# Surface Areas and Volumes – Exercise 13.7 – (MATHEMATICS) – 9th Class

**Exercise 13.7**

Assume π = 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.**

### Show Answer

**Sal.** (i) Radius (r) = 6 cm

Height (h) = 7 cm

Volume =

Volume =

(ii) Radius (r) = 3.5 ern

Height (h) = 12 cm

Volume =

**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.**

### Show Answer

**Sol.** (i) r = 7 cm, *l* = 25 cm.

∴

∴ Capacity = 1 22 x x 7 x 7 x 24 cm3

= 1232 cm³ = 1.232 *l*.

(ii) h = 12 cm, *l* = 13 cm.

∴

Capacity =

**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).**

### Show Answer

**Sol.** Let *r* be the required radius.

Height (*h*) = 15 cm

Volume = 1570 ⇒

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

### Show Answer

**Sol.** Height (h) = 9 cm, radius (r) = ?

Volume = 48π cm³ ⇒

⇒ r^{2} = 16 ⇒ r = 4 cm.

Diameter = 2r = 8 cm.

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

### Show Answer

**Sol.** Diameter (d) = 3.5 m

Deep, i.e., height (*h*) = 12 m

Capacity of the pit = πr²h

= 38.5 m³ = 38.5 kilolitres. [1 m³ = 1 kl]

**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.**

### Show Answer

**Sol.** Diameter = 28 cm, radius =

(i) Volume of cone = 9856 cm³

∴ Height (h) of the cone = 48 cm.

(ii) Slant height of cone (*l*)

(iii) Curved surface area of the cone = πrl

**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.**

### Show Answer

**Sol.** As sides are 5 cm, 12 cm, 13 cm. Hence, triangle is right angled.

We have *r* = 5 cm, *h* = 12 cm.

∴

× 5 × 12 cm^{3}

= 100π cm^{3}

**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.**

### Show Answer

**Sol.** In this case, *r* = 12 cm, *h* = 5 cm.

∴

= 240 π cm^{3}

Ratio of volumes obtained in Question 7 and Question 8:

**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.**

### Show Answer

**Sol.** Diameter (d) = 10.5 m ⇒ radius (r) =

Height (h) = 3 m

Volume of the heap of wheat =

= 86.625 m^{3}

Slant height of the cone =

=

Area of canvas required =

= 99.825 m²