Mensuration Test - 6

Side of the square field = 24 m

Perimeter of the square field = 4 × 24 = 96 m

Length of the rectangular field = 18 m

Breadth of the rectangular field = 16 m

Perimeter of the rectangular field = 2(18 + 16)

= 2(34) = 68 m

Distance covered by Piyush = 2 × 68 m = 136 m

Distance covered by Mohit = 3 × 96 m = 288 m

So, Mohit covers more distance than Piyush, the difference being 152 m.

Length of the rectangular field = 100 m
Breadth of the rectangular field = 50.5 m

Perimeter of the field = 2(100 m + 50.5 m) = 2(150.5) m = 301 m
Cost of ploughing = Rs. 15 per metre

Cost of ploughing 301 m = 301 × 15 = Rs. 4,515

Area of the land = 5.8 m × 5.2 m = 30.16 m²

Area of one flower bed = 1.5 m × 1.5 m = 2.25 m²

Area of 7 flower beds = 7 × 2.25 m² = 15.75 m²

So, area of the remaining part of the land = 30.16 m2 - 15.75 m2

= 14.41 m²

Length of each side of the park = 110 m

Perimeter of the equilateral triangular park = 3 × 110 m = 330 m
Distance covered by Mohan in 5 rounds = 5 × 330 m = 1650 m

Length of the rectangular park = 35 m

Breadth of the rectangular park = 25 m

Perimeter of the park = 2(35 + 25) = 2(60) = 120 m

So, distance travelled by the athlete in 18 rounds = 18 × 120 = 2160 m

(i) The perimeter of a rectangle is four times its sides. FALSE; perimeter of a rectangle = 2(l + b)
(ii) In a blackboard, the region available for writing is its area. TRUE
(iii) The distance covered along the boundary of a closed figure is called its perimeter. TRUE; the perimeter of a closed figure is the distance covered in one round along the boundary of the figure.

Area = 400 cm²
25 × P = 400
P = 16 cm

Perimeter = 2(16 + 25) = 82 cm
So, P = 16 cm, Q = 82 cm

Also, perimeter = 132
2(18 + R) = 132

18 + R = 66
R = 48 cm

Area = S = 48 × 18 = 864 cm²