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Volumes Practice Test 3

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Question : 18 of 40

Marks: +1, -0

The radius and the height of a right circular cone are in the ratio of 3 : 5. If its volume is 120π cu m, its slant height is

3 $\sqrt{34}$ m
2 $\sqrt{28}$ m
2 $\sqrt{44}$ m
2 $\sqrt{34}$ m

Solution:

Concept:

Use the cone volume formula and the slant height relation:

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

.

Explanation:

Let

r=3x\ \text{m}

and

h=5x\ \text{m}

.

Volume

=\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3x)^2 (5x) = 15\pi x^3\ \text{m}^3

.

Given volume

=120\pi\ \text{m}^3

, so

15\pi x^3=120\pi

.

\Rightarrow x^3=8

.

\Rightarrow x=2\ \text{m}

.

Thus

r=3\times2=6\ \text{m}

and

h=5\times2=10\ \text{m}

.

Slant height

l=\sqrt{r^2+h^2}=\sqrt{6^2+10^2}=\sqrt{36+100}=\sqrt{136}=2\sqrt{34}\ \text{m}

.

Answer:

2\sqrt{34}\ \text{m}

, which is option D.

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