SAT Math : Geometry

Study concepts, example questions & explanations for SAT Math

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Example Questions

Example Question #183 : Solid Geometry

Cone

In terms of \(\displaystyle r\), express the volume \(\displaystyle V\) of the above right circular cone.

Possible Answers:

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{ r^{2} - 576 }\)

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{ 576 - r^{2} }\)

\(\displaystyle V = \pi r^{2} + 24 \pi r\)

\(\displaystyle V =8 \pi r^{2}\)

\(\displaystyle V = \pi r^{2} + \pi r\sqrt{ 576 - r^{2} }\)

Correct answer:

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{ 576 - r^{2} }\)

Explanation:

The volume \(\displaystyle V\) of a cone can be calculated from its height \(\displaystyle h\) and the radius \(\displaystyle r\) of its base using the formula

\(\displaystyle V = \frac{1}{3} \pi r^{2} h\)

The slant height \(\displaystyle l\) is shown in the diagram to be 24. By the Pythagorean Theorem, 

\(\displaystyle h^{2}+ r ^{2} = l^{2}\)

Setting \(\displaystyle l = 24\) and solving for \(\displaystyle h\):

\(\displaystyle h^{2}+ r ^{2} = 24^{2}\)

\(\displaystyle h^{2}+ r ^{2} = 576\)

\(\displaystyle h^{2}+ r ^{2} - r^{2} = 576 -r^{2}\)

\(\displaystyle h^{2} = 576 - r^{2}\)

\(\displaystyle h =\sqrt{ 576 - r^{2} }\)

Substituting in the volume formula for \(\displaystyle h\):

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{ 576 - r^{2} }\).

 

Example Question #11 : Cones

Cone

In terms of \(\displaystyle r\), express the volume \(\displaystyle V\) of the provided right circular cone.

Possible Answers:

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{r^{2} - 400}\)

\(\displaystyle V = \pi r^{2}+ \pi r \sqrt{r^{2} + 400}\)

\(\displaystyle V = \frac{20}{3} \pi r^{2}\)

\(\displaystyle V = \frac{1}{3} \pi r^{2} \sqrt{r^{2} + 400}\)

\(\displaystyle V = \pi r^{2}+ 20 \pi r\)

Correct answer:

\(\displaystyle V = \frac{20}{3} \pi r^{2}\)

Explanation:

The volume \(\displaystyle V\) of a cone can be calculated from its height \(\displaystyle h\) and the radius \(\displaystyle r\) of its base using the formula

\(\displaystyle V = \frac{1}{3} \pi r^{2} h\)

The height of the cone is shown to be equal to 20, so substituting accordingly:

\(\displaystyle V = \frac{1}{3} \pi r^{2} \cdot 20\)

\(\displaystyle V = \frac{20}{3} \pi r^{2}\)

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