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Concept Detail
Maths / Surface Area and Volume / Volume of Cone
(A Brief Glimpse of ABHYAS Content - Have aLook !!!!)

# Volume of Cone

Cone: A cone is a three-dimensional figure with one circular base. A curved surface connects the base and the vertex.

Volume of Cone: .The volume V of a cone with radius r is one-third the area of the base B times the height h.

Volume of the cone of radius r and height h = $\dpi{100} \frac{1}{3}{\pi}r^{2}h$

Illustration: Find the volume of a right circular cone 1.02m high , if the radius of its base is 28 cm.

Solution: We know that the volume V of a right circular cone of radius r and height h is given $\dpi{100} V=\frac{1}{3}{\pi}r^{2}h$.   Here,   r = 28cm   and h = 1.02 m = 1.02 X 100 cm = 102 cm

$\therefore V=\frac{1}{3}\times {\frac{22}{7}}\times 28\times 28\times 102=83776\;cm^3$

Illustration: A conical tent is 9 m high and the radius of its base is 12 m.

(i) What is the cost of the canvas required to make it, if a square metre canvas costs Rs 10?

(ii) How many persons can be accommodated in the tent, if each person requires 2 square metre on the ground and $15\;m^{3}$ of space to breath in?

Solution: We have,

r = Radius of the base of the conical tent = 12 m

h = Height of the conical tent = 9 m

l = Slant height of the conical tent = $\sqrt{r^{2}+h^{2}}$

$=\sqrt{(12)^{2}+9^{2}}$

$=\sqrt{144+81}$$=\sqrt{225}\;m$

$=15\;m$

(i) Area of the lateral surface = $\pi rl=\frac{22}{7}\times 12\times 15\;m^{2}$

$=565.7\;m^{2}$

So, total cost of the canvas = Rs(565.2 X 10) = Rs 5652

(ii) Area of the base of the conical tent = $\pi r^{2}=\frac{22}{7}\times 12\times 12\;m^{2}$

$=452.16\;m^{2}$

As each person requires 2 sq. metres of floor area.

So, max number of persons who will have enough space on the ground $=\frac{452.16}{2}=226$

Also, Volume of the conical tent $=\frac{1}{3}\times Area\;of\;the\;base\times Height$

Volume of the conical tent$=\frac{1}{3}\times 452.16\times 9\;m^{3}$

Volume of the conical tent = 1356.48 $\;m^{3}$

We have, air space required person $=15\;m^{3}$

So, number of persons who will have enough air space to breath in $=\frac{1356.48}{15}=90$

Between 226 and 90, the smaller number is 90.

Hence 90 persons can be accomodated.

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