Surface Areas and Volumes NCERT Solution, study material, CBSE Notes

NCERT Solution: Surface Areas and Volumes

Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
= (2 × 22/7 × 2.1 × 4.5) m²
= 59.4 m²
(ii) Total surface area of the tank = 2πr(r + h) m²
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
= 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 – 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35) cm²
= 2200 cm²

The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr²
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
= (198 + 198/7) cm²
= 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
= 7920 cm²

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Radius (r) = 10.5/2 cm = 5.25 cm

Slant height (l) = 10 cm

Curved surface area of the cone = (πrl) cm²

=(22/7 × 5.25 × 10) cm²

=165 cm²

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m²
= 22/7 × 12 × (21 + 12) m²
= (22/7 × 12 × 33) m²
= 1244.57 m²

Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.

(i) Curved surface of a cone = 308 cm²
Slant height (l) = 14cm
Let r be the radius of the base
∴ πr�� = 308
⇒ 22/7 × r × 14 = 308
⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm²
= 22/7 × 7 ×(14 + 7) cm²
= (22 × 21) cm²
= 462 cm²

A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m²canvas is rs 70

Radius of the base (r) = 24 m
Height of the conical tent (h) = 10 m
Let l be the slant height of the cone.
∴ l²= h²+ r²
⇒ l = √h²+ r²
⇒ l = √10²+ 24²
⇒ l = √100+ 576
⇒ l = 26 m
(ii) Canvas required to make the conical tent = Curved surface of the cone
Cost of 1 m² canvas = rs70
∴ πrl = (22/7 × 24 × 26) m² = 13728/7 m²
∴ Cost of canvas = rs 13728/7 × 70 = rs 137280

What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π = 3.14).

Radius of the base (r) = 6 m
Height of the conical tent (h) = 8 m
Let l be the slant height of the cone.
∴ l = √h²+ r²
⇒ l = √8²+ 6²
⇒ l = √100
⇒ l = 10 m
CSA of conical tent = πrl
= (3.14 × 6 × 10) m² = 188.4 m²
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be x.
20 cm will be wasted in cutting.
So, the length will be (x – 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(x – 0.2 m) × 3] m = 188.4 m²
⇒ x – 0.2 m = 62.8 m
⇒ x = 63 m