Sectors - SAT Math

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Question

In the circle above, the length of arc BC is 100 degrees, and the segment AC is a diameter. What is the measure of angle ADB in degrees?

Answer

Since we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees.

Since angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees.

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Question

The length of an arc, , of a circle is $8\pi$ and the radius, , of the circle is . What is the measure in degrees of the central angle, $\theta$ , formed by the arc ?

Answer

The circumference of the circle is $2\pi r$ .

$2\pi(16)=32\pi$

The length of the arc S is $8\pi$ .

A ratio can be established:

$\frac{S}{\text{circumference}}=\frac{\theta}{360^o}$

$\frac{8\pi}{32\pi}=\frac{\theta}{360^o}$

Solving for _ $\theta$ _yields 90o.

Note: This makes sense. Since the arc S was one-fourth the circumference of the circle, the central angle formed by arc S should be one-fourth the total degrees of a circle.

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Question

In the figure above that includes Circle O, the measure of angle BAC is equal to 35 degrees, the measure of angle FBD is equal to 40 degrees, and the measure of arc AD is twice the measure of arc AB. Which of the following is the measure of angle CEF? The figure is not necessarily drawn to scale, and the red numbers are used to mark the angles, not represent angle measures.

Answer

The measure of angle CEF is going to be equal to half of the difference between the measures two arcs that it intercepts, namely arcs AD and CD.

$\angle CEF=\frac{1}{2}(AD-CD)$

Thus, we need to find the measure of arcs AD and CD. Let's look at the information given and determine how it can help us figure out the measures of arcs AD and CD.

Angle BAC is an inscribed angle, which means that its meausre is one-half of the measure of the arc that it incercepts, which is arc BC.

$\angle BAC=\frac{1}{2}BC$

$35^o=\frac{1}{2}BC\rightarrow BC=70^o$

Thus, since angle BAC is 35 degrees, the measure of arc BC must be 70 degrees.

We can use a similar strategy to find the measure of arc CD, which is the arc intercepted by the inscribed angle FBD.

$40^o=\frac{1}{2}CD\rightarrow CD=80^o$

Because angle FBD has a measure of 40 degrees, the measure of arc CD must be 80 degrees.

We have the measures of arcs BC and CD. But we still need the measure of arc AD. We can use the last piece of information given, along with our knowledge about the sum of the arcs of a circle, to determine the measure of arc AD.

We are told that the measure of arc AD is twice the measure of arc AB. We also know that the sum of the measures of arcs AD, AB, CD, and BC must be 360 degrees, because there are 360 degrees in a full circle.

Because AD = 2AB, we can substitute 2AB for AD.

This means the measure of arc AB is 70 degrees, and the measure of arc AD is 2(70) = 140 degrees.

Now, we have all the information we need to find the measure of angle CEF, which is equal to half the difference between the measure of arcs AD and CD.

$\angle CEF=\frac{1}{2}(AD-CD)$

$\angle CEF=\frac{1}{2}(140^o-80^o)=\frac{1}{2}(60^o)$

$\angle CEF=30^o$

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Question

A pie has a diameter of 12". A piece is cut out, having a surface area of 4.5π. What is the angle of the cut?

Answer

This is simply a matter of percentages. We first have to figure out what percentage of the surface area is represented by 4.5π. To do that, we must calculate the total surface area. If the diameter is 12, the radius is 6. Don't be tricked by this!

A = π * 6 * 6 = 36π

Now, 4.5π is 4.5π/36π percentage or 0.125 (= 12.5%)

To figure out the angle, we must take that percentage of 360°:

0.125 * 360 = 45°

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Question

Eric is riding a Ferris wheel. The Ferris wheel has 18 compartments, numbered in order clockwise. If compartment 1 is at 0 degrees and Eric enters compartment 13, what angle is he at?

Answer

12 compartments further means 240 more degrees. 240 is the answer.

360/12 = 240 degrees

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Question

What is the angle of a sector of area on a circle having a radius of ?

Answer

To begin, you should compute the complete area of the circle:

$A=\pi r^2$

For your data, this is:

$A=15^2\pi=225\pi$

Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:

$\frac{45}{225\pi}$

Now, multiply this by the total degrees in a circle:

$\frac{45}{225\pi}*360=22.918311805212$

Rounded, this is $22.92^{\circ}$ .

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Question

What is the angle of a sector that has an arc length of on a circle of diameter ?

Answer

The first thing to do for this problem is to compute the total circumference of the circle. Notice that you were given the diameter. The proper equation is therefore:

$C=\pi d$

For your data, this means,

$C=12\pi$

Now, to compute the angle, note that you have a percentage of the total circumference, based upon your arc length:

$\frac{13.5}{12\pi}*360=128.9155039044336$

Rounded to the nearest hundredth, this is $128.92^{\circ}$ .

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Question

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, $m \angle B = 90 ^{\circ }$ , and $\bigtriangleup ABC$ is a right triangle. The acute angles of $\bigtriangleup ABC$ are complementary, so

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

$11t\div 11 = 88 \div 11$

The measure of inscribed $\angle C$ is

$m \angle C = (4t+1 )^{\circ } = (4 \cdot 8+1 )^{\circ } = 33 ^{\circ }$ .

An inscribed angle of a circle intercepts an arc of twice its degree measure, so

$m \overarc {AB} = 2 \cdot m \angle C= 2 \cdot 33^{\circ } = 66 ^{\circ }$ .

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Question

Figure NOT drawn to scale

Refer to the above figure. $\overline{AB}$ is a diameter of the circle. Evaluate .

Answer

$\overline{AB}$ is a diameter, so $\overarc{BC A }$ is a semicircle, and

$m \overarc{BC A } = 180^{\circ }$ ,

or, equivalently,

$m \overarc{BC} + m \overarc {AC } = 180^{\circ }$

In terms of , since $m \overarc {AC } = t^{\circ }$ ,

$m \overarc{BC} + t ^{\circ }= 180^{\circ }$

$m \overarc{BC} + t^{\circ } - t ^{\circ }= 180 ^{\circ } - t^{\circ }$

$m \overarc{BC} =( 180 - t)^{\circ }$

$\overline{NB}$ and $\overline{NC }$ , being a secant segment and a tangent segment to a circle, respectively, intercept two arcs such that the measure of the angle that the segments form is equal to one-half the difference of the measures of the intercepted arcs - that is,

$\frac{1}{2} (m \overarc{BC} - m \overarc {AC }) = m \angle CNB$

Setting $m \overarc{BC} = \left (180 - t \right )^{\circ }$ , $m \overarc {AC } = t^{\circ }$ , and $m \angle CNB = 34^{\circ }$ :

$\frac{1}{2}\left [ (180 - t) - t \right ]= 34$

$\frac{1}{2} (180 - 2t) = 34$

$\frac{1}{2} \cdot 180 -\frac{1}{2} \cdot 2t = 34$

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Question

Refer to the above diagram. Evaluate the measure of $\angle BND$ .

Answer

The total measure of the arcs that comprise a circle is $360 ^{\circ }$ , so from the above diagram,

$m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

$8t \div 8 = 360 \div 8$

Therefore,

$m \overarc { AC}= 45^{\circ }$

and

$m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\overline{NB}$ and $\overline{ND}$ , which is $\angle BND$ , is half the difference of the measures of the arcs they intercept, so

$m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$m\angle BND = \frac{1}{2} (90^{\circ } )$

$m\angle BND = 45 ^{\circ }$

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle B$ and $\angle D$ are two such angles, so

$m \angle B + m \angle D = 180 ^{\circ }$

Setting $m \angle B = 88^{\circ }$ and $m \angle D =u ^{\circ }$ , and solving for :

,

the correct response.

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (180-t)^{\circ }$ , and solving for :

,

The statement turns out to be true regardless of the value of . Therefore, without further information, the value of cannot be determined.

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (t+12)^{\circ }$ , and solving for :

$2t \div 2 = 168 \div 2$

,

the correct response.

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Question

Figure NOT drawn to scale.

Refer to the above diagram. $\overline{AB}$ is a diameter. Evaluate $m \angle CBA$

Answer

$\overline{AB}$ is a diameter, so $\overarc{ACB}$ is a semicircle - therefore, $m \overarc{ACB} = 180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $t ^{\circ }= m \overarc{AC}$ , then

$t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$ ,

and

$m \overarc{CB} = (180 -t )^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NC}$ and $\overline{NB}$ are such segments intercepting $\overarc{AC}$ and $\overarc{CB}$ , it holds that

$\frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $m \overarc{AC} = t ^{\circ }$ , $m \overarc{CB} = (180 -t )^{\circ }$ , and $m \angle CNB = 28 ^{\circ }$ :

$\frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\frac{1}{2} \left ( 180-2 t \right ) = 28$

$m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\angle CBA$ , has half this measure:

$m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$ .

This is the correct response.

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Question

Figure NOT drawn to scale.

In the above figure, $\overline{AB}$ is a diameter. Also, the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5. Give the measure of $\angle TNA$ .

Answer

$\overline{AB}$ is a diameter, so $\overarc {ATB}$ is a semicircle, which has measure $180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc {AT } + m \overarc {BT} = m \overarc {ATB}$

If we let $t ^{\circ }= m \overarc {AT }$ , then, substituting:

$t ^{\circ }+ m \overarc {BT} = 180 ^{\circ }$ ,

and

$m \overarc {BT} =( 180 - t ) ^{\circ }$

the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5; this is also the ratio of their degree measures; that is,

$\frac{m \overarc{BT}}{m \overarc{AT}} = \frac{5}{3}$

Setting $m \overarc {AT } = t ^{\circ }$ and $m \overarc {BT} =( 180 - t ) ^{\circ }$ :

$\frac{180-t }{t} = \frac{7}{5}$

Cross-multiply, then solve for :

$\frac{12t }{12}= \frac{900}{12}$

$m \overarc {AT } =75^{\circ }$ , and $m \overarc {BT} =( 180 -75 ) ^{\circ } = 105 ^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NB}$ and $\overline{NT}$ are such segments whose angle $\angle TNA$ intercepts $\overarc{AT}$ and $\overarc{BT}$ , it holds that:

$m \angle TNA = \frac{1}{2}\left ( m \overarc{BT}- m \overarc{AT} \right )$

$m \angle TNA = \frac{1}{2}\left ( 105 ^{\circ } - 75^{\circ } \right )$

$m \angle TNA = \frac{1}{2}\left ( 30 ^{\circ } \right )$

$m \angle TNA = 15^{\circ }$

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Question

The radius of the circle above is and $\angle A=45^{\circ}$ . What is the area of the shaded section of the circle?

Answer

Area of Circle = πr2 = π42 = 16π

Total degrees in a circle = 360

Therefore 45 degree slice = 45/360 fraction of circle = 1/8

Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π

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Question

is a square.

$\overline{AB}=\overline{BC}=\overline{CD}$

The arc from to is a semicircle with a center at the midpoint of $\overline{BC}$ .

All units are in feet.

The diagram shows a plot of land.

The cost of summer upkeep is $2.50 per square foot.

In dollars, what is the total upkeep cost for the summer?

Answer

To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle.

The area of the square is straightforward:

30 * 30 = 900 square feet

Because each side is 30 feet long, AB + BC + CD = 30.

We can substitute BC for AB and CD since all three lengths are the same:

BC + BC + BC = 30

3BC = 30

BC = 10

Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet.

The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π.

The total area of the plot is the square less the semicircle: 900 - 12.5π square feet

The cost of upkeep is therefore 2.5 * (900 – 12.5π) = $(2250 – 31.25π).

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Question

A circular, 8-slice pizza is placed in a square box that has dimensions four inches larger than the diameter of the pizza. If the box covers a surface area of 256 in2, what is the surface area of one piece of pizza?

Answer

The first thing to do is calculate the dimensions of the pizza box. Based on our data, we know 256 = s2. Solving for s (by taking the square root of both sides), we get 16 = s (or s = 16).

Now, we know that the diameter of the pizza is four inches less than 16 inches. That is, it is 12 inches. Be careful! The area of the circle is given in terms of radius, which is half the diameter, or 6 inches. Therefore, the area of the pizza is π * 62 = 36π in2. If the pizza is 8-slices, one slice is equal to 1/8 of the total pizza or (36π)/8 = 4.5π in2.

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Question

In the figure, PQ is the arc of a circle with center O. If the area of the sector is $3\pi$ what is the perimeter of sector?

Answer

First, we figure out what fraction of the circle is contained in sector OPQ: $\frac{30^{\circ}}{360^{\circ}}= \frac{1}{12}$ , so the total area of the circle is $\dpi{100} \small 12\times 3\pi=36$ .

Using the formula for the area of a circle, ${\pi}r^{2}$ , we can see that $\dpi{100} \small r=6$ .

We can use this to solve for the circumference of the circle, $2{\pi}r$ , or $12{\pi}$ .

Now, OP and OQ are both equal to r, and PQ is equal to $\dpi{100} \small \frac{1}{12}$ of the circumference of the circle, or ${\pi}$ .

To get the perimeter, we add OP + OQ + PQ, which give us $12+{\pi}$ .

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Question

If B is a circle with line AC = 12 and line BC = 16, then what is the area formed by DBE?

Answer

Line AB is a radius of Circle B, which can be found using the Pythagorean Theorem:

$AB^2=AC^2+BC^2\rightarrow AB=\sqrt{AC^2+BC^2}=\sqrt{16^2+12^2}=\sqrt{400}=20$

Since AB is a radius of B, we can find the area of circle B via:

$Area=\pi R^2=\pi(20^2)=400\pi$

Angle DBE is a right angle, and therefore $\frac{1}{4}$ of the circle so it follows:

$Area(DBE)=\frac{400}{4}\pi=100\pi$

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