Question

The perimeters of the ends of a frustum of a cone are 36cm and 48cm. if the height of the frustum is 11cm, find it's volume.

Answer 1

Let the side of perimeter 36 cm have radius, r1 and the side of perimeter 48 cm have radius, r2 .

So,  perimeter = 36 cm

  => 2 π r1=36

  => r1=36/2π

  =>r1=18/π

Again,

  perimeter=48 cm

  => 2 π r2=48

        => r2= 24/π

Given height of frustum(h)= 11 cm

 

therefore, Volume of Frustum (v)= 1/3 π {(r1)2+ (r2)2 + (r1r2)} 

V=  1/3 x π x h {(18/π)2+(24/π)2+(18/π x 24/π)}V=  1/3 x π x h {324/π2+576/π2+432/π2}V=  1/3 x π x h x 1/π2 (324+576+432)V=  1/3 x π x h x 1/π2 (1332)V=  1/3 x h x 1/π (1332)V=  1/3 x 11 x 1/π (1332)  [since h=11]V=  1/3 x 11 x 7/22 (1332)  [π=22/7,  1/π= 7/22]V=  1/3 x 11 x 7/22 2  (1332­­­­­)666 222V=  7 x 222 V=  1554 cm3

 

Ans = volume of frustum = 1554 cm3.................t=wish u got your answer

Answer 2

$\huge\bold\green{AnSwer:-}$

According to the question Height of frustum = 11 cm

Also (Given)

Perimeter = 36 cm

2πr1 = 36

So,

$\tt{\rightarrow r1=\dfrac{36}{2\pi}}$

Or,

$\tt{\rightarrow r1=\dfrac{18}{\pi}}$

Now,

Perimeter = 48 cm

2πr2 = 48

$\tt{\rightarrow r2=\dfrac{24}{\pi}}$

Using Formula :-

Volume of Frustum

= 1/3 π {(r1)²+ (r2)² + (r1r2)} h

$\tt{\rightarrow\dfrac{1}{3}\pi[(r1)^2+(r2)^2+(r1r1)]h}$

$\tt{\rightarrow\dfrac{1}{3}\times\pi\times h[(\dfrac{18}{\pi})^2+(\dfrac{24}{\pi})^2+[\dfrac{18}{\pi}\times\dfrac{24}{\pi}]}$

$\tt{\rightarrow\dfrac{1}{3}\times\pi\times 11[\dfrac{324}{\pi^2}+\dfrac{576}{\pi^2}+\dfrac{432}{\pi^2}]}$

$\tt{\rightarrow\dfrac{1}{3}\times\pi \times 11\times\dfrac{1}{\pi^2}(324+576+432)}$

$\tt{\rightarrow\dfrac{1}{3}\times\pi \times 11\times\dfrac{1}{\pi^2}(1332)}$

$\tt{\rightarrow\dfrac{11}{3}\times\dfrac{1}{\pi}(1332)}$

$\tt{\rightarrow\dfrac{11}{3}\times\dfrac{7}{22}(1332)}$

$\tt{\rightarrow\dfrac{11\times 7\times 1332}{22\times 3}}$

Hence, Volume of frustum = 154 cm³