Sectors - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

What is the angle measure of in the figure above if the sector comprises 37% of the circle?

Answer

It is very easy to compute the angle of a sector if we know what it is as a percentage of the total circle. To do this, you merely need to multiply by ˚. This yields ˚.

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Question

What is the angle measure of in the figure above if the sector comprises % of the circle?

Answer

It is very easy to compute the angle of a sector if we know what it is as a percentage of the total circle. To do this, you merely need to multiply by ˚. This yields ˚.

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Question

What is the angle measure of in the figure if the sector comprises of the circle?

Answer

It is very easy to compute the angle of a sector if we know what it is as a percentage of the total circle. To do this, you merely need to multiply by ˚. This yields ˚

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Question

Which is the greater quantity?

(a) The degree measure of a 10-inch-long arc on a circle with radius 8 inches.

(b) The degree measure of a 12-inch-long arc on a circle with radius 10 inches.

Answer

(a) A circle with radius 8 inches has crircumference $C = 2\pi r = 2\pi \cdot 8 = 16\pi$ inches. An arc 10 inches long is $\frac{10}{16\pi } = \frac{5}{8\pi }$ of that circle. $\frac{5}{8\pi } \times 360 ^{\circ } = \frac{225}{\pi } ^{\circ }$ , the degree measure of this arc.

(b) A circle with radius 10 inches has crircumference $C = 2\pi r = 2\pi \cdot 10 = 20\pi$ inches. An arc 12 inches long is $\frac{12}{20\pi }=\frac{3}{5\pi }$ of that circle. $\frac{3}{5\pi } \times 360 ^{\circ } = \frac{216}{\pi } ^{\circ }$ , the degree measure of this arc.

(a) is the greater quantity.

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Question

Note: figure NOT drawn to scale

Refer to the above figure. Which is the greater quantity?

(a) $m \widehat{ACB}$

(b) $180 ^{\circ }$

Answer

Since

$7^{2} + 12 ^{2} = 49 + 144 = 169 = 13 ^{2}$ ,

the triangle is a right triangle with right angle $\angle ACB$ .

$\angle ACB$ is an inscribed angle on the circle, so the arc it intercepts is a semicricle. Therefore, $\widehat{ACB}$ is also a semicircle, and it measures $180 ^{\circ }$ .

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Question

Refer to the above figure. Which is the greater quantity?

(a)

(b) 55

Answer

The measure of an inscribed angle of a circle is one-half that of the arc it intercepts. Therefore, $x = \frac{1}{2} \cdot 110 = 55$ .

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Question

Note: Figure NOT drawn to scale

Refer to the above figure. Which is the greater quantity?

(a)

(b) 90

Answer

The measure of an arc intercepted by an inscribed angle of a circle is twice that of the angle. Therefore, $y = 2 \cdot 44 = 88 < 90$

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Question

$\odot A$ has twice the radius of $\odot B$ . Sector 1 is part of $\odot A$ ; Sector 2 is part of $\odot B$ ; the two sectors are equal in area.

Which is the greater quantity?

(a) Twice the degree measure of the central angle of Sector 1

(b) The degree measure of the central angle of Sector 2

Answer

$\odot A$ has twice the radius of $\odot B$ , so $\odot A$ has four times the area of $\odot B$ . This means that for a sector of $\odot A$ to have the same area as a sector of $\odot B$ , the central angle of the latter sector must be four times that of the former sector. This makes (b) greater than (a), which is only twice that of the former sector.

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Question

The area of the shaded sector in circle O is $4.5\pi$ . What is the angle measure ?

Answer

Remember that the angle for a sector or arc is found as a percentage of the total degrees of the circle. The proportion of to is the same as $4.5\pi$ to the total area of the circle.

The area of a circle is found by:

$A = \pi * r^2$

For our data, this means:

$A = 6^2\pi = 36\pi$

Now we can solve for using the proportions:

$\frac{x}{360} = \frac{4.5\pi}{36\pi}$

Cross multiply:

$36x\pi=1620\pi$

Divide both sides by $36\pi$ :

Therefore, is ˚.

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Question

The area of the shaded sector in circle O is $2.81\pi$ . What is the angle measure , rounded to the nearest hundredth?

Answer

Remember that the angle for a sector or arc is found as a percentage of the total degrees of the circle. The proportion of to is the same as $4.5\pi$ to the total area of the circle.

The area of a circle is found by:

$A = \pi * r^2$

For our data, this means:

$A = 7.5^2\pi = 56.25\pi$

Now we can solve for using the proportions:

$\frac{x}{360} = \frac{2.81\pi}{56.25\pi}$

Cross multiply:

$56.25x\pi=1011.6\pi$

Divide both sides by $56.25\pi$ :

Therefore, is ˚.

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Question

The arc-length for the shaded sector is . What is the value of , rounded to the nearest hundredth?

Answer

Remember that the angle for a sector or arc is found as a percentage of the total degrees of the circle. The proportion of to is the same as to the total circumference of the circle.

The circumference of a circle is found by:

$C = 2\pir$

For our data, this means:

$C = 29.1\pi=18.2\pi$

Now we can solve for using the proportions:

$\frac{x}{360} = \frac{9.31}{18.2\pi}$

Cross multiply:

$18.2x\pi=3351.6$

Divide both sides by $18.2\pi$ :

Therefore, is ˚.

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Question

The arc-length for the shaded sector is . What is the value of , rounded to the nearest hundredth?

Answer

Remember that the angle for a sector or arc is found as a percentage of the total degrees of the circle. The proportion of to is the same as to the total circumference of the circle.

The circumference of a circle is found by:

$C = 2\pir$

For our data, this means:

$C = 28.7\pi=17.4\pi$

Now we can solve for using the proportions:

$\frac{x}{360} = \frac{19.4}{17.4\pi}$

Cross multiply:

$17.4x\pi=6984$

Divide both sides by $17.4\pi$ :

Therefore, is ˚.

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Question

$\bigtriangleup ABC$ is inscribed in a circle. $\overarc {ABC }$ is a semicircle. $m \angle A = 40 ^{\circ }$ .

Which is the greater quantity?

(a) $m \overarc {BAC}$

(b) $m \overarc {BCA }$

Answer

The figure referenced is below:

$\overarc {ABC }$ is a semicircle, so $\overarc {ADC }$ is one as well; as a semicircle, its measure is $180 ^{\circ }$ . The inscribed angle that intercepts this semicircle, $\angle B$ , is a right angle, of measure $90 ^{\circ }$ . $m \angle A = 40 ^{\circ }$ , and the sum of the measures of the interior angles of a triangle is $180 ^{\circ }$ , so

$m \angle C = 180 ^{\circ } - (m \angle A + m \angle B )$

$= 180 ^{\circ } - (40 ^{\circ } +90 ^{\circ } )$

$= 180 ^{\circ } - 130 ^{\circ }$

$= 50 ^{\circ }$

$\angle C$ has greater measure than $\angle A$ , so the minor arc intercepted by $\angle C$ , which is $\overarc {BA}$ , has greater measure than that intercepted by $\angle A$ , which is $\overarc {BC}$ . It follows that the major arc corresponding to the latter, which is $\overarc {BA C}$ , has greater measure than that corresponding to the former, which is $\overarc {BC A }$ .

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Question

Refer to the above figure, Which is the greater quantity?

(a) The area of $\bigtriangleup ABC$

(b) The area of the orange semicircle

Answer

$\bigtriangleup ABC$ has angles of degree measure 30 and 60; the third angle must measure 90 degrees, making $\bigtriangleup ABC$ a right triangle.

For the sake of simplicity, let ; the reasoning is independent of the actual length. The smaller leg of a 30-60-90 triangle has length equal to $\frac{\sqrt{3}}{3}$ times that of the longer leg; this is about

$AB = \frac{\sqrt{3}}{3} \approx \frac{1.7}{3} \approx 0.57$

The area of a right triangle is half the product of its legs, so

$A = \frac{1}{2} \cdot 1 \cdot 0.57 = 0.5 \cdot 0.57 \approx 0.285$

Also, if , then the orange semicircle has diameter 1 and radius $\frac{1}{2}$ . Its area can be found by substituting $r = \frac{1}{2}$ in the formula:

$A = \frac{1}{2} \cdot \pi r^{2}$

$= \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}$

$= \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )$

$= \frac{1}{8} \cdot \pi$

$\approx 0. 125 \cdot 3.14$

$\approx 0. 3925$

The orange semicircle has a greater area than $\bigtriangleup ABC$

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Question

In the above figure, $\overline{AD}$ is a diameter of the circle.

Which is the greater quantity?

(a) $m \angle CAD$

(b) $m \angle CBD$

Answer

That $\overline{AD}$ is a diameter of the circle is actually irrelevant to the problem. Two inscribed angles of a circle that both intercept the same arc, as $\angle CAD$ and $\angle CBD$ both do here, have the same measure.

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Question

Trapezoid is inscribed in a circle, with $\overline{AD}$ a diameter.

Which is the greater quantity?

(a) $m \overarc{AC}$

(b) $m \overarc{DB}$

Answer

Below is the inscribed trapezoid referenced, along with its diagonals.

$\overline{AD} ||\overline{BC}$ , so, by the Alternate Interior Angles Theorem,

$\angle ACB \cong \angle CAD$ , and their intercepted angles are also congruent - that is,

$m \overarc{AB} = m \overarc{DC}$

$m \overarc{AB}+ m \overarc{BC} = m \overarc{DC} + m \overarc{CB}$

By the Arc Addition Principle,

$m \overarc{A C} = m \overarc{D B}$ .

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Question

In the above figure, is the center of the circle, and $m \overarc {AD} = 60 ^{\circ }$ . Which is the greater quantity?

(a)

(b)

Answer

Construct $\overline{OD}$ . The new figure is below:

$m \overarc {AD} = 60 ^{\circ }$ , so $m \overarc {AC} < 60 ^{\circ }$ . It follows that their respective central angles have measures

$m \angle AOD = 60 ^{\circ }$

and

$m \angle AOC < 60 ^{\circ }$ .

Also, since $m \overarc {AD} = 60 ^{\circ }$ and $m \overarc {ACB} = 180 ^{\circ }$ - $\overarc {ACB}$ being a semicircle - by the Arc Addition Principle, $m \overarc {BD} = 120^{\circ }$ . $\angle DAO$ , an inscribed angle which intercepts this arc, has half this measure, which is $60 ^{\circ }$ . The other angle of $\bigtriangleup ADO$ , which is $\angle ADO$ , also measures $60 ^{\circ }$ , so $\bigtriangleup ADO$ is equilateral.

, since all radii are congruent;

by reflexivity;

$m \angle AOC < m \angle AOD$

By the Side-Angle-Side Inequality Theorem (or Hinge Theorem), it follows that . Since $\bigtriangleup ADO$ is equilateral, , and since all radii are congruent, . Substituting, it follows that .

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Question

In the above figure, $\overline{AD}$ is a diameter of the circle. Which is the greater quantity?

(a) $m \angle ADB$

(b) $m \angle ACB$

Answer

Both $\angle ADB$ and $\angle ACB$ are inscribed angles of the same circle which intercept the same arc; they are therefore of the same measure. The fact that $\overline{AD}$ is a diameter of the circle is actually irrelevant to the problem.

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Question

Figure NOT drawn to scale.

Refer to the above diagram. $m \overarc {NO}$ is the arithmetic mean of $m \overarc {NP}$ and $m \overarc {OP}$ .

Which is the greater quantity?

(a) $m \angle NTO$

(b) $60 ^{\circ }$

Answer

$m \overarc {NO}$ is the arithmetic mean of $m \overarc {NP}$ and $m \overarc {OP}$ , so

$m \overarc {NO} = \frac{1}{2} \left ( m \overarc {NP} + m \overarc {OP} \right )$

By arc addition, this becomes

$m \overarc {NO} = \frac{1}{2} \cdot m \overarc {NPO}$

Also, $m \overarc {NO} + m \overarc {NPO}= 360 ^{\circ }$ , or, equivalently,

$m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO}$ , so

$m \overarc {NO} = \frac{1}{2} \cdot \left ( 360 ^{\circ } - m \overarc {NO} \right )$

Solving for $m \overarc {NO}$ :

$m \overarc {NO} = \frac{1}{2} \cdot 360 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO}$

$m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO} = 180 ^{\circ } - \frac{1}{2} \cdot m \overarc {NO} + \frac{1}{2} \cdot m \overarc {NO}$

$\frac{3}{2} \cdot m \overarc {NO} = 180 ^{\circ }$

$\frac{2} {3} \cdot \frac{3}{2} \cdot m \overarc {NO} = \frac{2} {3} \cdot 180 ^{\circ }$

$m \overarc {NO} = 120 ^{\circ }$

Also,

$m \overarc {NPO} = 360 ^{\circ } - m \overarc {NO} = 360 ^{\circ } -120 ^{\circ } = 240 ^{\circ }$

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

$m \angle NTO = \frac{1}{2}(m \overarc{NPO} - m \overarc{N O} ) = \frac{1}{2}(240^{\circ }- 120^{\circ } ) = \frac{1}{2}(120^{\circ } ) = 60^{\circ }$

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Question

Figure NOT drawn to scale

In the above diagram, $BA = AX \sqrt{2}$ .

Which is the greater quantity?

(a) $m \overarc{AD}$

(b) $m \overarc{BC}$

Answer

$\bigtriangleup AXB$ is a right triangle whose hypotenuse $\overline{AB}$ has length $\sqrt{2}$ times that of leg $\overline{AX}$ . This is characteristic of a triangle whose acute angles both have measure $45 ^{\circ }$ -and consequently, whose acute angles are congruent. Therefore,

$\angle A \cong \angle B$

These inscribed angles being congruent, the arcs they intercept, $\overarc{AD}$ and $\overarc{BC}$ , are also congruent.

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