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Common Core: High School - Geometry : Circles

Study concepts, example questions & explanations for Common Core: High School - Geometry

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Example Questions

Example Question #71 : Circles

Calculate the arc length of a circle that has a central angle of $\displaystyle 93$ degrees and an of 100 $\displaystyle 100$ .

Refer to the following figure to help calculate the solution.

Circlearc9 10

Possible Answers:

s=9.185 $\displaystyle s=9.185$

s=9.851 $\displaystyle s=9.851$

s=9.158 $\displaystyle s=9.158$

s=9.518 $\displaystyle s=9.518$

s=9.150 $\displaystyle s=9.150$

Correct answer:

s=9.158 $\displaystyle s=9.158$

Explanation:

To calculate the arc length of a circle that has a central angle of $\displaystyle 93$ degrees and a circumference of 100 $\displaystyle 100$ , refer to the figure and the algebraic formula for arc length.

Circlearc9 10

The algebraic formula for arc length is as follows.

$s=\frac{\theta}{360^\circ}\times C$ $\displaystyle s=\frac{\theta}{360^\circ}\times C$

where,

$\\s=\text{Arc Length} \\C=\text{Circumference} \\\theta=\text{Central Angle}\\A=Area$ $\displaystyle \\s=\text{Arc Length} \\C=\text{Circumference} \\\theta=\text{Central Angle}\\A=Area$

For this particular question the known information is,

$\\s=\text{Arc Length} \\A=100 \\\theta=93^\circ\\A=\pi r^2\\C=2\pi r$ $\displaystyle \\s=\text{Arc Length} \\A=100 \\\theta=93^\circ\\A=\pi r^2\\C=2\pi r$

First, calculate the radius and the circumference.

$\\100=\pi r^2 \\\\\frac{100}{\pi}=r^2 \\\\\sqrt{\frac{100}{\pi}}=r$ $\displaystyle \\100=\pi r^2 \\\\\frac{100}{\pi}=r^2 \\\\\sqrt{\frac{100}{\pi}}=r$

$C=2\pi\sqrt{\frac{100}{\pi}}$ $\displaystyle C=2\pi\sqrt{\frac{100}{\pi}}$

Substitute these values into the formula and solve for the arc length.

$s=\frac{\theta}{360^\circ}\times C$ $\displaystyle s=\frac{\theta}{360^\circ}\times C$

$s=\frac{93^\circ}{360^\circ}\times 2\pi\sqrt{\frac{100}{\pi}}$ $\displaystyle s=\frac{93^\circ}{360^\circ}\times 2\pi\sqrt{\frac{100}{\pi}}$

s=9.158 $\displaystyle s=9.158$

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Example Question #23 : Arc Length, Radii, Radian, And Sector Similarity And Proportionality: Ccss.Math.Content.Hsg C.B.5

Calculate the arc length of a circle that has a central angle of $\displaystyle 58$ degrees and an of $\displaystyle 15$ .

Refer to the following figure to help calculate the solution.

Circlearc11 12

Possible Answers:

s=2.212 $\displaystyle s=2.212$

s=2.122 $\displaystyle s=2.122$

s=2.207 $\displaystyle s=2.207$

s=2.218 $\displaystyle s=2.218$

s=2.221 $\displaystyle s=2.221$

Correct answer:

s=2.212 $\displaystyle s=2.212$

Explanation:

To calculate the arc length of a circle that has a central angle of $\displaystyle 58$ degrees and a circumference of $\displaystyle 15$ , refer to the figure and the algebraic formula for arc length.

Circlearc11 12

The algebraic formula for arc length is as follows.

$s=\frac{\theta}{360^\circ}\times C$ $\displaystyle s=\frac{\theta}{360^\circ}\times C$

where,

$\\s=\text{Arc Length} \\C=\text{Circumference} \\\theta=\text{Central Angle}\\A=Area$ $\displaystyle \\s=\text{Arc Length} \\C=\text{Circumference} \\\theta=\text{Central Angle}\\A=Area$

For this particular question the known information is,

$\\s=\text{Arc Length} \\A=15 \\\theta=58^\circ\\A=\pi r^2\\C=2\pi r$ $\displaystyle \\s=\text{Arc Length} \\A=15 \\\theta=58^\circ\\A=\pi r^2\\C=2\pi r$

First, calculate the radius and the circumference.

$\\15=\pi r^2 \\\\\frac{15}{\pi}=r^2 \\\\\sqrt{\frac{15}{\pi}}=r$ $\displaystyle \\15=\pi r^2 \\\\\frac{15}{\pi}=r^2 \\\\\sqrt{\frac{15}{\pi}}=r$

$C=2\pi\sqrt{\frac{15}{\pi}}$ $\displaystyle C=2\pi\sqrt{\frac{15}{\pi}}$

Substitute these values into the formula and solve for the arc length.

$s=\frac{\theta}{360^\circ}\times C$ $\displaystyle s=\frac{\theta}{360^\circ}\times C$

$s=\frac{58^\circ}{360^\circ}\times 2\pi\sqrt{\frac{15}{\pi}}$ $\displaystyle s=\frac{58^\circ}{360^\circ}\times 2\pi\sqrt{\frac{15}{\pi}}$

s=2.212 $\displaystyle s=2.212$

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Example Question #71 : Circles

Determine the area of the sector of a circle that has a central angle of $\displaystyle 58$ degrees and an area of $\displaystyle 15$ .

Refer to the following figure to help calculate the solution.

Circlearc11 12

Possible Answers:

$\text{sector area}= 2.61\overline{4}$ $\displaystyle \text{sector area}= 2.61\overline{4}$

$\text{sector area}= 2.46\overline{1}$ $\displaystyle \text{sector area}= 2.46\overline{1}$

$\text{sector area}= 2.41\overline{6}$ $\displaystyle \text{sector area}= 2.41\overline{6}$

$\text{sector area}= 2.14\overline{6}$ $\displaystyle \text{sector area}= 2.14\overline{6}$

$\text{sector area}= 2.41\overline{9}$ $\displaystyle \text{sector area}= 2.41\overline{9}$

Correct answer:

$\text{sector area}= 2.41\overline{6}$ $\displaystyle \text{sector area}= 2.41\overline{6}$

Explanation:

State the known information,

$\\\text{Circle Area}= 15 \\\text{Central Angle}=58^\circ$ $\displaystyle \\\text{Circle Area}= 15 \\\text{Central Angle}=58^\circ$

Since the question is asking for the area of the sector, a ratio will need to be constructed. Recall that a circle is composed of 360 degrees. Therefore, the following ratio can be made,

$\\\frac{\theta}{360^\circ}=\frac{\text{sector area}}{15} \\\\\frac{58^\circ}{360^\circ}=\frac{\text{sector area}}{15} \\\\\frac{29}{180}=\frac{\text{sector area}}{15} \\\\\frac{29}{180}\times 15=\text{sector area} \\\\\text{sector area}=2.41\overline{6}$ $\displaystyle \\\frac{\theta}{360^\circ}=\frac{\text{sector area}}{15} \\\\\frac{58^\circ}{360^\circ}=\frac{\text{sector area}}{15} \\\\\frac{29}{180}=\frac{\text{sector area}}{15} \\\\\frac{29}{180}\times 15=\text{sector area} \\\\\text{sector area}=2.41\overline{6}$

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Common Core: High School - Geometry : Circles

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #71 : Circles

Example Question #23 : Arc Length, Radii, Radian, And Sector Similarity And Proportionality: Ccss.Math.Content.Hsg C.B.5

Example Question #71 : Circles

All Common Core: High School - Geometry Resources

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