Exercise 11.3 Practice
Surface Area and Volume of a Right Circular Cone
Q1: Volume of a Cone
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Find the volume of a right circular cone with:
(i) radius $10.5 \text{ cm}$, height $18 \text{ cm}$
(ii) radius $7 \text{ cm}$, height $21 \text{ cm}$
(i) radius $10.5 \text{ cm}$, height $18 \text{ cm}$
(ii) radius $7 \text{ cm}$, height $21 \text{ cm}$
Formula: Volume of a cone $V = \frac{1}{3}\pi r^2 h$.
(i) Radius $r = 10.5 \text{ cm}$, height $h = 18 \text{ cm}$:
$V = \frac{1}{3} \times \frac{22}{7} \times (10.5)^2 \times 18$
$= \frac{1}{3} \times \frac{22}{7} \times 110.25 \times 18$
$= 22 \times 15.75 \times 6 = 2079 \text{ cm}^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times (10.5)^2 \times 18$
$= \frac{1}{3} \times \frac{22}{7} \times 110.25 \times 18$
$= 22 \times 15.75 \times 6 = 2079 \text{ cm}^3$.
(ii) Radius $r = 7 \text{ cm}$, height $h = 21 \text{ cm}$:
$V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 21$
$= 22 \times 7 \times 7 = 1078 \text{ cm}^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 21$
$= 22 \times 7 \times 7 = 1078 \text{ cm}^3$.
(i) $2079 \text{ cm}^3$, (ii) $1078 \text{ cm}^3
Q2: Capacity of a Conical Funnel
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Find the capacity in litres of a conical funnel with:
(i) radius $5 \text{ cm}$, slant height $13 \text{ cm}$
(ii) height $8 \text{ cm}$, slant height $10 \text{ cm}$
(i) radius $5 \text{ cm}$, slant height $13 \text{ cm}$
(ii) height $8 \text{ cm}$, slant height $10 \text{ cm}$
(i) $r = 5 \text{ cm}, l = 13 \text{ cm}$:
Height $h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 4 = 314 \text{ cm}^3$.
Capacity in litres $= \frac{314}{1000} = 0.314$ litres.
Height $h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 4 = 314 \text{ cm}^3$.
Capacity in litres $= \frac{314}{1000} = 0.314$ litres.
(ii) $h = 8 \text{ cm}, l = 10 \text{ cm}$:
Radius $r = \sqrt{l^2 - h^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ cm}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 6^2 \times 8 = 3.14 \times 12 \times 8 = 301.44 \text{ cm}^3$.
Capacity in litres $= \frac{301.44}{1000} = 0.30144$ litres.
Radius $r = \sqrt{l^2 - h^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ cm}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 6^2 \times 8 = 3.14 \times 12 \times 8 = 301.44 \text{ cm}^3$.
Capacity in litres $= \frac{301.44}{1000} = 0.30144$ litres.
(i) $0.314$ litres, (ii) $0.30144$ litres
Q3: Finding Height from Volume
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The volume of a right circular cone is $6280 \text{ cm}^3$ and the radius of its base is $20 \text{ cm}$. Find the height of the cone. (Use $\pi = 3.14$)
Given: Volume $V = 6280 \text{ cm}^3$, Radius $r = 20 \text{ cm}.
$V = \frac{1}{3}\pi r^2 h$
$6280 = \frac{1}{3} \times 3.14 \times 20^2 \times h$
$6280 = \frac{1}{3} \times 3.14 \times 400 \times h$
$h = \frac{6280 \times 3}{1256} = 15 \text{ cm}$.
$6280 = \frac{1}{3} \times 3.14 \times 20^2 \times h$
$6280 = \frac{1}{3} \times 3.14 \times 400 \times h$
$h = \frac{6280 \times 3}{1256} = 15 \text{ cm}$.
The height of the cone is $15 \text{ cm}.
Q4: Finding Diameter from Volume
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If the volume of a right circular cone of height $24 \text{ cm}$ is $1232\pi \text{ cm}^3$, find the diameter of its base.
Given: Height $h = 24 \text{ cm}$, Volume $V = 1232\pi \text{ cm}^3.
$V = \frac{1}{3}\pi r^2 h$
$1232\pi = \frac{1}{3}\pi r^2 (24)$
$1232 = 8r^2$
$r^2 = 154$.
$r = \sqrt{154} \approx 12.4 \text{ cm}$.
$1232\pi = \frac{1}{3}\pi r^2 (24)$
$1232 = 8r^2$
$r^2 = 154$.
$r = \sqrt{154} \approx 12.4 \text{ cm}$.
Diameter $= 2r = 2 \times 12.4 = 24.8 \text{ cm}$.
The diameter of the base is approximately $24.8 \text{ cm}.
Q5: Conical Tent Volume
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A conical tent has a base diameter of $7 \text{ m}$ and is $6 \text{ m}$ deep (high). What is its capacity in kilolitres?
Diameter $d = 7 \text{ m} \Rightarrow r = 3.5 \text{ m}$.
Depth (height) $h = 6 \text{ m}$.
Depth (height) $h = 6 \text{ m}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 6$
$= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 6 = 77 \text{ m}^3$.
$= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 6 = 77 \text{ m}^3$.
Since $1 \text{ m}^3 = 1$ kilolitre, the capacity is $77$ kilolitres.
The capacity of the tent is $77$ kilolitres.
Q6: Comprehensive Cone Analysis
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The volume of a right circular cone is $3140 \text{ cm}^3$. If the diameter of the base is $20 \text{ cm}$, find:
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
(Use $\pi = 3.14$)
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
(Use $\pi = 3.14$)
Given: Volume $V = 3140 \text{ cm}^3$, Diameter $d = 20 \text{ cm} \Rightarrow r = 10 \text{ cm}$.
(i) Height of the cone:
$V = \frac{1}{3}\pi r^2 h \Rightarrow 3140 = \frac{1}{3} \times 3.14 \times 10^2 \times h$
$3140 = \frac{314}{3}h$
$h = \frac{3140 \times 3}{314} = 30 \text{ cm}$.
$V = \frac{1}{3}\pi r^2 h \Rightarrow 3140 = \frac{1}{3} \times 3.14 \times 10^2 \times h$
$3140 = \frac{314}{3}h$
$h = \frac{3140 \times 3}{314} = 30 \text{ cm}$.
(ii) Slant height of the cone:
$l = \sqrt{r^2 + h^2} = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000} \approx 31.62 \text{ cm}$.
$l = \sqrt{r^2 + h^2} = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000} \approx 31.62 \text{ cm}$.
(iii) Curved surface area of the cone:
CSA $= \pi r l = 3.14 \times 10 \times 31.62 = 992.868 \text{ cm}^2$.
CSA $= \pi r l = 3.14 \times 10 \times 31.62 = 992.868 \text{ cm}^2$.
(i) $30 \text{ cm}$, (ii) $31.62 \text{ cm}$, (iii) $992.868 \text{ cm}^2
Q7: Ratio of Volumes
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Two right circular cones have their heights in the ratio $1:3$ and the radii of their bases in the ratio $3:1$. Find the ratio of their volumes.
Let the heights of the cones be $h_1 = h$ and $h_2 = 3h$.
Let the radii of their bases be $r_1 = 3r$ and $r_2 = r$.
Let the radii of their bases be $r_1 = 3r$ and $r_2 = r$.
Volume of the first cone $V_1 = \frac{1}{3}\pi r_1^2 h_1 = \frac{1}{3}\pi (3r)^2 h = 3\pi r^2 h$.
Volume of the second cone $V_2 = \frac{1}{3}\pi r_2^2 h_2 = \frac{1}{3}\pi r^2 (3h) = \pi r^2 h$.
Ratio of volumes $\frac{V_1}{V_2} = \frac{3\pi r^2 h}{\pi r^2 h} = 3$.
The ratio of their volumes is $3:1$.
Q8: Canvas for a Tent
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How many metres of canvas $2.5 \text{ m}$ wide will be required to make a conical tent whose base radius is $7 \text{ m}$ and height is $24 \text{ m}$?
Radius $r = 7 \text{ m}$, Height $h = 24 \text{ m}$.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ m}$.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ m}$.
Curved Surface Area (canvas required) $= \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2$.
Width of canvas $= 2.5 \text{ m}$.
Length of canvas required $= \frac{\text{Area}}{\text{Width}} = \frac{550}{2.5} = 220 \text{ m}$.
Length of canvas required $= \frac{\text{Area}}{\text{Width}} = \frac{550}{2.5} = 220 \text{ m}$.
$220 \text{ m}$ of canvas will be required.
Q9: Heap of Wheat
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A heap of wheat is in the form of a cone whose diameter is $10.5 \text{ m}$ and height is $3 \text{ m}$. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Diameter $d = 10.5 \text{ m} \Rightarrow r = 5.25 \text{ m}$.
Height $h = 3 \text{ m}$.
Height $h = 3 \text{ m}$.
Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3$
$= \frac{22}{7} \times 27.5625 = 86.625 \text{ m}^3$.
$= \frac{22}{7} \times 27.5625 = 86.625 \text{ m}^3$.
To find the canvas area, we need the slant height $l$.
$l = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05 \text{ m}$.
$l = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05 \text{ m}$.
Area of canvas (CSA) $= \pi r l = \frac{22}{7} \times 5.25 \times 6.05 = 99.825 \text{ m}^2$.
Volume is $86.625 \text{ m}^3$. Canvas area required is $99.825 \text{ m}^2$.