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Common Core: 7th Grade Math : Know and Use the Formulas for the Area and Circumference of a Circle: CCSS.Math.Content.7.G.B.4

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Example Questions

Example Question #2 : How To Find Circumference

The area of a circle is 25t^2 . Give the circumference of the circle in terms of .

Let $\pi=3.14$ .

Possible Answers:

20t

17.71 t^2

17.71t

20t^2

15t

Correct answer:

17.71t

Explanation:

The area of a circle can be calculated as $Area=\pi r^2$ , where is the radius of the circle.

$Area=\pi r^2\Rightarrow 25t^2=\pi r^2\Rightarrow r^2=\frac{25t^2}{\pi}$

$\Rightarrow r=\frac{5t}{\sqrt{\pi}}=\frac{5t}{\sqrt{3.14}}\approx 2.82t$

The circumference can be calculated as $Circumference =2\pi r$ , where is the radius of the circle.

$Circumference =2\pi r\Rightarrow Circumference =2\times 3.14\times 2.82t=17.71t$

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Example Question #3 : How To Find Circumference

The perimeter of a given rectangle is equal to the circumference of a given circle. The circle has radius inches; the width of the rectangle is $8 \pi$ inches. What is the length of the rectangle?

Possible Answers:

$7\pi$ inches

$(30 - 8 \pi )$ inches

$44 \pi$ inches

$22 \pi$ inches

$14 \pi$ inches

Correct answer:

$22 \pi$ inches

Explanation:

The circumference of a circle with radius inches is

$C = 2 \pi r = 2 \pi \cdot 30 = 60 \pi$ inches.

The perimeter of the rectangle is therefore $60 \pi$ inches. To find its length, substitute $W = 8\pi$ and $P = 60\pi$ into the equation and solve for :

2 L + 2W = P

$2 L + 2 \cdot 8 \pi = 60 \pi$

$2 L + 16 \pi = 60 \pi$

$2 L + 16 \pi - 16 \pi = 60 \pi - 16 \pi$

$2 L =44 \pi$

$2 L \div 2=44 \pi \div 2$

$L = 22\pi$ inches

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Example Question #78 : Geometry

A manufacturer makes wooden circles out of square blocks of wood. If the wood costs $0.25 per square inch, what is the minimum waste cost possible for cutting a circle with a radius of 44 in.?

Possible Answers:

5808 dollars

1936π dollars

1936 – 484π dollars

7744 – 1936π dollars

1936 dollars

Correct answer:

1936 – 484π dollars

Explanation:

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 88 inches for our problem. Its total area would be 88 * 88 or 7744 in².

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr² or π * 44² = 1936π in². Therefore, the area remaining would be 7744 – 1936π. The cost of the waste would be 0.25 * (7744 – 1936π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 4 from our subtraction. This would give us: 0.25 * 4 * (1936 – 484π). Since 0.25 is equal to 1/4, 0.25 * 4 = 1. Therefore, our final answer is: 1936 – 484π dollars.

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Example Question #1 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

A manufacturer makes wooden circles out of square blocks of wood. If the wood costs $0.20 per square inch, what is the minimum waste cost possible for cutting a circle with a radius of 25 in.?

Possible Answers:

500 - 125π dollars

625 dollars

500 dollars

2500 - 625π dollars

625 - 25π dollars

Correct answer:

500 - 125π dollars

Explanation:

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 50 inches for our problem. Its total area would be 50 * 50 or 2500 in².

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr² or π * 25² = 625π in². Therefore, the area remaining would be 2500 - 625π. The cost of the waste would be 0.2 * (2500 – 625π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 5 from our subtraction. This would give us: 0.2 * 5 * (500 – 125π). Since 0.2 is equal to 1/5, 0.2 * 625 = 125. Therefore, our final answer is: 500 – 125π dollars.

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Example Question #1 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

What is the circumference of the circle provided?

Possible Answers:

$62\pi\textup{ in}$

$61\pi\textup{ in}$

$64\pi\textup{ in}$

$60\pi\textup{ in}$

$63\pi\textup{ in}$

Correct answer:

$64\pi\textup{ in}$

Explanation:

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

$C=2r\pi$ or $C=d\pi$

The circle in this question provides us with the radius, so we can use the first formula to solve:

$C=2(32)\pi$

Solve:

$C=64\pi\textup{ in}$

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Example Question #1 : Area Of A Circle

What is the area of a circle that has a diameter of inches?

Possible Answers:

706.8583

153.938

176.7146

940

153.938

940

960

Correct answer:

176.7146

Explanation:

The formula for finding the area of a circle is $\pi r^{2}$ . In this formula, represents the radius of the circle. Since the question only gives us the measurement of the diameter of the circle, we must calculate the radius. In order to do this, we divide the diameter by .

$\frac{15}{2}=7.5$

Now we use 7.5 for in our equation.

$\pi (7.5)^{2}=176.7146 \: in^{2}$

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Example Question #122 : Area

What is the area of a circle with a diameter equal to 6?

Possible Answers:

$18\pi$

$3\pi$

$9\pi$

$36\pi$

Correct answer:

$9\pi$

Explanation:

First, solve for radius:

$r=\frac{d}{2}=\frac{6}{2}=3$

Then, solve for area:

$A=r^2\pi=3^2\pi=9\pi$

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Example Question #3 : Radius

The diameter of a circle is $4\ cm$ . Give the area of the circle.

Possible Answers:

$12.56\ cm^2$

$12 \ cm^2$

$11.56\ cm^2$

$13.56\ cm^2$

$13\ cm^2$

Correct answer:

$12.56\ cm^2$

Explanation:

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle, and $\pi$ is approximately 3.14 .

$Area=\frac{\pi d^2}{4}=\frac{\pi\times 4^2}{4}=4\pi \Rightarrow Area\approx 4\times 3.14\Rightarrow Area\approx 12.56 \ cm^2$

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Example Question #4 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

The diameter of a circle is . Give the area of the circle in terms of .

Possible Answers:

11.56 t

12.56 t^2

11.56 t^2

12.56 t

12 t^2

Correct answer:

12.56 t^2

Explanation:

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle and $\pi$ is approximately 3.14 .

$Area=\frac{\pi (4t)^2}{4}=\frac{16\pi t^2}{4}=4\pi t^2 \Rightarrow Area\approx 4\times 3.14\times t^2$

$\Rightarrow Area\approx 12.56t^2$

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Example Question #1 : Area Of A Circle

The radius of a circle is $\frac{2}{\sqrt{\pi }}$ . Give the area of the circle.

Possible Answers:

$2\pi$

$4\pi$

$\frac{4}{\pi }$

Correct answer:

Explanation:

The area of a circle can be calculated as $Area=\pi r^2$ , where is the radius of the circle, and $\pi$ is approximately 3.14 .

$Area=\pi r^2=\pi\times (\frac{2}{\sqrt{\pi}})^2=\pi\times \frac{4}{\pi}\Rightarrow Area=4$

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Common Core: 7th Grade Math : Know and Use the Formulas for the Area and Circumference of a Circle: CCSS.Math.Content.7.G.B.4

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

All Common Core: 7th Grade Math Resources

Example Questions

Example Question #2 : How To Find Circumference

Example Question #3 : How To Find Circumference

Example Question #78 : Geometry

Example Question #1 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

Example Question #1 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

Example Question #1 : Area Of A Circle

Example Question #122 : Area

Example Question #3 : Radius

Example Question #4 : Know And Use The Formulas For The Area And Circumference Of A Circle: Ccss.Math.Content.7.G.B.4

Example Question #1 : Area Of A Circle

All Common Core: 7th Grade Math Resources

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