SAT Math - How to find the volume of a cone

What is the volume of a right cone with a diameter of 6 cm and a height of 5 cm?

$15\pi \ cm^{3}$

$120\pi \ cm^{3}$

$180\pi \ cm^{3}$

$45\pi \ cm^{3}$

$60\pi \ cm^{3}$

Explanation

The general formula is given by $V = 1/3Bh = 1/3\pi r^{2}h$ , where = radius and = height.

The diameter is 6 cm, so the radius is 3 cm.

$V=\frac{\pi 3^2\times 5}{3}=15\pi$

There is a large cone with a radius of 4 meters and height of 18 meters. You can fill the cone with water at a rate of 3 cubic meters every 25 seconds. How long will it take you to fill the cone?

$41\ \textup{minutes}\ 53\ \textup{seconds}$

$3\ \textup{minutes}\ 20\ \textup{seconds}$

$800\ \textup{minutes}$

$600\ \textup{minutes}$

$60\ \textup{minutes}$

Explanation

First we will calculate the volume of the cone

$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\cdot 4^{2}\cdot 18=96\pi\ m^{3}$

Next we will determine the time it will take to fill that volume

$96\pi \ m^{3}\times \frac{25\ s}{3\ m^{3}}=800\pi \ s$

We will then convert that into minutes

$800\pi \ s\times \frac{1\ minute}{60\ seconds}=13.\overline{3}\pi \ minutes\approx 41.9\ minutes\approx 41\ minutes\ 53\ seconds$

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

$3000\pi$

$9000\pi$

$6000\pi$

$900\pi$

$300\pi$

Explanation

Write the formula to find the volume of a cone.

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

Substitute the known values and simplify.

$V=\frac{1}{3}\pi (10)^2 (90)= \frac{1}{3}\pi (100)(90)= \pi (100)(30)=3000\pi$

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?

$3\pi$

$\sqrt{27}$

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.

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

$27\pi = \pi r^3$

A cone has a base radius of 13 in and a height of 6 in. What is its volume?

None of the other answers

4394_π_ in3

1014_π_ in3

1352_π_ in3

338_π_ in3

Explanation

The basic form for the volume of a cone is:

V = (1/3)πr_2_h

For this simple problem, we merely need to plug in our values:

V = (1/3)π_132 * 6 = 169 * 2_π = 338_π_ in3

Cone

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

$294 \pi$

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

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

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

$175 \pi$

Explanation

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

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

and , so substitute and evaluate:

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

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

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

$= 294 \pi$

Cone

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

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

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

$294 \pi$

$175 \pi$

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

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 , but not .

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

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

Setting and , 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:

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

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

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

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

Cone

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

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

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

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

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

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

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 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 :

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

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

$15\pi$

$45\pi$

$30\pi$

$60\pi$

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.

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

$64\pi$