SSAT Upper Level Quantitative - How to find the volume of a cone

A cone has a diameter of and a height of . In cubic meters, what is the volume of this cone?

$48\pi:m^3$

$144\pi:m^3$

$36\pi:m^3$

$54\pi:m^3$

Explanation

First, divide the diameter in half to find the radius.

$\frac{d}{2}=r$

$\frac{12}{2}=6:m$

Now, use the formula to find the volume of the cone.

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

$\text{Volume}=\frac{1}{3}\pi \times 6^2\times4=\frac{1}{3}\pi \times 36\times4=\frac{144}{3}\pi=48\pi:m^3$

A right cone has a volume of $8\pi$ , a height of and a radius of the circular base of . Find .

$\sqrt[3]{2}$

$\sqrt[3]{3}$

$\sqrt{2}$

$2\sqrt[3]{2}$

Explanation

The volume of a cone is given by:

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

where is the radius of the circular base, and is the height; the perpendicular distance from the base to the vertex. Substitute the known values in the formula:

$Volume=\frac{1}{3}\pi r^2h\Rightarrow 8\pi=\frac{1}{3}\pi\times (2t)^2\times 3t$

$\Rightarrow 8\pi=4\pi t^3\Rightarrow t^3=2\Rightarrow t=\sqrt[3]{2}$

A cone has a radius of inches and a height of inches. Find the volume of the cone.

$36\pi:in^3$

$108\pi:in^3$

$12\pi:in^3$

$24\pi:in^3$

Explanation

The volume of a cone is given by the formula:

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

Now, plug in the values of the radius and height to find the volume of the given cone.

$\text{Volume}=\frac{1}{3}\pi\times3^2\times12=\frac{1}{3}\pi\times9\times12=\frac{108}{3}\pi=36\pi:in^3$

A cone has the height of 4 meters and the circular base area of 4 square meters. If we want to fill out the cone with water (density = $1000\frac{Kg}{m^3}$ ), what is the mass of required water (nearest whole kilogram)?

6333

Explanation

The volume of a cone is:

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

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

As the circular base area is $\pi r^2$ , so we can rewrite the volume formula as follows:

$Volume =\frac{A\times h}{3}$

where is the circular base area and known in this problem. So we can write:

$Volume =\frac{A\times h}{3}=\frac{4\times 4}{3}=\frac{16}{3}\approx 5.333 m^3$

We know that density is defined as mass per unit volume or:

$\rho =\frac{m}{v}$

Where $\rho$ is the density; is the mass and is the volume. So we get:

$m=\rho v=1000\times 5.333=5333 Kg$

The vertical height (or altitude) of a right cone is . The radius of the circular base of the cone is . Find the volume of the cone in terms of .

$\frac{2}{3}\pi t^3$

$\frac{1}{3}\pi t^3$

$\frac{2}{3}\pi t^2$

$\frac{1}{3}\pi t^2$

$\pi t^3$

Explanation

The volume of a cone is:

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

where is the radius of the circular base, and is the height (the perpendicular distance from the base to the vertex).

$Volume=\frac{1}3{}\pi r^2h =\frac{1}3{}\pi t^2\times 2t=\frac{2}{3}\pi t^3$

Chestnut wood has a density of about $0.45\frac{g}{cm^3}$ . A right circular cone made out of chestnut wood has a height of three meters, and a base with a radius of two meters. What is its mass in kilograms (nearest whole kilogram)?