Perimeter and Area of Plane Figures Test

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Perimeter and Area of Plane Figures Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

What is the perimeter of an isosceles triangle that has equal sides of length 15 cm each and an unequal side of length 8 cm?

A
38 cm

B
36 cm

C
34 cm

D
32 cm

Solution

Length of each equal side = 15 cm [Given]

Length of the unequal side = 8 cm [Given]

We know that perimeter is the distance along the boundary of a closed figure.

∴Perimeter of the given isosceles triangle = Sum of all three sides

= 15 cm + 15 cm + 8 cm

= 38 cm
Question 2
1 / -0

The perimeter of a regular octagon is 64 cm. What is the length of each of its sides?

A
10 cm

B
9 cm

C
8 cm

D
7 cm

Solution

We know that a regular octagon has 8 sides of equal lengths.

Therefore, perimeter of the octagon = (8 × Side)

Perimeter of the octagon = 64 cm

∴ 8 × Side = 64 cm

⇒ Side = 64cm/8 = 8 cm

Thus, the length of each side of the regular octagon is 8 cm.
Question 3
1 / -0

Use the following information to answer the next question.

A rectangular frame of length 50 cm and width 20 cm is to be decorated along its borders with a ribbon.

What is the length of the ribbon required for decorating the frame?

A
140 cm

B
100 cm

C
70 cm

D
50 cm

Solution

Length of frame = 50 cm

Breadth of frame = 20 cm

The length of the ribbon required for decorating the frame is equal to its perimeter.

∴ Perimeter of the frame = 2 (Length + Breadth)

= 2 (50 + 20) cm

= 2 (70) cm

= 140 cm

Thus, the length of the ribbon required for decorating the frame is 140 cm.
Question 4
1 / -0

Use the following information to answer the next question.

Rahul formed a square by bending a string of length 56 cm in such a way that both the ends of the string are joined. The string, which is bent in the form of square, is placed on a floor.

What is the area of the floor covered by the string?

A
112 cm²

B
196 cm²

C
218 cm²

D
234 cm²

Solution

Since the string is bent in the form of a square by joining both the ends of the string, the perimeter of the string is the length of the string.

∴ Perimeter of the square = 56 cm

⇒ 4 × Side of the square = 56 cm

⇒ Side of the square= 56cm/4 = 14cm

Area of the square = Side × Side = 14 cm × 14 cm = 196 cm²

Thus, the area of the floor covered by the string is 196 cm².
Question 5
1 / -0

Use the following information to answer the next question.

A rectangular frame of length 12 m is formed with a 40 m long wire. The wire is re-bent to form a square.

Which of the following alternatives represents the correct statement?

A
The area of the square is 2 sq. m less than that of rectangle.

B
The area of the square is 4 sq. m less than that of rectangle.

C
The area of the square is 2 sq. m more than that of rectangle.

D
The area of the square is 4 sq. m more than that of rectangle.

Solution

It is given that a rectangular frame of length 12 m is formed with a 40 m long wire.

∴ Perimeter of the rectangle = 40 m

⇒ 2 (Length + Breadth) = 40 m

⇒ 12 m + Breadth = 20 m

⇒ Breadth = 20 m − 12 m = 8 m

∴ Area of rectangle = Length × Breadth = 12 m × 8 m = 96 sq. m

The perimeter of square is also 40 m.

∴ 4 × Side of square = 40 m

⇒ Side of square = 40/4m = 10 m

∴ Area of square = Side × Side = 10 m × 10 m = 100 sq. m

Therefore, difference between the areas = (100 sq. m − 96 sq. m) = 4 sq. m

Thus, the area of the square is 4 sq. m more than that of rectangle.
Question 6
1 / -0

A rectangular jogging track is of length 60 m and breadth 40 m. How much time will a man take to cover three rounds of the track, jogging at the uniform speed of 80 m/min?

A
4.5 min

B
5 min

C
6 min

D
7.5 min

Solution

Length of the track = 60 m

Breadth of the track = 40 m

Distance covered by man in 1 round of the jogging track is equal to the perimeter of the rectangle.

Perimeter of a rectangle is given by, 2 × (length + breadth).

∴ Distance covered by man in 1 round = 2 × (60 m + 40 m) = 200 m

∴ Distance covered by man in 3 rounds = 3 × 200 m = 600 m

Speed = 80 m/min

∴ Time taken to cover 3 rounds of track = Distance/Speed =600/80 min = 7.5 min.

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Re-Attempt Weekly Quiz Competition

Perimeter and Area of Plane Figures Test - 1

Result

Perimeter and Area of Plane Figures Test - 1

Score

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Rank

-

SHARING IS CARING

Question 1

38 cm

36 cm

34 cm

32 cm

Solution

Question 2

10 cm

9 cm

8 cm

7 cm

Solution

Question 3

140 cm

100 cm

70 cm

50 cm

Solution

Question 4

112 cm2

196 cm2

218 cm2

234 cm2

Solution

Question 5

The area of the square is 2 sq. m less than that of rectangle.

The area of the square is 4 sq. m less than that of rectangle.

The area of the square is 2 sq. m more than that of rectangle.

The area of the square is 4 sq. m more than that of rectangle.

Solution

Question 6

4.5 min

5 min

6 min

7.5 min

Solution

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112 cm²

196 cm²

218 cm²

234 cm²