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High School Math : Non-Cubic Prisms

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Example Questions

Example Question #1 : Non Cubic Prisms

The length of a box is 3 times the width. Which of the following gives the length (L inches) in terms of the width (W inches) of the box?

Possible Answers:

L = W + 3

L = 3W

L = 3/W

L = ½ (3W)

Correct answer:

L = 3W

Explanation:

When reading word problems, there are certain clues that help interpret what is going on. The word “is” generally means “=” and the word “times” means it will be multiplied by something. Therefore, “the length of a box is 3 times the width” gives you the answer: L = 3 x W, or L = 3W.

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Example Question #1 : Prisms

The width of a box, in inches, is 5 inches less than three times its length. Which of the following equations gives the width, W inches, in terms of the length, L inches, of the box?

Possible Answers:

W=5-3L

W=5L-3

W=3-5L

W=3L-5

Correct answer:

W=3L-5

Explanation:

We notice the width is “5 inches less than three times its width,” so we express W as being three times its width (3L) and 5 inches less than that is 3L minus 5. In this case, W is the dependent and L is the independent variable.

W = 3L - 5

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Example Question #11 : Prisms

Angie is painting a 2 foot cube for a play she is in. She needs $25\hspace{1 mm}mL$ of paint for every square foot she paints. How much paint does she need?

Possible Answers:

$600\hspace{1 mm}mL$

$100\hspace{1 mm}mL$

None of the available answers

It is impossible to convert between metric units and feet.

$1.041\overline{6}\hspace{1 mm}mL$

Correct answer:

$600\hspace{1 mm}mL$

Explanation:

First we must calculate the surface area of the cube. We know that there are six surfaces and each surface has the same area:

$Area=6(2^2)=6\times 4=24\hspace{1 mm}feet^2$

Now we will determine the amount of paint needed

$24\hspace{1 mm}feet^2\times \frac{25\hspace{1 mm}mL}{1\hspace{1 mm}foot^2}=600\hspace{1 mm}mL$

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Example Question #3 : Non Cubic Prisms

Find the surface area of the following triangular prism.

Half_box

Possible Answers:

$144 + 52\sqrt{2} m^2$

$54 + 144\sqrt{3} m^2$

$54 + 144\sqrt{2} m^2$

$144 + 54\sqrt{3} m^2$

$144 + 54\sqrt{2} m^2$

Correct answer:

$144 + 54\sqrt{2} m^2$

Explanation:

The formula for the surface area of a triangular prism is:

$SA = 2 (base) + lateral\ area$

$SA = 2 \cdot \left(\frac{1}{2}\right)(l)(w) + (l+w+y)(h)$

Where is the length of the triangle, is the width of the triangle, is the hypotenuse of the triangle, and is the height of the prism

Use the formula for a 45-45-90 triangle to solve for the length of the hypotenuse:

$s-s-s\sqrt{2}$

$6-6-6\sqrt{2}$

Plugging in our values, we get:

$SA = (6m)(6m) + (6m+6m+6\sqrt{2}m)(9m)$

$SA = (6m)(6m) + (12m + 6\sqrt{2}m)(9m)$

$SA = 36m^2 + 108m^2 + 54\sqrt{2}m^2 = 144 + 54\sqrt{2} m^2$

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Example Question #1 : How To Find The Surface Area Of A Prism

Find the surface area of the following triangular prism.

Half_box

Possible Answers:

$256m^2 + 96\sqrt{2}m^2$

$96m^2 + 96\sqrt{2}m^2$

$256m^2 + 94\sqrt{2}m^2$

$256m^2 + 256\sqrt{2}m^2$

$96m^2 + 256\sqrt{2}m^2$

Correct answer:

$256m^2 + 96\sqrt{2}m^2$

Explanation:

The formula for the surface area of a triangular prism is:

SA = 2(base)+(perimeter)(height)

$SA = 2\left(\frac{1}{2}\right)(l)(w) + (l+w+y)(h)$

SA = (l)(w) + (l+w+y)(h)

Where is the length of the base, is the width of the base, is the hypotenuse of the base, and is the height of the prism

Use the formula for a 45-45-90 triangle to find the length of the hypotenuse:

$a-a-a\sqrt{2}$

$8m-8m-8\sqrt{2}m$

Plugging in our values, we get:

$SA = (8m)(8m) + (8m+8m+8\sqrt{2}m)(12m)$

$SA = 256m^2 + 96\sqrt{2}m^2$

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Example Question #3 : Non Cubic Prisms

Find the surface area of the following triangular prism.

Triangular_prism

Possible Answers:

$216 + 216\sqrt{3}m^2$

$216 + 16\sqrt{3}m^2$

$218 + 16\sqrt{3}m^2$

$216 + 18\sqrt{3}m^2$

$18 + 18\sqrt{3}m^2$

Correct answer:

$216 + 18\sqrt{3}m^2$

Explanation:

The formula for the surface area of an equilateral, triangular prism is:

SA = 2(base)+(perimeter)(height)

$SA = 2\left(\frac{s^2\sqrt{3}}{4}\right)+(3s)(h)$

Where is the length of the triangle side and is the length of the height.

Plugging in our values, we get:

$SA = 2\left(\frac{(6m)^2\sqrt{3}}{4}\right)+(3)(6m)(12m)$

$SA = 216 + 18\sqrt{3}m^2$

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Example Question #691 : Geometry

David wants to paint the walls in his bedroom. The floor is covered by a $10\ ft \times 16\ ft$ carpet. The ceiling is $8\ ft$ tall. He selects a paint that will cover $75\ ft^2$ per quart and $300\ ft^2$ per gallon. How much paint should he buy?

Possible Answers:

2 gallons and 1 quart

1 gallon and 1 quart

3 quarts

1 gallon

1 gallon and 2 quarts

Correct answer:

1 gallon and 2 quarts

Explanation:

Find the surface area of the walls: SA_walls = 2lh + 2wh, where the height is 8 ft, the width is 10 ft, and the length is 16 ft.

This gives a total surface area of 416 ft². One gallon covers 300 ft², and each quart covers 75 ft², so we need 1 gallon and 2 quarts of paint to cover the walls.

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Example Question #2 : Non Cubic Prisms

A box is 5 inches long, 5 inches wide, and 4 inches tall. What is the surface area of the box?

Possible Answers:

$140\ in^2$

$25\ in^2$

$100\ in^2$

$120\ in^2$

$130\ in^2$

Correct answer:

$130\ in^2$

Explanation:

The box will have six total faces: an identical "top and bottom," and identical "left and right," and an identical "front and back." The total surface area will be the sum of these faces.

Since the six faces consider of three sets of pairs, we can set up the equation as:

$SA=2(\text{top})+2(\text{left})+2(\text{front})$

Each of these faces will correspond to one pair of dimensions. Multiply the pair to get the area of the face.

SA=2(lw)+2(wh)+2(lh)

Substitute the values from the question to solve.

SA=2(5*5)+2(5*4)+2(5*4)

SA=50+40+40

$SA=130\ in^2$

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Example Question #3 : Non Cubic Prisms

What is the surface area of a rectangular brick with a length of 12 in, a width of 8 in, and a height of 6 in?

Possible Answers:

None of the answers are correct

$432\ in^2$

$576\ in^2$

$216\ in^2$

$382\ in^2$

Correct answer:

$432\ in^2$

Explanation:

The formula for the surface area of a rectangular prism is given by:

SA = 2LW + 2WH + 2HL

SA = 2(12 * 8) + 2(8 * 6) + 2(6 * 12)

SA = 2(96) + 2(48) + 2(72)

SA = 192 + 96 + 144

SA = 432 in²

216 in² is the wrong answer because it is off by a factor of 2

576 in³ is actually the volume, V = L * W * H

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Example Question #12 : How To Find Surface Area

What is the surface area of an equilateral triangluar prism with edges of 6 in and a height of 12 in?

Let $\sqrt{3}=1.732$ and $\sqrt{2}=1.414$ .

Possible Answers:

$247.176\ in^{2}$

$165.135\ in^{2}$

$189.239\ in^{2}$

$268.732\ in^{2}$

$216.414\ in^{2}$

Correct answer:

$247.176\ in^{2}$

Explanation:

The surface area of the prism can be broken into three rectangular sides and two equilateral triangular bases.

The area of the sides is given by: $A=lw=12\cdot 6=72\ in^{2}$ , so for all three sides we get $\dpi{100} 216\ in^{2}$ .

The equilateral triangle is also an equiangular triangle by definition, so the base has congruent sides of 6 in and three angles of 60 degrees. We use a special right traingle to figure out the height of the triangle: 30 - 60 - 90. The height is the side opposite the 60 degree angle, so it becomes $3\sqrt{3}$ or 5.196.

The area for a triangle is given by $A=\frac{1}{2}bh$ and since we need two of them we get $A=bh=6\cdot 3\sqrt{3}=31.176\ in^{2}$ .

Therefore the total surface area is $216\ in^{2}+31.176\ in^{2}=247.176\ in^{2}$ .

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High School Math : Non-Cubic Prisms

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Non Cubic Prisms

Example Question #1 : Prisms

Example Question #11 : Prisms

Example Question #3 : Non Cubic Prisms

Example Question #1 : How To Find The Surface Area Of A Prism

Example Question #3 : Non Cubic Prisms

Example Question #691 : Geometry

Example Question #2 : Non Cubic Prisms

Example Question #3 : Non Cubic Prisms

Example Question #12 : How To Find Surface Area

All High School Math Resources

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