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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

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

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:
Here, r = \(\left(\frac{10.5}{2}\right)\) cm = 5.25 cm and l = 10 cm.
Curved Surface area of the cone = πrl = (\(\frac{22}{7}\) × 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.
Answer:
Here, r = \(\left(\frac{24}{2}\right)\)m = 12 m and l = 21 m
Total surface area of the cone = (πrl + πr²) = πr(l + r)
= \(\frac{22}{7}\) × 12 × (21 + 12) m²
= (\(\frac{22}{7}\) × 12 × 33)m² = 1244.57 m² (approx.)

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
So, \(\frac{22}{7}\) × r × 14 = 308 ⇒ r = \(\frac{308 \times 7}{22 \times 14}\) = 7
Thus, the radius of the base is 7 cm.

(ii) Total surface area of the cone = πr(l + r)
= \(\frac{22}{7}\) × 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) Here, r = 24 m and h = 10 m.
Let l be the slant height of the cone.
Then, l² = h² + r² ⇒ l = \(\sqrt{h^{2}+r^{2}}\)
So, l = \(\sqrt{24^{2}+10^{2}}\) m = \(\sqrt{576+100}\) m
= \(\sqrt{676}\) m = 26 m

(ii) Canvas required to make the conical tent
= Curved surface of the cone
= πrl = (\(\frac{22}{7}\) × 24 × 26) m²
Rate of canvas per 1 m² is ₹ 70.
∴ Total cost of canvas = ₹ (\(\frac{22}{7}\) × 24 × 26 × 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:
Let r m be the radius, h m be the height and l m be the slant height of the tent. Then, r = 6 m and h = 8 m.
So, l = \(\sqrt{r^{2}+h^{2}}=\sqrt{6^{2}+8^{2}}\)m = \(\sqrt{36+64}\) m
= 7100 m = 10 m
Area of the canvas used for the tent = Curved surface area of the cone
= πrl = (3.14 × 6 × 10) m² = 188.4 m²

Now, this area is bought in the form of a rectangle of width 3 m.
Length of tarpaulin required = \(\frac{\text { Area of tarpaulin required }}{\text { Width of tarpaulin }}\)
= \(\left(\frac{188.4}{3}\right)\)m = 62.8 m.
The extra material required for stitching margins and cutting
= 20 cm = 0.2 m.
So, the total length of tarpaulin required = (62.8 + 0.2) m = 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.
Answer:
Here, l = 25 m and r = \(\left(\frac{14}{2}\right)\) m = 7 m.
Curved surface area = πrl = (\(\frac{22}{7}\) × 7 × 25) m² = 550 m²
Rate of whitewashing is ₹ 210 per 100 m².
∴ Cost of whitewashing 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.
Answer:
Let r be the radius, h be the height and l be the slant height of the joker’s cap. Then, r = 7 cm, h = 24 cm.
∴ l = \(\sqrt{h^{2}+r^{2}}=\sqrt{24^{2}+7^{2}}\) cm = \(\sqrt{576+49}\) cm
= \(\sqrt{625}\) cm = 25 cm.
Sheet required for one cap = Curved surface area of the cone
= πrl = (\(\frac{22}{7}\) × 7 × 25)cm² = 550 cm²
Sheet required for 10 such caps = 10 × 550 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 π = 3.14 and take \(\sqrt{1.04}\) = 1.02)
Answer:
Let r be the radius, h be the height and l be the slant height of a cone.
r = \(\left(\frac{40}{2}\right)\)cm = 20 cm = 0.2 m, and h = 1 m.
l = \(\sqrt{r^{2}+h^{2}}=\sqrt{0.04+1}\) m = \(\sqrt{1.04}\) m = 1.02 m

Curved surface area of 1 cone = πrl = (3.14 × 0.2 × 1.02) m²
Curved surface of such 50 cones = (50 × 3.14 × 0.2 × 1.02) m²
Cost of painting @ ₹ 12 per m² = ₹ (50 × 3.14 × 0.2 × 1.02 × 12)
= ₹ 384.34 (approx.)