Question

The volume of a frustum cone is $210\pi f{t}^3$, whose base and upper area of a circle is 10 and 250 $f{t}^2$. Find the height of the cone. (Use $\pi$ = 22/7).

• A. 5.38 ft
• B. 4.38 ft
• C. 3.38 ft
• D. 6.38 ft

Answer 1

The correct option is D 6.38 ft

Volume of a frustum cone = $\frac{1}{3}(A_1+A_2+\sqrt{A_1{A}_2})h$

$210\pi =\frac{1}{3}(10+250+\sqrt{10\times 250})\times h$

$210\pi =\frac{1}{3}(260+\sqrt{2,500})\times h$

$210\pi =\frac{1}{3}(260+50)\times h$

$210\times 22/7\times 3=310\times h$

$h=\frac{1,980}{310}$

$h=$ 6.38 ft