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SAT Math : Geometry

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Example Questions

Example Question #183 : Solid Geometry

Cone

In terms of , express the volume of the above right circular cone.

Possible Answers:

$V = \frac{1}{3} \pi r^{2} \sqrt{ r^{2} - 576 }$

$V = \frac{1}{3} \pi r^{2} \sqrt{ 576 - r^{2} }$

$V = \pi r^{2} + 24 \pi r$

$V =8 \pi r^{2}$

$V = \pi r^{2} + \pi r\sqrt{ 576 - r^{2} }$

Correct answer:

$V = \frac{1}{3} \pi r^{2} \sqrt{ 576 - r^{2} }$

Explanation:

The volume of a cone can be calculated from its height and the radius of its base using the formula

$V = \frac{1}{3} \pi r^{2} h$

The slant height is shown in the diagram to be 24. By the Pythagorean Theorem,

$h^{2}+ r ^{2} = l^{2}$

Setting l = 24 and solving for :

$h^{2}+ r ^{2} = 24^{2}$

$h^{2}+ r ^{2} = 576$

$h^{2}+ r ^{2} - r^{2} = 576 -r^{2}$

$h^{2} = 576 - r^{2}$

$h =\sqrt{ 576 - r^{2} }$

Substituting in the volume formula for :

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

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Example Question #881 : Geometry

Cone

In terms of , express the volume of the provided right circular cone.

Possible Answers:

$V = \pi r^{2}+ 20 \pi r$

$V = \frac{1}{3} \pi r^{2} \sqrt{r^{2} - 400}$

$V = \pi r^{2}+ \pi r \sqrt{r^{2} + 400}$

$V = \frac{20}{3} \pi r^{2}$

$V = \frac{1}{3} \pi r^{2} \sqrt{r^{2} + 400}$

Correct answer:

$V = \frac{20}{3} \pi r^{2}$

Explanation:

The volume of a cone can be calculated from its height and the radius of its base using the formula

$V = \frac{1}{3} \pi r^{2} h$

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

$V = \frac{1}{3} \pi r^{2} \cdot 20$

$V = \frac{20}{3} \pi r^{2}$

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SAT Math : Geometry

Study concepts, example questions & explanations for SAT Math

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Example Question #183 : Solid Geometry

Example Question #881 : Geometry

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