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

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.3 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

Exercise – 13.3

1. Diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. Find its curved surface area (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Diameter of the base of the cone, d = 10.5 cm
Radius of the base of cone, r = diameter/2 = 10.5/2 cm = 5.25cm
The slant height of the cone, l = 10 cm
NCERT Class 9 Solutions Maths
Curved surface area = πrl
= $\frac{22}{7}$ × 5.25 × 10
= 165 cm²
Thus, curved surface area of the cone = 165 cm².

2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Diameter of cone, d = 24
Radius of cone, r = d/2 = 24/2 m = 12m
Slant height, l = 21 m
Total Surface area of the cone = πr(l + r)
= $\frac{22}{7}$ × 12 m × (12 m + 21 m)

= $\frac{22}{7}$ × 12 m × 33 m
= 8712/7 m²
= 1244.57 m²
Thus, total surface area of the cone = 1244.57 m².

3. Curved surface area of a cone is 308 cm^2, and its slant height is 14 cm. Find
(i) the radius of the base and
(ii) the total surface area of the cone.
(Assume π = $\mathbf{\frac{22}{7}}$ )

Answer – Let the radius of the cone = r
The slant height of the cone, l = 14 cm
Curved surface area = 308 cm²

(i) the radius of the base
We know that the Curved surface area of cone = πrl
πrl = 308 cm²
$\frac{22}{7}$ × r × 14 = 308 cm²
22 × r × 2 = 308 cm²
44 r = 308 cm²
r = 308/44 = 7 cm
Therefore, the radius of the cone base is 7 cm.

(ii) Total surface area of cone = Curved surface area of cone + Area of base (πr²)
Total surface area of cone = 308 + $\frac{22}{7}$ × 7²
= 308 + 154
= 462 cm²Therefore, the total surface area of the cone is 462 cm².

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 Rs 70.
(Assume π = $\mathbf{\frac{22}{7}}$ )

Answer –
Let the Slant height of conical tent = l
Height of conical tent, h = 10 m
Radius of the conical tent, r = 24m
NCERT Class 9 Solutions Maths

(i) slant height of the tent.
Slant height, l = √r² + h²
= √(24)² + (10)²
= √576 + 100
= √676
= 26 m
Therefore, the slant height of the tent is 26 m.

(ii) Curved surface area of the tent = πrl
= $\frac{22}{7}$ × 24 × 26 m²Cost of 1 m² canvas = Rs 70
The cost of the canvas required to make the tent, at ₹ 70 per m^²= 70 × Curved surface area of the cone
= 70 × $\frac{22}{7}$ × 24 × 26
= 10 × 22 × 24 × 26
= ₹ 137280
Therefore, the cost of the canvas required to make such a tent is ₹ 137280.

5. What length of tarpaulin 3 m wide will be required to make a 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 –
Height of conical tent, h = 8m
The radius of the base of the tent, r = 6m
The slant height of the tent = l
l = √(r²+h²)
= √(6²+8²)
= √(36 + 64)
= √100
= 10 m
Therefore, the curved surface area = πrl
= 3.14 × 6m × 10m
= 188.4 m²Now, width of the tarpaulin = 3m
Area of the tarpaulin = 188.4 m²So, Area of the tarpaulin = width of the tarpaulin × length of the tarpaulin
188.4 m²= 3 × length of the tarpaulin
⇒ Length of the tarpaulin = 188.4 m²/3
= 62.8 m
Extra length of the material = 20cm = 20/100m = 0.2m
Actual length required = 62.8m + 0.2m = 63m
Thus, the required length of the tarpaulin is 63 m.

6. The slant height and base diameter of a conical tomb are 25m and 14 m, respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m². (Assume π = $\mathbf{\frac{22}{7}}$ )
Answer – The slant height of the conical tomb, l = 25m
Base Diameter, d = 14 m
Base Radius, r = diameter/2 = 14/2 m = 7m
Curved surface area = πrl
= $\frac{22}{7}$ × 7 m × 25 m
= 550m²Cost of the whitewashing at ₹ 210 per 100 m²= $\frac{210}{100}$ × 550
= ₹ 1155
Thus, the cost of whitewashing the conical tomb is ₹1155.

7. A joker’s cap is in the form of a right circular cone with a base radius of 7 cm and a height of 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)

Answer – The radius of the conical cap, r = 7 cm
Height of conical cap, h = 24cm
Slant height = l
l = √(r²+ h²)
= √(7²+ 24²)
= √(49 + 576)
= √625
= 25 cm
Area of the sheet required to make each cap = πrl
= $\frac{22}{7}$ × 7 cm × 25 cm
= 550 cm²
Area of the sheet required to make 10 such caps = 10 × 550 cm² = 5500 cm²Thus, the area of the sheet required to make 10 such caps is 5500 cm².

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 a height of 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)

Answer – Diameter, d = 40cm = 40/100 m = 0.4m
Radius of cone, r = diameter/2 = 0.4/2 cm = 0.2 m
Height of cone, h = 1 m
The slant height of the cone l = √(r²+ h²)
= √(0.2²+ 1²)
= √1.04
= 1.02 m (given)
The slant height of the cone is 1.02 m
The curved surface area = πrl
= 3.14 × 0.2m × 1.02m
= 0.64056 m²Curved surface area of 50 cones = 50 × 0.64056 m² = 32.028 m²Cost of painting of 50 cones at ₹ 12 per m² = 32.028 × 12
= ₹ 384.34 (approx.)
Thus, the cost of painting all the cones is ₹ 384.34 (approx.)

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