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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

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

Note: Assume π = \(\frac{22}{7}\), unless stated otherwise.

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:
For the given cone, diameter d = 10.5 cm.
Then, radius r = \(\frac{10.5}{2}\) cm and slant height
l = 10 cm.
Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × \(\frac{10.5}{2}\) × 10 cm²
= \(\frac{22}{7}\) × \(\frac{105}{2}\) × cm²
= 11 × 15 cm²
= 165 cm²
Thus, the curved surface area of the given cone is 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:
For the given cone, diameter d = 24 m.
Then, radius r = \(\frac{24}{2}\) = 12 m and slant height l = 21 m.
Total surface area of a cone
= πr (l + r)
= \(\frac{22}{7}\) × 12(21 + 12) m²
= \(\frac{22 \times 12 \times 33}{7}\) m²
= \(\frac{8712}{7}\) m²
= 1244.57 m²
Thus, the total surface area of the given cone is 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:
For the given cone, slant height l = 14 cm and curved surface area = 308 cm²
(i) Curved surface area of a cone = πrl
∴ 308 cm² = \(\frac{22}{7}\) × r × 14 cm
∴ \(\frac{308 \times 7}{22 \times 14}\) cm = r
∴ r = 7 cm
Thus, the radius of the base of the cone is 7 cm.

(ii) Total surface area of a cone
= πrl + πr²
= 308 + \(\frac{22}{7}\) × 7 × 7 cm²
= 308 + 154 cm²
= 462 cm²
Thus, the total surface area of the cone is 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:
For the conical tent,
radius r = 24 m and height h = 10 m.

(i) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{10^{2}+24^{2}}\)
= \(\sqrt{100+576}\)
= \(\sqrt{676}\)
∴ l = 26 m
Thus, the slant height of the tent is 26 m.

(ii) Area of the canvas used to make tent
= Curved surface area of the conical tent
= πrl
= \(\frac{22}{7}\) × 24 × 26 m²
= \(\frac{13728}{7}\) m²
Cost of 1 m² canvas = ₹ 70
∴ Cost of \(\frac{13728}{7}\) m² canvas
= ₹ \(\left(70 \times \frac{13728}{7}\right)\)
= ₹ 1,37,280
Thus, the cost of canvas required is ₹ 1,37,280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l² = h² + r² = 8² + 6² = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m²
= 188.4 m²
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = \(\frac{188.4}{3}\) m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 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 whitewashing its curved surface at the rate of ₹ 210 per 100 m².
Answer:
For the given conical tomb,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{14}{2}\) = 7 m
and slant height l = 25 m. .
Area of the region to be whitewashed
= Curved surface area of the conical tomb
= πrl
= \(\frac{22}{7}\) × 7 × 25 m²
= 550 m²
Cost of whitewashing 100 m² region = ₹ 210
∴ Cost of whitewashing 550 m² region
= ₹ \(\left(\frac{210 \times 550}{100}\right)\)
= ₹ 1155
Thus, the cost of whitewashing the curved surface of the tomb is ₹ 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:
For the conical cap, radius r = 7 cm and height h = 24 cm.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{24^{2}+7^{2}}\)
= \(\sqrt{576+49}\)
= \(\sqrt{625}\) = 25 cm
Area of the sheet required to make 1 conical cap
= Curved surface area of the conical cap
= πrl
= \(\frac{22}{7}\) × 7 × 25 cm²
= 550 cm²
Area of sheet required to make 1 cap = 550 cm²
∴ Area of sheet required to make 10 caps
= 550 × 10 cm²
= 5500 cm²
Thus, the area of the sheet required to make 10 caps is 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled carboard. 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:
For the given cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{40}{2}\) = 20 cm = 0.2 m and
height h = 1 m.
l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{1^{2}+0.2^{2}}\)
= \(\sqrt{1.04}\)
= 1.02 m
Curved surface area of a cone
= πrl
= 3.14 × 0.2 × 1.02 m²
∴ Curved surface area of 50 cones
= 50 × 3.14 × 0.2 × 1.02 m²
= 32.028 m²
Cost of painting 1 m² region = ₹ 12
∴ Cost of painting 32.028 m² region
= ₹ (12 × 32.028)
= ₹ 384.34 (approx.)
Thus, the cost of painting all the 50 cones is ₹ 384.34 (approx.)