Solutions: Mensuration - Vardaan Learning Institute

1. The area of a trapezium is $34 \text{ cm}^2$...

Solution:
Area = $\frac{1}{2}(a+b)h$.
$34 = \frac{1}{2}(10 + b) \times 4$.
$34 = 2(10 + b) \Rightarrow 17 = 10 + b \Rightarrow b = 7 \text{ cm}$.
Other side is 7 cm.

2. The diagonal of a quadrilateral shaped field is 24 m...

Solution:
Area = $\frac{1}{2} d (h_1 + h_2) = \frac{1}{2} \times 24 \times (8 + 13)$.
$= 12 \times 21 = 252 \text{ m}^2$.
Area = $252 \text{ m}^2$.

3. Find the total surface area of a closed cylindrical petrol storage tank...

Solution:
$r = 2.1 \text{ m}, h = 4.5 \text{ m}$.
TSA = $2 \pi r (r + h)$
$= 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5)$
$= 2 \times 22 \times 0.3 \times 6.6 = 13.2 \times 6.6 = 87.12 \text{ m}^2$.

4. A cuboid is of dimensions $60 \text{ cm} \times 54 \text{ cm} \times 30 \text{ cm}$...

Solution:
Vol of cuboid = $60 \times 54 \times 30$.
Vol of cube = $6 \times 6 \times 6$.
Number = $\frac{60 \times 54 \times 30}{6 \times 6 \times 6} = 10 \times 9 \times 5 = 450$.
450 small cubes.

5. Find the height of the cylinder whose volume is $1.54 \text{ m}^3$ and diameter...

Solution:
$d = 140 \text{ cm} = 1.4 \text{ m}, r = 0.7 \text{ m}$.
Vol = $\pi r^2 h$.
$1.54 = \frac{22}{7} \times (0.7)^2 \times h$.
$1.54 = 22 \times 0.1 \times 0.7 \times h = 1.54 h \Rightarrow h = 1 \text{ m}$.

6. A flooring tile has the shape of a parallelogram...

Solution:
Area of 1 tile = $24 \text{ cm} \times 10 \text{ cm} = 240 \text{ cm}^2 = 0.024 \text{ m}^2$.
Total Area = $1080 \text{ m}^2$.
No. of tiles = $\frac{1080}{0.024} = 45,000$.
45000 tiles.

7. An ant is moving around... (a) Semicircle...

Solution:
(a) Perimeter = $\pi r + d = \frac{22}{7} \times 1.4 + 2.8 = 4.4 + 2.8 = 7.2 \text{ cm}$.
(b) Semicircle arc + sum of lines.

8. Daniel is painting the walls and ceiling...

Solution:
Area = $2(l+b)h + lb$
$= 2(15+10) \times 7 + 15 \times 10$
$= 2(25) \times 7 + 150 = 350 + 150 = 500 \text{ m}^2$.
Cans = $500 / 100 = 5$.
5 cans required.

9. A godown is in the form of a cuboid...

Solution:
Vol of godown = $60 \times 40 \times 30 = 72000 \text{ m}^3$.
Vol of box = $0.8 \text{ m}^3$.
Number = $72000 / 0.8 = 90,000$.
90,000 boxes.

10. A rectangular paper of width 14 cm is rolled along its width...

Solution:
Width becomes height $h = 14 \text{ cm}$.
Radius $r = 20 \text{ cm}$.
Vol = $\pi r^2 h = \frac{22}{7} \times 20 \times 20 \times 14 = 22 \times 400 \times 2 = 17600 \text{ cm}^3$.

11. A rectangular piece of paper 11 cm x 4 cm is folded...

Solution:
Circumference = 11 cm (length forms circle).
$2 \pi r = 11 \Rightarrow 2 \times \frac{22}{7} \times r = 11 \Rightarrow r = \frac{7}{4}$.
$h = 4$.
Vol = $\pi r^2 h = \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 4 = 38.5$.
Volume = $38.5 \text{ cm}^3$.

12. The lateral surface area of a hollow cylinder is 4224...

Solution:
LSA = Area of Rectangle.
$4224 = l \times b = l \times 33$.
$l = 4224 / 33 = 128 \text{ cm}$.
Perimeter = $2(l+b) = 2(128 + 33) = 2(161) = 322 \text{ cm}$.

13. A road roller takes 750 complete revolutions...

Solution:
$r = 42 \text{ cm} = 0.42 \text{ m}, h = 1 \text{ m}$.
CSA = $2 \pi r h = 2 \times \frac{22}{7} \times 0.42 \times 1 = 2.64 \text{ m}^2$.
Area = $750 \times 2.64 = 1980 \text{ m}^2$.

14. A company packages its milk powder...

Solution:
Label height = $20 - 2 - 2 = 16 \text{ cm}$.
$r = 7 \text{ cm}$.
Area = $2 \pi r h = 2 \times \frac{22}{7} \times 7 \times 16 = 704 \text{ cm}^2$.

15. Water is pouring into a cubiodal reservoir...

Solution:
Rate = 60 L/min.
Vol = $108 \text{ m}^3 = 108000 \text{ L}$.
Time = $108000 / 60 = 1800 \text{ min}$.
Hours = $1800 / 60 = 30 \text{ hours}$.