Question

A cone whose height is $15$ cm and radius of base is $6$ cm, is trimmed sufficiently to reduce it to a pyramid whose base is an equilateral triangle. The volume of the portion of removed is

• A. $325c{m}^3$
• B. $328c{m}^3$
• C. $320c{m}^3$
• D. $332c{m}^3$

Answer 1

Answer: (D)

Height of a cone $\left(h\right)=15cm$

Radius of a cone $\left(r\right)=6cm$

$\Rightarrow$ Volume of a cone $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times (6{)}^2\times 15$

$=\frac{3960}{7}$

$=565.71c{m}^3$

Pyramid is an equilateral triangle of side $6\sqrt{3}cm$

$\Rightarrow$ Area of equilateral triangle $=\frac{\sqrt{3}}{4}\times (side{)}^2$

$=\frac{\sqrt{3}}{4}\times (6\sqrt{3}{)}^2$

$=27\sqrt{3}c{m}^2$

$\Rightarrow$ Volume of a pyramid of height $15cm$ $=\frac{1}{3}\times 27\sqrt{3}\times 15$

$=135\sqrt{3}c{m}^3$

$=233.82c{m}^3$

$\Rightarrow$ Difference between volume of cone and volume of pyramid $=(565.71-233.82)c{m}^3$

$=331.89c{m}^3$

$=332c{m}^3$