Question #75d5c

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Question #75d5c

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Answer 1 · 2018-02-11T10:05:15

$V 2 = \frac{3}{4} V 1$ $V2=3/4V1$ has been proved as shown below

Explanation:

A cone circumscribes a sphere and has its slant height equal to the diameter of its base. Show that the volume of the cone is 3/4 the volume of the sphere?

Given:
Slant height equal to diameter of the base.
ie,
the cone has 2 slant heights and one base which are all equal, and forming an equilateral angle
The angle at the vertex is 60 degrees
Semi vertex angle is $\frac{1}{2} (60) = 30 d e g r e e s$ $1/2(60)=30 degrees$
If the slant height is $l$ $l$
radius of the circumscribed sphere is $r$ $r$
Angle between the radii is $θ = 120 d e g r e e s$ $theta=120 degrees$
Thus $l^{2} = r^{2} + r^{2} - 2 r r cos (120)$ $l^2=r^2+r^2-2rrcos(120)$
$cos 120 = - \frac{1}{2}$ $cos120=-1/2$
Now, $l^{2} = r^{2} + r^{2} - 2 r r (- \frac{1}{2})$ $l^2=r^2+r^2-2rr(-1/2)$
$l^{2} = 3 r^{2}$ $l^2=3r^2$
$l = (\sqrt{3}) r$ $l=(sqrt3)r$
Volume of a sphere is V1

$V 1 = \frac{4}{3} π r^{3}$ $V1=4/3pir^3$
radius of the cone is $a = \frac{l}{2}$ $a=l/2$
$a^{2} = {(\frac{l}{2})}^{2}$ $a^2=(l/2)^2$
$a^{2} = \frac{l^{2}}{4}$ $a^2=l^2/4$
$l^{2} = 3 r^{2}$ $l^2=3r^2$
$a^{2} = 3 \frac{r^{2}}{4}$ $a^2=3r^2/4$
Height of the cone is given by
$l^{2} = {(\frac{l}{2})}^{2} + h^{2}$ $l^2=(l/2)^2+h^2$
$h^{2} = 3 \frac{l^{2}}{4}$ $h^2=3l^2/4$
$h = \frac{\sqrt{3}}{2} l$ $h=(sqrt3)/2l$
$h = \frac{\sqrt{3}}{2} \sqrt{3} r$ $h=(sqrt3)/2sqrt3r$
$h = \frac{3}{2} r$ $h=3/2r$
Volume of the cone is V2
$V 2 = \frac{1}{3} π a^{2} h$ $V2=1/3pia^2h$
$V 2 = \frac{1}{3} π \frac{3 r^{2}}{4} (\frac{3}{2} r)$ $V2=1/3pi(3r^2)/4(3/2r)$
$V 2 = π r^{3}$ $V2=pir^3$

Thus,
Volume of sphere is $V 1 = \frac{4}{3} π r^{3}$ $V1=4/3pir^3$
Volume of the cone is $V 2 = π r^{3}$ $V2=pir^3$
$V 2 = \frac{3}{4} (\frac{4}{3}) π r^{3}$ $V2=3/4(4/3)pir^3$
$V 2 = \frac{3}{4} V 1$ $V2=3/4V1$

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Question #75d5c

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