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Area of Squares and Rectangles - Problems with Solutions
By
Catalin David
Problem 1
A rectangle has a length of 6 inches and a width of 4 inches. The area is
in
^{2}
Solution:
Area of a rectangle is L × W where L is the length and W is the width
6 × 4 = 24 in
^{2}
.
Problem 2
The side of a square is 5 meters. The area of the square is
m
^{2}
Solution:
The sides of а square are equal size. The square's area is a × a where
a
is the side of the square.
5 × 5 = 25
Problem 3
The area of a square is 16 in
^{2}
. Its side is
in.
Solution:
Area = a × a where
a
is the length of a side
a × a = 16
4 × 4 = 16
a = 4
Problem 4
The area of a rectangle is 45 cm
^{2}
. If its length is 9 cm, then its width is
cm
Solution:
Area = L × W
45 = 9 × W
W = 45 ÷ 9 = 5
Problem 5
The perimeter of a square is 24 cm. The area of the square is
cm
^{2}
Solution:
Perimeter is 4 × a where
a
is the side of the square
4 × a =24
a = 24 ÷ 4 = 6
Area = a × a = 6 × 6 = 36 cm
^{2}
Problem 6
The length of a rectangle is 12 cm and its width is 5 cm smaller. The area of the rectangle is
cm
^{2}
.
Solution:
L = 12
W = 12 - 5 = 7
Area = L × W = 12 × 7 = 84 cm
^{2}
Problem 7
A rectangle has the length of 12 cm and the width 3 times smaller. Its area is
cm
^{2}
Solution:
L = 12
W = 12 ÷ 3 = 4
Area = L × W = 12 × 4 = 48 cm
^{2}
Problem 8
The length of a rectangle is 6 cm and the width is 4 cm. If the length is greater by 2 cm, what should the width be so that the new rectangle have the same area as the first one?
Solution:
The area of the first rectangle is L × W = 6 × 4 = 24 cm
^{2}
The new length is 6 + 2 = 8 cm
8 × W = 24 then W = 24 ÷ 8 = 3 cm
Problem 9
How many squares with the side of 2 cm cover the surface of a rectangle with a length of 24 cm and a width of 8 cm?
Solution:
The area of the square is 2 × 2 = 4 cm
^{2}
The area of the rectangle is L × W = 24 × 8 = 192 cm
^{2}
The number of squares is 192 ÷ 4 = 48
Problem 10
A square with a side of 6 cm and a rectangle with a width of 4 cm have the same area. What's the length of the rectangle?
Solution:
The area of the square is 6 × 6 =36 cm
^{2}
The area of the rectangle is L × W. Because the areas are equal L × W = 36
L × 4 = 36
L = 36 ÷ 4 = 9 cm
Problem 11
The side of a square is 5 cm. If its side is doubled, how many times is the area of the new square bigger than the area of the old square?
Answer:
times.
Solution:
The area of the first square is 5 × 5 = 25 cm
^{2}
The area of the new square is 10 × 10 = 100 cm
^{2}
The area is 4 times bigger
Problem 12
On a field whose length is 25 metres and width is 12 metres a house was built the shape of a square whose side is 9 metres. What's the area of the garden?
Solution:
The field is a rectangle whose length is 25 metres and width is 12 metres. The area is L × W = 25 × 12 = 25 × 4 × 3 = 100 × 3 = 300 m
^{2}
The field covered by the house is a square with the side of 9 metres. The area is 9 × 9 = 81 m
^{2}
The garden's area will be 300 - 81 = 219 m
^{2}
Problem 13
What is the area of the blue zone?
Solution:
The area of the blue zone is half of the rectangle's area.
Area = (L × W) ÷ 2 = (12 × 9) ÷ 2 = 108 ÷ 2 = 54 m
^{2}
Problem 14
What is the area of the red zone?
Solution:
[AF] = 24 m
[DE] = 15 m
[BC] = [AF] - [DE] = 9 m
[FE] = 20 m
[AB] = 9 m
[CD] = [FE] - [AB] = 11 m
The area of the red zone is equal to the difference between the area of the rectangle with a length of 24 metres and a width of 20 metres and the area of the rectangle with a length of 11 metres and a width of 9 metres
Area = (24 × 20) - ( 11 × 9) = 480 - 99 = 381 m
^{2}
Problem 15
A park shaped like a square with a side of 30 metres has an alley with the width of 2 metres. What's the area of the green zone?
Solution:
The area of the park is 30 × 30 = 900 m
^{2}
The alley is shaped like a rectangle with a length of 30 m and a width of 2 m. Area is L × W = 30 × 2 = 60 m
^{2}
The green zone has an area of 900 - 60 = 840 m
^{2}
Problem 16
A TV has a length of 80 cm and a width of 50 cm. If the border is 5 cm thick, the area of the screen is
Solution:
The length of the screen is 80 - (5 + 5) = 80 - 10 = 70 cm, the width of the screen is 50 - (5 + 5) = 50 - 10 = 40 cm
The area of the screen is 70 × 40 = 2800 cm
^{2}
Problem 17
A rectangle with a length of 14 cm and a width of 10 cm is covered by black squares and blue squares with the same area. What is the total area of the blue squares?
Solution:
Because there are 5 squares on the width of the rectangle and 7 squares on its length, then the side of the square is 2 cm. The area of a square is 4 cm
^{2}
. There are 18 blue squares, so their area will be 4 × 18 = 72 cm
^{2}
Problem 18
Rectangle ABCD has a length of 8 cm and a width of 6 cm. What is the total area of the blue rectangles?
Solution:
The area of the rectangle ABCD is 48 cm
^{2}
. Rectangle ABCD is split into 4 equal rectangles. Thus, the area of rectangle AMOQ is 12 cm
^{2}
. Rectangle AMOQ is split into 4 equal rectangles. The area of a blue rectangle is 3 cm
^{2}
. The total area of the 2 blue rectangles is 6 cm
^{2}
.
Problem 19
The square has a side of 2 cm and the rectangle has a length of 9 cm and a width of 6 cm. What is the area of the red zone?
Solution:
The area of the square is 4 cm
^{2}
. It's split in 4 equal squares. The area of a small square is 1 cm
^{2}
, the area of the rectangle is 54 cm
^{2}
, so the area of the red zone is 54 - 1 = 53 cm
^{2}
.
Problem 20
A box whose every side is a rectangle has a length of 10 cm, a width of 6 cm and a height of 8 cm. What is the total area of the box?
Solution:
The sides of the box are equal in pairs. Two of the sides have edges of 10 cm and 6 cm.
Their area is (10 × 6) × 2 = 60 × 2 = 120 cm
^{2}
Two of the sides have edges of 10 cm and 8 cm.
Their area is (10 × 8) × 2 = 80 × 2 = 160 cm
^{2}
Two of the sides have edges of 8 cm and 6 cm.
Their area is (8 × 6) × 2 = 48 × 2 = 96 cm
^{2}
The total area is 120 + 160 + 96 = 376 cm
^{2}
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