RS Aggarwal Solutions: Perimeter and Area (Exercise 12F) MCQs

Solution 1: (b) 28 cm
Ratio in the sides of a rectangle = 7 : 5 
Let, length (l) = 7x 
And breadth (b) = 5x 
perimeter = 96 cm ∴ 2(7x + 5x) = 96 
⇒ 7x + 5x = 96/2 
⇒ 12x = 48 
⇒ x = 48/12 
⇒ x = 4 
∴ length (l) = 7x = 7 × 4 = 28 cm

Solution 2: (d) 126 cm
Area of a rectangle = 650 cm² 
and breadth (b) = 13 cm 
Let, length = l cm 
Now, Area = 650 
⇒ l × b = 650 
⇒ l × 13 = 650 
⇒ l = 650/13 
⇒ l = 50
∴ Perimeter 
= 2(l + b) 
= 2 (50 + 13) cm 
= 2 × 63 cm 
= 126 cm

Solution 3: (b) ₹ 2340
Length of a rectangular field (l) = 34 m 
and breadth (b) = 18 m 
Perimeter = 2 (l + b) 
= 2 (34 + 18) m 
= 2 × 52 m 
= 104 m Rate of fencing 
= ₹ 22.50 per m 
∴ Total cost = 22.50 × 104 
= ₹ 2340

Solution 4: (b) 16 m
Total cost of fencing = ₹ 2400 
Rate of fencing = ₹ 30 per m 
Perimeter of the rectangular field = Total Cost/Rate = 2400/30 = 80 m 
∴ 2 (l + b) = 80 
⇒ (l + b) = 80/2 
⇒ 24 + b = 40 ∴ length (l) = 24 m 
⇒ b = 40 - 24 
⇒ b = 16 m

Solution 5: (a) 48 cm
Let width of a rectangle = x 
Then length = 3x 
and diagonal = 6√10 cm 
∴ length² + breadth² = diagonal² 
∴ (3x)² + (x)² = (6√10)² 
⇒ 9x² + x² = 6 × 6 × 10
⇒ 10x² = 360 
⇒ x² = 360/10 
⇒ x² = 36 
⇒ x² = (6)² ∴ x = 6
∴ perimeter = 2 (l + b) 
= 2(3x + x) 
= 2 × 4x 
= 8x 
= 8 × 6 
= 48

Solution 6: (b) 2 : 1
Ratio in length and perimeter of a rectangle = 1 : 3 
Let length = x, 
then perimeter = 3x 
Now, 2(l + b) = perimeter
⇒ 2(x + b) = 3x 
⇒ x + b = 3x/2 
⇒ b = 3x/2 - x 
⇒ b = (3x - 2x)/2 
⇒ b = x/2
∴ Ratio in length and breadth 
= x : x/2 
= x/(x/2) 
= 2/1 
= 2 : 1

Solution 7: (d) 16
Length of the sheet (l) = 72 cm 
and breadth (b) = 48 cm 
∴ Area of the sheet = l × b = 72 × 48 cm² 
Now, area of paper for one envelope = 18 × 12 cm²
∴ No. of envelopes = Area of the sheet / area of paper for one envelope 
= (72 × 48) / (18 × 12) 
= (6 × 48) / 18 
= 48/3 
= 16

Solution 8: (b) 24.3 m²
Length of a rectangular room (l) = 5 m 40 cm 
= 5.4 m 
and breadth (b) = 4 m 50 cm 
= 4.5 m 
∴ Area = l × b 
= 5.4 × 4.5 m² 
= 24.3 m²

Solution 9: (d) 80000
Length of the lane = 150 m 
Breadth of the lane = 9 m 
∴ Area of the lane = (150 × 9) m² 
= 1350 m²
Area of the brick = 22.5 cm × 7.5 cm 
= 168.75 cm² 
= 168.75/10000 m² 
= 0.016875 m²
∴ Number of bricks required = Area of the lane / Area of one brick 
= 1350 / 0.016875 
= 1350 × 1000000/16875 
= 80000

Solution 10: (b) 200 cm²
Length of diagonal of a square = 20 cm 
Area of the square = 1/2 × (diagonals)² 
= 1/2 × (20)² 
= 1/2 × 400 
= 200 cm²

Solution 11: (c) 20 m
Total cost of fencing around a square field = ₹ 2000 and rate = ₹ 25 per metre 
∴ Perimeter of the square field = Total cost/Rate = 2000/25 = 80 m
Now, 
Perimeter of the square field = 4 × side 
⇒ 80 = 4 × side 
⇒ side = 80/4 
⇒ side = 20 m

Solution 12: (b) 22 cm
Circumference = πd 
= 22/7 × 7 
= 22 cm

Solution 13: (a) 28 cm
Circumference = 88 ∴ 2πr = 88 
⇒ 2r × 22/7 = 88 
⇒ 2r = 88 × 7/22 
⇒ 2r = 28 
∴ Diameter (d) = 2r = 28 cm

Solution 14: (b) 110 m
Distance covered in one revolution = Circumference of the circle 
= π × diameter 
= 22/7 × 70 
= 220 cm 
= 220/100 m
∴ Distance covered in 50 revolutions 
= 220/100 × 50 
= 110 m