# Into Math Grade 4 Module 9 Lesson 4 Answer Key Solve Area Problems

We included HMH Into Math Grade 4 Answer Key PDF Module 9 Lesson 4 Solve Area Problems to make students experts in learning maths.

## HMH Into Math Grade 4 Module 9 Lesson 4 Answer Key Solve Area Problems

I Can find the area of a rectangular region that is formed by taking away rectangular pieces from its interior.

Step It Out

1. Josh runs a glass-bottom boat touring business. He wants to install a non-slip surface on the walkway of the deck. What is the area of the walkway that needs the non-slip surface? A. What do you know?
______________
______________

Total area of the deck = Base X Height = 18 X 10 = 180 square feet

Total area of the glass = Base X Height = 12 x 4 = 48 square feet

B. How can you solve the problem?
______________
______________ Total area of the deck = Base X Height = 18 X 10 = 180 square feet

Total area of the glass = Base X Height = 12 x 4 = 48 square feet

C. Subtract the area of the glass from the area of the deck.

Area of the deck – area of the glass = 180 – 48 = 132 square feet.

D. The area of the walkway that needs the non-slip surface is _______.

The area of the walkway that needs the non-slip surface is 132 square feet

Turn and Talk What did you do to help make sense of the problem?

Step It Out

2. Josh has a second glass-bottom boat with two viewing areas. He wants to install a non-slip surface on the walkway of the deck. What is the area of the walkway that needs the non-slip surface? A. What do you know?
_______________________
_______________________

Total Base of Deck = 25 ft

Total Height of Deck = 15 ft

The base of Glass A = 7 ft

Height of Glass A = 9 ft

The base of Glass B = 11 ft

Height of Glass B = 9 ft

B. How can you solve the problem?
_______________________
_______________________ Total Area of Deck = Base X Height = 25 X 15 = 375 square ft

Area of Glass A = Base X Height = 7 X 9 = 63 square ft

Area of Glass B = Base X Height = 11 X 9 = 99 square ft

C. Add the areas of the glass, then subtract the sum from the area of the deck.

Total area of glasses = 63 + 99 = 162 square ft

Remaining area = 375 – 162 = 213 square ft

D. The area of the walkway that needs the non-slip surface is ___

The area of the walkway that needs the non-slip surface is 213 square ft.

Check Understanding Math Board

Question 1.
Carter places a 3-inch by 2-inch sticker on the back of an envelope that measures 8 inches by 10 inches. What is the area of the back of the envelope that is not covered by the sticker?

Base of envelope = 8 inches

Height of envelope = 10 inches

Base of stick = 3 inch

Height of stick = 2 inch

Total area of envelope = Base X Height = 8 x 10 = 80 square inches

Area of stick = Base X Height = 3 X 2 = 6 square inches

Therefore, the area of the back of the envelope that is not covered by the sticker = 80 – 6 = 74 square inches.

Question 2.
Chloe is building a wooden bench around a tree. The space for the tree is a square with a side length of 4 feet. The bench will be a rectangle that measures 12 feet by 14 feet.
What is the area of the bench? _______

Base of bench = 12 feet

Height of bench = 14 feet

Side for space for tree = 4 feet

Total area of bench = base X height = 12 X 14 = 168 square feet

Area of space for tree =  side X side = 4 X 4 = 16 square feet

Therefore, remaining area = 168 – 16 = 152 square feet.

Use Structure Find the area of the shaded part of the rectangle.

Question 3. Area = ___

From the figure,

Total Area be ‘A’ = Length x width

A = 4 x 8 = 32 square cm

Area of Un Shaded region be ‘a’ = Length x Width

a = 1 x 5 = 5 square cm

Area of shaded region = A – a = 32 – 5 = 27 square cm

Therefore, Area of shaded region = 27 square cm.

Question 4. Area = ___

From the figure,

Total area be ‘A’ = Length x Width

A = 5 x 5 = 25 square cm

Area of Unshaded region be ‘a’ = length x Width

a = 2 x 2 = 4 square cm

Area of shaded region = A – a = 25 – 4 = 21 square cm

Therefore, Area of Shaded region is 21 square cm.

Question 5. Area = ___

From the figure,

Total area be ‘A’ = Length x width

A = 9 x 5 = 45 square m

Area of unshaded region be ‘a’ = Length x Width

a = 3 x 3 = 9 square m

Area of shaded region = A – a = 45 – 9 = 36 square m

Therefore, Area of shaded region = 36 square m.

Question 6.
The central chamber of the Lincoln Memorial is 74 feet long and 60 feet wide. Lincoln’s statue sits on a pedestal that is 17 feet long and 16 feet wide. What is the area of the central chamber that surrounds the pedestal?
______________

Given,

The central chamber of the Lincoln Memorial is 74 feet long and 60 feet wide.

Lincoln’s statue sits on a pedestal that is 17 feet long and 16 feet wide.

Area of central chamber = 74 x 60 = 4,440 square feet

Area of pedestal = 17 x 16 = 272

Remaining area = 4,440 – 272 = 4,168 square feet.

Therefore, 4,168 square feet is the area of the central chamber that surrounds the pedestal.

Question 7.
STEM An electrician is installing a wall plate for a double light switch like the one shown. The wall plate is 12 centimeters by 13 centimeters. Each hole for the light toggle measures 1 centimeters by 2 centimeters. What is the area of the wall plate, not including
the holes? ___ Given,

An electrician is installing a wall plate for a double light switch like the one shown.

The wall plate is 12 centimeters by 13 centimeters.

Each hole for the light toggle measures 1 centimeter by 2 centimeters.

Area of wall plate = 12 x 13 = 156 square cm

Area of each hole = 1 x 2 = 2 square cm

Number of holes = 2

So, area of holes = 2 x 2 = 4 square cm

Remaining area = 156 = 4 = 152 square cm

Therefore, the area of the wall plate not including holes is 152 square cm.

Question 8.
Brett wants to hang wallpaper on a rectangular wall that measures 8 feet by 14 feet. The wall has a window with an area of 4 square feet. What is the area of the wall that Brett wants to wallpaper?
______________

Given,

Brett wants to hang wallpaper on a rectangular wall that measures 8 feet by 14 feet.

The wall has a window with an area of 4 square feet.

The total area of the wall be A= length x width

A = 8 x 14 = 112 square feet.

area of the window be a = length x width

a = 4 x 4 = 16 square feet

remaining area = A – a = 112 – 16 = 96 square feet

Therefore, the area of the wallpaper that Brett wants wallpaper is 96 square feet.

Question 9.
Owen is painting a wall. The wall has a window and a door, as shown in the diagram. What is the area of the wall not including the window and the door? _______ Given,

Owen is painting a wall.

The wall has a window and a door, as shown in the diagram.

Total area be A = length x width

A = 9 x 14 =126 square feet

Area of the window be a = length x width

a = 3 x 2 = 6 square feet

Area of the door be b = length x width

b = 7 x 3 = 21 square feet

Remaining area = A – ( a + b ) = 126 – ( 6 + 21 ) = 126 – 27 = 99 square feet.

the area of the wall not including the window and the door is 99 square feet.

Question 10.
Reason A square fountain covers an area of 25 square meters. A plaza surrounds it. The total area of the plaza is 4 times the area of the fountain. How can you find the area of the plaza not covered by the fountain?

Given,

A square fountain covers an area of 25 square meters.

A plaza surrounds it.

The total area of the plaza is 4 times the area of the fountain.

Area of the plaza = 4 x 25 = 100 square meters

remaining area = 100 – 25 = 75 square meters

Therefore, the area of the plaza not covered by the fountain is 75 square meters.

Question 11.
Laura is designing a square flower garden. The outer edge will have blue flowers, then there will be purple flowers, and the middle will have yellow flowers. How can you find the area that will only have yellow and blue flowers? Given,

Laura is designing a square flower garden.

The outer edge will have blue flowers, then there will be purple flowers, and the middle will have yellow flowers.

The area that will have only yellow flowers be a = length x width

a = 2 x 2 = 4 square yards

The area that will have only purple flowers be b = length x width

b = 4 x 4 = 16 square meters

The area that will have only blue flowers be c = length x width

c = 6 x 6 = 36 square yards.

So, a + c – b = 4 + 36 – 16 = 24 square yards

Therefore, the area that will only have yellow and blue flowers is 24 square yards.

Use Structure Find the area of the shaded part of the rectangle.

Question 12. Area = ___

From the figure,

total area = length x width =  5 x 11 = 55 square yards.

area of one figure be a = length x width

a = 2 x 3 = 6 square yards

area of second figure be b = length x width

b = 2 x 2 = 4 square yards

So, total area – ( a + b ) = 55 – ( 6 + 4 ) = 44 square yards.

Therefore, area of the shaded region is 44 square yards.

Question 13. Area = ___

From the figure,

total area = length x width =  7 x 9  = 63 square meters.

area of one figure be a = length x width

a = 4 x 1 = 4 square meters

area of second figure be b = length x width

b = 1 x 5 = 5 square meters

So, total area – ( a + b ) = 63 – ( 4 + 5 ) = 54 square meters.

Therefore, area of the shaded region is 54 square meters.

Scroll to Top