Volume of a Three-Dimensional Figure - SSAT Upper Level Quantitative

Card 0 of 472

Can we determine the volume of this shape?

Tap to see back →

Volume is a characteristic of three-dimensional shapes. This is a cube, which is a three-dimensional shape. We can think of volume as space within an object or shape, so the volume would be the amount of space inside this cube.

Find the volume of a regular hexahedron with a side length of .

Tap to see back →

A regular hexahedron is another name for a cube.

To find the volume of a cube,

Plugging in the information given in the question gives

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

Tap to see back →

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.

Chestnut wood has a density of about . 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)?

Tap to see back →

First, convert the dimensions to cubic centimeters by multiplying by : the cone has height , and its base has radius .

$3m$\frac{100cm}{1m}$=300cm\ \ \2m$\frac{100cm}{1m}$=200cm$

Its volume is found by using the formula and the converted height and radius.

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

$V=\frac{1}{3}$ \pi $(200cm)^{2}$ ( 300cm )\approx 12,566,371 ; $\textrm{cm}^{3}$$

Now multiply this by to get the mass.

$m=V\rho$

Finally, convert the answer to kilograms.

$5654867g $\frac{1kg}{1,000g}$ \approx 5,655kg$

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 = ), what is the mass of required water (nearest whole kilogram)?

Tap to see back →

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 .

Tap to see back →

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$$

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

Tap to see back →

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

Tap to see back →

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 cube has six square faces, each with area 64 square inches. Using the conversion factor 1 inch = 2.5 centimeters, give the volume of this cube in cubic centimeters, rounding to the nearest whole number.

Tap to see back →

The volume of a cube is the cube of its sidelength, which is also the sidelength of each square face. This sidelength is the square root of the area 64:

$$\sqrt{64}$=8$ inches.

Multiply this by 2.5 to get the sidelength in centimeters:

$\times 2.5$ centimeters.

The cube of this is

cubic centimeters

A cube has a side length of 5 inches. Give the volume and surface area of the cube.

Tap to see back →

A cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself twice. As a formula:

where is the length of any edge of the cube.

The Surface Area of a cube can be calculated as .

So we get:

Volume

Surface area

A cheese seller has a 2 foot x 2 foot x 2 foot block of gouda and she wants to cut it into smaller gouda cubes that are 1.5 inches on a side. How many cubes can she cut?

Tap to see back →

First we need to determine how many of the small cubes of gouda would fit along one dimension of the large cheese block. One edge of the large block is 24 inches, so 16 smaller cubes $(24\div 1.5)$ would fit along the edge. Now we simply cube this one dimension to see how many cubes fit within the whole cube. .

Give the volume of a cube with surface area .

Tap to see back →

Let be the length of one edge of the cube. Since its surface area is , one face has one-sixth of this area, or $\frac{N}{6}$ . Therefore,

,

and

$s =$\sqrt{ \frac{N}{6}$}= $\frac{ \sqrt{ N}$}{$\sqrt{ 6}$}= $\frac{ \sqrt{ 6N}$}{6}$

Cube this sidelength to get the volume:

$s ^{3}= \left ( $\frac{ \sqrt{ 6N}$}{6} \right $)^{3}$ = $\frac{\left( \sqrt{ 6N}$ \right $)^{3}$$}{6^{3}$}= $\frac{ 6N\sqrt{ 6N}$ }{216}= $\frac{N\sqrt{ 6N}$ }{36}$

The distance from one vertex of a cube to its opposite vertex is . Give the volume of the cube.

Tap to see back →

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s $=$\sqrt{\frac{N^{2}$$}{3}} = $\frac{N}{\sqrt{3}$} = $\frac{N\sqrt{3}$}{3}$

Cube this sidelength to get the volume:

The length of a diagonal of one face of a cube is 10. Give the volume of the cube.

Tap to see back →

Since a diagonal of a square face of the cube is 10, each side of each square has length $$\frac{\sqrt{2}$}{2}$ times this, or $10 \cdot $\frac{\sqrt{2}$}{2} = 5$\sqrt{2}$$ .

Cube this to get the volume of the cube:

$\left (5$\sqrt{2}$ \right $)^{3}$= $5^{3}$ \left ( $\sqrt{2}$ \right $)^{3}$ = 125 \left (2 $\sqrt{2}$ \right ) = 250 $\sqrt{2}$$ .

The distance from one vertex of a cube to its opposite vertex is one foot. Give the volume of the cube in inches.

Tap to see back →

Since we are looking at inches, we will look at one foot as twelve inches.

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s= $\sqrt{48}$ = $\sqrt{16}$ \cdot $\sqrt{3}$ = 4$\sqrt{3}$$ inches.

Cube this sidelength to get the volume:

cubic inches.

Find the volume of a regular octahedron with side lengths of .

Tap to see back →

Use the following formula to find the volume of a regular octahedron:

Plugging in the information from the question,

An aquarium is shaped like a perfect cube; the area of each glass face is 1.44 square meters. If it is filled to the recommended 90% capacity, then, to the nearest hundred liters, how much water will it contain?

Note: 1 cubic meter = 1,000 liters.

Tap to see back →

A perfect cube has square faces; if a face has area 1.44 square meters, then each side of each face measures the square root of this, or 1.2 meters. The volume of the tank is the cube of this, or

cubic meters.

Its capacity in liters is $1.728 \times 1,000 = 1,728$ liters.

90% of this is

$1,728 \times 0.9 = 1,555.2$ liters.

This rounds to 1,600 liters, the correct response.

The distance from one vertex of a perfectly cubic aquarium to its opposite vertex is meters. Give the volume of the aquarium in liters.

cubic meter = liters.

Tap to see back →

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s = $\sqrt{\frac{27}{4}$} = $\frac{\sqrt{27}$}{$\sqrt{4}$}} = $\frac{ \sqrt{9}$ \cdot $\sqrt{3}$}{2}= $\frac{3\sqrt{3}$}{2}$ meters.

Cube this sidelength to get the volume:

$s ^{3 } =\left ( $\frac{3\sqrt{3}$}{2} \right $)^{3 }$= $$\frac{3^{3 }$$ \left ($\sqrt{3}\right $)^{3 }$$}{2^{3 }$ } = $\frac{ 27 \cdot 3\sqrt{3}$ }{8 } = $\frac{ 81 \sqrt{3}$ }{8 }$ cubic meters.

To convert this to liters, multiply by 1,000:

$\frac{ 81\sqrt{3}$ }{8 } \cdot 1,000 = $\frac{81\sqrt{3}$ }{8 } \cdot $\frac{1,000}{1}$ = $\frac{ 81\sqrt{3}$ }{1 } \cdot $\frac{125}{1}$= 10,125$\sqrt{3}$ liters.

Give the volume of a cube with surface area 3 square meters.

Tap to see back →

Let be the length of one edge of the cube. Since its surface area is 3 square meters, one face has one-sixth of this area, or $\frac{3}{6}$ = $\frac{1}{2}$ square meters. Therefore, , and $s = $\sqrt{ \frac{1}{2}$} = $\frac{\sqrt{1}$}{$\sqrt{2}$}=\frac{1 \cdot \sqrt{2}$}{$\sqrt{2}$\cdot $\sqrt{2}$}= $\frac{\sqrt{2}$}{2}$ meters.

The choices are in centimeters, so multiply this by 100 - the sidelength is

$\frac{\sqrt{2}$}{2} \cdot 100 = 50 $\sqrt{2}$ centimeters.

The volume is the cube of this, or $V= \left( 50 $\sqrt{2}$ \right $)^{3}$= $50^{3}$ \cdot \left( $\sqrt{2}$ \right $)^{3}$= 125,000 \cdot 2 $\sqrt{2}$= 250,000$\sqrt{2}$$ cubic centimeters.

The length of a diagonal of a cube is $15$\sqrt{2}$$ . Give the volume of the cube.

Tap to see back →

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s = $\sqrt{150 }$ = $\sqrt{25}$ \cdot $\sqrt{6}$ = 5$\sqrt{6}$$

Cube the sidelength to get the volume:

$s ^{3 }= \left ( 5$\sqrt{6}$ \right $)^{3 }$$

$s ^{3 }= $5^{3 }$ \left ($\sqrt{6}$ \right $)^{3 }$ = 125 \cdot 6 $\sqrt{6}$ = 750 $\sqrt{6}$$