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High School Math : High School Math

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Example Questions

Example Question #3 : Pyramids

What is the surface are of a pyramid with a square base length of 15 and a slant height (the height from the midpoint of one of the side lengths to the top of the pyramid) of 12?

Possible Answers:

555

565

575

585

Correct answer:

585

Explanation:

To find the surface area of a pyramid we must add the areas of all five of the shapes creating the pyramid together.

We have four triangles that all have the same area and a square that supports the pyramid.

To find the area of the square we take the side length of 15 and square it $15^{2}=225$

The area of the square is 225 .

To find the area of the triangle we must use the equation for the area of a triangle which is $\frac{base\times height}{2}$

Plug in the slant height 12 as the height of the triangle and use the side length of the square 15 as the base in our equation to get

$\frac{1}{2}*15*12=90$

The area of each triangle is .

We then multiply the area of each triangle by 4 to find the area of all four triangles 4*90=360 .

The four triangles have a surface area of 360 .

We add the surface area of the four triangles with the area of the square to get the answer for the surface area of the pyramid which is 360+225=585 .

The answer is 585 .

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Example Question #75 : Solid Geometry

Find the surface area of the following pyramid.

Pyramid

Possible Answers:

330m^2

320m^2

340m^2

360m^2

350m^2

Correct answer:

340m^2

Explanation:

The formula for the surface area of a pyramid is:

$SA = 4 \left(A_{triangle}\right)+Base$

$SA = 4 \left(\frac{1}{2}{h_{s}}w\right)+(lw)$

Where h_s is the length of the slant height, is the width of the base, and is the length of the base

In order to determine the areas of the triangle, you will need to use the Pythagorean Theorem to find the slant height:

A^2 + B^2 = C^2

A^2 + (5m)^2 = (13m)^2

A^2 = 144m^2

A = 12m

Plugging in our values, we get:

$SA = 4 \left(\frac{1}{2}\cdot (12m)(10m)\right)+((10m)(10m))$

SA = 240m^2+100m^2 = 340m^2

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Example Question #75 : Solid Geometry

Find the surface area of the following pyramid.

Pyramid

Possible Answers:

326m^2

316m^2

356m^2

336m^2

346m^2

Correct answer:

336m^2

Explanation:

The formula for the surface area of a pyramid is:

$SA = (base)+\frac{1}{2}(perimeter)(slant height)$

$SA = (l)(w) + \frac{1}{2}(l+l+w+w)(h_s)$

Where is the length of the base, is the width of the base, and h_s is the slant height

Use the Pythagorean Theorem to find the length of the slant height:

A^2 + B^2 = C^2

A^2 + (6m)^2 = (10m)^2

A = 8m

Plugging in our values, we get:

$SA = (12m)(12m) + \frac{1}{2}(12m+12m+12m+12m)(8m)$

SA = 336m^2

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Example Question #71 : Solid Geometry

What is the surface area of a square pyramid with a base side equal to 4 and a slant length equal to 6?

Possible Answers:

Correct answer:

Explanation:

The surface area of of a square pyramid can be determined using the following equation:

SA=2bs+b^2

SA=(2)(6)(4)+4^2

SA=48+16=64

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Example Question #1931 : High School Math

What is the surface area of a square pyramid with a height of 12 in and a base side length of 10 in?

Possible Answers:

$260 \textup{ in}^{2}$

$100 \textup{ in}^{2}$

$360 \textup{ in}^{2}$

$150 \textup{ in}^{2}$

$420 \textup{ in}^{2}$

Correct answer:

$360 \textup{ in}^{2}$

Explanation:

The surface area of a square pyramid can be broken into the area of the square base and the areas of the four triangluar sides. The area of a square is given by:

$A = s^{2} = 100 in^{2}$

The area of a triangle is:

$A = \frac{1}{2} bh$

The given height of 12 in is from the vertex to the center of the base. We need to calculate the slant height of the triangular face by using the Pythagorean Theorem:

$SH^{2} = H^{2} + B^{2}$

where H=12 and B= 5 (half the base side) resulting in a slant height of 13 in.

So, the area of the triangle is:

$A = \frac{1}{2} bh = \frac{1}{2} \cdot 10 \cdot 13 = 65 in^{2}$

There are four triangular sides totaling $260 in^{2}$ for the sides.

The total surface area is thus $360 in^{2}$ , including all four sides and the base.

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Example Question #1 : 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 = 3/W

L = W + 3

L = 3W

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 #2 : 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=5L-3

W=3L-5

W=5-3L

W=3-5L

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 #1932 : High School Math

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:

$100\hspace{1 mm}mL$

It is impossible to convert between metric units and feet.

None of the available answers

$600\hspace{1 mm}mL$

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

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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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Pyramids

Example Question #75 : Solid Geometry

Example Question #75 : Solid Geometry

Example Question #71 : Solid Geometry

Example Question #1931 : High School Math

Example Question #1 : Prisms

Example Question #2 : Prisms

Example Question #1932 : High School Math

Example Question #3 : Non Cubic Prisms

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

All High School Math Resources

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