Math in Focus

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.5 Understanding Scale Drawings to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.5 Answer Key Understanding Scale Drawings

Math in Focus Grade 7 Chapter 7 Lesson 7.5 Guided Practice Answer Key

Calculate the scale factor.

Question 1.
In the diagram, trapezoid B has been enlarged to produce trapezoid A. Find the scale factor.

Answer:
3 cm

Explanation:
Length of A = 7.8 cm
Lenght of B = 2.6 cm
Scaled length = 7.8 cm
Original lenght = 2.6 cm
Scale factor = \(\frac{7.8}{2.6}\)
= 3 cm

Hands-On Activity

Redraw A Given Figure On Grid Paper At A Different Scale

Materials:

• different sizes of grid paper
• ruler

Work in pairs.

The drawing shows a figure formed by nine squares and a triangle enclosed in a polygon. You can use different-sized grids to produce the same drawing at different scales.

Step 1.
Use two different-sized grids. Redraw the figure on one grid. Have your partner redraw the figure on another grid.

Step 2.
Use a ruler to measure a length in the original drawing. Measure the corresponding length in your scale drawings. Then find the scale factor for each drawing. You can use the following to calculate the scale factor:

Complete.

Question 2.
The scale of a map is 1 inch : 15 miles. If the distance on the map between John’s home and his school is 0.6 inch, find the actual distance in miles.

Answer:
The actual distance is 9 miles.

Explanation:

Complete.

Question 3.
The actual distance between Boston and New York is 220 miles. The scale on a particular map is 1 inch : 25 miles. How far apart on the map are the two cities? 1 inch : 25 miles means 1 inch on the map represents 25 miles on the ground.
Let x inches be the length on the map.

Answer:
8.8 inches

Question 4.
A model car is built using a scale of 1: 18. The length of the model car is 12 inches. Find the actual length of the car in feet.
Answer:
Let x be the actual length of the car in inches.
1: 18 = Scaled length of the car: Actual length of the car
We have:
Substitute values:
1 : 18 = 12 : x
Write ratios in fraction form:
\(\frac{1}{18}\) = \(\frac{12}{x}\)
Write cross products:
x = 18 × 12
Simplify the x.
x = 216
Now convert to feet:
12 in = 1 ft
216 in = \(\frac{1}{12}\) × 216
= 18 ft
The actual length of the car is 18 feet.

Hands-On Activity

Investigate The Relationship Between The Scale Factor And Its Corresponding Area

Work individually.

A square that has a side length of 1 centimeter has an area of 1 square centimeter. In this activity, you will explore how enlarging such a square by a scale factor affects its area.

Step 1.
Suppose you enlarge the square by a factor of 2. Find the side length and the area of the resulting square.

Length of square = cm
(twice the original length)
Area of square = cm²
(increased by a factor of 4)
Answer:
Length of square is 4 cm.
Area of square = 16 cm²

Explanation:
The original length is 2 cm.
Now the lenght of square is 2 + 2 = 4 cm.
Therefore the length of square is 4 cm.
And area of the square is = 4 × 4 = 16 cm²

Step 2.
Suppose you enlarge the square by a factor of 3. Find the side length and the area of the resulting square.
Length of square = cm
(three times the original length)

Area of square = cm²
(increased by a factor of 9)
Answer:
Length of square is 9 cm.
Area of square = 9 × 9 = 18 cm²

Explanation:
Original length of the square is 3 cm.
New length of the square is 3 + 3 = 6 cm.
Area of the square = 6 × 6 = 36 cm²

Step 3.
Suppose you enlarge the square by a factor of 4. Find the side length and the area of the resulting square.
Length of square = cm
(four times the original length)

Area of square = cm²
(increased by a factor of )
Answer:
16 cm, 256 cm²

Explanation:
Given, Original length is 4 cm.
Now the new length of the square is 16 cm.
Area of the square is 16 × 16 = 256 cm²

Step 4.
Copy and complete the table.

Copy and complete.

a) Increasing the length of a square by a factor of 5 increases the area by a factor of .
Answer:
Area by a factor is 25 cm²

b) Increasing the length of a square by a factor of k increases the area by a factor of .
Answer:
Area of the square is k² cm²  

Math Journal
Compare the side lengths and the areas for the various scale factors. What pattern do you observe?

From the above activity, the area of a square increases as the square of the scale factor. This property applies to other two-dimensional figures. If you enlarge a figure by a scale factor of 3, its area will be enlarged by a scale of 3² = 9

Complete.

Question 5.
Sylvia makes a map of her yard. On a map, 1 inch represents 8 feet. On the map, the area of a patch of grass is 12 square inches. Find the actual area of the patch of grass.

Answer:
1 inch : 8 feet means 1 inch on the map represents 8 feet on the ground.
Map length : Actual length = 1 in. : 8 ft
Map area: Actual area = 1 in.² : 8² ft²
Let y represent the actual area of the patch of grass in square feet.
Write a proportion:
\(\frac{\text { Area of patch of grass on map }}{\text { Actual area of patch of grass }}=\frac{1}{64}\)
Substitute:
\(\frac{12}{y}\) = \(\frac{1}{64}\)
Write the cross products:
y = 12 · 64
Simplify
y = 768
The actual area of the grass patch is 768 square feet.

Copy and complete.

Question 6.
A blueprint is a type of scale drawing used by architects. An architect is making a blueprint for a conference room that will have a floor area of 196 square feet. If the scale on the blueprint is 1 inch : 7 feet, find the area of the conference room floor on the blueprint.

Let y represent the area of the conference room on the blueprint in square inches.

Answer:
1 inch : 7 feet means 1 inch on the bLueprint represents 7 feet on the ground.
Blueprint length : ActuaL Length = 1 in. : 7 ft
Blueprint area : Actual area = 1 in.² : 7² ft²
Let y represent the area of the conference room on the blueprint in square inches.
Write a proportion:
\(\frac{\text { Area of room on blueprint }}{\text { Actual area of room }}=\frac{1}{49}\)
Substitute:
\(\frac{y}{196}\) = \(\frac{1}{49}\)
Write the cross products:
49y = 196
Divide both sides by 49:
\(\frac{49y}{49}\) = \(\frac{196}{49}\)
Simplify:
y = 4
The gloor area of the conference room on the blueprint is 4 square inches.

Math in Focus Course 2B Practice 7.5 Answer Key

Solve. Show your work.

Question 1.
A model of a ship is 6 inches long. The actual ship is 550 feet (6,600 inches). Find the scale factor used for the model.

Answer:
We are given:
Model length: 6 in.
Actual length: 6,600 in.
Determine the scale factor used for the model:
Scale factor: \(\frac{\text { Model length }}{\text { Actual length }}\) = \(\frac{6}{6,600}\)
= \(\frac{1}{1,100}\)

Question 2.
On a blueprint, the length of a wall is 5 inches. The actual length of the wall is 85 feet. What scale is used for the blueprint?
Answer:
We are given:
Blueprint length: 5 in.
Actual length: 85 ft
Convert the actual length from feet to inches:
1 ft = 12 in.
85 ft = (85 · 12) in. = 1020 in.
Determine the scale factor used for the blueprint:
Scale factor: \(\frac{\text { Blueprint length }}{\text { Actual length }}\) = \(\frac{5}{1020}\)
= \(\frac{1}{204}\)

Question 3.
An artist made a painting of a water pitcher. Then the artist reduced the size of the painting. Find the scale factor of the reduction.

Answer:
We are given:
Big size: 12 in.
Reduced size: 8 in.
Determine the scale factor of reduction:
Scale factor: \(\frac{\text { Reduced size }}{\text { Big size }}\) = \(\frac{8}{12}\)
= \(\frac{2}{3}\)

Question 4.
The height of a building in a drawing is 15 inches. If the actual height of the building is 165 feet, find the scale factor of the drawing.
Answer:
We are given:
Drawing height: 15 in.
Actual height: 165 ft
Convert feet to inches:
1 ft = 12 in.
165 ft = (165 · 12) in. = 1980 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing height }}{\text { Actual height }}\) = \(\frac{15}{1980}\)
= \(\frac{1}{132}\)

Question 5.
In a scale drawing, a sofa is 3 inches long. If the actual length of the sofa ¡s 5 feet long, find the scale factor.
Answer:
We are given:
Drawing length: 3 in.
Actual length: 5 ft
Convert feet to inches:
1ft = 12 in.
5 ft = (5 · 12) in. = 60 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing length }}{\text { Actual length }}\) = \(\frac{3}{60}\)
= \(\frac{1}{60}\)

Question 6.
Daniel is making a scale drawing of his classroom for a project. The length of his classroom is 30 feet long. In his drawing, the length of the classroom is 6 inches. Find the scale factor of Daniel’s drawing.
Answer:
We are given:
Drawing length: 6 in.
Actual length: 30 ft
Convert feet to inches:
1 ft = 12 in.
30 ft = (30 · 12) in. = 360 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing height }}{\text { Actual height }}\) = \(\frac{6}{360}\)
= \(\frac{1}{60}\)

Question 7.
Two cities are 7 inches apart on a map. If the scale of the map is 0.5 inch : 3 miles, what is the actual distance between the two cities?
Answer:
0.5 inch : 3 miles means 0.5 inches on the map represents 3 miles on the ground.
Let x miles be the actual distance.
We have:
0.5 inches : 3 miles = 7 inches: x miles
Write ratios in fraction form:
\(\frac{0.5 \mathrm{in}}{3 \mathrm{mi}}\) = \(\frac{7 \mathrm{in}}{x \mathrm{mi}}\)
Write without units:
\(\frac{0.5}{3}\) = \(\frac{7}{x}\)
Write cross products:
0.5x = 3 · 7
Simplify:
0.5x = 21
Multiply both sides by 2:
x = 42
The actual distance is 42 miles.

Question 8.
A road map of New Orleans uses a scale of 1 inch : 3 miles. If Carlton Avenue is 1 .3 inches long on the map, what is the actual length of the street?
Answer:
1 inch : 3 miles means 1 inch on the map represents 3 miles on the ground
Let x miles be the actual distance.
We have:
1 inch : 3 miles = 13 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{3 \mathrm{mi}}\) = \(\frac{1.3 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{3}\) = \(\frac{1.3}{x}\)
Write cross products:
x = 3 · 1.3
Simplify:
x = 3.9
The actual distance is 3.9 miles.

Question 9.
The scale of a map is 1 inch : 85 miles.
a) On the map, the river is 14 inches long. Find the actual length of the river in miles.
Answer:
1 inch : 85 miles means 1 inch on the map represents 85 miles on the ground.
Let x miles be the actuaL length.
We have:
1 inch : 85 miles = 14 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in}}{85 \mathrm{mi}}\) = \(\frac{14 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{85}\) = \(\frac{14}{x}\)
Write cross products:
x = 85 · 14
Simplify:
x = 1,190
The actual length is 1,190 miles.

b) The actual distance between two towns is 765 miles. Find the distance on the map between these towns.
Answer:
Let y inches be the distance on the map.
We have:
1 inch : 85 miles = y inches : 765 miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{85 \mathrm{mi}}\) = \(\frac{y \mathrm{in}}{765 \mathrm{mi}}\)
Write without units:
\(\frac{1}{85}\) = \(\frac{y}{765}\)
Write cross products:
85y = 765
Divide both sides by 85.
\(\frac{85y}{85}\) = \(\frac{765}{85}\)
Simplify:
y = 9
The distance on the map is 9 inches.

Question 10.
Goodhope River is 48 miles long. What is the length of the river on a map with a scale of 1 inch : 15 miles?
Answer:
1 inch : 15 miles means 1 inch on the map represents 15 miles on the ground.
Let x inches be the length on the map.
We have:
1 inch : 15 miles = x inches : 48 miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{15 \mathrm{mi}}\) = \(\frac{x \mathrm{in}}{48 \mathrm{mi}}\)
Write without units:
\(\frac{1}{15}\) = \(\frac{x}{48}\)
Write cross products:
15x = 4
Divide both sides by 15:
\(\frac{15x}{15}\) = \(\frac{48}{15}\)
Simplify:
x = 3.2
On the map the length is 3.2 inches.

Question 11.
A map is drawn using a scale of 1 inch : 165 miles. The length of a road on the map is 12 inches. Find the actual length of the road.
Answer:
1 inch : 165 miles means 1 inch on the map represents 165 mites on the ground.
Let x mites be the actual length.
We have:
1 inch : 165 miles = 12 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in}}{165 \mathrm{mi}}\) = \(\frac{12 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{165}\) = \(\frac{12}{x}\)
Write cross products:
x = 165 · 12
Simplify:
x = 1980
The actual length of the road is 1980 miles.

Question 12.
On a particular map, 2 inches represents an actual distance of 64 miles. Towns A and B are 608 miles apart. Find the distance between the two towns, in inches, on the map.
Answer:
2 inches : 64 miles means 2 inches on the map represents 64 mites on the ground.
Let x inches be the distance on the map.
We have:
2 inches : 64 miles = x inches : 608 miles
Write ratios in fraction form:
\(\frac{2 \mathrm{in}}{64 \mathrm{mi}}\) = \(\frac{x \text { in. }}{608 \mathrm{mi}}\)
Write without units:
\(\frac{2}{64}\) = \(\frac{x}{608}\)
Write cross products:
64x = 2 · 608
64x = 1,216
Divide both sides by 64.
\(\frac{64x}{64}\) = \(\frac{1,216}{64}\)
Simplify:
x = 19
On the map the two cities are 19 inches apart.

Question 13.
On a particular map, 1 inch represents an actual distance of 2.5 miles. The actual area of a lake is 12 square miles. Find the area of the lake on the map.
Answer:
1 inch : 2.5 miles means 1 inch on the map represents 2.5 miles on the ground.
Map length : Actual length = 1 in. : 2.5 mi.
Map area : Actual area = 1 in.² : 2.5² mi.²
Let y represent the area of the lake on map in square inches.
Write a proportion:
\(\frac{\text { Area of lake on map }}{\text { Actual area of lake }}=\frac{1}{6.25}\)
Substitute:
\(\frac{y}{12}\) = \(\frac{1}{6.25}\)
Write the cross products
6.25y = 12
Divide both sides by 6.25 and simplify:
\(\frac{6.25y}{6.25}\) = \(\frac{12}{6.25}\)
y = 1.92
The area of the lake on the map is 1.92 square inches.

Question 14.
On the map, the area of a nature preserve is 54.2 square inches. If the scale of the map is 1 inch : 8 miles, find the actual area of the nature preserve.
Answer:
1 inch : 8 miles means 1 inch on the map represents 8 miles on the ground.
Map Length : Actual length = 1 in. : 8 mi.
Map area: Actual area = 1 in.² : 8² mi.²
Let y represent the actual area of the nature preserve in square miles.
Write a proportion:
\(\frac{\text { Area of nature preserve on map }}{\text { Actual area of nature preserve }}=\frac{1}{64}\)
Substitute:
\(\frac{54.2}{y}\) = \(\frac{1}{64}\)
Write the cross products:
y = 54.2 × 64
Simplify:
y = 3,468.8
The actual area of the nature preserve is 3,468.8 square miles.

Question 15.
The map shows two roads labeled A and B.

a) Using a ruler, measure, in centimeters, the lengths of roads A and B.

Answer:
We measure the lengths of the roads A and B, in centimeters, using a ruler:
Length A = 24 cm
Length B = 3.3 cm

b) Using the scale given, find, in kilometers, the actual lengths of roads A and B.
Answer:
We are given the scale:
1 : 50,000
We determine the actual length x of Road A in centimeters:
\(\frac{2.4}{x}\) = \(\frac{1}{50,000}\)
x = 2.4 · 50,000
= 120,000 cm
Convert to kilometers:
1 km = 1000 m = 100,000 cm
120,000 cm = \(\frac{120,000}{100,000}\) km = 1.2 km
We determine the actual length y of Road B in centimeters:
\(\frac{3.3}{y}\) = \(\frac{1}{50,000}\)
y = 3.3 · 50,000
= 165,000 cm
Convert to kilometers:
1 km = 1000 m = 100,000 cm
165,000 cm = \(\frac{165,000}{100,000}\) km = 1.65 km

Question 16.
The map shows seven cities in Florida. Using the scale on the map, use a ruler to measure the distance between the following pairs of cities. Then find the actual distance between them in miles.

a) Orlando and West Palm Beach
Answer:
We measure the lengths of the distance between Orlando and West Palm Beach, in centimeters, using a ruler:
Length Orlando-West Pam Beach = 3 cm
Length Fort Myers-Miami Beach = 24 cm
We are given the scale:
1 cm : 50 miles
We determine the actual distance x between Orlando and West Palm Beach in centimeters:
\(\frac{3}{x}\) = \(\frac{1}{50}\)
x = 3 · 50
= 150 miles

b) Fort Myers and Miami Beach
Answer:
We determine the actual distance y between Fort Myers and Miami Beach in centimeters:
\(\frac{2.4}{y}\) = \(\frac{1}{50}\)
x = 2.4 · 50
= 120 miles

Question 17.
Use the scale on the floor plan of a house to find each of the following.
a) The actual length and width of room 1.
Answer:
2 cm : 5 m means 2 cm on the plan represent 5 m on the ground.
Let x, y meters be the actual length and width of room 1.
We have:
2 cm : 5 m = 1.8 cm : x m
Write ratios in fraction form:
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{1.8 \mathrm{~cm}}{x \mathrm{~m}}\)
Write without units:
\(\frac{2}{5}\) = \(\frac{1}{2}\)
Write cross products:
2x = 5 · 1.8
Divide by 2 and simplify:
\(\frac{2x}{x}\) = \(\frac{9}{2}\)
x = 4.5 meters
We do the same to find the actual width:
2 cm : 5 m = 1.4 cm : y m
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{1.4 \mathrm{~cm}}{y \mathrm{~m}}\)
\(\frac{2}{5}\) = \(\frac{1.4}{y}\)
2y = 5 · 1.4
\(\frac{2y}{2}\) = \(\frac{7}{2}\)
y = 3.5 meters

b) The width of the door on the floor plan if its actual width is 0.8 meter.
Answer:
We determine the width z of the door on the floor plan:
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{z \mathrm{~cm}}{0.8 \mathrm{~m}}\)
\(\frac{2}{5}\) = \(\frac{z}{0.8}\)
5z = 2 · 0.8
\(\frac{5z}{5}\) = \(\frac{1.6}{5}\)
z = 0.32 cm

c) The actual area of the floor of the house to the nearest square meter

Answer:
We have:
Length scale: 2 cm : 5 m
Area scale: 2² cm² : 5² m²
We determine the actual area s of the floor:
\(\frac{4}{25}\) = \(\frac{7.6 \cdot 3}{s}\)
4s = 25 · 22.8
\(\frac{4s}{4}\) = \(\frac{570}{4}\)
s ≈ 143 m²

Question 18.
A tower is drawn using a scale of 1 inch : 3 feet. The height of the tower in the drawing is 1 foot 5 inches. Then, an architect decides to make a new scale drawing of the tower. In the new scale, the scale is 1 inch : 5 feet. Find the height of the tower in the new drawing.
Answer:
1 inch : 3 feet means 1 inch on the drawing represents 3 feet on the ground.
Let x miles be the actual height
We have:
1 inch : 3 feet 1 ft 5 in. : x feet
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{3 \mathrm{ft}}\) = \(\frac{(12+5) \mathrm{in} .}{x \mathrm{ft}}\)
Write without units:
\(\frac{1}{3}\) = \(\frac{17}{x}\)
Write cross products:
x = 3 · 17
Simplify
x = 51 ft
The actual height of the tower is 51 ft.
The new scale is 1 inch : 5 feet
We determine the height y of the tower in the new drawing:
\(\frac{1 \mathrm{in} .}{5 \mathrm{ft}}\) = \(\frac{y \mathrm{in} .}{51 \mathrm{ft}}\)
\(\frac{1}{5}\) = \(\frac{y}{51}\)
5y = 51
\(\frac{5y}{5}\) = \(\frac{51}{5}\)
y = 10.2 inches

Question 19.
Each student walked in a straight line from one point to another. Use a centimeter ruler to measure distances on the map shown. Use the scale on the map to find the distance each student walked in meters.

We measure the lengths of the roads, in centimeters, using a ruler:
Length DA = 1.8 cm
Length AB = 3.3 cm
Length BF = 2.5 cm
Length CF = 1.4 cm
Length EF = 2.8 cm
Length CE = 3.5 cm

a) Ethan walked from the library to the school, and then to the gym.
Answer:
We are given the scale:
1 : 12,500
We determine the actual length x Ethan walked:
\(\frac{D A+A B}{x}\) = \(\frac{1}{12,500}\)
\(\frac{1.8+3.3}{x}\) = \(\frac{1}{12,500}\)
x = 5.1 · 12,500
= 63,750 cm
Convert to meters:
1 m = 100 cm
63,750 cm = \(\frac{63,750}{100}\) m = 637.5 m

b) Joshua walked from the motel to the restaurant, and then to the movie theater.
Answer:
We determine the actual length y joshua walked:
\(\frac{F C+C E}{y}\) = \(\frac{1}{12,500}\)
\(\frac{1.4+3.5}{y}\) = \(\frac{1}{12,500}\)
x = 4.9 · 12,500
= 61,250 cm
Convert to meters:
1 m = 100 cm
61,250 cm = \(\frac{61,250}{100}\) m = 612.5 m

c) Chloe walked from the gym to the motel, and then to the movie theater.
Answer:
We determine the actual length z Chloe walked:
\(\frac{B F+F E}{z}\) = \(\frac{1}{12,500}\)
\(\frac{2.5+2.8}{z}\) = \(\frac{1}{12,500}\)
x = 5.3 · 12,500
= 66,250 cm
Convert to meters:
1 m = 100 cm
66,250 cm = \(\frac{66,250}{100}\) m = 662.5 m

Brain @ Work

Question 1.
You know you can bisect any angle using a compass and a straightedge. Geometers have known for thousands of years that it is impossible to trisect any given angle using a compass and straightedge. (The word trisect means to divide into three equal parts.) But it is possible to trisect certain angles, including a right angle. Using only a compass and straightedge, show how you can draw a right angle that is trisected.
Answer:
We are given the right angle:
m∠AOB – 90°

Use the compass: with the center in O, draw an arc. Labet its intersection with \(\overline{O A}\) and \(\overline{O B}\) by C and D

With the same radius place the compass in center C, then D and draw arcs which intersect the arc drawn in the previous step in N and M.

Use a ruler to draw \(\overrightarrow{O M}\) and \(\overrightarrow{O N}\).

m∠AOM = m∠MON = m∠NOB
= \(\frac{m \angle A O B}{3}\) = \(\frac{90^{\circ}}{3}\) = 30°

Question 2.
You accidentally broke your mother’s favorite plate. You want to ask an artist to reproduce it. You ask your math teacher to help you find the original size of the plate. She suggests that you locate three points on the rim of the plate and use the points to draw two segments. The point where the two perpendicular bisectors of these segments intersect will be the center of the plate. From that you can measure the radius. Copy and complete the diagram below. Then measure to find the diameter of the plate.

Answer:
We consider 3 points on the rim of the plate:

The center of the circle/plate is situated at the intersection of the perpendicular bisectors of \(\overline{A B}\) and \(\overline{B C}\).

The radius of the circle/plate is OA.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.4 Constructing Quadrilaterals to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.4 Answer Key Constructing Quadrilaterals

Math in Focus Grade 7 Chapter 7 Lesson 7.4 Guided Practice Answer Key

Construct the quadrilateral from the given dimension. Use a ruler and protractor.

Question 1.
Rectangle ABCD measuring 7 centimeters by 5 centimeters.
Explanation:

Answer:
we have drawn the rectangle with the measurements of length=5cm and breadth=7cm.

Construct the quadrilateral from the given information. Use a compass, ruler, and protractor.

Question 2.
Parallelogram KLMN: KL = 6.4 cm, LM = 4.8 cm, and m∠KLM = 60°.
Explanation:

Answer:
The measurements of the parallelogram of length= 4.8cm and breadth=6.4cm. The angle of KLM=60.

Construct the quadrilateral from the given
information. Use a ruler, compass, and protractor.

Question 3.
Rhombus PQRS with diagonal PR = 6.2 cm and PQ = 4.5 cm.
Explanation:

Answer:
we drawn the rhombus with the measurement angels of PR=6.2cm and PQ=4.5cm.

Math in Focus Course 2B Practice 7.4 Answer Key

Construct each quadrilateral from the given information.

Question 1.
Rectangle KLMN measuring 5.3 centimeters by 4.7 centimeters.
Answer:
The measuring angles of rectangle is Length = 5.3 cm and breadth= 4.7 cm

Explanation:

Question 2.
Square with side lengths of 7 centimeters.
Explanation:

Answer:
The measuring angles of the square is 7cm.

Question 3.
Rhombus DEFG with diagonal DF = 6 cm and DE = 4.5 cm.
Explanation:

Answer:

The measuring angles of the rhombus DEFG diagonals DF=6cm and DE=4.5cm.

Question 4.
Parallelogram PQRS with PQ = 3.8 cm, QR = 5 cm, and m∠QPS = 70°.
Answer:
Explanation:

The measuring angles of the parallelogram PQRS
PQ=3.8cm and QR=5cm
we got the angle 70°.

Question 5.
Quadrilateral ABCD with AB = 6.7 cm, BC = 7.2 cm, AD = 4.9 cm, CD = 6.2 cm, and m∠ABC = 55°.
Answer:
Explanation:

The measurements of quadrilateral ABCD of AB = 6.7, BC = 7.2, CD = 6.2 and DA = 4.9. We got the angle ABC = 55°.

Solve. Show your work.

Question 6.
Construct quadrilateral ABCD with diagonal AC = 5 cm, AB = CD = 4 cm, BC = 6 cm, and AD = 6 cm.
Answer:
Explanation:

The measurements of the rectangular quadrilateral is AB = CD = 4cm , BC and AD = 6cm. The diagonal AC = 5cm. So this is a quadrilateral rectangular.
We sketch the quadrilateral:

First we construct △ABC.
Use a ruler to draw the segment 6 cm long. Label its endpoints by B and C.

Because AB = 4 cm, use the ruler to set the compass to a radius of 4 cm. Then using B as center draw an arc of radius 4 cm above \(\overline{B C}\).

Because AC = 5 cm, use the ruler to set the compass to a radius of 5 cm. Then using C as center draw an arc of radius 5 cm that intersects the first arc. Label the intersection as A.

Use the ruler to draw \(\overline{A B}\) and \(\overline{A C}\).

Then we construct △ACD.
Because AD = 6 cm, use the ruler to set the compass to a radius of 6 cm. Then using A as center draw an arc of radius 6cm above \(\overline{A C}\).

Because CD = 4 cm, use the ruler to set the compass to a radius of 4 cm. Then using C as center draw an arc of radius 4 cm that intersects the first arc. Label the intersection as D.

Use the ruler to draw \(\overline{A D}\) and \(\overline{C D}\).

a) What type of quadrilateral is ABCD? Explain your reasoning.
Answer:
The quadrilateral ABCD is rectangle because the angles AB and CD are equal with the  =4cm and the angles BC and AD are equal with the = 6cm. So this is rectangular quadrilateral.

b) Draw \(\overline{B D}\) to intersect \(\overline{A C}\) at point E. Find the lengths of \(\overline{A E}\), \(\overline{C E}\), \(\overline{B E}\), and \(\overline{D E}\).
Answer:
Draw \(\overline{B D}\)

We measure AE, CE, BE, DE:
AE = 2.5
CE = 2.5
BE = 4.4
DE = 4.4

c) Do the diagonals \(\overline{A C}\) and \(\overline{B D}\) bisect each other? Justify your answer.
Answer:
The diagonals \(\overline{A C}\) and \(\overline{B D}\) bisect each other This is explained by the fact that the quadrilateral is a parallelogram.

Question 7.
Construct a quadrilateral STUV by following Steps 1 to 5.
STEP1
Draw \(\overline{S T}\) so that ST = 5 cm.
STEP 2
With T as the center, draw \(\overline{T U}\) perpendicular to \(\overline{S T}\) , with TU = 4 cm.
STEP 3
With U as the center, draw an arc of radius 5 centimeters.
STEP 4
With S as the center, draw an arc of radius 4 centimeters to intersect the arc drawn in Step 3. Label this point of intersection as V.
STEP 5
Complete the construction of quadrilateral STUV.
Answer:
Use a ruler to draw \(\overline{S T}\) so that is 5 cm long.

Using a protractor, draw ∠T with a measure of 90°.

Because TU = 4 cm, set the compass to a radius of 4 cm. Then using T as the center, draw an arc intersecting the ray drawn in previous step. Label this point of intersection as U.

With U as center, draw an arc of radius 5 cm.
With S as center, draw an arc of radius 4 cm to intersect the arc drawn in the previous step. Label this point of intersection as V.

Draw \(\overline{V U}\) and \(\overline{V S}\).

a) Find each of the angles in quadrilateral STUV.
Answer:
We measure the angles of the quadrilateral:
m∠T = 90°
m∠U = 90°
m∠V = 90°
m∠S = 90°

b) Name the quadrilateral.
Answer:
As all angles are right angles, the quadrilateral is a rectangle.

c) Find the lengths of the diagonals. What do you notice?
Answer:
We find the lengths of the diagonals:
SU ≈ 6.4
VT ≈ 6.4

The diagonals are congruent.

Question 8.
Construct a quadrilateral ABCD with all sides of length 3 centimeters and diagonal BD = 5.2 cm.
a) What type of quadrilateral is ABCD? Explain your reasoning.
Answer:
Explanation:

The quadrilateral ABCD is a square because the sides of the square AB, BC, CD  and DA are equal with 3 cm and the diagonal BD=5.2cm.

b) Find the measure of each of the angles formed by the intersection of the diagonals.
Answer:
The measuring of each of the angles formed by the intersection of the diagonals is acute angle 60°.

Question 9.
Construct quadrilateral ABCD with diagonal AC = 6 cm, AB = 3 cm, BC = 4 cm, CD = 4.5 cm, and AD = 9 cm. What type of quadrilateral does ABCD seem to be? Explain your reasoning.
Answer:
Explanation:

The quadrilateral ABCD is trapezoid. AB=3cm ,BC=4cm,CD=4.5cm and DA=9cm.
The diagonal AC = 6cm.So this is trapezoid.

Question 10.
Construct the figure below using the given dimensions.

Answer:
Use a ruler to draw \(\overline{A B}\) so that is 6.6 cm long.

Using a protractor, draw ∠B with a measure of 90° and ∠A with a measure of 90°.

Because AD = BC = 3.4 cm, set the compass to a radius of 3.4 cm. Then using A and B as centers, draw arcs intersecting the rays drawn in previous step. Label this points of intersection as D and C.

Because AE = EC = 4 cm, set the compass to a radius of 4 cm. Then using C and D as centers, draw arcs above \(\overline{C D}\). Label the point of intersection of the two arcs as E.

Draw \(\overline{E D}\) and \(\overline{E C}\).

Question 11.
a) Construct quadrilateral ABCD with AB = 6 cm, BC = 3 cm, AD = 4 cm, m ∠BAD = 120°, and m ∠ABC = 100°.
Explanation:

Answer:
The measurements of the quadrilateral ABCD is AB=6cm , BC=3cm , AD =4cm. The angle BAD= 120° angle ABC =100°.

b) Find the length of \(\overline{C D}\).
Answer:
The length of overline CD=10cm.

c) Label the midpoints of the four sides of this quadrilateral as W, X, Y, and Z. Join them to form quadrilateral WXYZ.
Answer:
Explanation:

d) Compare the lengths of \(\overline{W X}\) and \(\overline{Y Z}\) . Compare the lengths of \(\overline{X Y}\) and \(\overline{W Z}\). What do you notice?
Answer:
Measure WX, YZ, XY, WZ:
WX ≈ 3.6 cm
YZ ≈ 3.6 cm
XY ≈ 4.4 cm
WZ ≈ 4.4 cm
We notice that we have:
WX = YZ
XY = WZ

Question 12.
Construct parallelogram PQRS with PQ = 6 cm, a height of 4.5 centimeters and interior angles 45° and 135°.
Explanation:

Answer:
The measurements of the parallelogram PQRS with PQ =6cm and height= 4.5cm.
The interior angles are  45° and 135°.

Question 13.
Draw rhombus ABCD with AC = 5 cm and AB = 6.5 cm. Also draw diagonal \(\overline{B D}\). How are the two diagonals related to each other? Explain.
Answer:
Explanation:

let ABCD be a rhombus whose diagonal AC and BD are intersecting at point O. We know that the diagram of parallelogram bisect at each other.
OA=OC and OB=OD
angle COB= angle COD
angle COB+ angle COD= 180°
angle COB=angle COD= 90°.

Question 14.
Jenny plans to make a trapezoidal bookmark for each of her teachers. The top will be 5 centimeters long and have right angles at either end. The right side will be 12 centimeters long, and the bottom of the bookmark will make a 50° angle with this side. Construct a template for Jenny’s bookmark. How long is the left side of Jenny’s template to the nearest centimeter?
Answer:
Explanation:

The measurements of the trapezoid length is 5cm and breadth is 12 cm. The bottom of the book mark angle is  50°. The left side of the jenny’s template is 4cm.

Question 15.
Martha plans to cut squares of paper from a roll of wrapping paper. She will package the squares to sell as origami papers to raise funds for a charity. If the area of the square paper is 64 square centimeters, construct the square Martha can use as a template.
Answer:
Determine the length of the square side:
AB² = 64
AB = \(\sqrt{64}\) = 8
Sketch the rectangle:

Use a ruler to draw \(\overline{A B}\) so that is 7 cm long.

Using a protractor, draw ∠B with a measure of 90°.

Because BC = 5 cm, set the compass to a radius of 5 cm. Then using B as the center, draw an arc intersecting the ray drawn in previous step. Label this point of intersection as C. Because AD = 5 cm, use the same compass setting. Using A as the center, draw an arc:

Because CD = 7 cm, set the compass to a radius of 7 cm. Then using C as the center draw an arc intersecting the ray drawn in previous step. Label this point of intersection as D.

Draw \(\overline{C D}\) and \(\overline{A D}\)

Question 16.
Jessie plans to make a patchwork pattern from colored paper by repeating a rhombus whose diagonals measure 4 centimeters and 5 centimeters. Use the given dimensions to construct a template for the rhombus.
Answer:
We have to construct the rhombus ABCD.
AC = 4
BD = 5
Sketch the rhombus:
Use a ruler to draw \(\overline{A C}\) so that is 4 cm long.

Draw the perpendicular bisector of \(\overline{A C}\):

Because OB = OD = \(\frac{5}{2}\) = 2.5 cm, use the compass setting 2.5 cm. Then using O as the center, draw an arc to intersect the perpendicular bisector in the points B and D.

Draw \(\overline{A B}\), \(\overline{B C}\), \(\overline{C D}\) and \(\overline{D A}\).

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.3 Constructing Triangles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.3 Answer Key Constructing Triangles

Math in Focus Grade 7 Chapter 7 Lesson 7.3 Guided Practice Answer Key

Construct the triangle from the given information. Use a compass and ruler.

Question 1.
Triangle PQR: PQ = 5.6 cm, QR = 4.5 cm, and PR = 8.2 cm.

Answer:
The measurements of the triangle are PQ=5.6cm ,PR=8.2cm and QR=4.5cm.

Eplanation:

Construct the triangle from the given information. Use a ruler and compass.

Question 2.
Triangle ABC: BC = 4 cm, m∠ABC = 25°, and m∠ACB = 120°.
Answer:
The measurements of the triangle are PQ=5.6cm ,PR=8.2cm and QR=4.5cm.

Explanation:

Question 3.
Triangle KLM: KL = 8.2 cm, KM = 6.9 cm, and m∠LKM = 75°.
Answer:
The measurements of the triangle KLM are KL=8.2cm ,KM=6.9cm and the angle LKM=75°.

Explanation:

Construct the triangle from the given information. Use a compass, ruler, and protractor.

Question 4.
Triangle KLM: KL = 7cm, KM = 9 cm, and m∠KLM = 125°.
Answer:
The measurements of the triangle KLM are KL=7cm , KM=9cm and m∠KLM= 125°.

Explanation:

Hands-On Activity

Materials

• protractor
• compass
• ruler

Decide Whether Given Measures Can Be Used To Construct One Triangle, More Than One Triangle. Or No Triangles

Work in pairs.

Step 1.
Try to construct triangle ABC with AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.
Answer:
The measurements of the triangle ABC are  AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.

Explanation:

Step 2.
Try to construct triangle DEF with DE = 4.6 cm, EF = 6 cm, and DF = 12 cm.
Answer:
The measurements of the triangle DEF are DE = 4.6 cm, EF = 6 cm, and DF = 12 cm.

Explanation:

Step 3.
Try to construct triangle GHI with GH = 6 cm, HI = 5 cm, and JK = 6 cm, JL = 4.7 cm, and m∠GHI = 50°.

Step 4.
Try to construct triangle JKL with JK = 6 cm, JL = 4.7 cm, and m∠JKL = 50°.
Answer:
The measurements of the triangle JKL are JK = 6 cm, JL = 4.7 cm, and m∠JKL = 50°.

Explanation:

Step 5.
Try to construct triangle MNP with MN = 7 cm, m∠MNP = 60°, and m∠PMN = 40°.
Answer:
The measurements of the triangle MNP are MN = 7 cm, m∠MNP = 60°, and m∠PMN = 40°.

Explanation:

Step 6.
Were there any triangles you could not construct? Were there any triangles that you could construct in more than one way? Explain.
Answer:
Yes, at step 3 we could not construct any triangle. At step 1 we can construct more than one triangle because the measurements of the triangle ABC are  AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.

Math Journal
Use your results to decide whether you can always construct exactly one triangle from the given information. Justify your answer.

a) Given three side lengths
Answer:

b) Given two side lengths and an angle measure
Answer:

Find the number of triangles that can be constructed. Try constructing the triangles to make your decision.

Question 5.
PQ = 4.8 cm, QR = 5.4 cm, and m∠PQR = 100°.
Answer:
The measurements of the triangle PQR = PQ = 4.8 cm, QR = 5.4 cm, and m∠PQR = 100°.

Explanation:

Question 6.
AB = 6.2 cm, BC = 4.8 cm, and m∠BAC = 75°.
Answer:
The measurements of the triangle ABC are AB = 6.2 cm, BC = 4.8 cm, and m∠BAC = 75°.

Explanation:

Question 7.
ST = 7.7 cm, SU = 5.2 cm, and m∠STU = 40°.
Answer:
The measurements of the triangle STU are ST = 7.7 cm, SU = 5.2 cm, and m∠STU = 40°.

Explanation:

Math in Focus Course 2B Practice 7.3 Answer Key

Use the given information to construct each triangle.

Question 1.
In triangle CDE, CD = 7 cm, DE = 4 cm, and CE = 6.5 cm.
Answer:
The measurements of the triangle CDE , CD = 7 cm, DE = 4 cm, and CE = 6.5 cm.

Explanation:

Question 2.
In triangle ABC, BC = 6 cm, m∠ABC = 30°, and m∠ACB = 60°. Find m∠BAC and AC.
Answer:
The measurements of the triangle ABC , BC = 6 cm, m∠ABC = 30°, and m∠ACB = 60°.
B= 30° and C= 60°
A=180°-B-C
A=180°- 30°- 60°
A=90°.
AC=3cm.

Explanation:

Question 3.
In an equilateral triangle, each side length is 6.5 centimeters long.
Answer:
The measurements of the equilateral triangle , each side length is 6.5 cm.

Explanation:

Question 4.
In triangle ABC, AB = 4 cm, AC = 5 cm, and m∠ABC = 40°.
Answer:
The measurements of the triangle ABC, AB = 4 cm, AC = 5 cm, and m∠ABC = 40°.

Explanation:

Question 5.
In triangle ABC, AB = 6 cm, BC = 8 cm, and AC = 10 cm. What kind of triangle is triangle ABC? Classify it by both sides and angles.
Answer:
The measurements of the triangle ABC, AB = 6 cm, BC = 8 cm, and AC = 10 cm. The triangle ABC is a right angle triangle. Right angle triangle has 3 sides and it have one angle  90°.

Explanation:

Question 6.
In triangle XYZ, XV = XZ = 4 cm and YZ = 5 cm. Find m∠XZY.
Answer:
The measurements of the triangle XYZ, XV = XZ = 4 cm and YZ = 5 cm.
The angle of the XZY is 50°.

Explanation:

Solve. Show your work.

Question 7.
Triangle POR has the dimensions shown in the diagram.

a) Construct triangle POR.
Answer:

b) Using a ruler, from your construction, measure the length of \(\overline{P R}\).
Answer:
Using a ruler we measure the length of \(\overline{P R}\).
PR ≈ 6.1 cm

c) Find the measures of ∠P and ∠R without using a protractor. Justify your answer.
Answer:
△PQR is Isosceles because PQ = QR Therefore the angles P and R we congruent:
m∠P = m∠R
Find m∠P and m∠R:
m∠P + m∠R – m∠Q = 180°
2m∠P + 100° = 180°
2m∠P = 180° – 100°
2m∠P = 80°
m∠P = m∠R = \(\frac{80^{\circ}}{2}\) = 40°

Question 8.
Math Journal
Is it possible to construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm? Explain.
Answer:
We cannot construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm because the lengths of the PR , QR are very small.

Explanation:

Question 9.
Three triangles have angle measures of 500 and 600. In one triangle, the side included between these angles is 2 centimeters. In the second triangle, the included side length is 3 centimeters, and in the third triangle, the included side length is 4 centimeters.

a) Construct the three triangles.
Answer:

b) In each triangle, what is the measure of the third angle?
Answer:
Sum of angles of triangle is 180 so the given triangles have angle measures of 50 degree and 60 degree another angle is 70 degree.

c) Using the triangles constructed to help you, what can you deduce about the number of triangles that can be constructed if you are given three angle measures of a triangle but not the measure of any side length?
Answer:
Given only three angle measures of a triangle we can construct an infinity of triangles, their sides being proportional.

Question 10.
Math Journal
Suppose you are given three angle measures whose sum is 180°, Can you form a triangle given this information? Are there other different triangles you can form? Explain.
Answer:
The angle sum property of a triangle states that the angles of a triangle always adds upto 180°.

Explanation:

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.2 Constructing Perpendicular Bisectors to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.2 Answer Key Constructing Perpendicular Bisectors

Math in Focus Grade 7 Chapter 7 Lesson 7.2 Guided Practice Answer Key

Hands-On Activity

Materials:

• ruler

Explore The Distance Between Points On The Perpendicular Bisector Of A Segment And The Endpoints Of The Segment

Work in pairs.

Line XY is the perpendicular bisector of line segment AB. Points W, X, Y, and Z are four points on the perpendicular bisector of \(\overline{\mathrm{AB}}\).

Measure and record each length to the nearest tenth of a centimeter.

Step 2.
Compare the lengths of \(\overline{A W}\) and \(\overline{B W}\). Then compare the lengths of each of the following pairs of segments: \(\overline{A X}\) and \(\overline{B X}\), \(\overline{A Y}\) and \(\overline{B Y}\), and \(\overline{A Z}\) and \(\overline{B Z}\). What do you notice about each pair of line segment lengths?
Answer:

Math Journal
Suppose you choose any point on the perpendicular bisector and measure the distances from that point to points A and B. What do you predict about the distances? What conclusion can you make?

From the activity, any point on the perpendicular bisector of a line segment is equidistant from the two endpoints of the line segment.

Complete.

Question 1.
Copy or trace triangle XYZ. Then draw the perpendicular bisector of line segment XY.

Answer:

Copy the triangle and complete.

Question 2.
A trucking company has most of its business in Cities P, Q, and R. Where should it locate its new facility so that it is equidistant from all three cities? Mark the location of the facility on a copy of the map.

Answer:

Math in Focus Course 2B Practice 7.2 Answer Key

Draw each line segment and construct its perpendicular bisector.

Question 1.
XY = 5 cm
Answer:

Question 2.
PQ = 6.8
Answer:

Question 3.
AB = 8.8 cm
Answer:

Question 4.
MN = 11 cm
Answer:

Question 5.
Draw a line segment between 4 inches and 5 inches and label the endpoints M and N. Construct the perpendicular bisector of \(\overline{\mathrm{MN}}\). Explain briefly if you could construct a different perpendicular bisector of \(\overline{\mathrm{MN}}\).
Answer:

Question 6.
Math Journal
In the diagram below, point M is the midpoint of \(\overline{\mathrm{AB}}\). Ben drew line segment XY through point M. He labeled it as the perpendicular bisector of \(\overline{\mathrm{AB}}\). Do you agree with Ben? Give a reason for your answer.

Answer:
Yes, I do agree with Ben.

Explanation:
In the given diagram, point M is the midpoint of \(\overline{\mathrm{AB}}\). Ben also drew line segment XY through point M. So Ben labeled it as the perpendicular bisector of \(\overline{\mathrm{AB}}\). A perpendicular bisector is a line that bisects another line at a right angle, through the intersection point. Hence I agree with Ben.

Question 7.

Answer:

Question 8.
Draw the perpendicular bisectors of \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{PR}}\) on a copy of each polygon. Label the point where the two perpendicular bisectors meet as W.

Answer:

Question 9.
The point that is equidistant from points W, X, and Y. On a copy of WXYZ, mark the points that are described, if possible. Otherwise, explain why you cannot.
Answer:
We are given the polygon:

The point equidistant from points W, X and Y is the intersection of the perpendicular bisectors of the segments \(\overline{W X}\), \(\overline{X Y}\) and \(\overline{Y W}\).

It is enough to draw two of the perpendicular bisectors as the three of them intersect in the same point.
We draw the perpendicuLar bisectors on \(\overline{X Y}\) and \(\overline{W X}\):

They intersect in point A. therefore A is the point that is equidistant from points W, X and Y.

Question 10.
The point that is equidistant from \(\overline{W X}\), \(\overline{W Z}\), and \(\overline{X Y}\).
Answer:

The points equidistant from \(\overline{W X}\) and \(\overline{W Z}\) are on the angle bisector of ∠XWZ.
The points equidistant from \(\overline{W X}\) and \(\overline{X Y}\) are on the angle bisector of ∠WXY.
Therefore the point equidistant from \(\overline{W X}\), \(\overline{W Z}\) and \(\overline{X Y}\) is the intersection of the two angle bisectors.
We draw the angle bisector of ∠XYZ:

We draw the angle bisector of ∠WXY:

They intersect in point B, therefore B is the point that is equidistant from \(\overline{W X}\), \(\overline{W Z}\) and \(\overline{X Y}\):

Question 11.
The point that is equidistant from \(\overline{W X}\) and \(\overline{W Z}\), and also from points X and Y.
Answer:
We are given the polygon:

The points equidistant from \(\overline{W X}\) and \(\overline{W Z}\) are on the angle bisector of ∠XYZ.
The points equidistant from the points X and Y are on the perpendicular bisector of \(\overline{X Y}\).
Therefore the point equidistant from \(\overline{W X}\) and \(\overline{W Z}\) and from points X and Y is the intersection between the angle bisector of ∠XWZ and the perpendicular bisector of \(\overline{XY}\).
We draw the angle bisector of ∠XWZ:

We draw the perpendicular bisector of \(\overline{XY}\):

They intersect in point C, therefore C is the point that ¡s equidistant from \(\overline{W X}\) and \(\overline{W Z}\) and points X, Y.

Solve.

Question 12.
Mark the point on \(\overline{Q R}\) that is equidistant from points Q and R on a copy of triangle POR.

Answer:

Question 13.
Mr. Smith wants to put a circular water sprinkler in his triangular-shaped garden. His garden has a tree at each vertex. The water sprinkler is to be equidistant from the trees. Copy or trace the given triangle. Mark point W to show where Mr. Smith should put the water sprinkler.

Answer:
We are given:

The point equidistant from the 3 trees is the intersection of the perpendicuLar bisectors of the triangles edges

It is enough to draw two of the perpendicular bisectors as the three of them intersect in the same point
We draw the perpendicular bisectors on two edges:

They intersect in point A. therefore A is the point that is equidistant from the 3 trees.

Question 14.
Math Journal
Melissa was asked to bisect a very long line segment using a compass and a straightedge. Melissa opened the compass to the widest possible setting. She found that it was not wide enough to do the standard construction that she had learned in math class. Is it possible for Melissa to bisect the very long line segment with only her compass and a straightedge? Explain and give a suggestion.
Answer:
Let \(\overline{A B}\) be the initial very long segment.
In order to be able to use an opening of the compass greater than half of a segment, she should ‘decrease the length of the given long segment as much as possible. For this, she places the compass, with its greatest opening, in point A and draw an arc which intersects \(\overline{A B}\) in X₁. With the same opening she places the compass in B and draw an arc which intersects \(\overline{A B}\) in Y₁.. She got a smaller segment \(\overline{X_{1} Y_{1}}\).

She continues placing the compass in X₁ and Y₁ and drawing arcs until she gets a segment \(\overline{X_{n} Y_{n}}\),  for which she can construct the perpendicular bisector.

She constructs the perpendicular bisector of \(\overline{X_{n} Y_{n}}\), which is also the perpendicular bisector of \(\overline{A B}\) because AX_n = BY_n.

Decrease the length of the segment with equal values until the procedure of constructing the perpendicular bisector can be applied.

Question 15.
Math Journal
Copy the rectangle shown. Use the least possible number of arcs and lines to construct the perpendicular bisectors of each side of the rectangle. Explain how you know you have used the least possible number.

Answer:

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Lesson 4.1 Understanding Equivalent Equations to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Lesson 4.1 Answer Key Understanding Equivalent Equations

Math in Focus Grade 7 Chapter 4 Lesson 4.1 Guided Practice Answer Key

Copy and complete to state whether each pair of equations are equivalent equations. Give a reason for your answer.

Question 1.

x – 3 + 4x = 5 and 5x = 2
x – 3 + 4x = 5
5x – 3 = 5 Group like terms.
5x – 3 + = 5 + Add to both sides.
5x = Simplify.
x – 3 + 4x = 5 be rewritten as 5x = 2.
So, the equations have solutions and are
Answer:
No, they are not equivalent equations.

Explanation:
Given 1st equation, x – 3 + 4x = 5

It can also be written as 5x – 3 =5

Add 3 to both sides, 5x – 3 + 3 = 5 + 3

5x = 8

So, equation x – 3 + 4x = 5  cannot be written as 5x = 2

Question 2.
x + 7 = 12 and 2x = 10
First solve x + 7 = 12.
x + 7 = 12
x + 7 – = 12 – Subtract from both sides.
x = Simplify.
Then check to see if is the solution of the equation 2x = 10.
If x = , 2x = 2 • Substitute for x.
= a solution.
Because the equations have the solution, they are equations.
Answer:
Yes, Both are equivalent equations.

Explanation:
First, let’s solve x + 7 = 12

Subtract 7 on both sides.

x + 7 – 7 = 12 -7

x = 5

Lets consider second equation 2x = 10

substitute x= 5 which we got by solving  the 1st equation in the 2nd equation.

2.(5) = 10

Hence proved.

Question 3.
1.2x = 2.4 and x – 6 = 8
First solve x – 6 = 8.
x – 6 = 8
x – 6 + = 8 + Add to both sides,
x = Simplify.
Then check to see if is the solution of the equation 1.2x = 2.4.
If x = , 1.2x = 1.2 • Substitute for x.
= a solution.
Because the equations have solutions, they are equations.
Answer:
No, they are not equivalent equations.

Explanation:
First, lets consider x – 6 = 8

Add 6 on both sides

x – 6 + 6 = 8 + 6

x = 14

Lets substitute x = 14  in the first equation.

1.2 x = 2.4

1.2(14) = 16.8

Both the equations are not equivalent equations.

Question 4.
\(\frac{2}{5}\)x = 4 and x = 10.
If x = 10, \(\frac{2}{5}\)x = \(\frac{2}{5}\) • 10 Substitute 10 for x.
= a solution.
Because the equations have the solution, they are equations.
Answer:
Yes, both the equations have the same solution, they are equivalent equations.

Explanation:
Let’s consider first equation, (2/5)x =4

Let us  substitute the second equation, x =10 in first equation.

(2/5) × 10 = 4

Hence proved.

Both the equations have the same solution, they are equivalent equations.

Math in Focus Course 2A Practice 4.1 Answer Key

Tell whether each pair of equations are equivalent. Give a reason for your answer.

Question 1.
2x = 4 and 4x + 5 = 13
Answer:
Yes, Both the equations are equivalent.

Explanation:
Let us consider second equation, 4x + 5 = 13

Subtract 5 on both sides.

4x + 5 – 5 = 13 – 5

4x = 8

x = 8 ÷ 4

x = 2

Substitute x=2 in the first equation, 2x = 4

2(2) = 4

Hence proved.

Both the equations are equivalent.

Question 2.
-2x + 9 = 7 and —2x = 2
Answer:
No, both the equations are not equivalent.

Explanation:
Lets solve the first equation, -2x + 9 = 7

Subtract 9  on both sides.

-2x + 9 – 9 = 7 – 9

-2x = -2

The second equation is -2x = 2

So, both the equations are not equivalent.

Question 3.
5x – 4 + 3x = 8 and 8x = 12
Answer:
Both are equivalent equations.

Explanation:
Lets us consider first equation, 5x – 4 + 3x = 8

8x – 4 = 8

8x = 8 + 4

8x = 12

As we derived second equation from the first equation, both are equivalent equations.

Question 4.
\(\frac{3}{4}\)x – 7 = 2 and x = 12
Answer:
Yes, both are equivalent equations.

Explanation:
Let us consider first equation, (3/4)x = 2

Substitute, the second equation, x = 12 in the first equation.

(3/4)(12) – 7 = (3).(3) – 7

=9 – 7

=2

Hence proved. Both are equivalent equations.

Match each equation with an equivalent equation.

Answer:
5 – d
6 – c
7 – e
8 – a
9 – g
10 – f
11 – b

Explanation:
5.  0.5x + 1 = 1.5

0.5x = 1.5 – 1 = 0.5

x = 0.5 ÷ 0.5

x = 1

We can match 5th equation with equation (d) on the other side.

6. 9 + 3.5x = 16

Let us solve the equation

3.5x = 16 – 9

3.5x = 7

x = 7 ÷ 3.5

x = 2

We can match the equation (3/2)x = 3

x = 3 × (2/3)

x = 2

Hence both are same.

7. (4/5)x =4

4x = 20

x =5

We can match 7th equation with equation (e) on the other side.

2x =10

x =5

Hence Matched.

8.  2x + (1/2) = (7/2)

2x = (7/2) – (1/2)

2x = (6/2)

2x = 3

x = 3/2

As option (a) is 6x =9

It can be also written as 3/2 as both are dividends of 3.

9. x – 8.3 = 1.3

x = 1.3 + 8.3

x = 9.6

Option (g) is  (1/2)x = 4.8

x =9.6

Hence matched.

10. 13.9 = 2.5x

x = 13.9 ÷ 2.5

x = 5.56

Option (f) is 1.2 + x = 6.76

x = 6.76 – 1.2

x =5.56

Hence proved.

11. 4x = (4/9)

x = 1/9

We can match with option (b).

(3/5)x = (1/15)

x = (1/9)

Solve.

Question 12.
Math Journal Max was asked to write to \(\frac{2}{3}\)x = 3 – x. He wrote the following:
\(\frac{2}{3}\)x = 3 – x
\(\frac{2}{3}\)x ∙ 3 = 3 ∙ 3 – x
2x = 9 – x
He concluded that \(\frac{2}{3}\)x = 3 – x and 2x = 9 \(\frac{2}{3}\) x are equivalent equations. Do you agree with his conclusion? Give a reason for your answer.
Answer:
No, I don’t agree with Max, both are not equivalent equations.

Explanation:
Let us consider first equation, (2/3)x = 3 – x

2x = 9 -3x (Max gave the equation 2x = 9 – x which is wrong, if we multiplied 3 on both sides not only it is multiplied with 3 it is also multiplies with x. so it is 2x = 9 -3x)

5x = 9

x = 9/5

x = 1.8

The second equation, 9(2/3)x =2x

substitute x =1.8 in the second equation

2 (1.8) = 9(2/3)(1.8)

3.6 = 10.8

Hence the both the equations are not equivalent.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Algebraic Equations and Inequalities to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Answer Key Algebraic Equations and Inequalities

Math in Focus Grade 7 Chapter 4 Quick Check Answer Key

Solve each equation.

Question 1.
x + 4 = 10
Answer:
6

Explanation:
x + 4 = 10
x = 10 -4

x = 6

Question 2.
x – \(\frac{1}{2}\) = 2
Answer:
2(1/2)

Explanation:
x – (1/2) = 2

x = 2 + (1/2)

x = 2 (1/2)

Question 3.
\(\frac{1}{5}\)x = 3
Answer:
15

Explanation:
(1/5)x = 3
x=  3 × 5

x = 15

Question 4.
1.2x = 2.4
Answer:
2

Explanation:
1.2x=2.4

x= 2.4 ÷ 1.2

x= 2

State whether each statement is True or False.

Question 5.
x = 1 gives the solution of the algebraic equation 3x + 5 = 8.
Answer:
True

Explanation:
3x + 5 = 8

Given, x = 1. Substitute x= 1 in the equation.

3(1) + 5  = 8

8 = 8

Hence proved.

Question 6.
y = 2 gives the solution of the algebraic equation 6y – 3 = 8.
Answer:
False

Explanation:
6y – 3 = 8

Given y = 2, substitute in the equation.

6(2) – 3 = 8

12 – 3 = 8

9 ≠ 8

So False.

Question 7.
z = 6 gives the solution of the algebraic equation \(\frac{z}{3}\) = 3.
Answer:
False

Explanation:
\(\frac{z}{3}\) = 3

(z/3) = 3

Given, z = 6, Substitute it in the equation.

(6/3) = 3

2 ≠ 3

So  false.

Question 8.
w = 3 gives the solution of the algebraic equation 2w = 6.
Answer:
True

Explanation:
Given equation, 2w = 6.

Given w=3, substitute it in the equation.

2(3) = 6

6 = 6

So the equation is true.

Draw a number line to represent each inequality.

Question 9.
x ≥ 3.5
Answer:

Explanation:
Given equation, x ≥ 3.5

In the equation , x can be any number which is equal to (or) more than 3.5.

It is represented in the number line as given above.

Question 10.
y < \(\frac{1}{2}\)
Answer:

Explanation:
Given equation, y < (1/2)

In the equation, y is a number which is less than (1/2)

It is represented in the number line as given above.

Complete each with =, >, or <.

Question 11.
11 12
Answer:
11 < 12

Question 12.
-9 -7
Answer:
-9  <  -7

Question 13.
25 ∙ (-1) (-1) ∙ 25
Answer:
25 ∙ (-1)  =  (-1) ∙ 25

Question 14.
3 ÷ (-1) (-1) ÷ 3
Answer:
3 ÷ (-1) = (-1) ÷ 3

Use x to represent the unknown quantity. Write an algebraic inequality for each statement.

Question 15.
The box can hold less than 70 pounds.
Answer:
x < 70

Explanation:
Let us consider the box be “x”.

As the box can hold less than 70 pounds.

Hence the equation x < 70 .

Question 16.
You have to be at least 17 years old to qualify for the contest.
Answer:
x ≥ 17

Explanation:
“At least” means it should be minimum of 17 or more than that.

Given, you have to be at least 17 years old to qualify for the contest.

Hence the equation, x ≥ 17.

Question 17.
The width of luggage that you can carry onto the plane is at most 17 inches.
Answer:
y ≤ 17

Explanation:
“At most” means it should be maximum of 17 nor less than that, but not more than 17.

As, The width of luggage that you can carry onto the plane is at most 17 inches.

Hence the equation, y ≤ 17

Question 18.
There are more than 120 people standing in line for the roller coaster.
Answer:
y > 120

Explanation:
The statement states “There are more than 120 people standing in line for the roller coaster” i.e, higher than 120.

Hence the equation, y > 120

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 3 Review Test to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 3 Review Test Answer Key

Concepts and Skills
Simplify each expression.
Question 1.
1.4w – 0.6w
Answer:
1.4w – 0.6w = 0.8w.

Explanation:
1.4w – 0.6w
= 0.8w.

Question 2.
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m
Answer:
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m = \(\frac{31}{20}\)m.

Explanation:
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m
LCD of 4 n 5 = 20.
= [(3 × 5) + (4 × 4)]m ÷ 20
= (15 + 16)m ÷ 20
= 31m ÷ 20 or \(\frac{31}{20}\)m.

Question 3.
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y
Answer:
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.

Explanation:
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.
LCD of 6 n 2 = 6.
= [(1 × 1) + (1 × 3)]y ÷ 6] +  \(\frac{1}{3}\)y
= [(1 + 3)y ÷ 6] + \(\frac{1}{3}\)y
= (4y ÷ 6) + \(\frac{1}{3}\)y
= \(\frac{2}{3}\)y+ \(\frac{1}{3}\)y
= (2 + 1)y ÷ 3
= 3y ÷ 3
= y.

Question 4.
1.8m – 0.2m – 7m
Answer:
1.8m – 0.2m – 7m = -5.4m.

Explanation:
1.8m – 0.2m – 7m
= 1.6m – 7m
= -5.4m.

Question 5.
1.3a – 0.8b + 2.2b – a
Answer:
1.3a – 0.8b + 2.2b – a = 0.3a + 1.4b.

Explanation:
1.3a – 0.8b + 2.2b – a
= 1.3a – a – 0.8b + 2.2b
= 0.3a – 0.8b + 2.2b
= 0.3a + 1.4b.

Question 6.
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a
Answer:
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a = 1 + a + \(\frac{3}{5}\)b.

Explanation:
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a
= 1 + \(\frac{1}{5}\)a + \(\frac{4}{5}\)a + \(\frac{3}{5}\)b
= 1 + [(1 + 4)a ÷ 5] + \(\frac{3}{5}\)b
= 1 + (5a ÷ 5) + \(\frac{3}{5}\)b
= 1 + a + \(\frac{3}{5}\)b.

Expand each expression. Then simplify when you can.
Question 7.
1.2(2p – 3)
Answer:
1.2(2p – 3) = 2.4p – 3.6.

Explanation:
1.2(2p – 3)
= (1.2 × 2p) – (3 × 1.2)
= 2.4p – 3.6.

Question 8.
\(\frac{1}{3}\)(12p + 9q)
Answer:
\(\frac{1}{3}\)(12p + 9q) = 4p + 3q.

Explanation:
\(\frac{1}{3}\)(12p + 9q)
= [12p × \(\frac{1}{3}\)] + [ 9q × \(\frac{1}{3}\)]
= 4p + 3q.

Question 9.
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)
Answer:
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\) = \(\frac{t}{15}\) + \(\frac{1}{10}\).

Explanation:
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)
= [\(\frac{t}{3}\) × \(\frac{1}{5}\)] + [\(\frac{1}{2}\) × \(\frac{1}{5}\)]
= \(\frac{t}{15}\) + \(\frac{1}{10}\).

Question 10.
-4(-2q + 2.5)
Answer:
-4(-2q + 2.5) = 8q – 10.

Explanation:
-4(-2q + 2.5)
= (-4 × -2q) + (2.5 × -4)
= 8q – 10.

Question 11.
–\(\frac{2}{3}\)(6x + 3)
Answer:
–\(\frac{2}{3}\)(6x + 3) = 4x – 2.

Explanation:
–\(\frac{2}{3}\)(6x + 3)
= (6x × –\(\frac{2}{3}\)) + (3 × –\(\frac{2}{3}\))
= (2x × 2) + (1 × -2)
= 4x – 2.

Question 12.
-0.5(2m – 4n)
Answer:
-0.5(2m – 4n) = -m + 0.5n.

Explanation:
-0.5(2m – 4n)
= (2m × -0.5) – (1n × -0.5)
= -1m – (-0.5n)
= -m + 0.5n.

Question 13.
3(a + 3) + 2a
Answer:
3(a + 3) + 2a = 5a + 9.

Explanation:
3(a + 3) + 2a
= [(3 × a) + (3 × 3)] + 2a
= 3a + 9 + 2a
= 3a + 2a + 9
= 5a + 9.

Question 14.
4(2p – 3) – 3(p + 2)
Answer:
4(2p – 3) – 3(p + 2) = 5p – 6.

Explanation:
4(2p – 3) – 3(p + 2)
= [(2p × 4) – (3 × 4)] – [(3 × p) + (2 × 3)]
= (8p – 12) – (3p + 6)
= 8p – 12 – 3p – 6
= 8p – 3p – 12 – 6
= 5p – 6.

Question 15.
2.5(m – 2) + 5.6m
Answer:
2.5(m – 2) + 5.6m = 8.1m – 5.

Explanation:
2.5(m – 2) + 5.6m
= [(m × 2.5) – (2 × 2.5)] + 5.6m
= (2.5m – 5) + 5.6m
= 2.5m – 5 + 5.6m
= 2.5m + 5.6m – 5
= 8.1m – 5.

Question 16.
4(0.6n – 3) – 0.2(2n – 3)
Answer:
4(0.6n – 3) – 0.2(2n – 3) = 2n – 12.6.

Explanation:
4(0.6n – 3) – 0.2(2n – 3)
= 2.4n – 12 – 0.4n + 0.6
= 2.4n – 0.4n – 12 – 0.6
= 2n – 12.6.

Factor each expression.
Question 17.
4t – 20s
Answer:
4t – 20s = 4(t – 5s).

Explanation:
4t – 20s
= 4(t – 5s).

Question 18.
-6p – 21q
Answer:
-6p – 21q = -3(2p + 7q).

Explanation:
-6p – 21q
= -3(2p + 7q).

Question 19.
8i + 12 + 4j
Answer:
8i + 12 + 4j = 4(2i + 3 + j).

Explanation:
8i + 12 + 4j
= 4(2i + 3 + j)

Question 20.
6a + 10b – 20
Answer:
6a + 10b – 20 = 2(3a + 5b – 10).

Explanation:
6a + 10b – 20
= 2(3a + 5b – 10).

Question 21.
-9m – 3n – 6
Answer:
-9m – 3n – 6 = -3(3m + n + 2).

Explanation:
-9m – 3n – 6
= -3(3m + n + 2).

Question 22.
-15x – 6 – 12y
Answer:
-15x – 6 – 12y = -3(5x + 2 + 4y).

Explanation:
-15x – 6 – 12y
= -3(5x + 2 + 4y).

Translate each verbal description into an algebraic expression. Then simplify when you can.
Question 23.
One-fourth x less than the sum of 7 and 2x.
Answer:
\(\frac{x}{4}\) – (7 + 2x) = – \(\frac{7}{4}\)x – 7.

Explanation:
One-fourth x less than the sum of 7 and 2x.
=> \(\frac{x}{4}\) – (7 + 2x)
=> \(\frac{x}{4}\) – 7 – 2x
=> \(\frac{x}{4}\) – 2x – 7.
=> [(x – 8x) ÷ 4 ] – 7.
=> (-7x ÷ 4) – 7
=> – \(\frac{7}{4}\)x – 7.

Question 24.
4 times 5y divided by 18.
Answer:
(4 × 5y ) ÷ 18 = \(\frac{10}{9}\)y.

Explanation:
4 times 5y divided by 18
=> (4 × 5y ) ÷ 18
=> 20y ÷ 18
=> 10y ÷ 9 or \(\frac{10}{9}\)y.

Question 25.
Five-ninths of (3p + 1) subtracted from one-third of (q + p).
Answer:
[\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)] = 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

Explanation:
Five-ninths of (3p + 1) subtracted from one-third of (q + p)
=> [\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)]
=> \(\frac{15}{9}\)p + \(\frac{5}{9}\) – \(\frac{1}{3}\)q + \(\frac{1}{3}\)p.
=> \(\frac{15}{9}\)p + \(\frac{1}{3}\)p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= LCD of 9 n 3 = 9.
= [(15 + 3)p ÷ 9 ] – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= (18p ÷ 9) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= (2p ÷ 1) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

Problem Solving
Solve. Show your work.
Question 26.
After 14 boys leave a concert, the ratio of boys to girls is 3 : 10. If there are p girls at the concert, write an algebraic expression for the number of boys at the beginning of the concert in terms of p.
Answer:
An algebraic expression for the number of boys at the beginning of the concert in terms of p =
0.3 p + 14.

Explanation:
Number of girls in the concert = p.
After 14 boys leave a concert, the ratio of boys to girls is 3 : 10.
Let total number of boys be b.
Ratio of boys after 14 boys leave:
=> \(\frac{b – 14}{p}\) = \(\frac{3}{10}\)
=> 3 × p = 10( b – 14)
=> 3 p = 10 b – 140
=>  10 b = 3 p + 140
= >  b = 0.3 p + 14

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.4 Interior and Exterior Angles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.4 Answer Key Interior and Exterior Angles

Math in Focus Grade 7 Chapter 6 Lesson 6.4 Guided Practice Answer Key

Hands-On Activity

Materials:

• scissors
• ruler

Explore The Sum Of The Angles In A Triangle

Work in pairs.

Step 1.
Draw and cut out a triangle. Label the three angles of the triangle as 1, 2, and 3.

Step 2.
Cut the triangle into three pieces, so that each piece contains a vertex.

Step 3.
Rearrange the cut pieces on a straight line. What do you notice about the sum of the measures of the three interior angles?

Complete.

Question 1.
Find the value of p.

Answer:
p° = 88°

Explanation:

∠RPQ + ∠QRP + ∠PQR = 180°
35° + p° + 57° = 180°
p° + 92° = 180°
p° = 180° – 92°
p° = 88°

Complete.

Question 2.
Triangle XYZ is an isosceles triangle and m∠XYZ = 55°. Find the value of x.

Answer:
∠XYZ + ∠XZY + ∠YXZ = 180°
55° + 55° + x° = 180°
110° +x° = 180°
x° = 180° – 110°
x° = 70°

Find the value of each variable.

Question 3.

Answer:
x° = 70°

Explanation:

x° + 62° = 132°
x° = 132° – 62°
x° = 70°

Question 4.

Answer:
x° = 76°

Explanation:

x° + 35°= 70° + 41°
x° + 35° = 111°
x° = 111° – 35°
x° = 76°

Complete.

Question 5.
Triangle ABC is an isosceles triangle. Find the measures of ∠1 and ∠2.
AB = AC
m∠ACB = m∠ABC

Answer:
∠1 = 255°, ∠2 = 95°

Explanation:
m∠ACB = m∠ABC
∠1 = 75° + 180°
∠1 = 255°
∠2 = 180° – 75°
∠2 = 95°

Question 6.
Triangle EFG is an isosceles triangle and \(\overleftrightarrow{E F}\) is parallel to \(\overleftrightarrow{H I}\). Find the value of x.

Answer:
x° = 120°

Explanation:

∠FEG = ∠IGJ
= 30°
∠FGE = ∠FEG
= 30°
∠FEG + ∠FGE + ∠EFG = 180°
30° + 30° + x° = 180°
x° + 60° = 180°
x° = 180° -60°
x° = 120°

Math in Focus Course 2B Practice 6.4 Answer Key

Find the value of y.

Question 1.

Answer:
y° = 61°

Explanation:
y° + 52° + 67° = 180°
y° + 119° = 180°
y° = 180° -119°
y° = 61°

Question 2.

Answer:
y° + 26° + 18° = 180°
y° + 44° = 180°
y° = 180° – 44°
y° = 136°

Question 3.

Answer:
20° + y° = 90° + 42°
20° + y° = 132°
y° = 132° -20°
y° = 112°

Question 4.

Answer:
90° + 27° + x° = 180°
117° + x° = 180°
x° = 180° -117°
x° = 63°
63° + 51° + y° = 180°
114° + y° = 180°
y° = 180° – 114°
y° = 66°

Question 5.

Answer:
y° = 122°

Explanation:
y° + 28° + 30° = 180°
y° + 58° = 180°
y° = 180° – 58°
y° = 122°

Question 6.

Answer:
x°  = 57° , y° = 67°

Explanation:
x°  + 123°  = 180°
x°  = 180° -123°
x°  = 57°
56°  + 57° + y° = 180°
y°  + 113° = 180°
y° = 180° – 113°
y° = 67°

Find m∠1 and m∠2 in each diagram.

Question 7.

Answer:
62° + 90° + m∠1 = 18o° ∠s sum in triangle:
152° + m∠1 = 180° Simplify:
152° + m∠1 — 152° = 180° — 152° Subtract 152° from both sides:
m∠1 = 28° Simplify:
m∠2 = 90° + m∠1 Exterior ∠ of triangle:
m∠2 = 90° + 28° substitute
m∠2 = 118° simplify
m∠1 = 28°
m∠2 = 118°

Question 8.

Answer:
∠1 = 76°, ∠2 = 114°

Explanation:
∠1 + 28° + 76° = 180°
∠1 + 104° = 180°
∠1 = 180° – 104°
∠1 = 76°
76° + 38° + x° = 180°
114° + x° = 180°
x° = 180° – 114°
x° = 66°
x° + ∠2 = 180°
66° + ∠2 = 180°
∠2 = 180° – 66°
∠2 = 114°

Question 9.

Answer:
50° + 70° + m∠1 = 180° Corresponding ∠s and ∠s sum in triangle:
120° + m∠1 = 180° Simplify:
120° + m∠1 – 120° = 180° – 120° Subtract 120° from both sides:
m∠1 = 60° Simplify:
m∠2 = 50° + 70° Exterior ∠ of triangle:
m∠2 = 120° simplify
m∠1 = 60°
m∠2 = 120°

Question 10.

Answer:
70° = 43° + m∠1 Alternate interior ∠s
70° – 43° = 43° + m∠1 – 43° Subtract 43° from both sides:
m∠1 = 27° Simplify:
m∠2 + m∠1 + 43° = 180° ∠s sum of triangle:
m∠2 + 27° + 43° = 180°
m∠2 + 70° = 180° Simplify:
m∠2 + 70° — 70° = 180° — 70° Subtract 70° from both sides:
m∠2 = 110° Simplify:
m∠1 = 27
m∠2 = 110°

Use an equation to find the value of y.

Question 11.

Answer::
3y° = y° + 80° Exterior ∠ of triangle:
3y = y + 80 The equation for y is:
3y – y = y + 80 – y Subtract y from both sides:
2y = 80 Simplify:
\(\frac{2 y}{2}\) = \(\frac{80}{2}\) Divide by 2
y = 40 simplify
3y = y + 80
y = 40

Question 12.

Answer:
Let’s note by x° the measure of the congruent angles of the isosceles triangle
x° = 2y° Alternate interior ∠s
x° + 3y° = 180° Supplementary ∠s:
2y° + 3y° = 180° Substitute:
5y° = 180° Simplify:
\(\frac{5 y^{\circ}}{5}\) = \(\frac{180^{\circ}}{5}\) Divide by 5
y = 36 simplify
2y + 3y = 180
y = 36

Find the value of x and name the type of triangle that is shown.

Question 13.

Answer:
x° = 60°, x° = 60°

Explanation:
x° + x° + 60° = 180°
2x° + 60° = 180°
2x° = 180° – 60°
2x° = 120°
x° = 120 ÷ 2
x = 60°

Question 14.

Answer:
x° = 36°, x° = 36°

Explanation:
x°+ x°+ 108° = 180°
2x° + 108° = 180°
2x° = 180° – 108°
2x° = 72°
x° = 72÷2
x° = 36°

Question 15.

Answer:
x° = 8°

Explanation:
We all know that all the angles in a triangle add up to 180 degrees.
In the given figure, each triangle has 3 angles.  Thus, we have the sum of three angles as shown:
A° + B° +C° = 180°
62°+ 90°+ x° = 180°
x°+ 172° = 180°
x° = 180° – 172°
x° = 8°

Find the measure of each numbered angle.

Question 16.

Answer:
m∠2 = 148° Corresponding ∠s:
m∠2 = m∠3 + 63° Vertical ∠s and exterior ∠ of triangle:
148° = m∠3 + 63° Substitute:
148° – 63° = m∠3 + 63° – 63° Subtract 63° from both sides:
m∠3 = 85° Simplify:
m∠1 + 63° = 180° supplementary ∠s
m∠1 + 63° – 63° = 180° – 63° subtract 63° from both sides
m∠1 = 117° simplify
m∠1 = 117°
m∠2 = 148°
m∠3 = 85°

Question 17.

Answer:
m∠1 + 140° = 180° Supplementary ∠s
m∠1 + 140° — 140° = 180° – 140° Subtract 140° from both sides:
m∠1 = 40° Simplify:
m∠2 + 15° + 40° = 180° Alternate interior ∠s and ∠s sum of triangle:
m∠2 + 55° = 180° Simplify:
m∠2 + 55° – 55° = 180° – 55° subtract 55° from both sides
m∠2 = 125° simplify
m∠1 = 40°
m∠2 = 125°

Find the measure of each numbered angle.

Question 18.

Answer:

We are given
m∠5 = 50° Corresponding ∠s
m∠3 = m∠5 = 50° Corresponding ∠s
m∠4 = 50° Alternate Interior ∠s
m∠4 + m∠2 + 77° = 180° substitute
m∠2 + 127° = 180° simplify
m∠2 + 127° – 127° = 180° — 127° Subtract 127° from both sides:
m∠2 = 53° Simplify:
114° = m∠1 + m∠4 Exterior ∠ of triangle:
114° = m∠1 + 50° Substitute:
114° – 50° = m∠1 + 50° + 50° Substitute 50° from both sides
m∠1 = 64° simplify
m∠1 = 64°
m∠2 = 53°
m∠3 = 50°

Question 19.

Answer:

we are given
m∠1 + 67° + 53° = 180° Sum of ∠s in triangle:
m∠1 + 120° = 180° Simplify:
m∠1 + 120° – 120° = 180° — 120° Subtract 120° from both sides:
m∠1 = 60° Simplify:
m∠3 = 67° Alternate interior ∠s
m∠2 = m∠1 + m∠3 vertical ∠s
m∠2 = 60° + 67° substitute
m∠2 = 127° simplify
m∠1 = 60°
m∠2 = 127°

Solve.

Question 20.
\(\overleftrightarrow{\mathrm{BE}}\) is parallel to \(\overleftrightarrow{\mathrm{FH}}\). Find the measure of A
∠CAD in terms of ∠1 and ∠2.

Answer:
m∠ADE = m∠2 Corresponding ∠s:
m∠ADE = m∠CAD + m∠1 Exterior ∠ of triangle:
m∠2 = m∠CAD + m∠1 Substitute:
m∠2 — m∠1 = m∠CAD + m∠1 — m∠1 subtract m∠1 from both sides:
m∠CAD = m∠2 – m∠1 Simplify:
m∠CAD = m∠2 – m∠1

Question 21.
Math Journal
Explain why each of the following statement is true.
a) A triangle cannot have two right angles.
Answer:
m∠A = m∠B = 90° a) Let’s assume that a triangle ABC has two right angles:
m∠A + m∠B + m∠C = 180° Sum of ∠s in triangle:
90° + 90° + m∠C = 180° Substitute:
180° + m∠C = 180° Simplify:
180° + m∠C – 180° = 180° – 180° Subtract 180° from both sides:
m∠C = 0° Simplify:
We got a false statement as the triangle does not exist in this case. Therefore we assumed a wrong statement So a triangle cannot have two right angles.

b) The interior angle measures of an isosceles triangle cannot be 96°, 43°, and 43°.
Answer:
Lets assume that a triangle has the measures of its angles 96°, 43°, 43°
96° + 43° + 43° = 182° We compute the sum of the angles’ measures:
But the sum of a triangle’s angles measures is 180°. We got 182° ≠ 180°, therefore the statement that we assumed is false. So the interior measures of a triangle cannot be 96°, 43°, 43°.
See proofs (sum of the measures of the triangle’s angles)

Question 22.
m∠1 = 2x°, m∠2 = (x – 5)°, and m∠3 = 100°. Use an equation to find the value of x and then find the measures of ∠1 and ∠2.

Answer:
∠1 = 80°, ∠2 = 35°

Explanation:
∠1 + ∠3 = 180°
2x° + 100° = 180°
2x° = 180° – 100°
2x° = 80°
x° = 80°÷2
x° = 40°
∠1 = 2×40 = 80°
∠2 = 40 – 5 =35°

Use an equation to find the value of x.

Question 23.

Answer:
x° = 56°

Explanation:
x° + 94°+ 30° = 180°
x° + 124° = 180°
x° = 180° – 124°
x° = 56°

Question 24.

Answer:
x° = 59°, 2x° = 118°

Explanation:
2x° + 62° = 180°
2x° = 180° – 62°
2x° = 118°
x° = 118°÷2
x° = 59°
x + 13 = 59° + 13° = 72°

Use an equation to find the value of x.

Question 25.

Answer:
x° = 50°, y° = 65°

Explanation:
y° + 115° = 180°
y° = 180° -115°
y° = 65°
65° + 65° + x° = 180°
x° + 130° = 180°
x° = 180° – 130°
x° = 50°

Question 26.

Answer:
x° + x° + 12° = 180°
2x° + 12° = 180°
2x° = 180° – 12°
2x° = 68°
x° = 68 ÷ 2
x° = 34°

Brain@Work

ABCD is a rhombus, and the measure of ∠BCD is 68°. BDE is a triangle where the measure of ∠BED is 36° and the measure of ∠BDE is 73°. Find m∠EBC. Show how you obtain your answer.

Answer:
m∠EBD + 36° + 73° = 180° Sum of ∠s in triangle:
m∠EBD + 109° = 180° Simplify:
m∠EBD + 109° — 109° = 180° — 109° Subtract 109° from both sides:
m∠EBD = 71° Simplify:
\(\overline{B C}\) = \(\overline{D C}\) ABCD rhombus
m∠CDB = m∠CBD ∆BCD isosceles:
68° + m∠CDB + m∠CBD = 180° Sum of ∠s in triangle:
68° + 2m∠CBD = 180°
68° + 2m∠CBD — 68° = 180° — 68° Subtract 68° from both sides:
2m∠CBD = 112° Simplify:

Divide by 2
m∠CBD = 56° Simplify:
m∠EBC = m∠EBD — m∠CBD Substitute:
m∠EBC = 71° — 56°
= 15°

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.3 Alternate Interior, Alternate Exterior, and Corresponding Angles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.3 Answer Key Alternate Interior, Alternate Exterior, and Corresponding Angles

Math in Focus Grade 7 Chapter 6 Lesson 6.3 Guided Practice Answer Key

Hands-On Activity

Materials:

• protractor

Explore The Angles Formed By Parallel Lines And A Transversal Using A Protractor

Work in pairs.

Step 1.
On a piece of paper, draw a pair of parallel lines, \(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\), as shown. Draw a transversal \(\overleftrightarrow{P Q}\) that intersects the pair of parallel lines. Use a protractor to measure the angles in the diagram.

Step 2.
Record your results in a table like the one shown.

Math Journal
What do you notice about the measures of these angle pairs: Alternate interior angles, alternate exterior angles, and corresponding angles? Make a conjecture about the angle measures for each pair.

Math Journal
Compare the conjecture that you have made for corresponding angles, alternate interior angles, and alternate exterior angles with the conjectures made by other students. What do you observe?

When two parallel lines are cut by a transversal,

• the alternate interior angles are always congruent (alt. int. ∠s, || lines).
• the alternate exterior angles are always congruent (alt. ext. ∠s, || lines).
• the corresponding angles are always congruent (corr. ∠s, || lines).

Use the diagram at the right to complete the following questions.

Question 1.
\(\overleftrightarrow{\mathrm{AD}}\), \(\overleftrightarrow{\mathrm{EH}}\), \(\overleftrightarrow{\mathrm{XY}}\), and \(\overleftrightarrow{\mathrm{WZ}}\) are straight lines and \(\overleftrightarrow{\mathrm{XY}}\) is parallel to \(\overleftrightarrow{\mathrm{WZ}}\). Identify all the pairs of angles formed by the intersection of \(\overleftrightarrow{\mathrm{AD}}\) with \(\overleftrightarrow{\mathrm{XY}}\) and \(\overleftrightarrow{\mathrm{WZ}}\).

Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) a) We are given:
∠ABX and ∠BCW Pairs of corresponding angles:
∠XBC and ∠WCD
∠EFB and ∠FGC
∠BFG and ∠CGH
∠XBC and ∠BCG b) Alternate interior angles:
∠FBC and ∠BCW

a) Corresponding angles: ∠ABX and ∠BCW;
Answer:
∠XBC and ∠WCD
∠EFB and ∠FGC
∠BFG and ∠CGH

b) Alternate interior angles: ∠XBC and ∠BCG;
Answer:
∠FBC and ∠BCW

Question 2.
Name another transversal of the parallel lines in the diagram.
Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) we are given
A transversal of the parallel lines \(\overline{X Y}\) and \(\overline{W Z}\) other than \(\overline{A D}\) is \(\overline{E H}\).
\(\overline{E H}\)

Question 3.
Identify one pair of each of the following angles formed by the intersection of \(\overleftrightarrow{\mathrm{EH}}\) with \(\overleftrightarrow{\mathrm{XY}}\) and \(\overleftrightarrow{\mathrm{WZ}}\).
a) Corresponding angles
Answer:
\(\overline{X Y}\) || \(\overline{W Z}\) We are given:
∠EFY and ∠FOZ a) Identify a pair of corresponding angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):

b) Alternate interior angles
Answer:
∠BFG and ∠FGZ b) Identify a pair of alterrate interior angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):

c) Alternate exterior angles
Answer:
∠EFY and ∠CGR c) Identity a pair of aiterrate exterior angles formed by the intersection of \(\overline{E H}\) with \(\overline{X Y}\) and \(\overline{W Z}\):
a) ∠EFY and ∠FGZ
b) ∠BFG and ∠FGZ
c) ∠EFY and ∠CGH

Complete.

Question 4.
In the diagram, \(\overleftrightarrow{\mathrm{MN}}\) is parallel to \(\overleftrightarrow{\mathrm{PQ}}\). Find the measures of ∠1, ∠2, and ∠3.

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 81° Alternate interior ∠s:
m∠2 = 81° : Corresponding ∠s:
m∠3 + m∠2 = 180° Supplementary ∠s:
m∠3 + 81° = Substitute m∠2 = 81°:
m∠3 + 81° — 81° = 180° — 81° Subtract 81° from bothsides:
m∠3 = 99° simplify
81° ; 81° ; m∠2 ; 81° ; 81° ; 81° ; 81° ; 81° ; 99°

Math in Focus Course 2B Practice 6.3 Answer Key

\(\overleftrightarrow{\mathrm{MN}}\) is parallel to \(\overleftrightarrow{\mathrm{PQ}}\). Identify each pair of angles as corresponding, alternate interior, alternate exterior angles, or none of the above.

Question 1.
∠3, ∠6
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠3 and ∠6 are alternate interior angles because they lie inside the pair of parallel lines and on the opposite sides of the transversal.
Alternate interior angles

Question 2.
∠5, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠5 and ∠7 are corresponding angles because they lie in matching corners on same side of the transversal \(\overline{S T}\).
corresponding angles

Question 3.
∠1, ∠2
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠1 and ∠2 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

Question 4.
∠1, ∠8
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠1 and ∠8 are alternate exterior angles because they lie outside the pair of parallel lines and on the opposite sides of the transversal \(\overline{S T}\).
Alternate exterior angles

Question 5.
∠8, ∠6
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠8 and ∠6 are corresponding angles because they lie in matching corners on same side of the transversal \(\overline{S T}\).
corresponding angles

Question 6.
∠4, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠4 and ∠7 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

Question 7.
∠2, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠2 and ∠7 are alternate interior angles because they lie inside the pair of parallel lines and on the opposite sides of the transversal \(\overline{S T}\).
Alternate interior angles

Question 8.
∠6, ∠7
Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
∠6 and ∠7 are neither alternate interior, nor alternate exterior or corresponding, therefore none of the above.
None of the above

\(\overleftrightarrow{\mathrm{AB}}\) is parallel to \(\overleftrightarrow{\mathrm{CD}}\). Use the diagram to answer the following.

Question 9.
Name two angles that have the same measure as ∠2.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
m∠7 = m∠2 ∠2 and ∠7 are alternate exterior angles because they lie outside the pair of parallel lines and on the opposite sides of the transversal \(\overline{E F}\).
m∠6 = m∠2 ∠2 and ∠6 are corresponding angles because they lie in matching corners on the same side of the transversal \(\overline{E F}\).
∠6, ∠7

Question 10.
Name an angle that is supplementary to ∠6.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
∠5, ∠8, ∠3, ∠1 We identify supplementary angles to ∠6:
∠5, ∠8, ∠3, ∠1

Question 11.
If m∠4 = 46°, find m∠5.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
m∠6 = m∠4 Alternate interior ∠s:
m∠6 = 46° substitute:
m∠5 + m∠6 = 180° supplementary ∠s:
m∠5 + 46° = 180° substitute
m∠5 + 46° – 46° = 180° – 46° subtract 46° from both sides
m∠5 = 134° simplify
134°

Question 12.
If m∠1 = 131°, find m∠7.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
Corresponding ∠s:
m∠5 = m∠1 corresponding ∠s:
m∠5 = 131° Substitute:
m∠5 + m∠7 = 180° Supplementary ∠s:
131° + m∠7 = 180° Substitute:
131° + m∠7 – 131° = 180° – 131° Subtract 131° from both sides:
m∠7 = 49° Simplify:
49°

Find the measure of each numbered angle.

Question 13.
\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 78° vertical ∠s:
m∠1 = m∠2 Alternate interior ∠s:
m∠2 = 78° Substitute:
m∠3 = m∠2 vertical ∠s:
m∠3 = 78° Substitute:
m∠1 = 78° ;m∠2 = 78° ; m∠3 = 78°

Question 14.
\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 107° vertical ∠s:
m∠1 = m∠2 corresponding ∠s:
m∠1 = m∠2 vertical ∠s:
m∠2 = 107° Substitute:
m∠1 = 107° ;m∠2 = 107°

Question 15.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
m∠1 = 87° Alternate interior ∠s:
m∠2 = 52° corresponding ∠s:
m∠1 = 87° ; m∠2 = 52°

Question 16.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
m∠1 = 48° Alternate exterior ∠s:
m∠1 + m∠2 = 180° supplementary ∠s:
48° + m∠2 = 180° substitute
48° + m∠2 – 48° = 180° – 48° subtract 48° from both sides
m∠2 = 132° simplify
m∠1 = 48° ; m∠2 = 132°

Question 17.
\(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 126° Alternate interior ∠s:(transversal \(\overline{A B}\))
\(\overline{A B}\) || \(\overline{C D}\) we are given
m∠2 = m∠1 Alternate exterior ∠s:(transversal \(\overline{P Q}\))
m∠2 = 126° substitute
m∠1 = 126° ; m∠2 = 126°

Question 18.
\(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\).

Answer:
\(\overline{M N}\) || \(\overline{P Q}\) We are given:
m∠1 = 32° Alternate interior ∠s:(transversal \(\overline{G F}\))
\(\overline{A B}\) || \(\overline{C D}\) we are given
m∠2 = 105° corresponding ∠s:(transversal \(\overline{M N}\))
m∠1 = 32° ; m∠2 = 105°

Find the value of each variable.

Question 19.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
a = 142 Alternate interior ∠s (transversal \(\overline{T U}\)):
a = 142
2b° + a° = 180° Supplementary ∠s
2b° + 142° = 180° Substitute:
2b° + 142° — 142° = 180° — 142° Subtract 142° from both sides:
2b° = 38°
\(\frac{2 b^{\circ}}{2}\) = \(\frac{38^{\circ}}{2}\) Divide by 2:
b = 19 simplify
a = 142
b = 19

Question 20.
\(\overleftrightarrow{P Q}\) is parallel to \(\overleftrightarrow{R S}\).

Answer:
\(\overline{P Q}\) || \(\overline{R S}\) We are given:
Let’s note by X the intersection between the lines \(\overline{T U}\) and \(\overline{R S}\).
m∠TXS = Corresponding ∠s (transversal \(\overline{T U}\)):
m∠TXS + m∠TXR = 180° Supplementary ∠s
5w° + 3w° = 180° Substitute:
8w° = 180° Simplify:
\(\frac{8 w^{\circ}}{8}\) = \(\frac{180^{\circ}}{8}\) Divide by 8
w = 22.5° simplify
w = 22.5°

MathJournal
Determine whether \(\overleftrightarrow{A B}\) is parallel to \(\overleftrightarrow{C D}\). Use the fact that two lines are parallel if a pair of corresponding angles formed by a transversal are congruent. Explain your answer.

Question 21.

Answer:
m∠BMN = 66° ∠BMN and ∠MNC are alternate interior ∠s We have:
m∠BMN ≠ m∠MNC As 66° ≠ 65°, we have

This means the lines \(\overline{A B}\) and \(\overline{C D}\) are not parallel:

Question 22.

Answer:
m∠MND = 90° ∠RMB and ∠MND are corresponding ∠s. We have:
m∠MND = 90°
m∠RMB = m∠MND We get:
\(\overline{A B}\) = \(\overline{C D}\) This means the lines \(\overline{A B}\) and \(\overline{C D}\) are parallel:
\(\overline{A B}\) || \(\overline{C D}\)

Question 23.

Answer:
m∠AMN + m∠NMB = 180° Supptementary ∠s
108° + m∠NMB = 180° Substitute:
108° + m∠NMB — 108° = 180° — 108° Subtract 108° from both sides:
m∠NMB = 72° Simplify:
m∠NMB = 72° ∠SND and ∠NMB are corresponding ∠s. We have:
m∠SND = m∠NMB we got
\(\overline{A B}\) = \(\overline{C D}\) This means the lines \(\overline{A B}\) and \(\overline{C D}\) are parallel
\(\overline{A B}\) || \(\overline{C D}\)

\(\overline{M N}\) is parallel to \(\overline{P Q}\). Find each unknown angle measure.

Question 24.

Answer:
m∠QMN = m∠PQM AS \(\overline{M N}\) || \(\overline{M N}\), ∠PQM and ∠QMN are alternate interior ∠s. We hava
m∠QMN = 34° Substitute
∠RMQ + m∠QMN + m∠1 = 180° ∠RMS straight angle
22° + 34° + m∠1 = 180°
56° + m∠1 = 180° Simplify
56° + m∠1 – 56° = 180° – 56° subtract 56° from both sides
m∠1 = 124° simplify
m∠1 = 124°

Question 25.

Answer:
m∠1 = m∠NPQ As \(\overline{M N}\) and \(\overline{P Q}\), ∠1 and ∠NPQ are alternate interior ∠s. We have:
m∠1 = 25° Substitute:
m∠MPN + m∠NPQ + m∠2 = 180° ∠MPS straight:
90° + 25° + m∠2 = 180° Substitute:
115° + m∠2 = 180° Simplify:
115° + m∠2 – 115° = 180° – 115° subtract 115° from both sides
m∠2 = 65° simplify
m∠1 = 25°
m∠2 = 65°

Question 26.

Answer:
m∠TRN = 114° As \(\overline{M N}\) || \(\overline{P Q}\), a pair of corresponding angles is:
62° + m∠3 = 114° Substitute:
62° + m∠3 – 62° = 114° — 62° Subtract 62° from both sides:
m∠3 = 52° simplify
m∠2 = m∠3 alternate interior ∠s
m∠2 = 52°
m∠1 = 114° Alternate exterior ∠s
m∠1 = 114°
m∠2 = 52°
m∠3 = 52°

Question 27.

Answer:
m∠NPQ = m∠MNP As \(\overline{M N}\) || \(\overline{P Q}\), ∠MNP and ∠NPQ are alternate interior angles
m∠NPQ = 101° Substitute:
m∠1 + m∠NPQ + m∠QPS = 180° ∠MPS is a straight line
m∠1 + 101° + 21° = 180° substitute
m∠1 + 122° = 180° simplify
m∠1 + 122° – 122° = 180° – 122° subtract 122° from both sides
m∠1 = 58° simplify
m∠1 = 58°

Question 28.

Answer:
m∠DCQ = m∠DBN As \(\overline{M N}\) || \(\overline{P Q}\), a pair of a corresponding angles is :
m∠DCQ = 76° Substitute
m∠1 + m∠DCQ = 180° ∠ACD is straight
m∠1 + 76° = 180° substitute
m∠1 + 76° – 76° = 180° – 76° subtract 76° from both sides
m∠1 = 104° simplify
m∠MFH + m∠HFN = 180° m∠MFH straight line
m∠MFH + 100° = 180° substitute
m∠MFH + 100° – 100° = 180° Subtract 100° from both sides:
m∠MFH = 80° Simplify:
m∠PGH = m∠MFH ∠MFH and ∠PGH corresponding angles:
m∠2 = 80° Substitute:
m∠1 = 104°
m∠2 = 80°

Question 29.

Answer:
m∠1 + 103° = 180° Straight ∠
m∠1 + 103° — 103° = 180° — 103° Subtract 103° from both sides:
m∠1 = 77° Simplify:
m∠1 = m∠2 + 16° As \(\overline{M N}\) || \(\overline{P Q}\), a pair of alternate interior angles is:
77° = m∠2 + 16° Substitute:
77° – 16° = m∠2 + 16 — 16° Subtract 16° from both sides:
m∠2 = 61° Simplify:
m∠1 = 77°
m∠2 = 61°

\(\overline{\mathbf{A B}}\) is parallel to \(\overline{\mathbf{C D}}\). Find the value of x.

Solve. Show your work.

Question 30.

Answer:
27° + x° = 42 As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
27° + x° — 27° = 42° — 27° Subtract 27° from both sides:
x° = 15° Simplify:
x = 15

Question 31.

Answer:

we are given
Let X be the intersection of the Lines \(\overline{A B}\) and \(\overline{P D}\), Y be the intersection of the lines \(\overline{C D}\) and \(\overline{B Q}\) and Z be the intersection of the lines \(\overline{P D}\) || \(\overline{B Q}\).
110° + m∠PDY = 180° ∠CDY is straight
110° + m∠PDY = 180° – 110° subtract 110° from both sides
m∠PDY = 70° simplify
m∠DZY + 154° = 180° ∠PZD is straight
m∠DZY + 154° – 154 = 180° — 134° Subtract 154° from both sides:
m∠DZY = 26° Simplify:
m∠DZY + m∠PDY + m∠BYD = 180° Sum of the angles in ∆ZDY:
26° + 70° + m∠BYD = 180° Substitute:
96° + m∠BYD = 180° Simplify:
96° + m∠BYD – 96° = 180° — 96° Subtract 96° from both sides:
m∠BYD = 84° Simplify:
m∠BYD + m∠DYQ = 180° ∠BYQ is straight:
84° + m∠DYQ = 180° Substitute:
84° + m∠DYQ – 84° = 180° – 84° Subtract 84° from both sides:
m∠DYQ = 96° Simplify:
m∠ABQ = m∠DYQ As \(\overline{A B}\) || \(\overline{C D}\), a pair of corresponding angles is:
x° = 96° Substitute:
x = 96

Question 32.

Answer:
m∠BCD = m∠ABC As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
m∠BCD = 108° Substitute:
m∠BCD + m∠DCQ + m∠QCP = 180° ∠BCP is straight:
108° + x° + 35° = 180° Substitute:
143° + x° = 180° Simplify:
143° + x° — 143° = 180° – 143° Substitute:
x = 37 Simplify:
x = 37

Question 33.

Answer:

we are given
Let X be the intersection of the lines \(\overline{A B}\) and \(\overline{P C}\), Y be the intersection of the lines \(\overline{C D}\) and \(\overline{A P}\).
150° + m∠XCY = 180° ∠YCD is straight:
150° + m∠XCY — 150° = 180° — 150° Subtract 150° from both sides:
m∠XCY = 30° Simplify:
m∠XAP + 132° = 180° ∠XAB is straight:
m∠XAP = 48° simplify
m∠XAP = m∠PYC As \(\overline{A B}\) || \(\overline{C D}\), a pair of alternate interior angles is:
m∠PYC = 48° Substitute:
m∠PYC + m∠XCY + m∠YPC = Sum of the ∠s in a triangle:
48° + 30° + m∠YPC = 180° Substitute:
78° + m∠YPC = Simplify:
78° + m∠YPC — 78° = 180° — 78° Subtract 78° from both sides:
m∠YPC = 102° Simplify:
m∠YPC + x° = 180° ∠APY straight:
102 + x° = 180° Substitute:
102° + x° – 102° = 180° — 102 Subtract 102° from both sides:
x° = 78° simplify
x = 78

Solve. Show your work.

Question 34.
In the diagram below, \(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\). m∠1 = (x + 28)° and m∠2 = (3x + 14)°. Write and solve an equation to find the measures of ∠1 and ∠2.

Answer:
m∠1 = m∠2 As \(\overline{M N}\) || \(\overline{P Q}\), a pair of alternate interior angles is:
x + 28 = 3x + 11 substitute
x + 28 – x = 3x + 14 – x Subtract x from both sides:
28 = 2x + 14 simplify
28 — 14 = 2x + 14 — 14 Subtract 14 from both sides:
2x = 14 Simplify:
\(\frac{2 x}{2}\) = \(\frac{14}{2}\) Divide by 2:
x = 7 Simplify:
m∠1 = m∠2 = (7 + 28)° = 35° Determine m∠1, m∠2:
x + 28 = 3x + 14
m∠1 = 35° ;m∠2 = 35°

Question 35.
Math Journal
In a plane, if a line is perpendicular to one of two parallel lines, is it also perpendicular to the other? Explain your reasoning.
Answer:
\(\overline{A B}\) || \(\overline{C D}\) We are given:
\(\overline{E F}\) ⊥ \(\overline{A B}\)
m∠EMB = 90°: \(\overline{E F}\) ⊥ \(\overline{A B}\)
m∠MND = m∠EMB As \(\overline{A B}\) || \(\overline{C D}\), a pair of corresponding angles is:
m∠MND = 90° Substitute:
\(\overline{E F}\) ⊥ \(\overline{C D}\) m∠MND = 90°
yes

Question 36.
The diagram below contains examples of parallel lines cut by transversals. Line MN is parallel to line PQ and line AB is parallel to line CO.

a) Name two pairs of corresponding angles.
Answer:
∠1 and ∠3 a) As \(\overline{A B}\) || \(\overline{P Q}\), pairs of corresponding angles are (transversal \(\overline{A B}\)):
∠2 and ∠4
∠5 and ∠7

b) Name all the angles that have the same measure as ∠1.
Answer:
m∠5 = m∠1 b) vertical ∠s:
m∠3 = m∠1 corresponding ∠s:
m∠8 = m∠3 = m∠1
m∠7 = m∠5 = m∠1
a) ∠1 and ∠3
∠2 and ∠4
∠5 and ∠7
b) ∠1, ∠3, ∠5, ∠7 = ∠8

Question 37.
The two mirrors used in a periscope are parallel to each other as shown. m∠1 = 3x°, m∠2 = (60 – x)°, and m∠3 = 90°. Write and solve an equation to find the value of x. Then find the measure of ∠4.

Answer:
m∠1 = 3x° We are given:
m∠2 = (60 – x)°
m∠3 = 90°
m∠1 = m∠2 Alternate interior ∠s
3x° = (60 – x)°
3x = 60 — x Substitute:
3x + x = 60 – x + x Add x to both sides:
4x = 60 Simplify:
\(\frac{4 x}{4}\) = \(\frac{60}{4}\) Divide by 4:
x = 15 Simplify:
m∠2 + m∠3 + m∠4 = 180° Straight ∠:
Substitute:
(60 – x)° + 90° + m∠4 = 180°
(60 – 15)° + 90° + m∠4 = 180°
45° + 90° + m∠4 = 180°
135° + m∠4 = 180° Simplify:
135° + m∠4 – 135° = 180° — 135° Subtract 135° from both sides:
m∠4 = 45° Simplify:
x = 15
m∠4 = 45°

Question 38.
MathJournal
Use a diagram to illustrate each of the following: transversal, corresponding angles, alternate exterior angles, and alternate interior angles. Label your diagram and explain which angles are congruent.
Answer:

We draw the diagram of 2 parallel lines cut by a transversal:
m∠1 = m∠5 Pairs of corresponding angles:
m∠2 = m∠6
m∠3 = m∠7
m∠4 = m∠8
m∠1 = m∠7 Pairs of alternate exterior aigles
m∠2 = m∠8
m∠3 = m∠5 Pairs of alternate Interior angles
m∠4 = m∠6
See diagram

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.2 Angles that Share a Vertex to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.2 Answer Key Angles that Share a Vertex

Math in Focus Grade 7 Chapter 6 Lesson 6.2 Guided Practice Answer Key

Complete.

Question 1.
Find the value of p in the diagram.

Answer:
p° = 20°

Explanation:

7 p°+ 75° + 145° = 360°
7 p° + 220° = 360°
7p° + 220° – 360° = 360°-360°
7p° = 140°
p°  = 140 ÷ 7
p° = 20°

Solve.

Question 2.
\(\overleftrightarrow{B E}\) and \(\overleftrightarrow{C A}\) are straight lines. Find the value of each variable.


Answer:
r°=28°, q°=28°, 3q°=84°

Explanation:


∠BOA + ∠AOE = 180°
152°+r°=180°
r = 180°-152°
r= 28°
∠BOC + ∠COD + ∠DOE = 180°
q°+ 68°+3q° = 180°
4q°+68° = 180°
4q° = 180°-68°
4q° = 112°
q°= 112÷4
q°= 28°
3q°= 84°

Question 3.
\(\overleftrightarrow{P Q}\) is a straight line. Find the value of each variable.

Answer:
m°=31°, n°=17°, 3m°=93°

Explanation:

m°+149° = 180°
m° = 180°-149°
m° = 31°
3m°= 93°
n°+3m°+70°=180°
n°+93°+70°=180°
n°=180°-163°
n°= 17°

Complete.

Question 4.
In the diagram at the right, the ratio a: b: c = 1: 3 : 5. Find the values of a, b, and c.
The ratio a : b : c = 1 : 3 : 5. So, b = 3 • a and c = 5 • a.


Answer:
a°= 40°, b°=120°, c°=200°

Explanation:
The ratio given is a : b : c = 1 : 3 : 5. Let a=1a, b=3a, c=5a

Technology Activity

Materials:

• geometry software

Explore The Relationship Among Vertical Angles Using Geometry Software

Work in pairs.

Step 1.
Construct intersecting line segments, \(\overline{A B}\) and \(\overline{C D}\), as shown.

Step 2.
Select ∠AOD and find its measure.

Step 3.
Select ∠COB and find its measure.

Step 4.
Then select ∠AOC and ∠BOD and find their measures.

Step 5.
Select point D and drag it so that you change the measures of ∠AOD and ∠AOC. As the measure of ∠AOD changes, what do you notice about the measure of ∠AOC?

Math Journal
Describe what you notice about the measures of the vertical angles.
Answer:
Vertical angles are always equal to one another and are always congruent. The four angles all together always sum to a full angle 360°.

Complete.

Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines. Find the value of y.

Answer:
3y°=120°

Explanation:

∠COB = ∠AOD
3y°=120°
y°= 120°÷3
y° = 40°

Complete.

Question 6.
In the diagram, two straight lines intersect to form angles 1, 2, 3, and 4. Find the value of each variable if m∠1 = 114°.

m∠1 + m∠2 = 180° Supp. ∠S

Answer:
r°=66°, p°=66°, q°=114°

Explanation:


114°+r°=180°
r°=180°-114
r°=66°
∠3=∠1
q°=114°
∠4=∠2
p°=r°
as r°=66°
p°=66°

Math in Focus Course 2B Practice 6.2 Answer Key

Find the value of each variable.

Question 1.

Answer:
x°=307°

Explanation:

x°+53°=360°
x°=360°-53°
x°=307°

Question 2.

Answer:
x°=257°

Explanation:

x°+33°=290°
x°=290°-33°
x°=257°

Question 3.

Answer:
x°=100°

Explanation:

x°+110°+150°=360°
x°+260°=360°
x°=100°

Question 4.

Answer:
x°=95°

Explanation:

∠AOD+∠BOC+∠AOD = 360°
x°+110°+95°+60°=360°
x°+265°=360°
x°=360°-265°
x°=95°

Question 5.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
p°= 34°

Explanation:

∠COA = ∠BOD
34° = p°
Hence p° = 34°

Question 6.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
r°=16°

Explanation:

In the image given below, we can observe that CD and AB are two straight lines. Here, ∠AOB and ∠COD are vertical angles.
a°= 164°
a°+r°=180°
164°+r°=180°
r°=180°-164°
r°=16°

Question 7.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
a° = 35°, b° = 106°

Explanation:

∠FOD = ∠COE
a° = 35° (vertically opposite angles)
∠EOB = ∠AOF
b° = 106° (Vertically opposite angles)

Question 8.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
b° = 126°, a° = 31°

Explanation:
∠AOF = ∠EOB
b° = 126° (vertically opposite angles)
∠AOC = ∠DOB
31° = a° (vertically opposite angles)

Question 9.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines.

Answer:
x° = 119°, y°= 122°

Explanation:

∠COE + ∠AOC = 180°
y°+ 58° = 180°
y° = 180°-58°
y° = 122°
∠DOB + ∠FOD = 180°
x° + 61° = 180°
x° = 180°-61°
x° = 119°

Question 10.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
w + 90° = 168° Vertical ∠s:
w° + 90° – 90° = 168° – 90° Subtract 90° from both sides:
w° = 78° Simplify:
w = 78

Find the value of k.

Question 11.
The ratio ∠1 : ∠2 : ∠3 : ∠4 = 3 : 2 : 1 : 3.

Answer:
k° = 40°

Explanation:

Let the assume given ratio be x
Now, the ratio is 3x: 2x: x: 3x
The sum of all angles at one point is 360°
Now add 3x+ 2x + x + 3x = 360°
9x = 360°
x = 360 ÷ 9
x° = 40°
3x° = 120°
2x° = 80°
3x° = 120°
Now we need to calculate the numbered angle 3 which is named as k° that is 40°

Question 12.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
4k°= 84°, 3k° = 63

Explanation:

∠COE +∠EOF = 174°
4k°+ 90° = 174°
4k° = 174-90
4k°= 84°
k° = 84÷4
k° = 21°
3k°= 21×3
3k° = 63°

Question 13.

Answer:
2k°+5k°+3k°+130°=180°
10k°+130°=360°
10k° = 360°-130°
10k° = 230°
k° = 23°
IF k°=23° then 2k° = 2×23 = 46°, 3k° = 3×23=69°
5k° = 5×23=115°

Name the pairs of vertical angles.

Question 14.

Answer:
AE and DB are a pair of non-adjacent angles that are formed when two lines intersect.

Explanation:
Vertical lines are a pair of non-adjacent angles formed when two lines intersect.

Question 15.

Answer:
Vertical angles are a pair of opposite angles formed by intersecting lines.
∠MKF and ∠NKG Lines \(\overline{M N}\) and \(\overline{F G}\) intersect in K.
∠FKN and ∠MKC We identify the pairs of vertical angles:
∠MKF and ∠NKG
∠FKN and ∠MKC

Question 16.
\(\overleftrightarrow{P S}\) and \(\overleftrightarrow{R N}\) are straight lines.

Answer:
Vertical angles are a pair of opposite angles formed by ¡ntersecting tines.
∠SOR ard ∠NOP Lines \(\overline{P S}\) and \(\overline{R N}\) intersect in O.
∠SON and ∠ROP We identify the pairs of vertical angles:
∠SOR and ∠NOP
∠SON ard ∠ROP

Find the value of each variable.

Question 17.

Answer:
64°+23°+3e°=360°
87°+3e°=360°
3e°=360°-87°
3e°=273°
e°=273°÷ 3°
e° = 91°

Question 18.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
2p°=88, q°= 92°, r°=2°

Explanation:

∠DOA = ∠BOC
2p°=88
p°= 88÷2
p°= 44
∠DOE+∠EOB+∠BOC= 180°
r°+2p°+88 = 180°
r°+88+88 = 180°
r°+178°=180°
r°=180°-178°
r°=2°
2p°+q°=180°
88+q°=180°
q°=180-88
q°= 92°

Question 19.
\(\overleftrightarrow{A B}\), \(\overleftrightarrow{C D}\) and \(\overleftrightarrow{E F}\) are straight lines. The ratio a : b : c = 1 : 2 : 2.

Answer:
a° = 36°, b = 72°, c = 72°

Explanation:

Let the ratio be x
The sum of the ratio be x + 2x + 2x = 5x
∠AOC = ∠DOB (vertically opposite angles are equal)
∠FOD + ∠DOB + ∠EOB = 180°
c°+a°+b° = 180°
2x + x° + 2x° = 180°
5x° = 180°
x° = 180÷5
x° = 36°
2x°= 72°

Question 20.
\(\overleftrightarrow{A B}\) and \(\overleftrightarrow{C D}\) are straight lines.

Answer:
f° = 40° Vertical ∠s:
f = 40
e° + f° = 180° Straight ∠:
e° + 40° = 1800 Substitute:
e° + 40° – 40° = 180° – 40° Subtract 40° from both sides:
e = 140 Simplify:
d° + 90° + 15° = e° Vertical ∠s:
d° + 105° = 140° Simplify:
d° + 105° — 105° = 140° – 105° Subtract 105° from both sides:
d = 35 Simplify:
f = 40
e = 140
d = 35

Answer each of the following.

Question 21.
In the diagram, the ratio p: q: r = 1: 2 : 3. Find the values of p, q, and r.

Answer:
p = 60°, q = 120°, r = 180°

Explanation:

p: q: r = 1: 2 : 3
x+2x+3x = 360°
6x = 360°
x = 360° ÷ 6
x = 60°
2x= 120°
3x = 180°
p = 60°, q = 120°, r = 180°

Question 22.
If ∠P and ∠N are angles at a point and m∠P = 149°, what is the m∠N?
Answer:
m∠N = 31°

Explanation:
The sum of angles at a point is 180°
Given angles are ∠P = 149°
Here we need to calculate ∠N
∠P + ∠N = 180°
149° + ∠N = 180°
∠N = 180°-149°
∠N = 31°

Question 23.
If 67°, 102°, 1 5°, and x° are angles at a point, what is the value of x?
Answer:
x° = 176°

Explanation:
67°+102°+15°+x°=360°
184°+x° = 360°
x°=360° -184°
x° = 176°

In the diagram below, \(\overleftrightarrow{M P}\) and \(\overleftrightarrow{Q R}\) are straight lines. Answer each of the following.

Question 24.
Name the angle that is vertical to ∠MNR.
Answer:
∠QNP

Explanation:
Vertical angles mean the angles that are opposite to each other. By observing from the given figure the angle ∠QNP is vertical to ∠MNR.

Question 25.
What kind of angles are ∠RNP and ∠PNS?
Answer:
∠RNP and ∠PNS are adjacent angles because they have a common vertex (N) and a common side (\(\overline{N P}\)) and they do not overlap.
Adjacent angles

Question 26.
Find the measure of ∠QNS.
Answer:
m∠QNS + m∠SNP = m∠MNR Vertical ∠s:
m∠QNS + 61° = Substitute:
m∠QNS + 61° – 61° = 138° – 61° Subtract 61° from both sides:
m∠QNS = 77° Simplify:
m∠QNS = 77°

Question 27.
Find the measure of ∠PNR.
Answer:
m∠MNP = 180° Straight ∠
m∠PNR + 138° = 180° Substitute:
m∠PNR + 138° — 138° = 180° — 138° Subtract 138° from both sides:
m∠PNR = 42° Simplify:
m∠PNR = 42°

Use an equation to find the value of each variable.

Question 28.
\(\overleftrightarrow{\mathrm{AB}}\) is a straight line.

Answer:
x°=30°, y°=150°, w°=120°

Explanation:

∠COE + ∠BOE =90°
2x°+x°= 90°
3x°=90
x°=30°
2x°=60°
∠AOC + ∠COE = 180°
w°+2x°=180°
w°+60°=180°
w°=180°-60°
w°=120°
∠AOE + ∠BOE = 180°
y°+x°=180°
y°+30°=180°
y°=180°-30°
y°=150°

Question 29.
\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) is a straight line.

Answer:
3b° = 90°, b° = 30°, c° = 60°

Explanation:

3b°=90°
b°=90÷3
b°=30°
∠DOB = ∠AOC (vertically opposite angles)
∠AOC = c
∠DOB = c°
∠AOD = 120°
∠DOB + ∠AOD = 180°
c°+120° = 180°
c° = 180°- 120°
c° = 60°

\(\overleftrightarrow{\mathrm{AB}}\) and \(\overleftrightarrow{\mathrm{CD}}\) are straight lines.

Question 30.
The ratio ∠1 : ∠2 = 3 : 1

Answer:
p°= 45°, q°= 135°

Explanation:
Let the given ratio be x
The sum of angles in a traingle is 180°
∠AOD +∠AOC = 180°
The ratio ∠1 : ∠2 = 3x : 1x
3x + x = 180°
4x = 180°
x°= 180÷4
x°= 45°
3x = 135°
∠1 = ∠3 vertically opposite angles
∠2 = ∠4 vertically opposite angles
∠3 = q°= 135°
∠4= p°= 45°

Question 31.

Answer:
p°=50°, q°= 80°, 2p° = 100°

Explanation:

∠AOC = ∠DOB
2p° = p+50
2p°-p° = 50
p°=50°
∠DOB = 100°
∠DOB+∠COB = 180°
P+50 = 50+50 = 100°
100°+q°=180°
q°=180°-100°
q°= 80°

Solve.

Question 32.
The diagram below shows the flag of the Philippines, m∠ADB = 60° and m∠ADC = m∠BDC. Find the measures of ∠ADC and ∠BDC.

Answer:
∠ADB = 60°, ∠BDC = 150°

Explanation:
∠ADB = 60°
∠ADC = ∠BDC
∠ADB + ∠ADC +∠BDC = 360°
60° + ∠ADC + ∠ADC = 360°
60 + 2 ∠ADC = 360°
2 ∠ADC = 360°-60°
2∠ADC = 300°
∠ADC = 300÷2
∠ADC = 150°
∠BDC = 150°

Question 33.
Math Journal
The diagram shows a pattern on a carpet.
a) Are ∠4 and ∠6 vertical angles? Explain why or why not.
Answer:
No.
When two lines cross then vertical angles are opposite to each other.

Explanation:
The angles ∠4 and ∠6 are not vertical angles. Because the vertical angles are opposite to each other but by observing the angles from the given figure the angles given are not opposite to each other.

b) Suppose m∠4 = m∠6. Are ∠4 and ∠5 supplementary angles? Explain your answer.

Answer:
Yes, angles ∠4 and ∠5 are supplementary angles.

Explanation:
Two angles are supplementary when if they add up to 180°. If we add up the given angles ∠4 and ∠5 then the angles add up to 180°