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**NCERT Class 9 Mathematics Chapter 11 Surface Areas and Volumes**

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**Solutions****NCERT Class 9 Mathematics Chapter 11 Surface Areas and Volumes****Surface Areas and Volumes**

**Surface Areas and Volumes****Chapter – 11**

Exercise 11.1 |

**Q.1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.**

Ans: ∴ Diameter of the base = 10.5cm

Slant height (l) = 10cm

∴ Curved surface area of the cone = πrl

**Q.2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.**

Ans: Slant height (l) = 21m

Diameter of base = 24m

∴ Total curved surface area of the cone = πr(l + r)

**Q.3. Curved surface area of a cone is 308cm² and its slant height is 14 cm. Find **

**(i) radius of the base and **

**(ii) total surface area of the cone.**

Ans: (i) Slant height (l) = 14cm

Curved surface area = 308cm²

Hence the radius of the base is 7 cm.

(ii) Total surface area of the cone = πr(l + r)

Hence the total surface area of the cone is 462cm²

**Q. 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**

Ans: (i) h = 10m,

r = 24m

Hence the slant height of the tent is 26 m.

(ii) Curved surface area of the tent = πrl

∴ Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70

Hence the cost of the canvas is Rs. 137280

**Q.5. What length of tarpaulin 3 m wide will be required to make a 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 wast- age in cutting is approximately 20 cm (Use π = 3.14 )**

Ans: For conical tent

∴ Curved surface area = πrl

= 3.14 x 6 x 10 = 188.4m²

Width of tarpaulin = 3m

Extra length of the material required = 20cm = 0.2m

∴ Actual length of tarpaulin required = 62.8m + 0.2m = 63m

**Q.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 Rs. 210 per 100m².**

Ans: Slant height (l) = 25 m

Base diameter = 14m

∴ Cost of white-washing the curved surface of the tomb at the rate of Rs. 210 per 100 m².

**Q.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.**

Ans: Base radius (r) = 7cm

Height (h) = 24cm

∴ Curved surface area of a cap

∴ Curved surface area of 10 caps = 550 x 10 = 5500 cm²

Hence the area of the sheet required to make 10 such caps is 5500cm².

**Q.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 Rs. 12 per m² what will be the cost of painting all these cones? (Use π = 3.14 and take**

√(**1.04) = 1.02 )**

Ans: Base diameter = 40cm

∴

Height (h) = 1m

∴ Curved surface area = πrl

= 3.14 x 0.2 x 1.02 = 0.64056m²

∴ Curved surface area of 50 cones

= 0.64056 x 50m² = 32.028m²

∴ Cost of painting all these cones = 32.028 x 12

= 384.336 = Rs. 384.34 (approximately).

Exercise 11.2 |

**Q.1. Find the surface area of a sphere of radius:**

**(i) 10.5 cm**

**(ii) 5.6 cm**

**(iii) 14 cm**

Ans: (i) r = 10.5cm

∴ Surface area = 4πr²

(ii) r = 5.6cm

(iii) r = 14cm

**Q.2. Find the surface area of a sphere of radius:**

**(i) 14 cm**

**(ii) 21 cm**

**(iii) 3.5 cm**

Ans: (i) Diameter = 14cm

(ii) Diameter = 21 cm

(iii) Diameter = 3.5 cm

**Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14**

Ans: r = 10cm

∴ Total surface area of the hemisphere

= 3πr² = 3 x 3.14 x (10)² = 942cm²

**Q.4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

Ans: Case I. r = 7cm

∴ Case II. r = 14cm

∴ Ratio of surface areas of the balloon

**Q.5. A Hemisphere bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the incident the rate of Rs. 16 per 100 cm²**

Ans: Inner diameter = 10.5cm

∴ Inner surface area

∴ Cost of tin-plating at the rate of Rs. 16 per 100 cm²

**Q.6. Find the radius of a sphere whose surface area is 154 cm². **

Ans: Let the radius of the sphere be r cm.

Surface area = 154 cm²

Hence the radius of the sphere is 3.5 cm.

**Q.7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.**

Ans: Let the diameter of the earth be 2r.

∴ Surface area of the earth = 4πr²

∴ Ratio of their surface area

**Q.8. A Hemisphere bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.**

Ans: Inner radius of the bowl = 5 cm

Thickness of steel = 0.25 cm

∴ Outer radius of the bowl = 5 + 0.25 = 5.25cm

∴ Outer curved surface of the bowl

**Q.9. A right circular cylinder just encloses a sphere of radius r. Find**

**(i) surface area of the sphere.**

**(ii) curved surface area of the cylinder.**

**(iii) ratio of the areas obtained in (i) and (ii).**

Ans: (i) Surface area of the sphere = 4πr²

(ii) For cylinder

Radius of the base = r

Height = 2r

∴ Curved surface area of the cylinder = 2π(r)(2r) = 4πr²

(iii) Ratio of the areas obtained in (i) and (ii)