Question

the perimeter of the ends of the frustum of a cone are 36 cm and 48cm. if the height of the frustum is 11 cm find its volume

Answer 1

The answer is 1554 cm³

$\bold{2\pi r = 36} \\ \\ \bold{r = \frac{36}{2\pi} = \frac{18}{\pi}}$

So,

Radius of smaller end is :-

$\dfrac{36}{2\pi}$

Now,

$\bold{2\pi r = 48} \\ \\ \bold{r = \frac{48}{2\pi} = \frac{24}{\pi} }$

Radius of larger end is 24/pie

Now ,

Volume of frustrum is as follows :-

$\boxed{ \frac{1}{3}\pi h( {r1}^{2} + {r2}^{2} + r1r2)}$

$\frac{1}{3}\pi \times 11( { (\frac{18}{\pi})}^{2} + { (\frac{24}{\pi})}^{2} + (\frac{18}{\pi} \times \frac{24}{\pi})) \\ \\ = \frac{1}{3} \pi \times 11( \frac{324 + 576 + 432}{ {\pi}^{2} }) \\ \\ \\ = \frac{1}{3} \times\frac{22}{7} \times 11 \times \frac{1332 \times 7 \times 7}{22 \times 22} \\ \\ \\ = \frac{1332 \times 7}{6} = 222 \times 7 \\ \\ = 1554 {cm}^{3}$

So

The volume of the frustrum is :-

$\boxed{1554 {cm}^{3}}$

Answer 2

Answer: 1554 cm³

Step-by-step explanation:

Let  the radius  of  smaller    end  be  r₁ and  radius  of  larger end  be  r₂.

Since, 48 is  greater than 36,  Perimeter of smaller  end is 36cm and  Perimeter of larger end  is 48cm

Perimeter  of smaller end = 36 cm

2πr₁ = 36

r₁ = 36/2π

r₁ = 18/π

Perimeter  of larger end = 48cm

2πr₂ = 48

r₂ = 48/2π

r₂ = 24/π

Height of the  frustum (h)  =  11cm

We know  that,

Volume of  a  frustum,

V = 1/3 × πh(r₁² + r₂² +r₁.r₂)

$V=\frac{\pi}{3}\times11[(\frac{18}{\pi})^2+(\frac{24}{\pi})^2+(\frac{18}{\pi}\times\frac{24}{\pi})]\\\;\\V=\frac{11\pi}{3}[\frac{324}{\pi{^2}}+\frac{576}{\pi^2}+\frac{432}{\pi^2}]\\\;\\V=\frac{11\pi}{3}[\frac{324+576+432}{\pi^2}]\\\;\\V=\frac{11}{3}[\frac{1332}{\pi}]\\\;\\V=\frac{11}{3}\times\frac{1332}{\frac{22}{7}}\\\;\\V=\frac{1}{3}\times\frac{1332\times7}{2}\\\;\\V=\frac{444\times7}{2}\\\;\\V=222\times7\\\;\\V=1554\;\;cm^3$

Note:- Answer may  be 1553.4 if we put π=3.14 which is  negligible