The document Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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

â‡’ Radius (r) = (10.5/2)cm

Slant height (l) = 10 cm

âˆ´ Curvered surface area of the cone = Ï€rl

**Question 2. Find the total surfce area of a cone, if its slant height is 21 m and diameter of its base is 24 m. Solution:** Here, diameter = 24 m

â‡’ Radius (r) = (24/2)m = 12 m

Slant height (l) = 21 m

âˆ´ Total surface area = Ï€r(r + l)

**Question 3. Curved surface area of a cone is 308 cm ^{2} 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 = 308 cm

Slant height (l) = 14 cm

(i) Let the radius of the base be â€˜râ€™ cm.

Thus, the required radius of the cone is 7 cm.

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

âˆ´ Total surface area = [Curved surface area] + [Base area]

= 308 cm^{2} + 154 cm^{2} = 462 cm^{2}

**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 **

Solution:

Radius of the base (r) = 24 m (i)

âˆµ The slant height,

âˆ´ The slant height of the tent =

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

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

âˆ´ Area of the canvas required =

**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 apÏ€roximately 20 cm (Use Ï€ = 3.14). Solution:** Here, Base radius (r) = 6 m

Height (h) = 8 m

âˆ´ Area of the canvas (tarpaulin) required to make the tent = 188.4 m^{2 }

Let the length of the tarpaulin = â€˜Lâ€™ m

âˆ´ Length x Breadth = 188.4

â‡’ L x 3 = 188.4

â‡’ L= (188.4/3) = 62.8 m

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

Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63.0 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 **

Solution:

Slant height (l) = 25 m

âˆ´ Curved surface area = Ï€rl = (22/7) x 7 x 25 m

= 22 x 25 m

Rate of whitewashing = 210 per 100 m

âˆ´ Cost of whitewashing for 550 m

**Question 7. A jokerâ€™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**: Here, Radius of the base (r) = 7 cm height (h) = 24 cm

â‡’ Lateral surface area of 10 caps = 10 x 550 cm^{2 }= 5500 cm^{2 }

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

**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 **â‚¹**12 per m ^{2}, what will be the cost of painting all these cones? (Use Ï€ = 3.14 and take âˆš1.04= 1.02) Solution: **Here, âˆµ Diameter of the base = 40 cm

âˆ´ Radius (r) = 40/2 cm = 20 cm = 10/100 m

= 2/10 m = 0.2 m

Height (h) = 1 m

= 1.02 m [âˆµ âˆš1.04= 1.02 (Given)]

Now, curved surface area = Ï€rl

â‡’ Curved surface area of 1 cone = 3.14 x 0.2 x 1.02 m^{2}

â‡’ Curved surface are of 50 cones

Cost of painting

âˆµ Rate of painting = â‚¹ 12 per m^{2}

**SURFACE AREA OF A SPHERE**

A solid generated by revolving a circular lamina about any of its diameters, is called a sphere. If â€˜râ€™ be its radius, then its surface area = 4Ï€r^{2}.

For the following Ï€roblems, assume Ï€ = (22/7), unless stated otherwise.

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