conclude that the measure of an interior angle of a regular pentagon is: 108°

c. Copy your table from Exploration 1 and add a row for the measure of one interior angle in a regular polygon with n sides. Complete the table. Use dynamic geometry software to check your results.
The completed table along with the column of “Measure of one interior angle in a regular polygon” is:

Question 3.
What is the sum of the measures of the interior angles of a polygon?
We know that,
The sum of the measures of the interior angles of a polygon is:
Sum = 180° (n – 2)
Where
n is the number of sides

Question 4.
Find the measure of one interior angle in a regular dodecagon (a polygon with 12 sides).
We know that,
The measure of one interior angle in a regular polygon = $$\frac{180° (n – 2)}{n}$$
Where
n is the number of sides
So,
For a regular dodecagon, i.e., a polygon with 12 sides
The measure of one interior angle in a regular dodecagon = $$\frac{180° (12 – 2)}{12}$$
= $$\frac{180° (10)}{12}$$
= 15 × 10
= 150°
Hence, from the above,
We can conclude that the measure of one interior angle in a regular dodecagon is: 150°

### Lesson 7.1 Angles of Polygons

Monitoring Progress

Question 1.
The Coin shown is in the shape of an 11-gon. Find the sum of the measures of the interior angles.

It is given that the coin shown is in the shape of an 11-gon
So,
The number of sides in a coin (n) = 11
We know that,
The sum pf the measures of the interior angles = 180° (n – 2)
= 180° (11 – 2)
= 180° (9)
= 1620°
Hence, from the above,
We can conclude that the sum of the measures of the interior angles in a coin is: 1620°

Question 2.
The sum of the measures of the interior angles of a convex polygon is 1440°. Classify the polygon by the number of sides.
It is given that the sum of the measures of the interior angles of a convex polygon is 1440°
We know that,
The sum of the measures of the interior angles of a polygon = 180° (n – 2)
Where
n is the number of sides
So,
1440° = 180° (n – 2)
n – 2 = $$\frac{1440}{180}$$
n – 2 = 8
n = 8 + 2
n = 10
Hence, from the above,
We can conclude that the polygon with 10 sides is called “Decagon”

Question 3.
The measures of the interior angles of a quadrilateral are x°, 3x°. 5x°. and 7x° Find the measures of all the interior angles.
It is given that the measures of the interior angles of a quadrilateral are x°, 3x°, 5x°, and 7x°
We know that,
The sum of the measures of the interior angles of a quadrilateral is: 360°
So,
x° + 3x° + 5x° + 7x° = 360°
16x° = 360°
x° = $$\frac{360}{16}$$
x° = 22.5°
So,
The measures of all the interior angles of a quadrilateral are:
x° = 22.5°
3x° = 3 (22.5)° = 67.5°
5x° = 5 (22.5)° = 112.5°
7x° = 7 (22.5)° = 157.5°
Hence, from the above,
We can conclude that the measures of the internal angles of a quadrilateral are:
22.5°, 67.5°, 112.5°, and 157.5°

Question 4.
Find m∠S and m∠T in the diagram.

The given figure is:

From the given figure,
We can observe that the number of the sides is 5
We know that,
The sum of the measures of the interior angles of the pentagon = 540°
Let
∠S = ∠T = x°
So,
93° + 156° + 85° + x° + x° = 540°
2x° + 334° = 540°
2x° = 540° – 334°
2x° = 206°
x° = $$\frac{206}{2}$$
x° = 103°
Hence, from the above,
We can conclude that
∠S = ∠T = 103°

Question 5.
Sketch a pentagon that is equilateral but not equiangular.
e know that,
The “Equilateral” means all the sides are congruent
The “Equiangular” means all the angles are congruent
Hence,
A pentagon that is equilateral but not equiangular is:

Question 6.
A convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75°. What is the measure of an exterior angle at the sixth vertex?
It is given that a convex hexagon has exterior angles with measures 34°, 49°, 58°, 67°, and 75°
We know that,
A hexagon has 6 number of sides
We know that,
The sum of the measures of the exterior angles of any polygon is: 360°
Let the exterior angle measure at the sixth vertex of a convex hexagon be: x°
So,
34° + 49° + 58° + 67° + 75° + x° = 360°
283° + x° = 360°
x° = 360° – 283°
x° = 77°
Hence, from the above,
We can conclude that the measure of the exterior angle at the sixth vertex of the convex hexagon is: 77°

Question 7.
An interior angle and an adjacent exterior angle of a polygon form a linear pair. How can you use this fact as another method to find the measure of each exterior angle in Example 6?
It is given that an interior angle and an adjacent exterior angle of a polygon form a linear pair i.e, it forms a supplementary angle
So,
Interior angle measure + Adjacent exterior angle measure = 180°
So,
By using the above property,
The sum of the angle measure of the exterior angle of any polygon = 180° × 4 = 360°
So,
The measure of each exterior angle = $$\frac{360}{4}$$
= 90°

### Exercise 7.1 Angles of Polygons

Question 1.
VOCABULARY
Why do vertices connected by a diagonal of a polygon have to be nonconsecutive?

Question 2.
WHICH ONE DOESNT BELONG?
Which sum does not belong with the other three? Explain your reasoning.

 The sum of the measures of the interior  angles of a quadrilateral The sum of the measures of the exterior angles of a quadrilateral The sum of the measures of the interior  angles of a pentagon The sum of the measures of the exterior angles of a pentagon

The given statements are:
a. The sum of the measures of the interior angles of a quadrilateral
b. The sum of the measures of the exterior angles of a quadrilateral
c. The sum of the measures of the interior angles of a pentagon
d. The sum of the measures of the exterior angles of a pentagon
We know that,
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
The sum of the angle measures of the exterior angles of any polygon is: 360°
We know that,
The number of sides of a quadrilateral is: 4
The number of sides of a pentagon is: 5
So,
The sum of the angle measures of the interior angles of a quadrilateral = 180° ( 4 – 2)
= 180° (2)
= 360°

The sum of the angle measures of the interior angles of a pentagon = 180° ( 5 – 2)
= 180° (3)
= 540°
Hence, from the above,
We can conclude that the statement (c) does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the sum of the measures of the interior angles of the indicated convex po1gon.

Question 3.
nonagon

Question 4.
14-gon
The given convex polygon is: 14-gon
So,
The number of sides of 14-gon is: 14
We know that,
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
So,
The sum of the angle measures of the interior angles of 14-gon = 180° (14 – 2)
= 180° (12)
= 2160°

Question 5.
16-gon

Question 6.
20-gon
The given convex polygon is: 20-gon
So,
The number of sides of 20-gon is: 20
We know that,
The sum of the angle measures of the interior angles of a polygon = 180° (n – 2)
So,
The sum of the angle measures of the interior angles of 20-gon = 180° (20 – 2)
= 180° (18)
= 3240°

In Exercises 7-10, the sum of the measures of the interior angles of a convex polygon is given. Classify the polygon by the number of sides.

Question 7.
720°

Question 8.
1080°
It is given that
The sum of the angle measures of the interior angles of a convex polygon is: 1080°
We know that,
The sum of the measures of the interior angles of a polygon = 180° (n – 2)
So,
1080° = 180° (n – 2)
n – 2 = $$\frac{1080}{180}$$
n – 2 = 6
n = 6 + 2
n = 8
Hence, from the above,
We can conclude that the number of sides is: 8
Hence,
The given polygon is: Octagon

Question 9.
2520°

Question 10.
3240°
It is given that
The sum of the angle measures of the interior angles of a convex polygon is: 3240°
We know that,
The sum of the measures of the interior angles of a polygon = 180° (n – 2)
So,
3240° = 180° (n – 2)
n – 2 = $$\frac{3240}{180}$$
n – 2 = 18
n = 18 + 2
n = 20
Hence, from the above,
We can conclude that the number of sides is: 20
Hence,
The given polygon is: 20-gon or Icosagon

In Exercises 11-14, find the value of x.

Question 11.

Question 12.

The given figure is:

From the given figure,
We can observe that
The polygon has 4 sides
The given angle measures of a polygon with 4 sides are:
103°, 133°, 58°, and x°
So,
The sum of the angle measures of the interior angles of a polygon with 4 sides = 180° (4 – 2)
= 180° (2)
= 360°
So,
103° + 133° + 58° + x° = 360°
294° + x° = 360°
x° = 360° – 294°
x° = 66°
Hence, from the above,
We can conclude that the value of x is: 66°

Question 13.

Question 14.

The given figure is:

From the given figure,
We can observe that
The polygon has 4 sides
The given angle measures of a polygon with 4 sides are:
101°, 68°, 92°, and x°
So,
The sum of the angle measures of the interior angles of a polygon with 4 sides = 180° (4 – 2)
= 180° (2)
= 360°
So,
101° + 68° + 92° + x° = 360°
261° + x° = 360°
x° = 360° – 261°
x° = 99°
Hence, from the above,
We can conclude that the value of x is: 99°

In Exercises 15-18, find the value of x.

Question 15.

Question 16.

The given figure is:

From the given figure,
We can observe that
The polygon has 5 sides
The given angle measures of a polygon with 5 sides are:
140°, 138°, 59°, x°, and 86°
So,
The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2)
= 180° (3)
= 540°
So,
140° + 138° + 59° + x° + 86° = 540°
423° + x° = 540°
x° = 540° – 423°
x° = 117°
Hence, from the above,
We can conclude that the value of x is: 117°

Question 17.

Question 18.

The given figure is:

From the given figure,
We can observe that
The polygon has 8 sides
The given angle measures of a polygon with 8 sides are:
143°, 2x°, 152°, 116°, 125°, 140°, 139°, and x°
So,
The sum of the angle measures of the interior angles of a polygon with 8 sides = 180° (8 – 2)
= 180° (6)
= 1080°
So,
143° + 2x° + 152° + 116° + 125° + 140° + 139° + x° = 1080°
815° + 3x° = 1080°
3x° = 1080° – 815°
3x° = 265°
x° = $$\frac{265}{3}$$
x° = 88.6°
Hence, from the above,
We can conclude that the value of x is: 88.6°

In Exercises 19 – 22, find the measures of ∠X and ∠Y.

Question 19.

Question 20.

The given figure is:

From the given figure,
We can observe that
The polygon has 5 sides
We know that,
If the angles are not mentioned in a polygon, the consider that angles as equal angles
Now,
The given angle measures of a polygon with 5 sides are:
47°, 119°, 90°, x°, and x°
So,
The sum of the angle measures of the interior angles of a polygon with 5 sides = 180° (5 – 2)
= 180° (3)
= 540°
So,
47° + 119° + 90° + x° + x° = 540°
256° + 2x° = 540°
2x° = 540° – 256°
2x° = 284°
x° = $$\frac{284}{2}$$
x° = 142°
Hence, from the above,
We can conclude that
∠X = ∠Y = 142°

Question 21.

Question 22.

The given figure is:

From the given figure,
We can observe that
The polygon has 6 sides
We know that,
If the angles are not mentioned in a polygon, the consider that angles as equal angles
Now,
The given angle measures of a polygon with 6 sides are:
110°, 149°, 91°, 100°,  x°, and x°
So,
The sum of the angle measures of the interior angles of a polygon with 6 sides = 180° (6 – 2)
= 180° (4)
= 720°
So,
110° + 149° + 91° + 100 +  x° + x° = 720°
440° + 2x° = 720°
2x° = 720° – 440°
2x° = 280°
x° = $$\frac{280}{2}$$
x° = 140°
Hence, from the above,
We can conclude that
∠X = ∠Y = 140°

In Exercises 23-26, find the value of x.

Question 23.

Question 24.

The given figure is:

From the given figure,
We can observe that
The polygon has 7 sides
The given angle measures of a polygon with 7 sides are:
50°, 48°, 59°, x°, x°, 58°, and 39°
So,
The sum of the angle measures of the exterior angles of any polygon is: 360°
So,
50° + 48° + 59° +  x° + x° + 58° + 39° = 360°
254° + 2x° = 360°
2x° = 360° – 254°
2x° = 106°
x° = $$\frac{106}{2}$$
x° = 53°
Hence, from the above,
We can conclude that the value of x is: 53°

Question 25.

Question 26.

The given figure is:

From the given figure,
We can observe that
The polygon has 5 sides
The given angle measures of a polygon with 5 sides are:
45°, 40°, x°, 77°, and 2x°
So,
The sum of the angle measures of the exterior angles of any polygon is: 360°
So,
45° + 40° +  x° + 77° + 2x° = 360°
162° + 3x° = 360°
3x° = 360° – 162°
3x° = 198°
x° = $$\frac{198}{3}$$
x° = 66°
Hence, from the above,
We can conclude that the value of x is: 66°

In Exercises 27-30, find the measure of each interior angle and each exterior angle of the indicated regular polygon.

Question 27.
pentagon

Question 28.
18-gon
The given polygon is: 18-gon
So,
The number of sides of 18-gon is: 18
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
So,
The measure of each interior angle of 18-gon = $$\frac{180° (18 – 2)}{18}$$
= $$\frac{180° (16)}{18}$$
= 160°
The measure of each exterior angle of 18-gon = $$\frac{360°}{18}$$
= 20°
Hence, from the above,
We can conclude that
The measure of each interior angle of 18-gon is: 160°
The measure of each exterior angle of 18-gon is: 20°

Question 29.
45-gon

Question 30.
90-gon
The given polygon is: 90-gon
So,
The number of sides of 90-gon is: 90
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
So,
The measure of each interior angle of 90-gon = $$\frac{180° (90 – 2)}{90}$$
= $$\frac{180° (88)}{90}$$
= 176°
The measure of each exterior angle of 90-gon = $$\frac{360°}{90}$$
= 4°
Hence, from the above,
We can conclude that
The measure of each interior angle of 90-gon is: 176°
The measure of each exterior angle of 90-gon is: 4°

ERROR ANALYSIS
In Exercises 31 and 32, describe and correct the error in finding the measure of one exterior angle of a regular pentagon.

Question 31.

Question 32.

It is given that there are 10 exterior angles, two at each vertex
So,
The number of sides based on 10 exterior angles is: 10
We know that,
The measure of each exterior angle of any polygon = $$\frac{360°}{n}$$
So,
The measure of each exterior angle of a polygon with 10 sides = $$\frac{360°}{10}$$
= 36°
Hence, from the above,
We can conclude that the measure of each exterior angle in a polygon of 10 sides is: 36°

Question 33.
MODELING WITH MATHEMATICS
The base of a jewelry box is shaped like a regular hexagon. What is the measure of each interior angle of the jewelry box base?

Question 34.
MODELING WITH MATHEMATICS
The floor of the gazebo shown is shaped like a regular decagon. Find the measure of each interior angle of the regular decagon. Then find the measure of each exterior angle.

It is given that the floor of the gazebo shown above is shaped like a regular decagon
Now,
We know that,
The number of sides of a regular decagon is: 10
Now,
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
So,
We know that,
The measure of each interior angle of a regular decagon = $$\frac{180° (10 – 2)}{10}$$
= $$\frac{180° (8)}{10}$$
= 144°
The measure of each exterior angle of a regular decagon = $$\frac{360°}{10}$$
= 36°
Hence, from the above,
We can conclude that
The measure of each interior angle of a regular decagon is: 144°
The measure of each exterior angle of a regular decagon is: 36°

Question 35.
WRITING A FORMULA
Write a formula to find the number of sides n in a regular polygon given that the measure of one interior angle is x°.

Question 36.
WRITING A FORMULA
Write a formula to find the number of sides n in a regular polygon given that the measure of one exterior angle is x°.
It is given that the measure of one exterior angle is: x°
We know that,
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
Where
n is the number of sides
So,
x° = $$\frac{360°}{n}$$
n = $$\frac{360°}{x°}$$
Hence, from the above,
We can conclude that
The formula to find the number of sides n in a regular polygon given that the measure of one exterior angle is:
n = $$\frac{360°}{x°}$$

REASONING
In Exercises 37-40, find the number of sides for the regular polygon described.

Question 37.
Each interior angle has a measure of 156°.

Question 38.
Each interior angle has a measure of 165°.
It is given that each interior angle has a measure of 165°
We know that,
The measure of each interior angle of a polygon = $$\frac{180° (n – 2)}{n}$$
So,
165° = $$\frac{180° (n – 2)}{n}$$
165n = 180 (n – 2)
165n = 180n – 360
180n – 165n = 360
15n = 360
n = $$\frac{360}{15}$$
n = 24
Hence, from the above,
We can conclude that the number of sides with each interior angle 165° is: 24

Question 39.
Each exterior angle has a measure of 9°.

Question 40.
Each exterior angle has a measure of 6°.
It is given that each exterior angle has a measure of 6°
We know that,
The measure of each exterior angle of a polygon = $$\frac{360°}{n}$$
So,
6° = $$\frac{360°}{n}$$
6n = 360
n = $$\frac{360}{6}$$
n = 60
Hence, from the above,
We can conclude that the number of sides with each exterior angle 6° is: 60

Question 41.
DRAWING CONCLUSIONS
Which of the following angle measures are possible interior angle measures of a regular polygon? Explain your reasoning. Select all that apply.
(A) 162°
(B) 171°
(C) 75°
(D) 40°

Question 42.
PROVING A THEOREM
The Polygon Interior Angles Theorem (Theorem 7.1) states that the sum of the measures of the interior angles of a convex n-gon is (n – 2) • 180°. Write a paragraph proof of this theorem for the case when n = 5.

Polygon Interior Angles Theorem:
Statement:
The sum of the measures of the interior angles of a convex n-gon is: 180° (n – 2)

Proof of the Polygon Interior Angles Theorem:

ABCDE is an “n” sided polygon. Take any point O inside the polygon. Join OA, OB, OC.
For “n” sided polygon, the polygon forms “n” triangles.
We know that the sum of the angles of a triangle is equal to 180 degrees
Therefore,
The sum of the angles of n triangles = n × 180°
From the above statement, we can say that
Sum of interior angles + Sum of the angles at O = 2n × 90° ——(1)
But, the sum of the angles at O = 360°
Substitute the above value in (1), we get
Sum of interior angles + 360°= 2n × 90°
So, the sum of the interior angles = (2n × 90°) – 360°
Take 90 as common, then it becomes
The sum of the interior angles = (2n – 4) × 90°
Therefore,
The sum of “n” interior angles is (2n – 4) × 90°
Hence,
When n= 5,
The sum of interior angles = ([2 × 5] – 4) × 90°
= (10 – 4) × 90°
= 6 × 90°
= 540°

Question 43.
PROVING A COROLLARY
Write a paragraph proof of the Corollary to the Polygon Interior Angles Theorem (Corollary 7. 1).

Question 44.
MAKING AN ARGUMENT
Your friend claims that to find the interior angle measures of a regular polygon. you do not have to use the Polygon Interior Angles Theorem (Theorem 7. 1). You instead can use the Polygon Exterior Angles Theorem (Theorem 7.2) and then the Linear Pair Postulate (Postulate 2.8). Is your friend correct? Explain your reasoning.

Explanation:
It is given that your friend claims that to find the interior angle measures of a regular polygon. you do not have to use the Polygon Interior Angles Theorem. You instead can use the Polygon Exterior Angles Theorem and then the Linear Pair Postulate.
We know that
In a polygon,
The sum of the angle measures of exterior angles + The sum of the angle measures of interior angles = 180°
So,
The sum of the angle measures of interior angles = 180° – (The sum of the angle measures of exterior angles)
So,
From the above,
We know that
According to Linear Pair Postulate
The sum of the exterior angle and interior angle measures is: 180°
The angle measures of the exterior angles can be found out by using the “Polygon Exterior Angles Theorem”
Hence, from the above,
We can conclude that the claim of your friend is correct

Question 45.
MATHEMATICAL CONNECTIONS
In an equilateral hexagon. four of the exterior angles each have a measure of x°. The other two exterior angles each have a measure of twice the sum of x and 48. Find the measure of each exterior angle.

Question 46.
THOUGHT-PROVOKING
For a concave polygon, is it true that at least one of the interior angle measures must be greater than 180°? If not, give an example. If so, explain your reasoning.
We know that,
For a concave polygon,
The angle measure of at least one interior angle should be greater than 180°
Now,
The word “Concave” implies that at least 1 interior angle is folding in and so this “Folding in” should be greater than 180° to produce the required shape
Hence, from the above,
We can conclude that it is true that at least one of the interior angle measures must be greater than 180°

Question 47.
WRITING EXPRESSIONS
Write an expression to find the sum of the measures of the interior angles for a concave polygon. Explain your reasoning.

Question 48.
ANALYZING RELATIONSHIPS
Polygon ABCDEFGH is a regular octagon. Suppose sides $$\overline{A B}$$ and $$\overline{C D}$$ are extended to meet at a point P. Find m∠BPC. Explain your reasoning. Include a diagram with your answer.
It is given that polygon ABCDEFGH is a regular polygon and $$\overline{A B}$$ and $$\overline{C D}$$ are extended to meet at a point P.
So,
The representation of the regular octagon is:

We know that,
The angle measure of each exterior angle = $$\frac{360°}{n}$$
Where
n is the number of sides
It is given that a polygon is: Octagon
So,
The number of sides of the Octagon is: 8
So,
The angle measure of each exterior angle = $$\frac{360°}{8}$$
= 45°
We know that,
All the angles in the Octagon are equal
So,
From the given figure,
We can observe that ΔBPC is an Isosceles triangle
So,
From ΔBPC,
∠B = ∠C = 45°, ∠P = x°
We know that,
the sum of the angle measures of a given triangle is: 180°
So,
45° + x° + 45° = 180°
x° + 90° = 180°
x° = 180° – 90°
x° = 90°
Hence, from the above,
∠BPC = 90°

Question 49.
MULTIPLE REPRESENTATIONS
The formula for the measure of each interior angle in a regular polygon can be written in function notation.

a. Write a junction h(n). where n is the number of sides in a regular polygon and h(n) is the measure of any interior angle in the regular polygon.
b. Use the function to find h(9).
c. Use the function to find n when h(n) = 150°.
d. Plot the points for n = 3, 4, 5, 6, 7, and 8. What happens to the value of h(n) as n gets larger?

Question 50.
HOW DO YOU SEE IT?
Is the hexagon a regular hexagon? Explain your reasoning.

Question 51.
PROVING A THEOREM
Write a paragraph proof of the Polygon Exterior Angles Theorem (Theorem 7.2). (Hint: In a convex n-gon. the sum of the measures of an interior angle and an adjacent exterior angle at any vertex is 180°.)

Question 52.
ABSTRACT REASONING
You are given a convex polygon. You are asked to draw a new polygon by increasing the sum of the interior angle measures by 540°. How many more sides does our new polygon have? Explain your reasoning.
It is given that you are given a convex polygon and you are asked to draw a new polygon by increasing the sum of the interior angle measures by 540°
So,
We know that
The sum of the angle measures of the interior angles in a polygon = 180° (n – 2)
Let the number of sides of a new polygon be x
So,
180° (x – 2) + 540° = 180° (n – 2)
180x – 360° + 540° = 180n – 360°
180x – 180n = -540°
180 (x – n) = -540°
x – n = $$\frac{540}{180}$$
x – n = -3
n – x = 3
n = x + 3
Hence, from the above,
We can conclude that
We have to add 3 more sides to the original convex polygon

Maintaining Mathematical Proficiency

Find the value of x.

Question 53.

Question 54.

The given figure is:

From the given figure,
We can observe that
113° and x° are the corresponding angles
We know that,
According to the “Corresponding Angles Theorem”, the corresponding angles are congruent i.e., equal
So,
x° = 113°
Hence, from the above,
We can conclude that the value of x is: 113°

Question 55.

Question 56.

The given figure is:

From the given figure,
We can observe that
(3x + 10)° and (6x – 19)° are the corresponding angles
We know that,
According to the “Corresponding angles Theorem”, the corresponding angles are congruent i.e., equal
So,
(3x + 10)° = (6x – 19)°
6x – 3x = 19° + 10°
3x° = 29°
x° = $$\frac{29}{3}$$
x° = 9.6°
Hence, from the above,
We can conclude that the value of x is: 9.6°

### 7.2 Properties of Parallelograms

Exploration 1

Discovering Properties of Parallelograms

Work with a partner: Use dynamic geometry software.

a. Construct any parallelogram and label it ABCD. Explain your process.
Sample

The representation of parallelogram ABCD is:

b. Find the angle measures of the parallelogram. What do you observe?
The representation of parallelogram ABCD with the angles is:

Hence,
From the parallelogram ABCD,
We can observe that
∠A = 105°, ∠B = 75°, ∠D = 105°, and ∠C = 75°

c. Find the side lengths of the parallelogram. What do you observe?
The representation of the parallelogram ABCD along with the side lengths is:

Hence,
From the parallelogram ABCD,
We can observe that
AB = CD = 5cm, and AC = BD = 2.8cm

d. Repeat parts (a)-(c) for several other parallelograms. Use your results to write conjectures about the angle measures and side lengths of a parallelogram.
The representation of parallelogram ABCD along with its angles and the side lengths is:

Hence,
From the parallelogram ABCD,
We can conclude that
a. The opposite sides (i.e., AB and CD; AC and BD) are congruent i.e., equal
b. The opposite angles (i.e., A and C; B and D) are congruent i.e, equal

Exploration 2

Discovering a Property of Parallelograms

Work with a partner: Use dynamic geometry software.

a. Construct any parallelogram and label it ABCD.
The representation of the parallelogram ABCD is:

b. Draw the two diagonals of the parallelogram. Label the point of intersection E.

The representation of the parallelogram ABCD along with its diagonals is:

From the parallelogram ABCD,
We can observe that
The diagonals of parallelogram ABCD are: AC and BD
The intersection point of AC and BD is: E

c. Find the segment lengths AE, BE, CE, and DE. What do you observe?
The representation of the parallelogram ABCD along the segment lengths is:

Hence,
From the parallelogram ABCD,
We can observe that
AE = DE = 1.9 cm
CE = BE = 2.7 cm

d. Repeat parts (a)-(c) for several other parallelograms. Use your results to write a conjecture about the diagonals of a parallelogram.
MAKING SENSE OF PROBLEMS
To be proficient in math, you need to analyze givens, constraints, relationships, and goals.
The representation of the parallelogram ABCD along with the length of the diagonals is:

Hence, from the above,
We can conclude that
The diagonals bisect each other
The lengths of the diagonals are:
AD = 3.8 cm and BC = 5.4 cm

Question 3.
What are the properties of parallelograms?
The properties of parallelograms are:
a. The opposite sides are parallel.
b. The opposite sides are congruent.
c. The opposite angles are congruent.
d. Consecutive angles are supplementary.
e. The diagonals bisect each other.

### Lesson 7.2 Properties of Parallelograms

Monitoring progress

Question 1.
Find FG and m∠G.

The given figure is:

It is given that
EH = 8 and ∠E = 60°
We know that,
In a parallelogram,
a. The opposite sides are congruent
b. The opposite angles are congruent
So,
From the given figure,
We can say that
FG = HE and GH = FE
∠G = ∠E and ∠H = ∠F
Hence, from the above,
We can conclude that
FG = 8 and ∠G = 60°

Question 2.
Find the values of x and y.

The given figure is:

From the given parallelogram,
We can observe that
JK = 18 and LM = y + 3
∠J = 2x° and ∠L = 50°
We know that,
In a parallelogram,
a. The opposite sides are congruent
b. The opposite angles are congruent
So,
JK = LM and ∠J = ∠L
So,
y + 3 =18
y = 18 – 3
y = 15
2x° = 50°
x° = $$\frac{50}{2}$$
x° = 25°
Hence, from the above,
We can conclude that the values of x and y are: 25° and 15

Question 3.
WHAT IF?
In Example 2, find in m∠BCD when m∠ADC is twice the measure of ∠BCD.
From Example 2,
It is given that ABCD is a parallelogram
In the parallelogram ABCD,
It is given that
We know that,
In a parallelogram,
The sum of the angle measure of the consecutive angles is supplementary
So,
From Example 2,
We can observe that
So,
∠BCD = 180° – 110°
= 70°
Hence, from the above,
We can conclude that
∠BCD = 70°

Question 4.
Using the figure and the given statement in Example 3, prove that ∠C and ∠F are supplementary angles.

Question 5.
Find the coordinates of the intersection of the diagonals of STUV with vertices S(- 2, 3), T(1, 5), U(6, 3), and V(3, 1).
The given coordinates of the parallelogram STUV are:
S (-2, 3), T (1, 5), U (6, 3), and V (3, 1)
Compare the given points with (x1, y1), (x2, y2)
We know that,
The opposite vertices form a diagonal
So,
In the parallelogram STUV,
SU and TV are the diagonals
So,
We know that,
The intersection of the diagonals means the midpoint of the vertices of the diagonals because diagonals bisect each other
So,
The midpoint of SU = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{6 – 2}{2}$$, $$\frac{3 + 3}{2}$$)
= ($$\frac{4}{2}$$, $$\frac{6}{2}$$)
= (2, 3)
Hence, from the above,
We can conclude that the coordinates of the intersection of the diagonals of parallelogram STUV is: (2, 3)

Question 6.
Three vertices of ABCD are A(2, 4), B(5, 2), and C(3, – 1). Find the coordinates of vertex D.
The given vertices of parallelogram ABCD are:
A (2, 4), B (5, 2), and C (3, -1)
Let the fourth vertex of the parallelogram ABCD be: (x, y)

We know that,
In a parallelogram,
The diagonals bisect each other i.e., the angle between the diagonals is 90°
So,
We can say that the diagonals are the perpendicular lines
So,
In the given parallelogram,
AC and BD are the diagonals
Now,
Slope of AC = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{-1 – 4}{3 – 2}$$
= $$\frac{-5}{1}$$
= -5
Slope of BD = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 2}{x – 5}$$
We know that,
AC and BD are the perpendicular lines
SO,
The product of the slopes of the perpendicular lines is equal to -1
So,
(Slope of AC) × (Slope of BD) = -1
-5 × $$\frac{y – 2}{x – 5}$$ = -1
$$\frac{y – 2}{x – 5}$$ = $$\frac{1}{5}$$
Equate the numerator and denominator of both expreesions
We get,
y – 2 = 1                    x – 5 = 5
y = 1 + 2                   x = 5 + 5
y = 3                          x = 10
Hence, from the above,
We can conclude that the coordinates of the vertex D are: (10, 3)

Exercise 7.2 Properties of Parallelograms

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Why is a parallelogram always a quadrilateral, but a quadrilateral is only sometimes a parallelogram?

Question 2.
WRITING
You are given one angle measure of a parallelogram. Explain how you can find the other angle measures of the parallelogram.
The parallelogram is:

It is given that you have one angle measure of a polygon
Let the one angle measure of the given parallelogram be ∠A
We know that,
In a parallelogram,
The opposite angles are congruent i.e., equal
The consecutive angles have the sum 180°
So,
From the given figure,
We can say that
∠A = ∠C and ∠B = ∠D
Hence, from the above,
We can measure the other angles of the parallelogram by using the property of “The opposite angles are congruent”

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the value of each variable in the parallelogram.

Question 3.

Question 4.

The given figure is:

We know that,
According to the Parallelogram Opposite sides Theorem,
AB = CD and AD = BC
So,
From the given figure,
We can say that
n = 12          m + 1 = 6
n = 12          m = 6 – 1
n = 12          m = 5
Hence, from the above,
We can conclude that the values of m and n are: 5 and 12

Question 5.

Question 6.

The given figure is:

We know that,
According to the parallelogram Opposite sides Theorem,
AB = CD and AD = BC
According to the parallelogram Opposite Angles Theorem,
∠A = ∠C and ∠B = ∠D
Hence, from the figure,
(g + 4)° = 65°             16 – h = 7
g° = 65° – 4°                h = 16 – 7
g° = 61°                       h = 9
Hence, from he above,
We can conclude that the values of g and h are: 61° and 9

In Exercises 7 and 8. find the measure of the indicated angle in the parallelogram.

Question 7.
Find m∠B.

Question 8.
Find m ∠ N.
The given figure is:

We know that,
According to the parallelogram Consecutive angles Theorem,
The sum of the consecutive angle measures is equal to: 180°
So,
From the given figure,
∠M + ∠N = 180°
95° + ∠N = 180°
∠N = 180° – 95°
∠N = 85°
Hence, from the above,
We can conclude that the value of ∠N is: 85°

In Exercises 9-16. find the indicated measure in LMNQ. Explain your reasoning.

Question 9.
LM

Question 10.
LP
The given figure is:

From the given figure,
We can observe that
LN = 7
We know that,
In the parallelogram,
The diagonals bisect each other. So, the length of each diagonal will be divided into half of the value of the diagonal length
So,
LN can be divided into LP and PN
So,
LP = $$\frac{LN}{2}$$
LP = $$\frac{7}{2}$$
LP = 3.5
Hence, from the above,
We can conclude that the value of LP is: 3.5

Question 11.
LQ

Question 12.
MQ
The given figure is:

Hence,
From the given figure,
We can observe that
MQ = 8.2
Hence, from the above,
We can conclude that the value of MQ is: 8.2

Question 13.
m∠LMN

Question 14.
m∠NQL
The given figure is:

We know that,
According to the parallelogram Consecutive angles Theorem,
The sum of the consecutive angle measures is: 180°
So,
From the figure,
∠L + ∠M = 180°
100° + ∠M = 180°
∠M = 180° – 100°
∠M = 80°
We know that,
According to the parallelogram Opposite angles Theorem,
The opposite angles are congruent i.e., equal
So,
From the figure,
∠M = ∠Q
So,
∠Q = 80°
Hence, from the above,
We can conclude that the value of ∠NQL is: 80°

Question 15.
m∠MNQ

Question 16.
m∠LMQ
The given figure is:

We know that,
According to the parallelogram Consecutive angles Theorem,
The sum of the consecutive angle measures is: 180°
So,
From the figure,
∠L + ∠M = 180°
100° + ∠M = 180°
∠M = 180° – 100°
∠M = 80°
Hence, from the above,
We can conclude that the value of ∠LMQ is: 80°

In Exercises 17-20. find the value of each variable in the parallelogram.

Question 17.

Question 18.

The given figure is:

We know that,
According to the parallelogram Opposite Angles Theorem,
∠A = ∠C and ∠B = ∠D
According to the parallelogram Consecutive angles Theorem,
∠A + ∠B = 180° and ∠C + ∠D = 180° and ∠A + ∠D = 180° and ∠B + ∠C = 180°
So,
From the given figure,
(b – 10)° + (b + 10)° = 180°
2b° = 180°
b° = $$\frac{180}{2}$$
b° = 90°
Hence, from the above,
We can conclude that the value of b is: 90°

Question 19.

Question 20.

The given figure is:

We know that,
In the parallelogram,
The diagonals bisect each other
So,
From the given figure,
We can say that
$$\frac{v}{3}$$ = 6                       2u + 2 = 5u – 10
v = 6 (3)                                                     5u – 2u = 10 + 2
v = 18                                                        3u = 18
v = 18                                                         u = $$\frac{18}{3}$$
v = 18                                                          u = 6
Hence, from the above,
We can conclude that the values of u and v are: 6 and 18

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in using properties of parallelograms.

Question 21.

Question 22.

We know that,
According to the properties of the parallelogram,
GJ and HK are the diagonals of the parallelogram and F is the perpendicular bisector of the diagonals
So,
Snce F is the perpendicular bisector,
We know that,
GF = FJ and KF = FH
Hence, from the baove,
We can conclude that because quadrilateral GHJK is a parallelogram,
$$\overline{G F}$$ = $$\overline{F J}$$

PROOF
In Exercises 23 and 24, write a two-column proof.

Question 23.
Given ABCD and CEFD are parallelograms.
Prove $$\overline{A B} \cong \overline{F E}$$

Question 24.
Given ABCD, EBGF, and HJKD are parallelograms.
Prove ∠2 ≅∠3

Given:
ABCD, EBGF, and HJKD are parallelograms
Prove:
∠2 ≅ ∠3

In Exercises 25 and 26, find the coordinates of the intersection of the diagonals of the parallelogram with the given vertices.

Question 25.
W(- 2, 5), X(2, 5), Y(4, 0), Z(0, 0)

Question 26.
Q(- 1, 3), R(5, 2), S(1, – 2), T(- 5, – 1)
The given coordinates of the parallelogram are:
Q (-1, 3), R (5, 2), S (1, -2), and T (-5, -1)
Compare the given coordinates with (x1, y1), and (x2, y2)
We know that,
By the parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other
So,
From the given coordinates,
The diagonals of the parallelogram are: QS and RT
So,
The midpoint of QS = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{1 – 1}{2}$$, $$\frac{3 – 2}{2}$$)
= (0, $$\frac{1}{2}$$)
The midpoint of RT = ($$\frac{x1 + x2}{2}$$, $$\frac{y1 + y2}{2}$$)
= ($$\frac{5 – 5}{2}$$, $$\frac{2 – 1}{2}$$)
= (0, $$\frac{1}{2}$$)
Hence from the above,
We can conclude that the coordinates of the intersection of the diagonals of the given parallelogram are:
(0, $$\frac{1}{2}$$)

In Exercises 27-30, three vertices of DEFG are given. Find the coordinates of the remaining vertex.

Question 27.
D(0, 2), E(- 1, 5), G(4, 0)

Question 28.
D(- 2, – 4), F(0, 7), G(1, 0)
The given vertices of parallelogram are:
D (-2, -4), F (0, 7), and G (1, 0)
Let the fourth vertex of the parallelogram be: (x, y)
We know that,
In a parallelogram,
The diagonals bisect each other i.e., the angle between the diagonals is 90°
So,
We can say that the diagonals are the perpendicular lines
So,
In the given parallelogram,
DF and EG are the diagonals
Now,
Slope of DF = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{7 + 4}{0 + 2}$$
= $$\frac{11}{2}$$
Slope of EG = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 0}{x – 1}$$
= $$\frac{y}{x – 1}$$
We know that,
DF and EG are the perpendicular lines
So,
The product of the slopes of the perpendicular lines is equal to -1
So,
(Slope of DF) × (Slope of EG) = -1
$$\frac{11}{2}$$ × $$\frac{y}{x – 1}$$ = -1
$$\frac{y}{x – 1}$$ = –$$\frac{2}{11}$$
Equate the numerator and denominator of both expreesions
We get,
y = -2                        x – 1 = 11
y = -2                        x = 11 + 1
y = -2                        x = 12
Hence, from the above,
We can conclude that the coordinates of the fourth vertex are: (12, -2)

Question 29.
D(- 4, – 2), E(- 3, 1), F(3, 3)

Question 30.
E (1, 4), f(5, 6), G(8, 0)
The given vertices of a parallelogram are:
E (1, 4), F (5, 6), and G (8, 0)
Let the fourth vertex of the parallelogram be: (x, y)
We know that,
In a parallelogram,
The diagonals bisect each other i.e., the angle between the diagonals is 90°
So,
We can say that the diagonals are the perpendicular lines
So,
In the given parallelogram,
FH and EG are the diagonals
Now,
Slope of FH = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{y – 6}{x – 5}$$
Slope of EG = $$\frac{y2 – y1}{x2 – x1}$$
= $$\frac{0 – 4}{8 – 1}$$
= $$\frac{-4}{7}$$
= –$$\frac{4}{7}$$
We know that,
FH and EG are the perpendicular lines
So,
The product of the slopes of the perpendicular lines is equal to -1
So,
(Slope of FH) × (Slope of EG) = -1
–$$\frac{4}{7}$$ × $$\frac{y – 6}{x – 5}$$ = -1
$$\frac{y – 6}{x – 4}$$ = $$\frac{7}{4}$$
Equate the numerator and denominator of both expreesions
We get,
y – 6 = 7                    x – 4 = 4
y = 7 + 6                   x = 4 + 4
y = 13                        x = 8
Hence, from the above,
We can conclude that the coordinates of the fourth vertex are: (8, 13)

MATHEMATICAL CONNECTIONS
In Exercises 31 and 32. find the measure of each angle.

Question 31.
The measure of one interior angle of a parallelogram is 0.25 times the measure of another angle.

Question 32.
The measure of one interior angle of a parallelogram is 50 degrees more than 4 times the measure of another angle.
It is given that the measure of one interior angle of a parallelogram is 50 degrees more than 4 times the measure of another angle.
So,
The measure of one interior angle is: x°
The measure of another interior angle is: 50° + 4x°
We know that,
The opposite angles of the parallelogram are equal
The sum of the angles of the parallelogram is: 360°
So,
x° + 4x + 50° + x° + 4x + 50° = 360°
10x° + 100° = 360°
10x° = 360° – 100°
10x° = 260°
x° = $$\frac{260}{10}$$
x° = 26°
So,
The angle measures of the parallelogram are:
x° = 26°
4x° + 50° = 4 (26°) + 50°
= 104° + 50°
= 154°
Hence, from the above,
We can conclude that the angle measures are: 26° and 154°

Question 33.
MAKING AN ARGUMENT
m∠B = 124°, m∠A = 56°, and m∠C = 124°.

Question 34.
ATTENDING TO PRECISION
∠J and ∠K are Consecutive angles in a parallelogram. m∠J = (3t + 7)°. and m∠K = (5t – 11)°. Find the measure of each angle.
It is given that ∠J and ∠K are the consecutive angles in a parallelogram
So,
∠J + ∠K = 180°
Now,
It is given that
∠J = (3t + 7)° and ∠K = (5t – 11)°
So,
(3t + 7)° + (5t – 11)° = 180°
8t° – 4 = 180°
8t° = 180° + 4°
8t° = 184°
t° = $$\frac{184}{8}$$
t° = 23°
So,
∠J = (3t + 7)°
= 3 (23)° + 7
= 69° + 7°
= 76°
∠K = (5t – 11)°
= 5 (23)° – 11
= 115° – 11°
= 104°
Hence, from the above,
We can conclude that the measure of each angle is: 76° and 104°

Question 35.
CONSTRUCTION
Construct any parallelogram and label it ABCD. Draw diagonals $$\overline{A C}$$ and $$\overline{B D}$$. Explain how to use paper folding to verify the Parallelogram Diagonals Theorem (Theorem 7.6) for ABCD.

Question 36.
MODELING WITH MATHEMATICS
The feathers on an arrow from two congruent parallelograms. The parallelograms are reflections of each other over the line that contains their shared side. Show that m ∠ 2 = 2m ∠ 1.

Question 37.
PROVING A THEOREM
Use the diagram to write a two-column proof of the Parallelogram Opposite Angles Theorem (Theorem 7.4).

Given ABCD is a parallelogram.
Prove ∠A ≅ ∠C, ∠B ≅ ∠D

Question 38.
PROVING A THEOREM
Use the diagram to write a two-column proof of the Parallelogram Consecutive Angles Theorem (Theorem 7.5).

Given PQRS is a parallelogram.
Prove x° + y° = 180°
Given:
PQRS is a parallelogram
Prove:
x° + y° = 180°

Question 39.
PROBLEM-SOLVING
The sides of MNPQ are represented by the expressions below. Sketch MNPQ and find its perimeter.
MQ = – 2x + 37 QP = y + 14
NP= x – 5 MN = 4y + 5

Question 40.
PROBLEM SOLVING
In LMNP, the ratio of LM to MN is 4 : 3. Find LM when the perimeter of LMNP is 28.
It is given that
In the parallelogram LMNP,
The ratio of LM to MN is: 4 : 3
It is also given that
The perimeter of the parallelogram LMNP is: 28
So,
Let the length of LM be 4x
Let the length of MN be 3x
We know that,
The opposite sides of the parallelogram are equal
The perimeter is the sum of all the sides
So,
4x + 3x + 4x + 3x = 28
8x + 6x = 28
14x = 28
x = $$\frac{28}{14}$$
x = 2
So,
The length of LM = 4 x
= 4 (2)
= 8
Hence, from the above,
We can conclude that the length of LM is: 8

Question 41.
ABSTRACT REASONING
Can you prove that two parallelograms are congruent by proving that all their corresponding sides are congruent? Explain your reasoning.

Question 42.
HOW DO YOU SEE IT?
The mirror shown is attached to the wall by an arm that can extend away from the wall. In the figure. points P, Q, R, and S are the vertices of a parallelogram. This parallelogram is one of several that change shape as the mirror is extended.

a. What happens to m∠P as m∠Q increases? Explain.
From the given figure,
We can observe that
∠P and ∠Q are the consecutive angles
So,
∠P + ∠Q = 180°
Now,
To make the sum 180°, if one angle measure increases, then the other angle measure has to decrease
Hence, from the above,
We can conclude that when ∠Q increases, ∠P has to decrease

b. What happens to QS as m∠Q decreases? Explain.
From the given figure,
QS is a diagonal of the parallelogram
Q and S are the opposite angles
We know that,
The opposite angles are equal
So,
As ∠Q decreases, the length of QS may also decrease or may also increase

c. What happens to the overall distance between the mirror and the wall when m∠Q decreases? Explain.
From the given figure,
We can observe that,
As the angle between Q and the wall increases,
The overall distance between the mirror and the wall increase

Question 43.
MATHEMATICAL CONNECTIONS
In STUV m∠TSU = 32°, m∠USV = (x2)°, m∠TUV = 12x°, and ∠TUV is an acute angle. Find m∠USV.

Question 44.
THOUGHT-PROVOKING
Is it possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram? Explain your reasoning.
Yes, it is possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram

Explanation:
We know that,
The diagonals bisect each other and the angles may or may not be 90° in the diagonals
So,
After the bisecting with the diagonals in a quadrilateral,
We can observe that the quadrilateral is divided into four triangles
Hence, from the above,
We can conclude that it is possible that any triangle can be partitioned into four congruent triangles that can be rearranged to form a parallelogram

Question 45.
CRITICAL THINKING
Points W(1. 2), X(3, 6), and Y(6, 4) are three vertices of a parallelogram. How many parallelograms can be created using these three vertices? Find the coordinates of each point that could be the fourth vertex.

Question 46.
PROOF
In the diagram. $$\overline{E K}$$ bisects ∠FEH, and $$\overline{F J}$$ bisects ∠EFG. Prove that $$\overline{E K}$$ ⊥ $$\overline{F J}$$. (Hint: Write equations using the angle measures of the triangles and quadrilaterals formed.)

Question 47.
PROOF
Prove the congruent Parts of Parallel Lines Corollary: If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.

Given
Prove $$\overline{H K}$$ ≅ $$\overline{K M}$$
(Hint: Draw $$\overline{K P}$$ and $$\overline{M Q}$$ such that quadrilatcral GPKJ and quadrilateral JQML are parallelorams.)

Maintaining Mathematical Proficiency

Determine whether lines l and m are parallel. Explain your reasoning.

Question 48.

The given figure is:

From the given figure,
We can observe that the given angles are the corresponding angles i.e., an interior angle and an exterior angle
Hence,
According to the Corresponding Angles Theorem,
We can conclude that l and m are parallel lines

Question 49.

Question 50.

The given figure is:

From the given figure,
We can observe that the given angles are the consecutive interior angles
We know that,
The sum of the angle measures of the consecutive interior angles is: 180°
But, from the given figure,
The sum of the angle measures is not 180°
Hence, from the above,
We can conclude that l is not parallel to m

### 7.3 Proving That a Quadrilateral is a Parallelogram

Exploration 1

Proving That a Quadrilateral is a Parallelogram

Work with a partner: Use dynamic geometry software.

a. Construct a quadrilateral ABCD whose opposite sides are congruent.

We know that,
If a quadrilateral which has opposite sides congruent and each angle measure not equal to 90° and the diagonals bisected each other, then that quadrilateral is called the “Parallelogram”
Hence,
The representation of the quadrilateral ABCD as a parallelogram is:

c. Repeat parts (a) and (b) for several other quadrilaterals. Then write a conjecture based on your results.

If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal

d. Write the converse of your conjecture. Is the Converse true? Explain.

REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of objects.
From part (c),
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal
The converse of your conjecture is:
If the opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal
Now,
From the below figure,

We can observe that
When ∠B and ∠D are 90°,
Hence, from the above,
We can conclude that the converse of the conjecture from part (c) is true

Exploration 2

Proving That a Quadrilateral Is a Parallelogram

Work with a partner: Use dynamic geometry software.

a. Construct any quadrilateral ABCD whose opposite angles are congruent.

We know that,
For a quadrilateral to be a parallelogram,
The opposite sides are equal and the opposite angles are equal
Hence, from the above figure,
We can conclude that the given quadrilateral ABCD is a parallelogram

c. Repeat parts (a) and (b) for several other quadrilaterals. Then write a conjecture based on your results.
From parts (a) and (b),
If the opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal

d. Write the converse of your conjecture. Is the converse true? Explain.
From part (c),
If the opposite angles of a quadrilateral are equal, then the opposite sides of a quadrilateral are equal
Hence,
The converse of the conjecture of the quadrilaterals is given as:
If the opposite sides of a quadrilateral are equal, then the opposite angles of a quadrilateral are equal

Question 3.
How can you prove that a quadrilateral is a parallelogram?
The number of ways to prove that a quadrilateral is a parallelogram are:
a. The opposite sides are congruent
b. The opposite angles are congruent
c. The opposite sides are parallel
d. The consecutive angles are supplementary
e. An angle is supplementary to both its consecutive angles

Question 4.
From the given figure,
We can observe that the opposite angles are equal
So,
From the conjecture of the quadrilateral,
If the opposite angles of the quadrilateral are equal, then the opposite sides of the quadrilateral are equal
We know that,
if the opposite angles are equal and the angles are not 90°, then the quadrilateral is called the “parallelogram”
Hence, from the above,
We can conclude that the quadrilateral ate the left is the “Parallelogram”

### Lesson 7.3 Proving that a Quadrilateral is a Parallelogram

Monitoring Progress

Question 1.
In quadrilateral WXYZ, m∠W = 42°, m∠X = 138°, and m∠Y = 42°. Find m∠Z. Is WXYZ a parallelogram’? Explain your reasoning.
It is given that
∠W = 42°, ∠X = 138°, and ∠Y = 42°
Let
∠Z = x°
We know that,
The sum of the angles of a quadrilateral is 360°
So,
∠W + ∠X + ∠Y + ∠Z = 360°
42° + 138° + 42° + x° = 360°
84° + 138° + x° = 360°
x° = 360° – 222°
x° = 138°
So,
∠Z = 138°
Now,
We know that,
If the opposite angles of a quadrilateral are equal and the angle is not equal to 90°, then that quadrilateral is called the “Parallelogram”
Hence, from the above,
We can conclude that WXYZ is a parallelogram

Question 2.
For what values of x and y is quadrilateral ABCD a parallelogram? Explain your reasoning.

The given figure is:

We know that,
For a quadrilateral to be a parallelogram,
The opposite angles are equal
So,
From the given figure,
4y° = (y + 87)°                                    2x° = (3x – 32)°
4y° – y° = 87°                                      3x° – 2x° = 32°
3y° = 87°                                              x° = 32°
y° = 87 / 3                                            x° = 32°
y° = 29°                                                x° = 32°
Hence, from the above,
We can conclude that for
x° = 32° and y° = 29°
The quadrilateral ABCD is a parallelogram

State the theorem you can use to show that the quadrilateral is a parallelogram.

Question 3.

The given figure is:

From the given figure,
We can observe that the opposite sides are equal and are parallel
Hence,
According to the “Opposite sides parallel and congruent Theorem”,
The given quadrilateral is a parallelogram

Question 4.

The given figure is:

From the given figure,
We can observe that the opposite sides are congruent and are parallel
Hence,
According to the “Opposite sides parallel and congruent Theorem”,
The given quadrilateral is a parallelogram

Question 5.

The given figure is:

From the given figure,
We can observe that the opposite angles are equal and are parallel
Hence,
According to the “Opposite angles parallel and congruent Theorem”,
The given quadrilateral is a parallelogram

Question 6.
For what value of x is quadrilateral MNPQ a parallelogram? Explain your reasoning.

The given figure is:

From the “Parallelogram Diagonals Converse Theorem”,
MP and NQ bisect each other
So,
MP = NQ
From the given figure,
It is given that
MP = 10 – 3x
NQ = 2x
So,
2x = 10 – 3x
2x + 3x = 10
5x = 10
x = $$\frac{10}{5}$$
x = 2
Hence, from the above,
We can conclude that for x = 2, the quadrilateral MNPQ is a parallelogram

Question 7.
Show that quadrilateral JKLM is a parallelogram.

From the given coordinate plane,
The coordinates of the quadrilateral JKLM are:
J (-5, 3), K (-3, -1), L (2, -3), and M (2, -3)
Now,
For the quadrilateral JKLM to be a parallelogram,
The opposite sides of the quadrilateral JKLM must be equal
So,
JL = KM
We know that,
The distance between the 2 points = $$\sqrt{(x2 – x1)² + (y2 – y1)²}$$
So,
JL = $$\sqrt{(5 + 2)² + (3 + 3)²}$$
= $$\sqrt{(7)² + (6)²}$$
= $$\sqrt{49 + 36}$$
= $$\sqrt{85}$$
= 9.21
KM = $$\sqrt{(5 + 2)² + (3 + 3)²}$$
= $$\sqrt{(7)² + (6)²}$$
= $$\sqrt{49 + 36}$$
= $$\sqrt{85}$$
= 9.21
Hence, from the above,
We can conclude that
The quadrilateral JKLM is a parallelogram according to the “Opposite sides parallel and congruent Theorem”

Question 8.
Refer to the Concept Summary. Explain two other methods you can use to show that quadrilateral ABCD in Example 5 is a parallelogram.
Concept Summary
Ways to Prove a Quadrilateral is a Parallelogram

The other ways to prove a quadrilateral parallelogram are:
a. Prove that the sum of the consecutive angles is supplementary
b. Prove that an angle is supplementary to both the consecutive angles

### Exercise 7.3 Proving that a Quadrilateral is a Parallelogram

Vocabulary and Core Concept Check

Question 1.
WRITING

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

 Construct a quadrilateral with opposite sides congruent. Construct a quadrilateral with one pair of parallel sides. Construct a quadrilateral with opposite angles congruent, Construct a quadrilateral with one pair of opposite sides congruent and parallel.

The given statements are:
a. Construct a quadrilateral with opposite sides congruent
b. Construct a quadrilateral with opposite angles congruent
c. Construct a quadrilateral with one pair of parallel sides
d. Construct a quadrilateral with one pair of opposite sides congruent and parallel
Now,
We know that,
For constructing a parallelogram,
We need
a. 2 pairs of opposite sides congruent and parallel
b. 2 pairs of opposite angles congruent and parallel
Now,
From the given sattements,
Statements a., b., and d are needed for onstructing a parallelogram whereas statement c. is unnecessary

Monitoring Progress and Modeling with Mathematics

In Exercises 3-8, state which theorem you can use to show that the quadrilateral is a parallelogram.

Question 3.

Question 4.

The given figure is:

From the given figure,
We can observe that the opposite sides are congruent and parallel
Hence,
By using the “Parallelogram Opposite sides Congruent and Parallel Theorem”,
We can say that the given quadrilateral is a parallelogram

Question 5.

Question 6.

The given figure is:

From the given figure,
We can observe that the opposite sides are congruent and parallel
Hence,
By using the “Parallelogram Opposite sides Congruent and Parallel Theorem”,
We can say that the given quadrilateral is a parallelogram

Question 7.

Question 8.

The given figure is:

From the given figure,
We can observe that the diagonals of the quadrilateral bisected each other
Hence,
By using the “Parallelogram Diagonals Converse Theorem”,
We can say that the given quadrilateral is a parallelogram

In Exercises 9-12, find the values of x and y that make the quadrilateral a parallelogram.

Question 9.

Question 10.

The given figure is:

It is given that the given quadrilateral is a parallelogram
So,
We know that,
According to the “Parallelogram Opposite sides congruent and parallel Theorem”,
The lengths of the opposite sides of the given quadrilateral are equal
So,
x = 16 and y = 9
Hence, from the above,
We can conclude that for the given quadrilateral to be a parallelogram,
The alues of x and y are:
x = 16 and y = 9

Question 11.

Question 12.

The given figure is:

It is given that the given quadrilateral is a parallelogram
Now,
We know that,
According to the “Opposite angles Theorem”,
The opposite angles of the parallelogram are equal
So,
(4x + 13)° = (5x – 12)°
4x° – 5x° = -12° – 13°
x° = -25°
x° = 25°
(3x – 8)° = (4y + 7)°
3x – 8 – 7 = 4y°
4y° = 3 (25°)  15
4y° = 75 – 15
4y° = 60°
y° = $$\frac{60}{4}$$
y° = 15°
Hence, from the above,
We can conclude that for a given quadrilateral to be a parallelogram,
The values of x and y are:
x = 25° and y = 15°

In Exercises 13-16, find the value of x that makes the quadrilateral a parallelogram.

Question 13.

Question 14.

The given figure is:

It is given that the given quadrilateral is a parallelogram
Now,
We know that,
According to the “Parallelogram Opposite angles parallel and congruent Theorem”,
The lengths of the opposite sides of the parallelogram are equal
So,
2x + 3 = x + 7
2x – x = 7 – 3
x = 4
Hence, from the above,
We can conclude that for a given quadrilateral to be a parallelogram,
The value of x is:
x = 4

Question 15.

### Page 6

#### Question 3030.Write a ratio for the number of red balls to blue balls. Red balls = 10 Blue balls = 8 Green balls = 6

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#### Prentice Hall Geometry Chapter 7: Similarity Chapter Exam Instructions

Choose your answers to the questions and click 'Next' to see the next set of questions. You can skip questions if you would like and come back to them later with the "Go To First Skipped Question" button. When you have completed the practice exam, a green submit button will appear. Click it to see your results. Good luck!

Prentice Hall Geometry: Online Textbook Help  /  Math Courses

This quiz is incomplete! To play this quiz, please finish editing it.
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• Q. Are the triangles similar?
• Q. Find the height of the tree.
• Q. The distance between two cities on a map is 4.125 inches.  Find the actual distance if the scale on the map is 2 inches = 40 miles.
• Q. Similarity Ratio of JKL~SRQ: _
• Q. Determine the length of QT
• Q. How long is segment AB?
• Q. Identify the length of line segment FH
• Q. What can we conclude about MN and KL?
They are parallel by Triangle Proportionality Theorem
They are parallel by the Converse of the Triangle Proportionality Theorem
They are congruent by SSS
They will intersect 10 units down from N
• Q. The triangles are similar by...
• Q. The triangles are similar by...
• Q. The triangles are similar by...
• Q. Find length of segment AB.
• Q. How is triangle ACB congruent to ECB, if it is.
• Q. To find the height of a very tall pine tree, you place a mirror on the ground and stand where you can see the top of the pine tree.  How tall is the tree?
• Q. To measure the distance across a pond, a surveyor locates points A, B, C, D, and E as shown.  What is AB to the nearest meter?
• Q. To measure the distance across a pond, a surveyor locates points A, B, C, D, and E as shown.  What is AB to the nearest meter?
• Q. The triangles are similar. Find the missing length.
• Q. What is a true similarity statement?
• Q. #2 Triangle MNR is similar to triangle MBA.  Which segment corresponds to MN?
• Q. Find the perimeter of the large triangle.
• Q. How are these triangles similar?
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Chapter 7 Practice Test Answer Key

We are a different matter. Compared. And yet Sasha for you.

## Geometry test 7 chapter practice

After getting dressed, I knocked on the door. A minute later the attendant opened the door. Shiknuv at me, he quickly led me down the corridor to the exit. - Go home slut, - he said, - come the day after tomorrow, if you like it - I'll be on duty again.

Ch 7 Practice Test

At night, Utah arranged a sex marathon for the girl. All her holes were subjected to the stretching of his indefatigable trunk. He managed to enjoy her body two or even three times. The ass stretched out quickly and easily took in his cock. Natasha was just beginning to worry that upon arriving home, there would not be a trunk that would not dangle in her developed holes.

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It is a pity that she could not be present - the rules allowed the presence of only one of the owners, and in one. Of the four competitions it was forbidden to use strapons and dildos. And so I never participated in the punishment competition, and this greatly reduced the total number of points that I could score.

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