Sectors - ISEE Upper Level Mathematics Achievement

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Question

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Answer

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Use the following formula and solve for x:

$\frac{x}{360^{\circ}}100=43%$

Begin by dividing over the 100

$\frac{x}{360^{\circ}}=.43$

Then multiply by 360

$x=.43*360^{\circ}=154.8^{\circ}$

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Question

If sector AJL covers 45% of circle J, what is the measure of sector AJL's central angle?

Answer

If sector AJL covers 45% of circle J, what is the measure of sector AJL's central angle?

To find an angle measure from a percentage, simply convert the percentage to a decimal and then multiply it by 360 degrees.

$45%\rightarrow 0.45$

$0.45 * 360^{\circ}=162^{\circ}$

So, our answer is 162 degrees.

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Question

A giant clock has a minute hand four feet long. Since noon, the tip of the minute hand has traveled $10 \pi$ feet. What time is it now?

Answer

The circumference of the path traveled by the tip of the minute hand over the course of one hour is:

$C = 2 \pi r = 2 \pi \cdot 4 = 8 \pi$ feet.

Since the tip of the minute hand has traveled $10 \pi$ feet since noon, the minute hand has made

$\frac{10 \pi }{8 \pi } = \frac{5}{4} = 1 \frac{1}{4}$ revolutions. Therefore, $1 \frac{1}{4}$ hours have elapsed since noon, making the time 1:15 PM.

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Question

Note: Figure NOT drawn to scale

Refer to the above diagram. $\overarc {ABC}$ is a semicircle. Evaluate $m \overarc {AB}$ .

Answer

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\angle B$ , which intercepts the semicircle $\overarc{ABC}$ , is such an angle. Consequently,

$m \angle B = 90 ^{\circ }$

$4x \div 4 = 108 \div 4$

$m \angle C = (x+10)^{\circ } = (27+10)^{\circ }= 37^{\circ }$

Inscribed $\angle C$ intercepts an arc with twice its angle measure; this arc is $\overarc{AB}$ , so

$m \widehat{AB} = 2 \cdot m \angle C = 2 \cdot 37 ^{\circ } = 74 ^{\circ }$ .

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Question

Figure NOT drawn to scale

Refer to the above diagram. $\overarc {ABC}$ is a semicircle. Evaluate $m \overarc {BCA}$ given $\measuredangle B=90^\circ$ .

Answer

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\angle B$ , which intercepts the semicircle $\overarc {ABC}$ , is such an angle. Consequently, $\bigtriangleup ABC$ is a right triangle, and $\angle A$ and $\angle C$ are complementary angles. Therefore,

$m \angle A + m \angle C = 90 ^{\circ }$

$y + \left ( \frac{y}{3}+2\right ) = 90$

$y + \frac{1}{3}y +2 = 90$

$\frac{4}{3}y +2 = 90$

$\frac{4}{3}y +2 - 2 = 90 - 2$

$\frac{4}{3}y = 88$

$\frac{3}{4} \cdot \frac{4}{3}y = \frac{3}{4} \cdot 88$

$m \angle C = \left (\frac{y}{3}+ 2 \right ) ^{\circ }=\left ( \frac{66}{3}+ 2 \right )^{\circ }=( 22 + 2 )^{\circ }= 24^{\circ }$

Inscribed $\angle C$ intercepts an arc with twice its angle measure; this arc is $\overarc {AB}$ , so

$m \overarc {AB} = 2 \cdot m \angle C = 2 \cdot 24 ^{\circ } = 48 ^{\circ }$ .

The major arc corresponding to this minor arc, $\overarc {BCA}$ , has measure

$m \overarc {BCA} = 360 ^{\circ } - m \overarc {AB} = 360 ^{\circ } - 48^{\circ } = 312^{\circ }$

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Question

In the above diagram, radius .

Calculate the length of $\overarc {AB}$ .

Answer

Inscribed $\angle ACB$ , which measures $72 ^{\circ }$ , intercepts an arc with twice its measure. That arc is $\overarc {AB}$ , which consequently has measure

$72 ^{\circ } \times 2 = 144 ^{\circ }$ .

This makes $\overarc {AB}$ an arc which comprises

$\frac{144}{360} = \frac{144 \div 72}{360 \div 72} = \frac{2}{5}$

of the circle.

The circumference of a circle is $2 \pi$ multiplied by its radius, so

$C = 2 \pi r = 2\pi \cdot 20 = 40 \pi$ .

The length of $\overarc {AB}$ is $\frac{2}{5}$ of this, or $\frac{2}{5} \cdot 40 \pi = 16 \pi$ .

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Question

Figure NOT drawn to scale.

The circumference of the above circle is 100. $\overarc{AB }$ and $\overarc{CD}$ have lengths 20 and 15, respectively. Evaluate .

Answer

The length of $\overarc{AB }$ comprises $\frac{20}{100} = \frac{1}{5}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{5} \cdot 360 ^{\circ } = 72 ^{\circ }$ . Similarly, The length of $\overarc{CD}$ comprises $\frac{15}{100} = \frac{3}{20}$ of the circumference of the circle. Therefore, its degree measure is $\frac{3}{20} \cdot 360 ^{\circ } = 54 ^{\circ }$ .

If two chords cut each other inside the circle, as $\overline{AC }$ and $\overline{BD}$ do, and one pair of vertical angles are examined, then the degree measure of each angle is half the sum of those of the arcs intercepted - that is,

$t = \frac{1}{2} ( m \overarc{AB}+ m \overarc {CD})$

$t = \frac{1}{2} ( 72 ^{\circ } + 54 ^{\circ })$

$t = \frac{1}{2} ( 126^{\circ })$

$t =63 ^{\circ }$

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Question

Figure NOT drawn to scale.

The circumference of the above circle is 120. $\overarc{AB }$ and $\overarc{CD}$ have lengths 10 and 20, respectively. Evaluate .

Answer

The length of $\overarc{AB }$ comprises $\frac{10}{120} = \frac{1}{12}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{12} \cdot 360 ^{\circ } = 30^{\circ }$ . Similarly, The length of $\overarc{CD}$ comprises $\frac{20}{120} = \frac{1}{6}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{6} \cdot 360 ^{\circ } = 60 ^{\circ }$ .

If two secants are constructed to a circle from an outside point, the degree measure of the angle the secants form is half the difference of those of the arcs intercepted - that is,

$t = \frac{1}{2} ( m \overarc {CD} - m \overarc{AB})$

$t = \frac{1}{2} ( 60 - 30 ) = \frac{1}{2} \cdot 30 = 15$ .

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Question

Figure NOT drawn to scale.

Refer to the above diagram. $\overarc{AB }$ and $\overarc{ABC}$ have lengths 80 and 160, respectively. Evaluate .

Answer

The circumference of the circle is the sum of the two arc lengths:

The length of $\overarc{AB }$ comprises $\frac{80}{240} = \frac{1}{3}$ of the circumference of the circle. Therefore, its degree measure is $\frac{1}{3} \cdot 360 ^{\circ } = 120 ^{\circ }$ . Consequently, $\overarc{ACB }$ is an arc of degree measure $360 ^{\circ } - 120 ^{\circ } = 240^{\circ }$ .

The segments shown are both tangents from to the circle. Consequently, the degree measure of the angle they form is half the difference of the angle measures of the arcs they intercept - that is,

$t = \frac{1}{2} ( m \overarc {ACB} - m \overarc{AB})$

$t = \frac{1}{2} ( 240 ^{\circ } - 120^{\circ }) = \frac{1}{2} \cdot 120^{\circ } = 60 ^{\circ }$

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Question

The circumference of the above circle is $30 \pi$ . Give the area of the shaded region.

Answer

The radius of a circle is found by dividing the circumference $C = 30 \pi$ by $2 \pi$ :

$r= \frac{C}{2 \pi}= \frac{30 \pi}{2 \pi} = 15$

The area of the entire circle can be found by substituting for in the formula:

$A = \pi r ^{2} = \pi \cdot 15 ^{2} = 225 \pi$ .

The area of the shaded $80 ^{\circ }$ sector is $\frac{80 }{360 }$ of the total area:

$\frac{80 }{360 } \times 225 \pi = \frac{2}{9} \times 225 \pi = 50 \pi$

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Question

Give the area of the white region of the above circle if $\overarc{AB}$ has length $12 \pi$ .

Answer

If we let be the circumference of the circle, then the length of $\overarc{AB}$ is $\frac{80}{360} = \frac{2}{9 }$ of the circumference, so

$\frac{2}{9 } C = 12 \pi$

$\frac{9 }{2}\cdot \frac{2}{9 } C =\frac{9 }{2}\cdot 12 \pi$

$C =54\pi$

The radius is the circumference divided by $2 \pi$ :

$r= 54 \pi \div 2 \pi = 27$

Use the formula to find the area of the entire circle:

$A = \pi r ^{2} = \pi \left (27 \right )^{2} = 729 \pi$

The area of the white region is $1 - \frac{2}{9} = \frac{7}{9}$ of that of the circle, or

$\frac{7}{9} \cdot 729 \pi = 567 \pi$

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Question

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the area of the highlighted wedge?

Answer

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the area of the highlighted wedge?

To begin, let's recall our formula for area of a sector.

$A_{sector}=\frac{\theta}{360^{\circ}}\pi r^2$

Now, we have theta and r, so we just need to plug them in and simplify!

$A_{sector}=\frac{65^{\circ}}{360^{\circ}}\pi (4in)^2=2.89\pi in^2$

So our answer is $2.89\pi in^2$

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Question

A giant clock has a minute hand three feet long. How far, in inches, did the tip move between noon and 12:20 PM?

Answer

The distance that the tip of the minute hand moves during one hour is the circumference of a circle with radius feet. This circumference is $2\pi r = 2 \pi \times 3 = 6\pi$ feet. minutes is one-third of an hour, so the tip of the minute hand moves $\frac{1}{3} \times 6\pi = 2\pi$ feet, or $2\pi \times 12 = 24\pi$ inches.

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Question

A giant clock has a minute hand six feet long. How far, in inches, did the tip move between noon and 1:20 PM?

Answer

The distance that the tip of the minute hand moves during one hour is the circumference of a circle with radius 6 feet. This circumference is $2\pi r = 2 \pi \times 6 = 12\pi$ feet. One hour and twenty minutes is $1 \frac{1}{3}$ hours, so the tip of the hand moved $1 \frac{1}{3} \times 12 \pi = 16 \pi$ feet, or $16 \pi \times 12 = 192 \pi$ inches.

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Question

In the above figure, express in terms of .

Answer

The measure of an arc - $\overarc{AB}$ - intercepted by an inscribed angle - $\angle ACB$ - is twice the measure of that angle, so

$m \overarc{AB} = 2 \cdot m \angle ACB$

$\frac{4y}{4} = \frac{4x- 27}{4}$

$y = x- \frac{27}{4}$

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Question

In the above diagram, radius .

Give the length of $\overarc {AB}$ .

Answer

The circumference of a circle is $2 \pi$ multiplied by its radius , so

$C = 2 \pi \cdot OA = 2 \pi \cdot 12 = 24 \pi$ .

$\angle ACB$ , being an inscribed angle of the circle, intercepts an arc $\overarc {AB}$ with twice its measure:

$m \overarc {AB} = 2 \cdot m\angle ACB = 2 \cdot 72 ^{\circ } = 144 ^{\circ }$

The length of $\overarc {AB}$ is the circumference multiplied by $\frac{144}{360}$ :

$\frac{144}{360} \times 24 \pi = \frac{48 \pi }{5}$ .

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Question

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the length of the arc of the highlighted wedge?

Answer

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . If the screen has a radius of 4 inches, what is the length of the arc of the highlighted wedge?

To begin, let's recall our formula for length of an arc.

$Arc=\frac{\theta}{360^{\circ}}2 \pi r$

Now, just plug in and simplify

$Arc=\frac{65^{\circ}}{360^{\circ}}2 \pi (4in)=\frac{13^{\circ}}{9^{\circ}}\pi in\approx 4.54in$

So, our answer is 4.54in

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Question

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Answer

Sector SOW has a central angle of $45^{\circ}$ . What percentage of the circle does it cover?

Recall that there is a total of 360 degrees in a circle. SOW occupies 45 of them. To find the percentage, simply do the following:

$\frac{45^{\circ}}{360^{\circ}}100=.125100=12.5%$

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Question

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . What percentage of the circle is highlighted?

Answer

While visiting a history museum, you see a radar display which consists of a circular screen with a highlighted wedge with an angle of $65^{\circ}$ . What percentage of the circle is highlighted?

To find the percentage of a sector, simply put the degree measure of the angle over 360 and multiply by 100.

$Sector=\frac{65^{\circ}}{360^{\circ}}*100\approx 18.06 %$

So, our answer is 18.06%

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