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SAT Math : Solid Geometry

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Example Question #875 : Geometry

Find the volume of a cone with radius 3 and height 5.

Possible Answers:

$15\pi$

$30\pi$

$60\pi$

$45\pi$

Correct answer:

$15\pi$

Explanation:

To solve, simply use the formula for the volume of a cone. Thus,

$V=\frac{1}{3}\pi{r^2}h=\frac{1}{3}\pi*{3^2}*5=\frac{1}{3}\pi*9*5=15\pi$

To remember the formula for volume of a cone, it helps to break it up into it's base and height. The base is a circle and the height is just h. Now, just multiplying those two together would give you the formula of a cylinder (see problem 3 in this set). So, our formula is going to have to be just a portion of that. Similarly to volume of a pyramid, that fraction is one third.

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Example Question #871 : Geometry

Find the area of a cone whose radius is 4 and height is 3.

Possible Answers:

$64\pi$

$48\pi$

$24\pi$

$16\pi$

Correct answer:

$64\pi$

Explanation:

To solve, simply use the formula for the area of a cone. Thus,

$V=\frac{4}{3}\pi{r^2}h=\frac{4}{3}*\pi*4^2*3=4*\pi*16=64\pi$

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Example Question #876 : Geometry

The volume of a right circular cone is $9\pi$ . If the cone's height is equal to its radius, what is the radius of the cone?

Possible Answers:

$3\pi$

$\sqrt{27}$

Correct answer:

Explanation:

The volume of a right circular cone with radius and height is given by:

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

Since the height of this cone is equal to its radius, we can say:

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

Now, we can substitute our given volume into the equation and solve for our radius.

$9\pi = \frac{1}{3}\pi r^3$

$27\pi = \pi r^3$

27 = r^3

r = 3

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Example Question #877 : Geometry

Cone

The above is a right circular cone. Give its volume.

Possible Answers:

$294 \pi$

$\frac{245}{3} \pi \sqrt{11}$

$49 \pi+35 \pi\sqrt{11}$

$\frac{1,925}{3} \pi$

$175 \pi$

Correct answer:

$\frac{245}{3} \pi \sqrt{11}$

Explanation:

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

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

We are given r = 7 , but not .

, , and the slant height of a right circular cone are related by the Pythagorean Theorem:

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

Setting l = 18 and r = 7 , substitute and solve for :

$h^{2}+ 7^{2}= 18^{2}$

$h^{2}+49= 324$

$h^{2}+49 - 49 = 324 - 49$

$h^{2} =275$

Taking the square root of both sides and simplifying the radical:

$h =\sqrt{275} = \sqrt{25} \cdot \sqrt{11} = 5 \sqrt{11}$

Now, substitute for and and evaluate:

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

$V = \frac{1}{3} \pi \cdot 7^{2} \cdot 5 \sqrt{11}$

$V = \frac{1}{3} \pi \cdot 49 \cdot 5 \sqrt{11}$

$V = \frac{245}{3} \pi \sqrt{11}$

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Example Question #181 : Solid Geometry

Cone

The above is a right circular cone. Give its volume.

Possible Answers:

$\frac{245}{3} \pi \sqrt{11}$

$175 \pi$

$294 \pi$

$49 \pi+35 \pi\sqrt{11}$

$\frac{1,925}{3} \pi$

Correct answer:

$294 \pi$

Explanation:

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

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

r = 7 and h = 18 , so substitute and evaluate:

$V = \frac{1}{3} \pi \cdot 7^{2} \cdot 18$

$= \frac{1}{3} \pi \cdot 49 \cdot 18$

$=\left ( \frac{1}{3} \cdot 49 \cdot 18 \right ) \cdot \pi$

$= 294 \pi$

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Example Question #182 : Solid Geometry

Cone

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

Possible Answers:

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

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

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

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

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

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 #183 : Solid Geometry

Cone

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

Possible Answers:

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

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

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

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

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

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 : Solid Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #875 : Geometry

Example Question #871 : Geometry

Example Question #876 : Geometry

Example Question #877 : Geometry

Example Question #181 : Solid Geometry

Example Question #182 : Solid Geometry

Example Question #183 : Solid Geometry

All SAT Math Resources

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