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Common Core: 7th Grade Math : Solve Problems Involving Area, Volume and Surface Area of Two- and Three-Dimensional Objects: CCSS.Math.Content.7.G.B.6

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Example Question #1 : Solve For Surface Area

The length of the side of a cube is x -5 . Give its surface area in terms of .

Possible Answers:

x^3-15 x^2+75 x-125

x^3-15 x^2-75 x+125

$6x^{2} -10x + 25$

$6x^{2} -60x + 150$

$6x^{2} -60x - 150$

Correct answer:

$6x^{2} -60x + 150$

Explanation:

Substitute s = x - 5 in the formula for the surface area of a cube:

$A = 6s^{2}$

$A = 6 \left ( x - 5\right )^{2}$

$A = 6 \left ( x^{2} - 2 \cdot 5 \cdot x + 5^{2}\right )$

$A = 6 \left ( x^{2} - 10x +25\right )$

$A = 6 x^{2} - 60x +150$

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Example Question #1 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

If a cube has one side measuring cm, what is the surface area of the cube?

Possible Answers:

Correct answer:

Explanation:

To find the surface area of a cube, use the formula $6s^{2}$ , where represents the length of the side. Since the side of the cube measures , we can substitute in for .

$6(4)^{2}=96\: cm^{2}$

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Example Question #3 : Solve For Surface Area

Find the surface area of a non-cubic prism with the following measurements:

l=7;w=6;h=2

Possible Answers:

168

136

Correct answer:

136

Explanation:

The surface area of a non-cubic prism can be determined using the equation:

SA=2lw+2wh+2lh

SA=2(7)(6)+2(6)(2)+2(7)(2)=84+24+28=136

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Example Question #71 : Geometry

A small rectangular jewelry box has two square ends with areas of 36 square centimeters, and a width of 10 centimeters. What is the surface area of the outside of the jewelry box.

Possible Answers:

$72 \; cm^2$

$240 \; cm^2$

$312 \; cm^2$

$216 \; cm^2$

$360 \; cm^2$

Correct answer:

$312 \; cm^2$

Explanation:

To find the surface area of the rectangular box we just need to add up the areas of all six sides. We know that two of the sides are 36 square centimeters, that means we need to find the areas of the four mising sides. To find the area of the missing sides we can just multiply the side of one of the squares (6 cm) by the width of the box:

$6 \times 10 = 60 \; cm^2$

But remember we have four of these rectangular sides:

$4 \times 60 = 240 \; cm^2$

Now we just add the two square sides and four rectangular sides to find the total surface area of the jewelry box:

$36 \; cm^2 + 36 \; cm^2 + 240 \; cm^2 = 312 \; cm^2$

That is the total surface area!

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

Alice is wrapping a rectangular box that measures $5\textup {in}\times 6\textup {in}\times 12\textup {in}$ . How many square feet of wrapping paper does she need?

Possible Answers:

$2.75 \textup{ ft}^{2}$

$3.25\textup{ ft}^{2}$

$2.25\textup{ ft}^2$

$1.75\textup{ ft}^{2}$

$2.00 \textup{ ft}^{2}$

Correct answer:

$2.25\textup{ ft}^2$

Explanation:

The surface area of a rectangular prism is given by

$\textup {SA=2lw+2lh+2wh}$ where is the length, is the width, and is the height.

Let $\textup l=5\textup { in}$ , $\textup {w}=6\textup{ in}$ , and $\textup {h}=12\textup { in}$

So the equation to solve becomes $\textup SA=2\cdot 5\cdot 6+2\cdot 5\cdot 12+2\cdot 6\cdot 12$ or $324\textup{ in}^{2}$

However the question asks for an answer in square feet. Knowing that $144\textup { in}^{2}=1\textup { ft}^{2}$ we can convert square inches to square feet. It will take $2.25\textup{ ft}^{2}$ of paper to wrap the present.

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Example Question #1 : Solve For Area

Calculate the area of the provided figure.

Possible Answers:

$141\pi\textup{ in}^2$

$140\pi\textup{ in}^2$

$144\pi\textup{ in}^2$

$143\pi\textup{ in}^2$

$142\pi\textup{ in}^2$

Correct answer:

$144\pi\textup{ in}^2$

Explanation:

In order to solve this problem, we need to recall the area formula for a circle:

$A=r^2\pi$

Now that we have the correct formula, we can substitute in our known values and solve:

$A=12^2\pi$

$A=144\pi\textup{ in}^2$

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Example Question #2 : Solve For Area

Calculate the area of the provided figure.

Possible Answers:

$364\pi\textup{ in}^2$

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

$361\pi\textup{ in}^2$

$362\pi\textup{ in}^2$

$363\pi\textup{ in}^2$

Correct answer:

$361\pi\textup{ in}^2$

Explanation:

In order to solve this problem, we need to recall the area formula for a circle:

$A=r^2\pi$

Now that we have the correct formula, we can substitute in our known values and solve:

$A=19^2\pi$

$A=361\pi\textup{ in}^2$

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Example Question #1 : Solve For Area Of A Rectangle

Calculate the area of the provided figure.

Possible Answers:

$120\textup{ in}^2$

$100\textup{ in}^2$

$110\textup{ in}^2$

$112\textup{ in}^2$

$102\textup{ in}^2$

Correct answer:

$120\textup{ in}^2$

Explanation:

In order to solve this problem, we need to recall the area formula for a rectangle:

$A=l\times w$

Now that we have the correct formula, we can substitute in our known values and solve:

$A=12\times10$

$A=120\textup{ in}^2$

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Example Question #2 : Solve For Area Of A Rectangle

Calculate the area of the provided figure.

Possible Answers:

$40\textup{ in}^2$

$42\textup{ in}^2$

$44\textup{ in}^2$

$43\textup{ in}^2$

$41\textup{ in}^2$

Correct answer:

$44\textup{ in}^2$

Explanation:

In order to solve this problem, we need to recall the area formula for a rectangle:

$A=l\times w$

Now that we have the correct formula, we can substitute in our known values and solve:

$A=11\times4$

$A=44\textup{ in}^2$

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Example Question #1 : Solve For Area Of A Rectangle

Abcdrectangle

In rectangle ABCD, the perimeter is 48 and side BC measures 4. What is the area of rectangle ABCD?

Possible Answers:

Correct answer:

Explanation:

The perimeter of a rectangle is 2L + 2W, or 2 times the length plus 2 times the width. Here you're given that side BC is 4, which means that the opposite side, AD, is also 4. So since that is two widths, you now have:

8 + 2L = 48

So 2L = 40

That means that the length is 20.

Since the area is LW, you can calculate the area as 20 * 4 = 80.

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Common Core: 7th Grade Math : Solve Problems Involving Area, Volume and Surface Area of Two- and Three-Dimensional Objects: CCSS.Math.Content.7.G.B.6

Study concepts, example questions & explanations for Common Core: 7th Grade Math

All Common Core: 7th Grade Math Resources

Example Questions

Example Question #1 : Solve For Surface Area

Example Question #1 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

Example Question #3 : Solve For Surface Area

Example Question #71 : Geometry

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

Example Question #1 : Solve For Area

Example Question #2 : Solve For Area

Example Question #1 : Solve For Area Of A Rectangle

Example Question #2 : Solve For Area Of A Rectangle

Example Question #1 : Solve For Area Of A Rectangle

All Common Core: 7th Grade Math Resources

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