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