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

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 13 Surface Areas and Volumes Ex 13.3.

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

**Ex
13.3 Class 9 Maths 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.

**Solution:
**Here, diameter of the base = 10.5 cm

⇒ Radius of the base (r) = 10.5/2 cm and slant height (l) = 10 cm

Curved surface area of the cone = Ï€rl

= 22/7 × 10.5/2 × 10 cm

^{2}

= 11 × 15 × 1 cm

^{2}= 165 cm

^{2}

**Ex
13.3 Class 9 Maths 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.

**Solution:
**Here, diameter of the base = 24 m

∴ Radius of the base (r) = 24/2 m = 12 m and slant height (l) = 21 m

∴ Total surface area of a cone = Ï€r(r + 1)

**Ex 13.3 Class 9 Maths 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.

**Solution:
**Here, curved surface area = Ï€rl = 308 cm

^{2}

Slant height (l) = 14 cm

**(i)**
Let the radius of the base be ‘r’ cm

∴ Ï€rl = 308 ⇒ 22/7 × r ×
14 = 308

r = 308 × 7/22 × 1/14 = 7 cm

Thus, the radius of the cone is 7 cm.

**(ii)**
Base area = Ï€r^{2} = 22/7 × 7^{2} cm^{2}

= 22 × 7 cm^{2} = 154 cm^{2 }

and curved surface area = 308 cm^{2} [Given]

∴ Total surface area of the cone

= [Curved surface area] + [Base area] = 308 cm^{2} + 154 cm^{2}

= 462 cm^{2}

**Ex 13.3 Class 9 Maths 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 m

^{2}canvas is ₹70.

**Solution:
**Here, height of the tent (h) = 10 m

Radius of the base (r) = 24 m

**(i)**
The slant height, l^{2} = r^{2 }+ h^{2 }

= 24^{2 }+ 10^{2 }= 576 + 100 = 676

Therefore,
l = √676 = 26 m

Thus, the required slant height of the tent is 26 m.

**(ii)** Curved surface area of the cone = Ï€rl

∴ Area of the canvas required

**Ex
13.3 Class 9 Maths 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 Ï€ = 3.14)

**Solution:
**Here, base radius (r) = 6 m

Height (h) = 8 m

∴ Slant height l

^{2}= r

^{2 }+ h

^{2 }

= 6^{2 }+ 8^{2 }= 36 + 64 = 100

Therefore,
l = √100 = 10 m

Now, curved surface area = Ï€rl

= 3.14 × 6 × 10 m^{2}

= 188.4 m^{2 }

Thus, area of the tarpaulin required to make the tent = 188.4 m^{2}

Let the length of the tarpaulin be L m.

Length × Breadth = 188.4

⇒ L × 3 = 188.4

⇒ L = 188.4/3 =
62.8 m

Extra length of tarpaulin required for margins = 20 cm = 20/100 m = 0.2 m

Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63 m

**Ex
13.3 Class 9 Maths 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 m

^{2}.

**Solution:
**Here, base radius (r) = 14/2 m = 7 m

Slant height (l) = 25 m

∴ Curved surface area = Ï€rl

= 22/7 × 7 × 25 m

^{2}= 550 m

^{2 }

Cost of white-washing for 100 m

^{2}area = ₹ 210

∴ Cost of white-washing for 550 m

^{2}area

= ₹ 210/100 × 550 = ₹ 1155

**Ex
13.3 Class 9 Maths 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.

**Solution:
**Radius of the base (r) = 7 cm and height (h) = 24 cm

Slant height, l

^{2}= h

^{2 }+

^{ }r

^{2 }= 24

^{2 }+ 7

^{2}

^{
}= 576 + 49 = 625

Therefore,
l = 25 cm

∴ Lateral surface area = Ï€rl = 22/7 × 7 × 25 cm^{2} = 550 cm^{2}

∴ Lateral surface area of 10 caps = 10 × 550 cm^{2} =
5500 cm^{2}

Thus, the required area of the sheet = 5500 cm^{2}

**Ex 13.3 Class 9 Maths 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 Ï€ = 3.14 and
take √1.04 = 1.02)

**Solution:****
**Diameter of the base = 40 cm

= ₹
384.336 = ₹ 384.34 (approx.)

Thus, the required cost of painting is ₹ 384.34
(approx.).

**You can also like these:**

**NCERT Solutions for Maths Class 10**