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Volumes Practice Test 3

Question : 18 of 40

Marks: +1, -0

Solution:

Suppose the base radius and the height of the right circular cone are 3x m and 5x m respectively.

∴ Volume of the cone =

\frac{1}{3}\pi r^{2}h

\frac{1}{3}\pi (3x)^{2}\times 5x

cu m

Now,

\frac{1}{3}\pi \times 9x^{2}\times 5x

= 120π

⇒

x^{3}

\frac{120\times 3}{9\times 5}

⇒

x^{3}

= 8 ⇒

x^{3}

(2)^{3}

⇒ x = 2 m

∴ The radius and the height of the cone will be 3 × 2 = 6 m and 5 × 2 = 10 m respectively.

∴ Slant height of the cone =

\sqrt{r^{2}+h^{2}}

\sqrt{(6)^{2}+(10)^{2}}

\sqrt{36+100}

\sqrt{136}

2\sqrt{34}

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