Cubes

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GMAT Quantitative › Cubes

Questions 1 - 10

is a cube and face has an area of . What is the length of diagonal of the cube ?

$3\sqrt{3}$

$2\sqrt{3}$

$3\sqrt{2}$

Explanation

To find the diagonal of a cube we can apply the formula $d=e\sqrt{3}$ , where is the length of the diagonal and where is the length of an edge of the cube.

Since we are given an area of a face of the cube, we can find the length of an edge simply by taking its square root.

$9=s^2 \rightarrow \sqrt 9 =s \rightarrow 3=s$

Here the length of an edge is 3.

Thefore the final andwer is $d=e\sqrt3=3\sqrt{3}$ .

is a cube and face has an area of . What is the length of diagonal of the cube ?

$3\sqrt{3}$

$2\sqrt{3}$

$3\sqrt{2}$

Explanation

To find the diagonal of a cube we can apply the formula $d=e\sqrt{3}$ , where is the length of the diagonal and where is the length of an edge of the cube.

Since we are given an area of a face of the cube, we can find the length of an edge simply by taking its square root.

$9=s^2 \rightarrow \sqrt 9 =s \rightarrow 3=s$

Here the length of an edge is 3.

Thefore the final andwer is $d=e\sqrt3=3\sqrt{3}$ .

is a cube with diagonal . What is the length of an edge of the cube?

$\sqrt{3}$

$\sqrt{2}$

$2\sqrt{2}$

$3\sqrt{2}$

Explanation

We are given the length of the diagonal of the cube.

Therefore we can find the length of an edge by using the formula

$d=e\sqrt{3}$ or $e=\frac{d}{\sqrt{3}}$ , and $\frac{3}{\sqrt{3}}=\frac{3}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}$ .

Therefore, the final answer is $\sqrt{3}$ .

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

$\frac{64 \sqrt{3}}{9} \textup{ yd}^{3}$

$\frac{64 \sqrt{2}}{6} \textup{ yd}^{3}$

$\frac{64 }{9} \textup{ yd}^{3}$

$32\sqrt{2} \textup{ yd}^{3}$

$32\sqrt{3} \textup{ yd}^{3}$

Explanation

Since we are looking at yards, we will look at twelve feet as four yards.

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

$s^{2}+s^{2}+s^{2}=4^{2}$

$3s^{2}=16$

$s^{2}= \frac{16}{3}$

$s^{2}= \sqrt{\frac{16}{3}} = \frac{\sqrt{16}}{\sqrt{3}} = \frac{4}{\sqrt{3}}$ yards.

Cube this sidelength to get the volume:

$s^{3}= \left ( \frac{4}{\sqrt{3}} \right )^{3} = \frac{4^{3}}{(\sqrt{3})^{3}} = \frac{64 }{3\sqrt{3} }= \frac{64 \cdot \sqrt{3}}{3\sqrt{3}\cdot \sqrt{3} } = \frac{64 \sqrt{3}}{9}$ cubic yards.

A sphere with surface area $40 \pi$ is inscribed inside a cube. Give the volume of the cube.

$80 \sqrt{10}$

$10 \sqrt{10}$

$40 \sqrt{10}$

Explanation

The sidelength of the cube is the diameter of the inscribed sphere, which is twice that sphere's radius. The sphere has surface area $40 \pi$ , so the radius is calculated as follows:

$4 \pi r^{2} = SA$

$4 \pi r^{2} = 40 \pi$

$r^{2} = 40 \pi \div 4 \pi = 10$

$r = \sqrt{10}$

The diameter of the sphere - and the sidelength of the cube - is twice this, or $2 \sqrt{10}$ .

Cube this sidelength to get the volume of the cube:

$V= \left ( 2 \sqrt{10} \right )^{3} = 2^{3} \cdot\left ( \sqrt{10 } \right )^{3} = 8 \cdot 10 \sqrt{10} = 80 \sqrt{10}$

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

Note: cubic meter liters.

$12,500\textup{ L}$

$15,625\textup{ L}$

$8,000\textup{ L}$

$10,000\textup{ L}$

The correct answer is not given among the other choices.

Explanation

A perfect cube has square faces; if a face has area 6.25 square meters, then each side of each face measures the square root of this, or 2.5 meters. The volume of the tank is the cube of this, or

$2.5^{3} = 15.625$ cubic meters.

Its capacity in liters is $15.625 \times 1,000 = 15,625$ liters.

80% of this is

$15,625 \times 0.8 = 12,500$ liters.

Which choice comes closest to the volume of a cube with surface area square centimeters?

$3.5 \textup{ m}^{3}$

$3 \textup{ m}^{3}$

$4.5 \textup{ m}^{3}$

$4 \textup{ m}^{3}$

$5 \textup{ m}^{3}$

Explanation

The suface area of a cube is six times the square of the length of one side, so solve for in the following:

$6s^{2} =135,000$

$s^{2} = 22,500$

This is the sidelength in centimeters; since we are looking at meters, divide this by 100 to convert to $150 \div 100 = 1.5$ meters.

Cube this to get volume

$V =1.5^{3}= 3.375$ cubic meters.

Of the given choices, 3.5 cubic meters comes closest.

The length of a diagonal of a cube is $\sqrt{17}$ . Give the volume of the cube.

$\frac{ 17 \sqrt{ 51} }{ 9} \right )$

$\frac{ 17 \sqrt{ 34} }{ 6} \right )$

$\frac{ 17 \sqrt{ 51} }{ 3} \right )$

$\frac{ 17 \sqrt{ 34} }{ 2} \right )$

Explanation

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

$s^{2}+s^{2}+s^{2}= ( \sqrt{17})^{2}$

$3s^{2}= 17$

$s^{2}= \frac{17}{3}$

$s =\sqrt{ \frac{17}{3}}$

Cube the sidelength to get the volume:

$s^{3} =\left (\sqrt{ \frac{17}{3}} \right )^{3}$

$=\frac{ \left (\sqrt{ 17} \right )^{3}}{\left (\sqrt{3} \right )^{3}} \right )$

$=\frac{ 17 \sqrt{ 17} }{ 3 \sqrt{3} } \right )$

$=\frac{ 17 \sqrt{ 17} \cdot \sqrt{3}}{ 3 \sqrt{3} \cdot \sqrt{3}} \right )$

$=\frac{ 17 \sqrt{ 51} }{ 9} \right )$

The length of a diagonal of one face of a cube is $5\sqrt{2}$ . Give the volume of the cube.

$125 \sqrt{2}$

$\frac{125 \sqrt{2}}{2}$

The correct answer is not among the other responses.

Explanation

A diagonal of a square has length $\sqrt{2}$ times that of a side, so each side of each square face of the cube has length $5 \sqrt{2} \div \sqrt{2}= 5$ . Cube this to get the volume:

$V= 5^{3} = 125$

The length of a diagonal of one face of a cube is $5\sqrt{2}$ . Give the volume of the cube.

$125 \sqrt{2}$

$\frac{125 \sqrt{2}}{2}$

The correct answer is not among the other responses.

Explanation

A diagonal of a square has length $\sqrt{2}$ times that of a side, so each side of each square face of the cube has length $5 \sqrt{2} \div \sqrt{2}= 5$ . Cube this to get the volume:

$V= 5^{3} = 125$

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