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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 11.3

Exercise 11.3

Assume π = 227, unless stated otherwise.

1. Find the volume of the right circular cone with:

(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

Instructions

We know: Volume of cone =
(i) radius 6 cm, height 7 cm : V = × × × = cm3
(ii) radius 3.5 cm, height 12 cm: V = × × × = cm3
We have found all the answers.

2. Find the capacity in litres of a conical vessel with:

(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

Instructions

Volume of the cone, V =
(i) radius 7 cm, slant height 25 cm:
h = l2−r2 = 252−72 =625−49 = cm
V = × × × = cm3
Capacity of the conical vessel = liters (as 1L = cm3) = L (Round off to three decimal places)
(ii) height 12 cm, slant height 13 cm
r = l2−h2 = 132−122 =169−144 = cm
V = × × × = cm3 (Round off to one decimal place)
Capacity of the conical vessel = L (Round off to three decimal places)

3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)

Instructions

We know: volume of the cone = where h = cm and volume of cone = cm3
Thus, 13πr2h = 1570 ⇒ × × r2 × = 1570
r2 = ⇒ r = cm
Thus, radius of the base of the cone 10 cm.

4. If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.

Instructions

Given: h = cm and volume = 48π cm3 with formula for finding volume =
Putting values: 13πr29 = 48 π ⇒ r2 = ⇒ r = cm.
Thus, Diameter = × Radius = cm
Thus, diameter of the base is 8 cm.

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Instructions

Finding volume of pit

  • From the given data, we get: Radius of pit (r) = m
  • We know volume of pit = where r and h are the radius and height,respectively.
  • Substituting values
  • We get volume of pit = m3
  • We know 1 m3 = L = kL
  • Thus, the capacity of the pit is kiloliters (kL).
  • We have found the desired answer.

6. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find:

(i) height of the cone (ii) slant height of the cone

(iii) curved surface area of the cone

Instructions

(i) Radius of cone (r) = cm with volume of cone = = cm3
So, 13πr2h = 9856 ⇒ × × × h = 9856
Thus, h = cm.
Thus, height is equal to 48 cm.
(ii) l = r2+h2 = 142+482 = 196+2304 = cm.
Thus, slant height of the cone is 50 cm.
(iii) Curved surface area of cone =
Substituting: CSA = × × = cm2
Curved surface area of the cone is 2200 cm2.

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Instructions

Finding volume of solid

  • Upon revolving the right triangle, we get
  • We know volume of solid = where r and h are the radius and height,respectively.
  • Substituting values
  • We get volume of solid = × π cm3
  • We have found the desired answer.

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Instructions

Finding volume of solid

  • In this case, we obtain a cone with height = cm and radius = cm which further gives the slant height = cm.
  • Substituting values in the volume formula
  • We get the volume = π cm3
  • Finding the ratio of PreviousCaseCurrentCase = or =
  • We have found the desired answer.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Instructions

We have: r = m with h = m. We also know: Volume of heap =
Thus, Volume = × × × = m3 (Round off to three decimal places)
The volume of the heap of wheat is 86.625 m3.
Area of canvas =
Area of canvas = 227 × 5.25 × 5.252+32 = 227 × 5.25 × = m2 (Round off to two decimal places)
Therefore, the area of the canvas is 99.83 m2.