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ACT Math : Other Polyhedrons

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Example Question #11 : Other Polyhedrons

A pyramid is placed inside a cube so that they share a base and height. If the surface area of the cube is $486\textup{ft}^{2}$ , what is the volume of the pyramid, in square feet?

Possible Answers:

$243\textup{ft}^{2}$

$729\textup{ft}^{2}$

$225\textup{ft}^{2}$

$216\textup{ft}^{2}$

Correct answer:

$243\textup{ft}^{2}$

Explanation:

First, we need to find the length of a side for the cube.

Recall that the surface area of the cube is given by the following equation:

$\text{Surface Area}=6s^2$ , where is the length of a side.

Plugging in the surface area given by the equation, we can then find the side length of the cube.

486=6s^2

s^2=81

s=9

Now, because the pyramid and the cube share a base, we know that the pyramid must be a square pryamid.

Recall how to find the volume of a pyramid:

$\text{Volume}=\text{Area of base}\times \text{height}\times \frac{1}{3}$

$\text{Area of Base}=9\times9=81$

Now, since the pyramid is the same height as the cube, the height of the pyramid is also .

$\text{Volume}=81\times 9\times \frac{1}{3}=243$

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Example Question #12 : Other Polyhedrons

In cubic feet, find the volume of the pentagonal prism illustrated below. The pentagon has an area of $16\textup{ft}^{2}$ , and the prism has a height of $8\textup{ft}$ .

Possible Answers:

$224\textup{ft}^{3}$

$96\textup{ft}^{3}$

$128\textup{ft}^{3}$

$20\textup{ft}^{3}$

Correct answer:

$128\textup{ft}^{3}$

Explanation:

For any prism, the volume is given by the following equation:

$\text{Volume}=\text{Area of base}\times\text{height}$

The question gives us the area of the base and the height.

$\text{Volume}=16\times8=128$

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Example Question #13 : Other Polyhedrons

If the side lengths of a cube are tripled, what effect will it have on the volume?

Possible Answers:

The volume will be times as large.

Correct answer:

The volume will be times as large.

Explanation:

Start by taking a cube that is $1\times1\times1$ . The volume of this cube is .

Next, triple the sides of this cube so that it becomes $3\times3\times3$ . The volume of this cube is .

The volume of the new cube is times as large as the original.

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Example Question #14 : Other Polyhedrons

The surface area of a cylinder is given by $2\pi rh+2\pi r^{2}$ , where is the radius and is the height. If a cylinder has a surface area of $20\pi$ $cm^{2}$ and a height of , what is its radius?

Possible Answers:

2 cm

5 cm

1cm

Correct answer:

2 cm

Explanation:

The fastest way to solve this problem is to plug all of the answer choices in for and look for an output of $20\pi$ .

Alternatively, one could set the surface area formula equal to $20\pi$ (knowing that h=3cm ), but this would require solving a quadratic.

$20\pi =2\pi r(3)+2\pi r^2$

$0=6\pi r +2\pi r^2-20\pi$

Rearranging this we get

$2\pi r^2+6 \pi r-20\pi=0$

Now factoring out a $2\pi$ we get:

$2\pi (r^2+3r-10)=0$

$2\pi(r+5)(r-2)=0$

$r+5=0 \rightarrow r=-5$

$r-2=0 \rightarrow r=2$

Since we cannot have a negative length our answer is 2 cm .

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Example Question #15 : Other Polyhedrons

If a half cylinder with a height of 5 and semicircular bases with a radius of 2, what is the surface area?

Possible Answers:

$20\pi+10$

$20+10\pi$

$20+14\pi$

$34\pi$

$20\pi+14$

Correct answer:

$20+14\pi$

Explanation:

The surface area of the half cylinder will consist of the lateral area and the base area.

There are three parts to the surface area:

Rectangular region, semi-circular bases of the half cylinder, and outer face of the cylinder.

The cross section of the half cylinder is a rectangle. Find the area of the rectangle. The diameter of the semicircle represents the width of the rectangle, which is double the radius. The length of the rectangle is the height of the cylinder.

$A_{\textup{rectangle}}=LW=(4)(5)=20$

Next, find the area of a semicircular bases.

$A_{\textup{semicircular base}} = \frac{\pi r^2}{2} = \frac{\pi (2^2)}{2}= 2\pi$

Since there are two semicircular bases of the semi-cylinder, the total area of the semicircular bases is $4\pi$ .

$A_{\textup{ TWO semicircular bases}} = 4\pi$

Find the area of the outer region. The area of the outer region is the half circumference multiplied by the height of the cylinder.

$A_{\textup{outer region}}= \frac{2\pi r}{2} \times h = 2\pi (5)= 10\pi$

Sum the areas of the rectangle, the two circular bases, and the outer region to find the lateral area.

$\sum A=20 + 4\pi + 10\pi = 20+14\pi$

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ACT Math : Other Polyhedrons

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #11 : Other Polyhedrons

Example Question #12 : Other Polyhedrons

Example Question #13 : Other Polyhedrons

Example Question #14 : Other Polyhedrons

Example Question #15 : Other Polyhedrons

All ACT Math Resources

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