## Surface Areas and Volumes

__EXERCISE 13.3__

**Q1: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.**

Answer: Given r = 10.5/2 = 5.25 cm, l = 10 cm.

Curved surface area (CSA) of the cone =

**πrl**

= (22/7) × 5.25 × 10 =

**165 cm² (Answer)**

**Q2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.**

Answer: Given, l = 21m, r = 24/2 = 12 m

TSA (cone) =

**πr(l + r)**= (22/7) × 12 (21 + 12)

=

**1244.57 m² (Answer)**

**Q3: 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:

(i) Given, l = 14cm, CSA = 308 cm²

CSA =

*πrl*= 308

⇒ r = (308 × 7) / (22 × 2) =

**7cm (Answer)**

(ii) TSA(cone) =

*πr (l + r)*

= (22/7) × 7 (14 + 7)

= 22 × 21

=

**462 cm² (Answer)**

**Q4: 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.**

Answer: Given, h = 10 m, r = 24 m

(i) Using Pythagoras theorem,

*l² = h² + r²*

= (10)² + (24)²

= 100 + 576 = 676

⇒ l = √676 =

**26m (Answer)**

(ii) CSA of conical tent =

*πrl*

= (22/7) × 24 × 26

Cost of 1 m² canvas = ₹ 70

∴ Cost of canvas = 70 × (22/7) × 24 × 26

=

**₹137280 (Answer)**

**Q5: 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 29 cm (use π = 3.14)**

Answer: Given h = 6 m, r = 8 m

∵

*l² = h² + r²*

⇒ l² = 36 + 64 = 100

⇒ l = √100 = 10m

Curved surface area of the tent =

*πrl*

= 3.14 × 6 × 10

∴ required length of tarpaulin = (3.14 × 6 × 10)×⅓ + 0.20m

= 62.8 m + 0.2 m =

**63 m (Answer)**

**Q6: 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 100 m².**

Answer: Given, h = 25m, r = 14/2 = 7m

CSA of tomb =

*πrl*

= (22/7) × 7 × 25

= 550 m²

Cost of white washing 100 m² = ₹ 210

∴ Cost of white washing 550 m² = (210/100) × 550 =

**₹ 1155 (Answer)**

**Q7: 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: Given, r = 7 cm, h = 24 cm

∵

*l² = r² + h²*

⇒ l² = 7² + 24² = 49 + 576 = 625

⇒ l = √(625) = 25cm

Total curved surface area of 1 cap =

*πrl*

= (22/7) × 7 × 25

= 550 cm²

Area of sheet required to make 10 such caps = 10 × 550 cm² =

**5500 cm² (Answer)**

**Q8: 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 m**²

**, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02)**

Answer: Given, r = 40/2 = 20cm = 0.20m, h = 1m

∵

*l² = r² + h²*= 1² + 0.2² = 1.04

⇒ l = √(1.04) = 1.02

CSA (1 cone) =

*πrl*

CSA (50 cones) = 50πrl

= 50 × 3.14 × 0.2 × 1.02

= 32.028 m²

Cost of painting an area of 1 m² = ₹12

∴ Cost of painting an area of 32.028 m² = Rs 12 × 32.028

=

**Rs 384.34 (Answer)**

## No comments:

## Post a Comment

We love to hear your thoughts about this post!

Note: only a member of this blog may post a comment.