ACT Math - Prisms | Practice Hub

Find the diagonal of a right rectangular prism if the length, width, and height are 3,4, and 5, respectively.

$5\sqrt2$

$\sqrt{23}$

$4\sqrt2$

$3\sqrt{2}$

Explanation

Write the diagonal formula for a rectangular prism.

$d=\sqrt{L^2+W^2+H^2}$

Substitute and solve for the diagonal.

$d=\sqrt{3^2+4^2+5^2}= \sqrt{9+16+25}= \sqrt{50}= 5\sqrt2$

Find the diagonal of a right rectangular prism if the length, width, and height are 3,4, and 5, respectively.

$5\sqrt2$

$\sqrt{23}$

$4\sqrt2$

$3\sqrt{2}$

Explanation

Write the diagonal formula for a rectangular prism.

$d=\sqrt{L^2+W^2+H^2}$

Substitute and solve for the diagonal.

$d=\sqrt{3^2+4^2+5^2}= \sqrt{9+16+25}= \sqrt{50}= 5\sqrt2$

Sturgis is in charge of designing a new exhibit in the shape of a rectangular prism for a local aquarium. The exhibit will hold alligator snapping turtles and needs to have a volume of . Sturgis knows that the exhibit will be long and go back into the wall.

What will the height of the new exhibit be?

Explanation

This sounds like a geometry problem, so start by drawing a picture so that you know exactly what you are dealing with.

Because we are dealing with rectangular prisms and volume, we will need the following formula:

We are solving for height, so you can begin by rearranging the equation to get by itself:

$\frac{V}{l*w}=h$

Then, plug in our knowns (, and )

$h=\frac{150m^3}{15m*5m}=\frac{150m}{75}=2:m$

Here is the problem worked out with a corresponding picture:

Prism

A rectangular prism has the following dimensions:

Length:

Width:

Height:

Find the volume.

Explanation

Given that the dimensions are: , , and and that the volume of a rectangular prism can be given by the equation:

$v=l \cdot w \cdot h$ , where is length, is width, and is height, the volume can be simply solved for by substituting in the values.

This final value can be approximated to .

What will the height of the new exhibit be?

Explanation

This sounds like a geometry problem, so start by drawing a picture so that you know exactly what you are dealing with.

Because we are dealing with rectangular prisms and volume, we will need the following formula:

We are solving for height, so you can begin by rearranging the equation to get by itself:

$\frac{V}{l*w}=h$

Then, plug in our knowns (, and )

$h=\frac{150m^3}{15m*5m}=\frac{150m}{75}=2:m$

Here is the problem worked out with a corresponding picture:

Prism

A rectangular prism has the following dimensions:

Length:

Width:

Height:

Find the volume.

Explanation

Given that the dimensions are: , , and and that the volume of a rectangular prism can be given by the equation:

$v=l \cdot w \cdot h$ , where is length, is width, and is height, the volume can be simply solved for by substituting in the values.

This final value can be approximated to .

If the dimensions of a right rectangular prism are 1 yard by 1 foot by 1 inch, what is the diagonal in feet?

$\frac{\sqrt{1441}}{12}$

$\sqrt3$

$\frac{\sqrt3}{3}$

$\frac{\sqrt{1306}}{3}$

$\frac{1}{9}$

Explanation

Convert the dimensions into feet.

$1 \textup{ yard}= 3\textup{ feet}$

$\textup{1 inch }= \frac{1}{12}\textup{ foot}$

The new dimensions of rectangular prism in feet are: $3\times1\times \frac{1}{12}$

Write the formula for the diagonal of a right rectangular prism and substitute.

$d=\sqrt{L^2+W^2+H^2}$

$d=\sqrt{3^2+1^2+\frac{1}{12}^2}=\sqrt{9+1+\frac{1}{144}}=\sqrt{\frac{1440}{144}+\frac{1}{144}}$

$d=\sqrt{\frac{1441}{144}} =\frac{\sqrt{1441}}{12}\textup{ ft}$

If three-quarters of the exhibit's volume will be water, how high up the wall will the water come?

Cannot be determined with the information provided

Explanation

The trickiest part of this question is the wording. This problem is asking for the height of the water in the exhibit if the exhibit is three-quarters full. We can find this at least two different ways.

The longer way requires that we begin by finding three quarters of the total volume:

$150:m^3*\frac{3}{4}=112.5:m^3$

Now we go back to our volume equation, and since we are again looking for height, we want it solved for :

Becomes

$\frac{V}{l*w}=h$

$h=\frac{112.5:m^3}{15:m*5:m}=1.5:m$

The easier way requires that we recognize a key detail. If we take three-quarters of the volume without changing our length or width, our new height will just be three-quarters of the total height. We can solve for the total height of the exhibit by using the volume equation and rearranging it to solve for :

$\frac{V}{l*w}=h$

At this point, we can substitute in our given values and solve for :

$\frac{150:m^3}{15:m*5:m}=2:m$

So, the total height of the exhibit is . We can now easily solve for three-quarters of the total height:

$h{}'=h*\frac{3}{4}$

$h{}'=2:m*\frac{3}{4}=1.5:m$

If three-quarters of the exhibit's volume will be water, how high up the wall will the water come?

Cannot be determined with the information provided

Explanation

The longer way requires that we begin by finding three quarters of the total volume:

$150:m^3*\frac{3}{4}=112.5:m^3$

Now we go back to our volume equation, and since we are again looking for height, we want it solved for :

Becomes

$\frac{V}{l*w}=h$

$h=\frac{112.5:m^3}{15:m*5:m}=1.5:m$

The easier way requires that we recognize a key detail. If we take three-quarters of the volume without changing our length or width, our new height will just be three-quarters of the total height. We can solve for the total height of the exhibit by using the volume equation and rearranging it to solve for :

$\frac{V}{l*w}=h$

At this point, we can substitute in our given values and solve for :

$\frac{150:m^3}{15:m*5:m}=2:m$

So, the total height of the exhibit is . We can now easily solve for three-quarters of the total height:

$h{}'=h*\frac{3}{4}$

$h{}'=2:m*\frac{3}{4}=1.5:m$

If the dimensions of a right rectangular prism are 1 yard by 1 foot by 1 inch, what is the diagonal in feet?

$\frac{\sqrt{1441}}{12}$

$\sqrt3$

$\frac{\sqrt3}{3}$

$\frac{\sqrt{1306}}{3}$

$\frac{1}{9}$

Explanation

Convert the dimensions into feet.

$1 \textup{ yard}= 3\textup{ feet}$

$\textup{1 inch }= \frac{1}{12}\textup{ foot}$

The new dimensions of rectangular prism in feet are: $3\times1\times \frac{1}{12}$

Write the formula for the diagonal of a right rectangular prism and substitute.

$d=\sqrt{L^2+W^2+H^2}$

$d=\sqrt{3^2+1^2+\frac{1}{12}^2}=\sqrt{9+1+\frac{1}{144}}=\sqrt{\frac{1440}{144}+\frac{1}{144}}$

$d=\sqrt{\frac{1441}{144}} =\frac{\sqrt{1441}}{12}\textup{ ft}$

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