Surface Areas & Volumes
NCERT Solutions • Class 9 Maths • Chapter 11Exercise 11.3
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
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
(i) Given: $r = 6 \text{ cm}, h = 7 \text{ cm}$.
Formula: Volume $V = \frac{1}{3} \pi r^2 h$
$V = \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7$
$= 22 \times 2 \times 6 =$ $264 \text{ cm}^3$.
Formula: Volume $V = \frac{1}{3} \pi r^2 h$
$V = \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7$
$= 22 \times 2 \times 6 =$ $264 \text{ cm}^3$.
(ii) Given: $r = 3.5 \text{ cm}, h = 12 \text{ cm}$.
$V = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12$
$= \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times 3.5 \times 12$
$= 11 \times 3.5 \times 4 =$ $154 \text{ cm}^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12$
$= \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times 3.5 \times 12$
$= 11 \times 3.5 \times 4 =$ $154 \text{ cm}^3$.
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
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
(i) Step 1: Find height ($h$)
$h = \sqrt{l^2 – r^2} = \sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576} = 24 \text{ cm}$.
Step 2: Find Volume
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 22 \times 7 \times 8 = 1232 \text{ cm}^3$.
Step 3: Convert to Litres ($1000 \text{ cm}^3 = 1 \text{ L}$)
Capacity $= \frac{1232}{1000} =$ $1.232 \text{ L}$.
$h = \sqrt{l^2 – r^2} = \sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576} = 24 \text{ cm}$.
Step 2: Find Volume
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 22 \times 7 \times 8 = 1232 \text{ cm}^3$.
Step 3: Convert to Litres ($1000 \text{ cm}^3 = 1 \text{ L}$)
Capacity $= \frac{1232}{1000} =$ $1.232 \text{ L}$.
(ii) Step 1: Find radius ($r$)
$r = \sqrt{l^2 – h^2} = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5 \text{ cm}$.
Step 2: Find Volume
$V = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{2200}{7} \text{ cm}^3$.
Step 3: Convert to Litres
Capacity $= \frac{2200}{7 \times 1000} = \frac{11}{35} \text{ L} \approx$ $0.314 \text{ L}$.
$r = \sqrt{l^2 – h^2} = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5 \text{ cm}$.
Step 2: Find Volume
$V = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{2200}{7} \text{ cm}^3$.
Step 3: Convert to Litres
Capacity $= \frac{2200}{7 \times 1000} = \frac{11}{35} \text{ L} \approx$ $0.314 \text{ L}$.
3. The height of a cone is 15 cm. If its volume is $1570 \text{ cm}^3$, find the radius of the base. (Use $\pi = 3.14$)
Calculation:
$V = \frac{1}{3} \pi r^2 h = 1570$
$\frac{1}{3} \times 3.14 \times r^2 \times 15 = 1570$
$15.7 \times r^2 = 1570$
$r^2 = \frac{1570}{15.7} = 100$
$r = \sqrt{100} =$ $10 \text{ cm}$.
$V = \frac{1}{3} \pi r^2 h = 1570$
$\frac{1}{3} \times 3.14 \times r^2 \times 15 = 1570$
$15.7 \times r^2 = 1570$
$r^2 = \frac{1570}{15.7} = 100$
$r = \sqrt{100} =$ $10 \text{ cm}$.
4. If the volume of a right circular cone of height 9 cm is $48 \pi \text{ cm}^3$, find the diameter of its base.
Calculation:
$\frac{1}{3} \pi r^2 h = 48 \pi$
$\frac{1}{3} \times r^2 \times 9 = 48$
$3 r^2 = 48 \Rightarrow r^2 = 16 \Rightarrow r = 4 \text{ cm}$.
Diameter $= 2r = 2 \times 4 =$ $8 \text{ cm}$.
$\frac{1}{3} \pi r^2 h = 48 \pi$
$\frac{1}{3} \times r^2 \times 9 = 48$
$3 r^2 = 48 \Rightarrow r^2 = 16 \Rightarrow r = 4 \text{ cm}$.
Diameter $= 2r = 2 \times 4 =$ $8 \text{ cm}$.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Given: $d = 3.5 \text{ m} \Rightarrow r = 1.75 \text{ m}$, $h = 12 \text{ m}$.
Volume:
$V = \frac{1}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 12$
$= 22 \times 0.25 \times 1.75 \times 4$
$= 38.5 \text{ m}^3$.
Since $1 \text{ m}^3 = 1 \text{ kilolitre (kL)}$:
Capacity $=$ $38.5 \text{ kL}$.
Volume:
$V = \frac{1}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 12$
$= 22 \times 0.25 \times 1.75 \times 4$
$= 38.5 \text{ m}^3$.
Since $1 \text{ m}^3 = 1 \text{ kilolitre (kL)}$:
Capacity $=$ $38.5 \text{ kL}$.
6. The volume of a right circular cone is $9856 \text{ cm}^3$. If the diameter of the base is 28 cm, find (i) height (ii) slant height (iii) curved surface area.
(i) Find height ($h$):
$r = 14 \text{ cm}$. $V = 9856$.
$\frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h = 9856$
$\frac{1}{3} \times 616 \times h = 9856$
$h = \frac{9856 \times 3}{616} =$ $48 \text{ cm}$.
$r = 14 \text{ cm}$. $V = 9856$.
$\frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h = 9856$
$\frac{1}{3} \times 616 \times h = 9856$
$h = \frac{9856 \times 3}{616} =$ $48 \text{ cm}$.
(ii) Find slant height ($l$):
$l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500}$
$l =$ $50 \text{ cm}$.
$l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500}$
$l =$ $50 \text{ cm}$.
(iii) Find CSA:
CSA $= \pi r l = \frac{22}{7} \times 14 \times 50$
$= 22 \times 2 \times 50 =$ $2200 \text{ cm}^2$.
CSA $= \pi r l = \frac{22}{7} \times 14 \times 50$
$= 22 \times 2 \times 50 =$ $2200 \text{ cm}^2$.
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.
When revolved about the side 12 cm, the solid formed is a cone with:
Height ($h$) = $12 \text{ cm}$
Radius ($r$) = $5 \text{ cm}$
$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5 \times 5 \times 12$
$= 100 \pi \text{ cm}^3 \approx$ $314.28 \text{ cm}^3$.
Height ($h$) = $12 \text{ cm}$
Radius ($r$) = $5 \text{ cm}$
$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5 \times 5 \times 12$
$= 100 \pi \text{ cm}^3 \approx$ $314.28 \text{ cm}^3$.
8. If the triangle ABC in Question 7 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.
Step 1: Calculate Volume
Revolved about 5 cm $\Rightarrow h = 5 \text{ cm}, r = 12 \text{ cm}$.
$V_2 = \frac{1}{3} \times \pi \times 12 \times 12 \times 5 = 240 \pi \text{ cm}^3$.
$V_2 \approx 754.28 \text{ cm}^3$.
Revolved about 5 cm $\Rightarrow h = 5 \text{ cm}, r = 12 \text{ cm}$.
$V_2 = \frac{1}{3} \times \pi \times 12 \times 12 \times 5 = 240 \pi \text{ cm}^3$.
$V_2 \approx 754.28 \text{ cm}^3$.
Step 2: Calculate Ratio
Ratio $= \frac{V_1}{V_2} = \frac{100 \pi}{240 \pi} = \frac{10}{24} = \frac{5}{12}$.
Ratio is $5:12$.
Ratio $= \frac{V_1}{V_2} = \frac{100 \pi}{240 \pi} = \frac{10}{24} = \frac{5}{12}$.
Ratio is $5:12$.
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.
Volume
$d = 10.5 \Rightarrow r = 5.25 \text{ m}$. $h = 3 \text{ m}$.
$V = \frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3$
$= 22 \times 0.75 \times 5.25 =$ $86.625 \text{ m}^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3$
$= 22 \times 0.75 \times 5.25 =$ $86.625 \text{ m}^3$.
Canvas Area (CSA)
$l = \sqrt{r^2 + h^2} = \sqrt{5.25^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05 \text{ m}$.
CSA $= \pi r l = \frac{22}{7} \times 5.25 \times 6.05$
$= 22 \times 0.75 \times 6.05 =$ $99.825 \text{ m}^2$.
CSA $= \pi r l = \frac{22}{7} \times 5.25 \times 6.05$
$= 22 \times 0.75 \times 6.05 =$ $99.825 \text{ m}^2$.