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Common Core: 7th Grade Math : Geometry

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

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All Common Core: 7th Grade Math Resources

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Example Questions

Example Question #71 : Geometry

Solve for :

Possible Answers:

$143^\circ$

$154^\circ$

$43^\circ$

$123^\circ$

$57^\circ$

Correct answer:

$123^\circ$

Explanation:

A pair of intersecting angles is illustrated by the provided image.

$123^\circ$ and are vertical angles because they share the same vertex, or corner spot. Vertical angles are always equal; thus, $x=123^\circ$

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Example Question #72 : Geometry

Solve for :

Possible Answers:

$142^\circ$

$104^\circ$

$120^\circ$

$14^\circ$

$76^\circ$

Correct answer:

$104^\circ$

Explanation:

A pair of intersecting angles is illustrated by the provided image.

$104^\circ$ and are vertical angles because they share the same vertex, or corner spot. Vertical angles are always equal; thus, $x=104^\circ$

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Example Question #73 : Geometry

Calculate the area of the provided figure.

Possible Answers:

$31.5\textup{ in}^2$

$34\textup{ in}^2$

$36.5\textup{ in}^2$

$32\textup{ in}^2$

$33.5\textup{ in}^2$

Correct answer:

$31.5\textup{ in}^2$

Explanation:

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

$A=\frac{b\times h}{2}$

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

$A=\frac{7\times9}{2}$

$A=31.5\textup{ in}^2$

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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

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

Possible Answers:

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

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

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

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

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

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 #2 : 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 #74 : Geometry

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 #1 : Solve For Surface Area

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:

$312 \; cm^2$

$360 \; cm^2$

$72 \; cm^2$

$216 \; cm^2$

$240 \; cm^2$

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 #75 : Geometry

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:

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

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

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

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

$2.25\textup{ ft}^2$

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:

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

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

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

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

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

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:

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

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

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

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

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

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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Common Core: 7th Grade Math : Geometry

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

All Common Core: 7th Grade Math Resources

Example Questions

Example Question #71 : Geometry

Example Question #72 : Geometry

Example Question #73 : Geometry

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

Example Question #74 : Geometry

Example Question #1 : Solve For Surface Area

Example Question #75 : Geometry

Example Question #1 : Solve For Area

Example Question #2 : Solve For Area

All Common Core: 7th Grade Math Resources

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