Question

The perimeter of ends of frustum are 48cm and 36cm.if the height of the frustum be 11 cm . find the volume

Answer 1

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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