JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

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Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:
Radius (r) = \(\frac{10.5}{2}\) cm = 5.25 cm

Slant height (l) = 10 cm
Curved surface area of the cone = (πrl) cm²
= (\(\frac{22}{7}\) × 5.25 × 10) cm²
= 165 cm²

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Ans. Radius (r) = \(\frac{24}{2}\) m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m²
= \(\frac{22}{7}\) × 12 × (21 + 12) m²
= \(\frac{22}{7}\) × 12 × 33 m²= 1244.57 m²

Question 3.
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.
Answer:
(i) Curved surface of a cone = 308 cm²
Slant height (l) = 14 cm
Let r be the radius of the base
∴ πrl = 308
⇒ \(\frac{22}{7}\) × r × 14 = 308
⇒ 44r = 308
⇒ r = \(\frac{308}{\frac{22}{7} \times 14}\) = 7 cm.

(ii) TSA of the cone = πr(l + r) cm²
= \(\frac{22}{7}\) × 7 × (14 + 7) cm²
= (22 × 21) cm²
= 462 cm²

Question 4.
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 m² canvas is ₹ 70.
Answer:
(i) 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 = \(\sqrt{\mathrm{h}^2+\mathrm{r}^2}=\sqrt{10^2+24^2}\)
= \(\sqrt{100 +576}\) = 26 m

(ii) Canvas required to make the conical tent = Curved surface of the cone
= πrl = \(\frac{22}{7}\) × 24 × 26 m² = \(\frac{13728}{7}\) m

Cost of 1 m² canvas = ₹ 70
∴ Cost of canvas = ₹ \(\frac{13728}{7}\) × 70
= ₹ 137280

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14)
Answer:
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 = \(\sqrt{\mathrm{h}^2+\mathrm{r}^2}\) = \(\sqrt{10^2+24^2}\)
= \( \sqrt{100} \) = 10 m

CSA of conical tent = πrl = (3.14 x 6 x 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
Area of sheet = CSA of tent
⇒ [(x – 0.2) m × 3] m² = 188.4 m²
⇒ x – 0.2 = 62.8
⇒ x = 63 m
∴ Length of tarpaulin sheet required = 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².

Ans. Radius (r) = \(\frac{14}{2}\) m = 7 m
Slant height of the tomb (l) = 25 m
Curved surface area = πrl m²
= \(\frac{22}{7}\) × 25 × 7 m²= 550 m²

Rate of white-washing = ₹ 210 per 100 m²
Total cost of white-washing the tomb = ₹ (550 × \(\frac{210}{100}\)) = ₹ 1155

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Ans.
Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = \(\sqrt{\mathrm{h}^2+\mathrm{r}^2}\) = \(\sqrt{24^2+7^2}\)
= \( \sqrt{625} \) = 25 m

Sheet required for one cap = Curved surface of the cone
= πrl cm²
= \(\frac{22}{7}\) × 7 × 25 cm²
= 550 cm²

Sheet required for 10 caps = 550 × 10 cm² = 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m2, what will be the cost of painting all these cones?
(Use n = 3.14 and take Vl.04 = 1.02)
Answer:
Radius of the cone (r) = \(\frac{40}{2}\) cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l be the slant height of a cone.
l = \(\sqrt{\mathrm{h}^2+\mathrm{r}^2}\) = \(\sqrt{1^2+0.2^2}\)
= \( \sqrt{1.04} \) = 1.02 m

Rate of painting = ₹ 12 per m²
Curved surface of 1 cone = πrl m²
= (3.14 × 0.2 × 1.02) m² = 0.64056 m²

Curved surface of such 50 cones= (50 × 0.64056) m²= 32.028 m²
Cost of painting all these cones = ₹ (32.028 × 12) = ₹ 384.34