RS Aggarwal Class 6 Math Twenty-one Chapter Concept Of Perimeter And Area Exercise 21E Solution

EXERCISE 21E

OBJECTIVE QUESTIONS

Mark (√) against the correct answer in each of the following:

(1) The sides of a rectangle are in the ratio 7 : 5 and its perimeter is 96 cm. The length of the rectangle is

Ans: (b) 28 cm

Let, the length of the rectangle = 7x cm and its breadth = 5x cm.

(2) The area of a rectangle is 650 cm² and its breadth is 13 cm. The perimeter of the rectangle is

Ans: (d) 126 cm

(3) The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 22.50 per meter is

Ans: (b) Rs 2340

Perimeter of rectangle = 2(34 + 18) = 104 m

Rate of fencing = Rs 22.50

∴ Cost of fencing = Rs (104 × 22.50) =Rs 2340

(4) The cost of fencing a rectangular field at Rs 30 per meter is Rs 2400. If the length of the field is 24 m, then its breadth is

Ans: (b) 16 m

The breadth of the field = 16 m.

(5) The area of a rectangular carpet is 120 m² and its perimeter is 46 m. The length of its diagonal is

Ans: (c) 17 m

Here, l + b = 23 and lb = 120.

(6) The length of a rectangle is three times its width and the length of its diagonal is 6√10 cm. The perimeter of the rectangle is

Ans: (a) 48 cm

Let the width of rectangle = x cm and its length = 3x cm.

Hence, Width = 6 cm and length = 6 × 3 = 18 cm

So, the perimeter = 2(18+6) = 48 cm

(7) If the ratio between the length and perimeter of a rectangular plot is 1 : 3, then the ratio between the length and breadth of the plot is

Ans: (b) 2 : 1

Here, let the length be x cm. Then, its perimeter is 3x cm.

(8) The length of the diagonal of a square is 20 cm. Its area is

Ans: (b) 200 cm²

(9) The cost of putting a fence around a square field at Rs 25 per meter is Rs 2000. The length of each side of the field is

Ans: (c) 20 m

(10) The diameter of a circle is 7 cm. Its circumference is

Ans: (b) 22 cm

(11) The circumference of a circle is 88 cm. Its diameter is

Ans: (a) 28 cm

C = 2r

Hence, the diameter = 2 × 14 = 28 cm

(12) The diameter of a wheel of a car is 70 cm. How much distance will it cover in making 50 revolutions?

Ans: (b) 110 m

In revolution, the wheel covers a distance equal to its circumference.

So, distance covered by the wheel in 1 revolution = 2.20 m

And, distance covered by the wheel in 500 revolution = (2.20 50) m = 110 m

(13) A lane 150 m long and 9 m wide is to paved with bricks, each measuring 22.5 cm by 7.5 cm. How many bricks are required?

Ans: (d) 80000

Here, the area of the lane = (15000 × 900) = 13500000 cm²

And, the area of the bricks = (22.5 × 7.5) = 168.75 cm²

Hence, the number of the bricks of the lane

(14) A room is 5 m 40 cm long and 4 m 50 cm broad. Its area is

Ans: (b) 24.3 m²

Here, Length = 5 m 40 cm = 5.40 m and Breadth = 4 m 50 cm = 4.50 m

Area of the room = (5.40 × 4.50) = 24.3 m²

(15) How many envelopes can be made out of a sheet of paper 72 cm by 48 cm, if each envelope requires a paper of size 18 cm by 12 cm?

Ans: (d) 16

Area of the sheet = (72 × 48) = 3456 cm²

And, Length of the each envelope = 18 cm and Breadth of the each envelope = 12 cm.

Area of the each envelope = (18 × 12) = 216 cm²

∴ number of the envelopes of the sheet