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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

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All Common Core: 7th Grade Math Resources

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Example Questions

Example Question #332 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Give the volume of a cube with surface area 240 square inches.

Possible Answers:

$80 \textup{ in}^{3}$ $\displaystyle 80 \textup{ in}^{3}$

$120 \sqrt{10} \textup{ in}^{3}$ $\displaystyle 120 \sqrt{10} \textup{ in}^{3}$

$40 \sqrt{10} \textup{ in}^{3}$ $\displaystyle 40 \sqrt{10} \textup{ in}^{3}$

$120 \textup{ in}^{3}$ $\displaystyle 120 \textup{ in}^{3}$

$80 \sqrt{10} \textup{ in}^{3}$ $\displaystyle 80 \sqrt{10} \textup{ in}^{3}$

Correct answer:

$80 \sqrt{10} \textup{ in}^{3}$ $\displaystyle 80 \sqrt{10} \textup{ in}^{3}$

Explanation:

Let $\displaystyle s$ be the length of one edge of the cube. Since its surface area is 240 square inches, one face has one-sixth of this area, or $\frac{240}{6} = 40$ $\displaystyle \frac{240}{6} = 40$ square inches. Therefore, $s^{2} = 40$ $\displaystyle s^{2} = 40$, and $s = \sqrt{40} = 2\sqrt{10}$ $\displaystyle s = \sqrt{40} = 2\sqrt{10}$.

The volume is the cube of this, or $V= \left ( 2\sqrt{10} \right )^{3}= 2^{3} \cdot \left ( \sqrt{10} \right )^{3}= 8 \cdot 10 \sqrt{10}= 80 \sqrt{10}$ $\displaystyle V= \left ( 2\sqrt{10} \right )^{3}= 2^{3} \cdot \left ( \sqrt{10} \right )^{3}= 8 \cdot 10 \sqrt{10}= 80 \sqrt{10}$ cubic inches.

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Example Question #1 : Finding Volume Of A Cube

Calculate the volume of the provided figure.

Possible Answers:

$342\textup{ in}^3$ $\displaystyle 342\textup{ in}^3$

$345\textup{ in}^3$ $\displaystyle 345\textup{ in}^3$

$344\textup{ in}^3$ $\displaystyle 344\textup{ in}^3$

$343\textup{ in}^3$ $\displaystyle 343\textup{ in}^3$

$341\textup{ in}^3$ $\displaystyle 341\textup{ in}^3$

Correct answer:

$343\textup{ in}^3$ $\displaystyle 343\textup{ in}^3$

Explanation:

In order to solve this problem, we need to recall the volume formula for a cube:

$V=l\times w\times h$ $\displaystyle V=l\times w\times h$

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

$V=7\times7\times7$ $\displaystyle V=7\times7\times7$

$V=343\textup{ in}^3$ $\displaystyle V=343\textup{ in}^3$

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Example Question #31 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

Calculate the volume of the provided figure.

Possible Answers:

$214\textup{ in}^3$ $\displaystyle 214\textup{ in}^3$

$212\textup{ in}^3$ $\displaystyle 212\textup{ in}^3$

$215\textup{ in}^3$ $\displaystyle 215\textup{ in}^3$

$213\textup{ in}^3$ $\displaystyle 213\textup{ in}^3$

$216\textup{ in}^3$ $\displaystyle 216\textup{ in}^3$

Correct answer:

$216\textup{ in}^3$ $\displaystyle 216\textup{ in}^3$

Explanation:

In order to solve this problem, we need to recall the volume formula for a cube:

$V=l\times w\times h$ $\displaystyle V=l\times w\times h$

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

$V=6\times6\times6$ $\displaystyle V=6\times6\times6$

$V=216\textup{ in}^3$ $\displaystyle V=216\textup{ in}^3$

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Example Question #31 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

Calculate the volume of the provided figure.

Possible Answers:

$1\textup,335\textup{ in}^3$ $\displaystyle 1\textup,335\textup{ in}^3$

$1\textup,334\textup{ in}^3$ $\displaystyle 1\textup,334\textup{ in}^3$

$1\textup,332\textup{ in}^3$ $\displaystyle 1\textup,332\textup{ in}^3$

$1\textup,333\textup{ in}^3$ $\displaystyle 1\textup,333\textup{ in}^3$

$1\textup,331\textup{ in}^3$ $\displaystyle 1\textup,331\textup{ in}^3$

Correct answer:

$1\textup,331\textup{ in}^3$ $\displaystyle 1\textup,331\textup{ in}^3$

Explanation:

In order to solve this problem, we need to recall the volume formula for a cube:

$V=l\times w\times h$ $\displaystyle V=l\times w\times h$

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

$V=11\times11\times11$ $\displaystyle V=11\times11\times11$

$V=1\textup,331\textup{ in}^3$ $\displaystyle V=1\textup,331\textup{ in}^3$

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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 #332 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #1 : Finding Volume Of A Cube

Example Question #31 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

Example Question #31 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6

All Common Core: 7th Grade Math Resources

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