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Chapter 11: Surface Areas and Volumes

    Surface Area of a Right Circular Cone

    Surface Area of a Sphere

    Volume of a Right Circular Cone

    Volume of a Sphere

    Exercise 11.1

    Exercise 11.2

    Exercise 11.3

    Exercise 11.4

    Summary

    Enhanced Curriculum Support

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Chapter 11: Surface Areas and Volumes > Exercise 11.1

Exercise 11.1

Assume π = 227, unless stated otherwise.

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

Instructions

We have Radius = Diameter ÷ 2 = 10.52 = cm
Applying the formula for the curved surface area (CSA) of a cone: = × 5.25 × = cm2
The curved surface area of the cone is 165 cm2.

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

Instructions

Diameter of the base = 24 m. Thus, Radius (r) = Diameter ÷ 2 = 242 = m
Total Surface Area =
Total Surface Area = × × ( + ) = m2 (Round off to two decimal places)
The total surface area of the cone is 1244.57 m2.

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

Instructions

We know that Curved Surface Area (CSA) of the cone = = cm2
Thus, 308 = × × r ⟹ r = 7 cm
Total surface area of the cone = πrl + πr2 = + 227 × 72 = cm2
We get: r = 7 cm and total surface area of the cone = 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

Instructions

Let ABC be a conical tent where h = m and r = m.
Let the slant height of the tent be l.
(i) In the right triangle ABO, we have: AB2 = AO2 + BO2(using Pythagoras’ theorem) i.e. l2 = h2 + r2
l2 = 102 + 242 = + = i.e. l = m
Therefore, the slant height of the tent is 26 m.
(ii) CSA of tent =
Curved Surface Area = 227 × × = m2.
We also have been given: Cost of 1 m2 canvas = Rs 70. So, cost of canvas = Rs 137287 × 70 = Rs
Therefore, the cost of the canvas required to make such a tent is Rs 137280.

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m?

Note: 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)

Instructions

Finding length of material

  • Using given values to calculate value of slant height
  • We find the value to be m.
  • We know that the curved surface of tent is equal to where l - slant height and r - radius.
  • Substituting the values
  • We find curved surface of tent = m2(Round off to one decimal place)
  • We have been given that 20 cm is wastage provision for the cloth.Let the length of the tarpaulin sheet required is L. Then, the effective length used for the tent is m
  • We have been given: breadth of tarpaulin = 3m. We also know: Area of material sheet = of the tent
  • Putting the values
  • Upon equating we get L = m
  • Thus, length of the required tarpaulin sheet will be 63 m.

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 100 m2

Instructions

Finding cost of whitewashing

  • The base radius (r) of the tent is = m
  • We know that the curved surface of tent is equal to where l - slant height and r - radius.
  • Substituting the values, we get: m2
  • Using the cost of the wahitewashing per 100m2, we can calculate the total cost.
  • We find the total cost to be Rs.
  • Thus, we have found the desired answer.

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.

Instructions

We have: r = am and h = cm. So, slant height, l2 = r2+h2 = 72+242 = = ( + ) = . Thus, l = cm.
CSA of 1 conical cap = = 227 × 7 × 25 = cm2
CSA of 10 caps = 10 × 550 cm2 = cm2
Therefore, 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 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)

Instructions

We have: r = diameter2 = 402 cm = cm = m and h = m
So, slant height of cone : l2 = r2+h2 ⟹ l2 = (0.22+12) = which gives: l = m
Slant height of the cone is 1.02 m.
CSA of each cone = = (3.14 × × ) = m
CSA of 50 such cones = (50 × 0.64056) = m2 (Round off to two decimal places)
We have: Cost of painting 1 m2 area = Rs 12. Cost of painting 32.03 m2 area = Rs (32.03 × 12) = Rs. (Round off to two decimal places)
Therefore, the cost of painting all these cones is Rs. 384.36.