Circles, Arc Lengths, and Sector Areas

31 soru

Soru 1Soru

Sector S1S_1 belongs to a circle with radius rr and has a central angle of measure θ\theta^\circ, where 0<θ<3600 < \theta < 360. Sector S2S_2 belongs to a circle with radius 2r2r and has a central angle of measure (θ2)\left(\frac{\theta}{2}\right)^\circ. Which of the following statements must be true? Select all such statements.

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Cevap: The area of sector S2S_2 is twice the area of sector S1S_1.; The arc length of sector S2S_2 is equal to the arc length of sector S1S_1.; The ratio of the area of sector S1S_1 to its arc length is half the ratio of the area of sector S2S_2 to its arc length.

Cevap

The true statements are that the area of sector S2S_2 is twice the area of sector S1S_1, the arc length of sector S2S_2 is equal to the arc length of sector S1S_1, and the ratio of area to arc length for sector S1S_1 is half that of sector S2S_2.
The area of sector S2S_2 is twice that of S1S_1 because quadrupling r2r^2 combined with halving the central angle results in a factor of 2. The arc length of sector S2S_2 equals that of S1S_1 because doubling rr and halving the angle cancel each other out. The area-to-arc-length ratio of any sector reduces to R2\frac{R}{2}, so sector S1S_1 with radius rr has ratio r2\frac{r}{2}, which is half the ratio rr of sector S2S_2.

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1
Calculate and compare the sector areas.
Area(S1)=θ360πr2\text{Area}(S_1) = \frac{\theta}{360}\pi r^2 and Area(S2)=θ/2360π(2r)2=θ720π(4r2)=2θ360πr2=2Area(S1)\text{Area}(S_2) = \frac{\theta/2}{360}\pi (2r)^2 = \frac{\theta}{720}\pi (4r^2) = \frac{2\theta}{360}\pi r^2 = 2 \cdot \text{Area}(S_1).
The sector area formula is Area=angle360πR2\text{Area} = \frac{\text{angle}}{360^\circ} \pi R^2. Doubling the radius quadruples R2R^2, while halving the angle reduces the fraction by half, producing a net doubling of area.
2
Calculate and compare the arc lengths.
Arc(S1)=θ360(2πr)\text{Arc}(S_1) = \frac{\theta}{360}(2\pi r) and Arc(S2)=θ/2360(2π2r)=θ360(2πr)=Arc(S1)\text{Arc}(S_2) = \frac{\theta/2}{360}(2\pi \cdot 2r) = \frac{\theta}{360}(2\pi r) = \text{Arc}(S_1).
The arc length formula is Arc=angle360(2πR)\text{Arc} = \frac{\text{angle}}{360^\circ} (2\pi R). Doubling the radius doubles RR, while halving the angle halves the fraction, leaving the product unchanged.
3
Calculate and compare the sector perimeters.
Perimeter(S1)=2r+Arc(S1)\text{Perimeter}(S_1) = 2r + \text{Arc}(S_1) and Perimeter(S2)=4r+Arc(S2)=4r+Arc(S1)\text{Perimeter}(S_2) = 4r + \text{Arc}(S_2) = 4r + \text{Arc}(S_1).
The perimeter of a sector consists of two straight radii and the curved arc length. 2Perimeter(S1)=4r+2Arc(S1)Perimeter(S2)2 \cdot \text{Perimeter}(S_1) = 4r + 2\text{Arc}(S_1) \neq \text{Perimeter}(S_2).
4
Evaluate the area-to-arc-length ratios for both sectors.
For S1S_1, Area(S1)Arc(S1)=θ360πr2θ3602πr=r2\frac{\text{Area}(S_1)}{\text{Arc}(S_1)} = \frac{\frac{\theta}{360}\pi r^2}{\frac{\theta}{360}2\pi r} = \frac{r}{2}. For S2S_2, Area(S2)Arc(S2)=2r2=r\frac{\text{Area}(S_2)}{\text{Arc}(S_2)} = \frac{2r}{2} = r.
The ratio of area to arc length simplifies to R2\frac{R}{2} for any sector, so doubling the radius doubles this ratio.

Anahtar Kavram

Geometric properties of circle sectors, including proportional relationships between radius, central angle, arc length, sector area, and total sector perimeter.
Tahmini Süre:2m 0s
Soru 2Soru

In circle OO, sector AOBAOB has a central angle of 6060^\circ and a radius of 1212. A smaller circle CC is inscribed within sector AOBAOB such that it is tangent to radii OAOA and OBOB, as well as to arc ABAB. What is the area of the region inside sector AOBAOB that lies outside circle CC?

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Cevap: 8π8\pi

Cevap

The area of the region inside sector AOBAOB outside circle CC is 8π8\pi.
The correct answer is derived by first finding the area of sector AOBAOB using 60360π(122)=24π\frac{60}{360} \pi (12^2) = 24\pi. Then, analyzing the geometry of the inscribed circle reveals that the line from OO to the center of circle CC bisects the 6060^\circ angle. In the resulting 30609030^\circ-60^\circ-90^\circ right triangle, the hypotenuse length is 2r2r, making the total radius of sector AOBAOB equal to 2r+r=3r=122r + r = 3r = 12, which yields r=4r = 4. The area of circle CC is π(42)=16π\pi (4^2) = 16\pi. Subtracting the circle area from the sector area gives 24π16π=8π24\pi - 16\pi = 8\pi.

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1
Calculate the area of sector AOBAOB
Sector area =60360×π(122)=16×144π=24π= \frac{60^\circ}{360^\circ} \times \pi (12^2) = \frac{1}{6} \times 144\pi = 24\pi
The area of a sector with central angle θ\theta and radius RR is θ360πR2\frac{\theta}{360^\circ} \pi R^2.
2
Find the radius rr of the inscribed circle CC
Distance from OO to center of circle CC is OP=2rOP = 2r, so total radius R=OP+r=3r=12    r=4R = OP + r = 3r = 12 \implies r = 4
The line segment connecting center OO to center PP of circle CC bisects the 6060^\circ angle into two 3030^\circ angles. A perpendicular dropped from PP to radius OAOA forms a 30609030^\circ-60^\circ-90^\circ right triangle where sin(30)=rOP=12\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}, giving OP=2rOP = 2r.
3
Calculate the area of circle CC
Area of circle C=πr2=π(42)=16πC = \pi r^2 = \pi (4^2) = 16\pi
The area of a circle with radius rr is πr2\pi r^2.
4
Subtract the area of circle CC from the area of sector AOBAOB
24π16π=8π24\pi - 16\pi = 8\pi
The desired region is the difference between the full sector area and the enclosed circle's area.

Anahtar Kavram

Inscribed circles in sectors and central angle sector area calculations
Tahmini Süre:2m 30s
Soru 3Soru

An equilateral triangle ABCABC with side length 636\sqrt{3} is inscribed in a circle with center OO. What is the area of sector AOBAOB, divided by π\pi?

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Cevap: 12

Cevap

The area of sector AOBAOB divided by π\pi is 12.
Since triangle ABCABC is equilateral, its three vertices divide the circle into three congruent arcs of 120120^\circ each. Thus, central angle AOB=120\angle AOB = 120^\circ. The relationship between the side length ss of an inscribed equilateral triangle and the radius RR of its circumscribed circle is s=R3s = R\sqrt{3}. Given s=63s = 6\sqrt{3}, we solve for RR to find R=6R = 6. The area of sector AOBAOB is 120360πR2=13π(62)=12π\frac{120^\circ}{360^\circ} \pi R^2 = \frac{1}{3} \pi (6^2) = 12\pi. Dividing this area by π\pi yields 1212.

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1
Find the central angle AOB\angle AOB corresponding to side ABAB of the inscribed equilateral triangle.
AOB=120\angle AOB = 120^\circ
An inscribed equilateral triangle divides the 360360^\circ circle into three equal central angles.
2
Calculate the radius RR of the circumscribed circle from the given side length s=63s = 6\sqrt{3}.
R=6R = 6
In an inscribed equilateral triangle, s=R3s = R\sqrt{3}. Substituting 63=R36\sqrt{3} = R\sqrt{3} gives R=6R = 6.
3
Compute the area of sector AOBAOB and divide by π\pi.
12
Sector Area=120360×π×62=12π\text{Sector Area} = \frac{120^\circ}{360^\circ} \times \pi \times 6^2 = 12\pi. Dividing by π\pi leaves 1212.

Anahtar Kavram

Relationship between inscribed shapes, circle radii, and sector area
Soru 4Soru

In circle OO, a sector with a central angle of 6060^\circ has an area of 6π6\pi. In circle PP, a sector with a central angle of 120120^\circ has an arc length equal to the arc length of the sector in circle OO. What is the area of the sector in circle PP?

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Cevap: 3π3\pi

Cevap

3π3\pi
The sector area of 3π3\pi is correctly calculated by first determining the radius of circle OO (r1=6r_1 = 6), using it to find the arc length (2π2\pi), setting that as the arc length for the sector in circle PP to find its radius (r2=3r_2 = 3), and finally computing its sector area 120360π(32)=3π\frac{120^\circ}{360^\circ} \cdot \pi (3^2) = 3\pi.

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1
Find the radius of circle OO
Radius r1=6r_1 = 6
The sector area formula is A=θ360πr2A = \frac{\theta}{360^\circ} \cdot \pi r^2. Given A=6πA = 6\pi and θ=60\theta = 60^\circ, we have 60360πr12=6π16r12=6r12=36r1=6\frac{60^\circ}{360^\circ} \cdot \pi r_1^2 = 6\pi \Rightarrow \frac{1}{6} r_1^2 = 6 \Rightarrow r_1^2 = 36 \Rightarrow r_1 = 6.
2
Calculate the arc length of the sector in circle OO
Arc length L=2πL = 2\pi
The arc length formula is L=θ3602πrL = \frac{\theta}{360^\circ} \cdot 2\pi r. Substituting θ=60\theta = 60^\circ and r1=6r_1 = 6, we get L=162π(6)=2πL = \frac{1}{6} \cdot 2\pi (6) = 2\pi.
3
Find the radius of circle PP
Radius r2=3r_2 = 3
The sector in circle PP has an arc length equal to 2π2\pi and a central angle of 120120^\circ. Using L=1203602πr2L = \frac{120^\circ}{360^\circ} \cdot 2\pi r_2, we get 2π=132πr2r2=32\pi = \frac{1}{3} \cdot 2\pi r_2 \Rightarrow r_2 = 3.
4
Calculate the area of the sector in circle PP
Sector area =3π= 3\pi
Using the sector area formula for circle PP: A2=120360πr22=13π(32)=3πA_2 = \frac{120^\circ}{360^\circ} \cdot \pi r_2^2 = \frac{1}{3} \cdot \pi (3^2) = 3\pi.

Anahtar Kavram

Relationship between central angles, radii, arc lengths, and sector areas in circles
Soru 5Soru

In a circle centered at point OO, sector AOBAOB has a radius of length rr, a central angle of θ\theta degrees, an arc length of LL, an area of AsA_s, and a perimeter of PP. If the ratio of the sector area to the arc length is AsL=4\frac{A_s}{L} = 4, and the ratio of the sector area to the sector perimeter is AsP=43\frac{A_s}{P} = \frac{4}{3}, which of the following statements must be true? Select all that apply.

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Cevap: The radius rr of the circle is 88.; The area AsA_s of sector AOBAOB is 3232.; The perimeter PP of sector AOBAOB is 2424.

Cevap

The true statements are that the radius of the circle is 8, the area of the sector is 32, and the perimeter of the sector is 24.
The ratio of sector area to arc length simplifies directly to half the radius, r2=4\frac{r}{2} = 4, giving a radius of 88. Expressing the sector perimeter as the sum of the arc length and two radii (L+16L + 16) and using AsP=4LL+16=43\frac{A_s}{P} = \frac{4L}{L + 16} = \frac{4}{3} yields L=8L = 8. From this, the sector area is As=32A_s = 32 and the sector perimeter is P=24P = 24.

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1
Relate sector area AsA_s and arc length LL to find the radius rr.
As=θ360πr2A_s = \frac{\theta}{360}\pi r^2 and L=θ3602πrL = \frac{\theta}{360}2\pi r, so AsL=r2=4    r=8\frac{A_s}{L} = \frac{r}{2} = 4 \implies r = 8.
Dividing the sector area formula by the arc length formula cancels the central angle fraction and π\pi.
2
Set up the ratio equation for AsP\frac{A_s}{P} using P=L+2rP = L + 2r.
Since r=8r = 8, P=L+16P = L + 16. Given As=4LA_s = 4L, we have 4LL+16=43    3L=L+16    2L=16    L=8\frac{4L}{L + 16} = \frac{4}{3} \implies 3L = L + 16 \implies 2L = 16 \implies L = 8.
Expressing both sector area and perimeter in terms of arc length allows direct solution for LL.
3
Calculate sector area AsA_s and perimeter PP.
As=4(8)=32A_s = 4(8) = 32 and P=8+2(8)=24P = 8 + 2(8) = 24.
Substituting L=8L = 8 and r=8r = 8 gives the exact values for area and perimeter.
4
Determine the central angle θ\theta.
8=θ3602π(8)    8=16πθ360    θ=(90π)28.658 = \frac{\theta}{360} \cdot 2\pi(8) \implies 8 = \frac{16\pi \theta}{360} \implies \theta = \left(\frac{90}{\pi}\right)^\circ \approx 28.65^\circ.
Plugging L=8L = 8 and r=8r = 8 into the arc length formula gives θ28.65\theta \approx 28.65^\circ, which is less than 6060^\circ.

Anahtar Kavram

Relating sector area, arc length, and perimeter using proportional ratios and fundamental circle formulas.
Soru 6Soru

A sector of a circle with a radius of 1212 units has an area of 24π24\pi square units. What is the measure, in degrees, of the central angle of the sector?

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Cevap: 60

Cevap

The measure of the central angle of the sector is 6060^\circ.
The area of the full circle is πr2=π(12)2=144π\pi r^2 = \pi (12)^2 = 144\pi. The sector area represents a fraction of the total area, specifically 24π144π=16\frac{24\pi}{144\pi} = \frac{1}{6}. Multiplying this fraction by the full circle's central angle of 360360^\circ gives 16×360=60\frac{1}{6} \times 360^\circ = 60^\circ.

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1
Find the total area of the circle
The total area of the circle is π×122=144π\pi \times 12^2 = 144\pi.
The area of a full circle with radius rr is given by A=πr2A = \pi r^2.
2
Relate the sector area to the total circle area
The fraction of the circle represented by the sector is 24π144π=16\frac{24\pi}{144\pi} = \frac{1}{6}.
The area of a sector is proportional to the fraction of the total central angle (360360^\circ) it subtends.
3
Calculate the central angle in degrees
\theta = \frac{1}{6} \times 360^\circ = 60^\circ.
Multiply the fraction of the circle by 360360^\circ to get the central angle measure.

Anahtar Kavram

Relationship between central angle measure, total circle area, and sector area
Soru 7Soru

In a circle, an arc of length 4π4\pi corresponds to a central angle of 4040^\circ. If the area of the sector formed by this central angle is kπk\pi, what is the value of kk?

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Cevap: 36

Cevap

The value of kk is 36.
Using the arc length equation 4π=403602πr4\pi = \frac{40}{360} \cdot 2\pi r, we solve for the radius r=18r = 18. Substituting r=18r = 18 into the sector area formula A=40360π(18)2A = \frac{40}{360} \cdot \pi (18)^2 results in A=36πA = 36\pi. Thus, k=36k = 36.

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1
Calculate the radius of the circle using the arc length formula
Radius r=18r = 18
Arc length is related to central angle and radius by L=θ3602πrL = \frac{\theta}{360^\circ} \cdot 2\pi r. Substituting L=4πL = 4\pi and θ=40\theta = 40^\circ gives 4π=192πr    r=184\pi = \frac{1}{9} \cdot 2\pi r \implies r = 18.
2
Calculate the area of the sector using the radius and central angle
Sector Area A=36πA = 36\pi
Sector area is calculated using A=θ360πr2A = \frac{\theta}{360^\circ} \cdot \pi r^2. Substituting θ=40\theta = 40^\circ and r=18r = 18 gives A=19π(182)=36πA = \frac{1}{9} \cdot \pi (18^2) = 36\pi.
3
Extract the coefficient kk from kπk\pi
k=36k = 36
Comparing 36π36\pi to kπk\pi directly yields k=36k = 36.

Anahtar Kavram

Relationship between central angle, arc length, radius, and sector area
Soru 8Soru

In a circle centered at point OO, sector OABOAB has a central angle of 6060^\circ and a radius of 1212. A smaller circle is inscribed inside sector OABOAB such that it is tangent to radius OAOA, radius OBOB, and arc ABAB. If the area of the region inside sector OABOAB that lies outside the inscribed circle is expressed in the form kπk\pi, what is the value of kk?

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Cevap: 8

Cevap

The correct value of kk is 8.
By using the geometry of the 3030^\circ-6060^\circ-9090^\circ right triangle formed by the angle bisector and the radius of tangency, the radius of the inscribed circle is found to be r=4r = 4. Subtracting its area (16π16\pi) from the sector's area (24π24\pi) gives 8π8\pi, so k=8k = 8.

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1
Determine the relationship between the radius of the larger circle RR and the radius of the inscribed circle rr.
OP=2rOP = 2r and R=3rR = 3r.
The center PP of the inscribed circle lies on the angle bisector of AOB=60\angle AOB = 60^\circ, creating a 3030^\circ angle with radius OAOA. The perpendicular distance from PP to radius OAOA is rr, so sin(30)=rOP=12\sin(30^\circ) = \frac{r}{OP} = \frac{1}{2}, giving OP=2rOP = 2r. Since the inscribed circle touches arc ABAB, OP+r=ROP + r = R, so 3r=R3r = R.
2
Calculate the radius rr of the inscribed circle.
r=4r = 4.
Given R=12R = 12, solving 3r=123r = 12 yields r=4r = 4.
3
Compute the area of sector OABOAB.
Areasector=24π\text{Area}_{\text{sector}} = 24\pi.
The formula for the area of a sector is θ360πR2\frac{\theta}{360^\circ} \pi R^2. Here, 60360π(122)=16×144π=24π\frac{60^\circ}{360^\circ} \pi (12^2) = \frac{1}{6} \times 144\pi = 24\pi.
4
Compute the area of the inscribed circle.
Areacircle=16π\text{Area}_{\text{circle}} = 16\pi.
The area of a circle with radius r=4r = 4 is πr2=π(42)=16π\pi r^2 = \pi (4^2) = 16\pi.
5
Subtract the area of the inscribed circle from the area of sector OABOAB to find kk.
k=8k = 8.
Arearegion=24π16π=8π\text{Area}_{\text{region}} = 24\pi - 16\pi = 8\pi, which means k=8k = 8.

Anahtar Kavram

Inscribed shapes within sectors, arc length, and sector area relations
Soru 9Soru

Two concentric circles centered at point OO have radii of 66 units and 1010 units. A central angle θ\theta defines a sector that intersects the region between the concentric circles, creating a region bounded by an outer arc, an inner arc, and two straight line segments. If the area of this bounded region is 16π16\pi square units, what is the perimeter of the region?

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Cevap: 8+8π8 + 8\pi

Cevap

The perimeter of the region is 8+8π8 + 8\pi units.
The region between the concentric circles has an area equal to the fraction of the central angle times the area of the ring: θ360π(10262)=64π(θ360)\frac{\theta}{360^\circ} \cdot \pi (10^2 - 6^2) = 64\pi \left(\frac{\theta}{360^\circ}\right). Setting this equal to 16π16\pi gives θ360=14\frac{\theta}{360^\circ} = \frac{1}{4}. The outer arc length is 14(2π10)=5π\frac{1}{4}(2\pi \cdot 10) = 5\pi, and the inner arc length is 14(2π6)=3π\frac{1}{4}(2\pi \cdot 6) = 3\pi. The region is bounded on the sides by two segments of length 106=410 - 6 = 4 units each. The total perimeter is 5π+3π+4+4=8+8π5\pi + 3\pi + 4 + 4 = 8 + 8\pi.

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1
Express the area of the bounded region in terms of the central angle fraction f=θ360f = \frac{\theta}{360^\circ}.
Area=fπ(R2r2)=fπ(10262)=64πf\text{Area} = f \cdot \pi(R^2 - r^2) = f \cdot \pi(10^2 - 6^2) = 64\pi f.
The area of the region between two concentric sector arcs is the difference between the outer sector area and the inner sector area.
2
Solve for the fraction ff using the given area of 16π16\pi.
64πf=16π    f=16π64π=1464\pi f = 16\pi \implies f = \frac{16\pi}{64\pi} = \frac{1}{4}.
Equating the algebraic expression for the region's area to 16π16\pi allows finding the proportion of the circle represented by the central angle.
3
Calculate the lengths of the outer arc, inner arc, and straight boundary segments.
Outer Arc=14(2π10)=5π\text{Outer Arc} = \frac{1}{4}(2\pi \cdot 10) = 5\pi, Inner Arc=14(2π6)=3π\text{Inner Arc} = \frac{1}{4}(2\pi \cdot 6) = 3\pi, Segment Length=106=4\text{Segment Length} = 10 - 6 = 4.
Arc lengths use the formula f2πrf \cdot 2\pi r, and each straight boundary segment is the radial distance between the two circles.
4
Sum all boundary components to find the total perimeter.
Perimeter=5π+3π+4+4=8+8π\text{Perimeter} = 5\pi + 3\pi + 4 + 4 = 8 + 8\pi.
The complete perimeter of the bounded region consists of the outer arc, inner arc, and two radial segments.

Anahtar Kavram

Annular Sector Area and Perimeter
Tahmini Süre:1m 30s
Soru 10Soru

A circle has a radius of 66. A central angle of 6060^\circ intercepts an arc on the circle. What is the perimeter of the sector defined by this central angle?

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Cevap: 2π+122\pi + 12

Cevap

2π+122\pi + 12
The sector's perimeter includes the curved arc length and the two straight radii bounding it. The central angle of 6060^\circ represents 60360=16\frac{60}{360} = \frac{1}{6} of the full circle. The arc length is 16×2π(6)=2π\frac{1}{6} \times 2\pi(6) = 2\pi. Adding the two radii of length 66 gives 2π+6+6=2π+122\pi + 6 + 6 = 2\pi + 12.

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1
Calculate the arc length of the sector
Arc length = 2π2\pi
The arc length formula is Arc Length=θ360×2πr\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r. Substituting θ=60\theta = 60^\circ and r=6r = 6 yields 60360×12π=16×12π=2π\frac{60}{360} \times 12\pi = \frac{1}{6} \times 12\pi = 2\pi.
2
Calculate the perimeter of the sector by adding the arc length to the two radii
Perimeter = 2π+122\pi + 12
The perimeter of a sector consists of the arc length plus two radii (2r2r). Thus, Perimeter=2π+2(6)=2π+12\text{Perimeter} = 2\pi + 2(6) = 2\pi + 12.

Anahtar Kavram

Perimeter of a sector equals arc length plus twice the radius (L+2rL + 2r).
Tahmini Süre:45s
Soru 11Soru

A circle has a radius of 1010 units. A sector within this circle is defined by a central angle of 7272^\circ. Which of the following statements about this sector are true? Select all such statements.

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Cevap: The arc length of the sector is 4π4\pi units.; The area of the sector is 20π20\pi square units.; The ratio of the sector's area to the circle's total area is 11 to 55.

Cevap

The statements confirming an arc length of 4π4\pi units, a sector area of 20π20\pi square units, and a sector-to-total area ratio of 11 to 55 are correct.
The central angle fraction is 72360=15\frac{72^\circ}{360^\circ} = \frac{1}{5}. Multiplying the full circumference 20π20\pi by 15\frac{1}{5} gives an arc length of 4π4\pi. Multiplying the total area 100π100\pi by 15\frac{1}{5} gives a sector area of 20π20\pi. The ratio of sector area to total area is also equal to 15\frac{1}{5}.

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1
Find the central angle fraction of the circle
The fraction is 72360=15\frac{72^\circ}{360^\circ} = \frac{1}{5}
Arc length and sector area are proportional to the central angle relative to a full 360360^\circ turn.
2
Calculate the arc length of the sector
Arc Length =15×2π(10)=4π= \frac{1}{5} \times 2\pi(10) = 4\pi units
The arc length is the circle's circumference multiplied by the central angle fraction.
3
Calculate the area of the sector
Sector Area =15×π(10)2=20π= \frac{1}{5} \times \pi(10)^2 = 20\pi square units
The sector area is the circle's total area multiplied by the central angle fraction.
4
Determine the area ratio
Ratio =Sector AreaTotal Area=20π100π=15= \frac{\text{Sector Area}}{\text{Total Area}} = \frac{20\pi}{100\pi} = \frac{1}{5}
The ratio of the sector area to the total area is identical to the central angle fraction.

Anahtar Kavram

Arc Length and Sector Area Formulas
Tahmini Süre:1m 0s
Soru 12Soru

A circle has a circumference of 20π20\pi. A sector of this circle is defined by a central angle that intercepts an arc of length 5π5\pi. What is the area of this sector?

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Cevap: 25π25\pi

Cevap

The area of the sector is 25π25\pi.
First, the radius is found using 2πr=20π2\pi r = 20\pi, which gives r=10r = 10. The total area of the circle is π(10)2=100π\pi (10)^2 = 100\pi. The fraction of the circle defined by the sector is 5π20π=14\frac{5\pi}{20\pi} = \frac{1}{4}. Multiplying the total area by this fraction gives 14×100π=25π\frac{1}{4} \times 100\pi = 25\pi.

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1
Find the radius of the circle from the circumference.
Since C=2πr=20πC = 2\pi r = 20\pi, solving for rr gives r=10r = 10.
The radius is required to calculate the total circle area.
2
Determine the fraction of the circle that the sector represents.
\text{Fraction} = \frac{\text{Arc Length}}{\text{Circumference}} = \frac{5\pi}{20\pi} = \frac{1}{4}.
The ratio of arc length to circumference gives the proportion of the total circle occupied by the sector.
3
Calculate the total area of the circle.
A = \pi r^2 = \pi (10)^2 = 100\pi.
The area of a circle with radius 10 is 100π100\pi.
4
Multiply the total area by the sector fraction.
\text{Sector Area} = \frac{1}{4} \times 100\pi = 25\pi.
Applying the proportional fraction yields the sector area.

Anahtar Kavram

The ratio of a sector's arc length to the full circumference is equal to the ratio of the sector's area to the full circle's area.
Soru 13Soru

Circle KK has a radius of 1010. Points PP and QQ lie on circle KK such that central angle POQ\angle POQ measures 7272^\circ, where point OO is the center of circle KK. Which of the following statements are true? Select all that apply.

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Cevap: The length of minor arc PQPQ is 4π4\pi.; The area of sector POQPOQ is 20π20\pi.

Cevap

The correct statements are that the length of minor arc PQPQ is 4π4\pi, and the area of sector POQPOQ is 20π20\pi.
The central angle of 7272^\circ corresponds to 72360=15\frac{72^\circ}{360^\circ} = \frac{1}{5} of the entire circle. The circle has a circumference of 2π(10)=20π2\pi(10) = 20\pi and an area of π(10)2=100π\pi(10)^2 = 100\pi. Taking 15\frac{1}{5} of the circumference gives an arc length of 4π4\pi, and taking 15\frac{1}{5} of the total area gives a sector area of 20π20\pi. Both of these statements are mathematically accurate.

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1
Find the central angle fraction relative to the full circle
Fraction=72360=15\text{Fraction} = \frac{72^\circ}{360^\circ} = \frac{1}{5}
A complete circle subtends 360360^\circ, so the arc and sector comprise one-fifth of the circle.
2
Calculate the circumference and length of minor arc PQPQ
\text{Circumference} = 2\pi(10) = 20\pi, \quad \text{Arc Length } PQ = \frac{1}{5} \times 20\pi = 4\pi
Arc length is the central angle fraction multiplied by the total circumference.
3
Calculate the total circle area and area of sector POQPOQ
\text{Total Area} = \pi(10)^2 = 100\pi, \quad \text{Sector Area } POQ = \frac{1}{5} \times 100\pi = 20\pi
Sector area is the central angle fraction multiplied by the total area of the circle.
4
Evaluate sector perimeter and area ratio for remaining options
\text{Perimeter} = 4\pi + 2(10) = 4\pi + 20; \quad \text{Ratio} = \frac{20\pi}{100\pi} = \frac{1}{5}
Sector perimeter requires adding the two straight radii to the arc length, and the ratio of sector area to total area is 1:51:5.

Anahtar Kavram

Arc Length and Sector Area of a Circle
Tahmini Süre:1m 30s
Soru 14Soru

A sector of a circle has an area of 15π15\pi square units and a perimeter of 10+6π10 + 6\pi units. What is the measure of the central angle of the sector, in degrees?

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Cevap: 216216^\circ

Cevap

The correct answer is 216216^\circ, corresponding to the option stating 216216^\circ.
The perimeter of a sector is defined as 2r+s=10+6π2r + s = 10 + 6\pi, which yields a radius of r=5r = 5 and an arc length s=6πs = 6\pi. Checking with the sector area formula 12rs=12(5)(6π)=15π\frac{1}{2}rs = \frac{1}{2}(5)(6\pi) = 15\pi confirms these measurements. The total area of the circle is π(5)2=25π\pi (5)^2 = 25\pi. The sector thus constitutes 15π25π=35\frac{15\pi}{25\pi} = \frac{3}{5} of the circle. Multiplying this fraction by 360360^\circ gives a central angle of 216216^\circ.

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1
Express the sector perimeter formula and solve for radius and arc length.
Radius r=5r = 5 and arc length s=6πs = 6\pi.
The perimeter of a sector equals two radii plus its arc length: Perimeter=2r+s=10+6π\text{Perimeter} = 2r + s = 10 + 6\pi. Equating standard and π\pi-termed components gives 2r=10    r=52r = 10 \implies r = 5 and arc length s=6πs = 6\pi.
2
Verify consistency using the sector area formula.
Area=15π\text{Area} = 15\pi, matching the given information.
The area of a sector can also be calculated as 12rs=12(5)(6π)=15π\frac{1}{2} r s = \frac{1}{2}(5)(6\pi) = 15\pi.
3
Find the total area of the circle and the fraction of the circle occupied by the sector.
Total area =25π= 25\pi, area fraction =35= \frac{3}{5}.
The full circle area is πr2=π(52)=25π\pi r^2 = \pi (5^2) = 25\pi. The sector represents 15π25π=35\frac{15\pi}{25\pi} = \frac{3}{5} of the entire circle.
4
Calculate the central angle θ\theta in degrees.
θ=216\theta = 216^\circ.
Multiply the fraction by 360360^\circ: θ=35×360=216\theta = \frac{3}{5} \times 360^\circ = 216^\circ.

Anahtar Kavram

Perimeter, Arc Length, and Area of a Circle Sector
Soru 15Soru

In a circle centered at point OO with radius RR, radii OAOA and OBOB bound a sector AOBAOB with a central angle of 120120^\circ. Point PP lies on minor arc ABAB such that the ratio of the length of arc APAP to the length of arc PBPB is 1:31:3. Segment PQPQ is drawn perpendicular to radius OAOA, intersecting OAOA at point QQ. What is the ratio of the area of the region bounded by line segment PQPQ, line segment AQAQ, and minor arc APAP to the area of sector AOBAOB?

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Cevap: 4π3316π\frac{4\pi - 3\sqrt{3}}{16\pi}

Cevap

4π3316π\frac{4\pi - 3\sqrt{3}}{16\pi}
The correct answer 4π3316π\frac{4\pi - 3\sqrt{3}}{16\pi} is obtained by finding the central angle of sector AOPAOP (3030^\circ), subtracting the area of right triangle OQPOQP (316R2\frac{\sqrt{3}}{16}R^2) from the area of sector AOPAOP (π12R2\frac{\pi}{12}R^2), and dividing this bounded area by the total sector AOBAOB area (π3R2\frac{\pi}{3}R^2).

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1
Determine the central angle AOP\angle AOP of sector AOPAOP
AOP=11+3×120=30\angle AOP = \frac{1}{1+3} \times 120^\circ = 30^\circ
Arc length is directly proportional to central angle. A 1:31:3 ratio means sector AOPAOP represents 14\frac{1}{4} of the 120120^\circ central angle.
2
Calculate the area of sector AOPAOP and the total area of sector AOBAOB
Area(Sector AOP)=30360πR2=π12R2\text{Area(Sector } AOP) = \frac{30^\circ}{360^\circ} \pi R^2 = \frac{\pi}{12} R^2, and Area(Sector AOB)=120360πR2=π3R2\text{Area(Sector } AOB) = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{\pi}{3} R^2
The area of a sector with central angle θ\theta is θ360πR2\frac{\theta}{360^\circ} \pi R^2.
3
Calculate the area of right triangle OQPOQP
Area(OQP)=316R2\text{Area}(\triangle OQP) = \frac{\sqrt{3}}{16} R^2
In OQP\triangle OQP, QOP=30\angle QOP = 30^\circ and hypotenuse OP=ROP = R. Thus OQ=Rcos(30)=R32OQ = R \cos(30^\circ) = \frac{R\sqrt{3}}{2} and PQ=Rsin(30)=R2PQ = R \sin(30^\circ) = \frac{R}{2}. Area=12×OQ×PQ=12(R32)(R2)=316R2\text{Area} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \left(\frac{R\sqrt{3}}{2}\right)\left(\frac{R}{2}\right) = \frac{\sqrt{3}}{16} R^2.
4
Calculate the area of the bounded region and take the ratio to the area of sector AOBAOB
Ratio=π12R2316R2π3R2=4π3316π\text{Ratio} = \frac{\frac{\pi}{12}R^2 - \frac{\sqrt{3}}{16}R^2}{\frac{\pi}{3}R^2} = \frac{4\pi - 3\sqrt{3}}{16\pi}
Subtract the triangle area from sector AOPAOP area to get the bounded region area, then divide by sector AOBAOB area.

Anahtar Kavram

Sector Area, Arc Length Proportions, and Geometric Region Subdivision
Tahmini Süre:3m 0s
Soru 16Soru

Two concentric circles centered at point OO have radii rr and RR, where r<Rr < R. A sector bounded by radii OAOA and OBOB of the outer circle has central angle θ\theta^\circ. Region SS is the region lying inside sector AOBAOB but outside the inner circle. The area of region SS is equal to 33 times the area of the sector of the inner circle bounded by central angle θ\theta^\circ. If the perimeter of region SS is equal to 116\frac{11}{6} times the length of arc ABAB, what is the value of θ\theta?

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Cevap: 540π\frac{540}{\pi}

Cevap

540π\frac{540}{\pi}
The area of region SS is θ360π(R2r2)\frac{\theta}{360}\pi(R^2 - r^2), which is equal to 33 times the area of the inner sector θ360πr2\frac{\theta}{360}\pi r^2. This simplifies to R2r2=3r2    R=2rR^2 - r^2 = 3r^2 \implies R = 2r. The perimeter of region SS consists of outer arc length LouterL_{\text{outer}}, inner arc length Linner=12LouterL_{\text{inner}} = \frac{1}{2} L_{\text{outer}}, and two straight line segments each of length Rr=rR - r = r. Setting the total perimeter 32Louter+2r\frac{3}{2} L_{\text{outer}} + 2r equal to 116Louter\frac{11}{6} L_{\text{outer}} yields 2r=13Louter2r = \frac{1}{3} L_{\text{outer}}, or Louter=6rL_{\text{outer}} = 6r. Substituting Louter=θ360(2πR)=4πrθ360L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{4\pi r \theta}{360} gives 4πrθ360=6r\frac{4\pi r \theta}{360} = 6r, which solves to θ=540π\theta = \frac{540}{\pi}.

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1
Relate outer radius RR to inner radius rr using sector areas.
R=2rR = 2r
The area of region SS is θ360π(R2r2)\frac{\theta}{360}\pi(R^2 - r^2) and the inner sector area is θ360πr2\frac{\theta}{360}\pi r^2. Setting θ360π(R2r2)=3θ360πr2\frac{\theta}{360}\pi(R^2 - r^2) = 3 \cdot \frac{\theta}{360}\pi r^2 gives R2r2=3r2R^2 - r^2 = 3r^2, so R2=4r2R^2 = 4r^2 and R=2rR = 2r.
2
Express the perimeter of region SS in terms of outer arc length LouterL_{\text{outer}} and radius rr.
Perimeter(S)=32Louter+2r\text{Perimeter}(S) = \frac{3}{2} L_{\text{outer}} + 2r
Region SS is bounded by outer arc ABAB (LouterL_{\text{outer}}), inner arc CDCD (LinnerL_{\text{inner}}), and two straight segments ACAC and BDBD of length Rr=2rr=rR - r = 2r - r = r. Since R=2rR = 2r, Linner=12LouterL_{\text{inner}} = \frac{1}{2} L_{\text{outer}}, making total perimeter Louter+12Louter+2(r)=32Louter+2rL_{\text{outer}} + \frac{1}{2} L_{\text{outer}} + 2(r) = \frac{3}{2} L_{\text{outer}} + 2r.
3
Use the given perimeter relationship to express LouterL_{\text{outer}} in terms of rr.
Louter=6rL_{\text{outer}} = 6r
Setting 32Louter+2r=116Louter\frac{3}{2} L_{\text{outer}} + 2r = \frac{11}{6} L_{\text{outer}} yields 2r=(11696)Louter=13Louter2r = \left(\frac{11}{6} - \frac{9}{6}\right) L_{\text{outer}} = \frac{1}{3} L_{\text{outer}}, so Louter=6rL_{\text{outer}} = 6r.
4
Solve for θ\theta using the definition of outer arc length.
θ=540π\theta = \frac{540}{\pi}
Louter=θ360(2πR)=θ360(4πr)L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{\theta}{360}(4\pi r). Setting 4πrθ360=6r\frac{4\pi r \theta}{360} = 6r simplifies to πθ90=6\frac{\pi \theta}{90} = 6, giving θ=540π\theta = \frac{540}{\pi}.

Anahtar Kavram

Annular sector area and perimeter relations combining arc length formulas and concentric circle geometry.
Tahmini Süre:2m 30s
Soru 17Soru

A circle has a radius of 99 inches. What is the area, in square inches, of a sector of this circle formed by a central angle of 4040^\circ?

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Cevap: 9π9\pi

Cevap

The area of the sector is 9π9\pi square inches.
The area of a sector with radius rr and central angle θ\theta is calculated as Area=πr2(θ360)\text{Area} = \pi r^2 \left(\frac{\theta}{360^\circ}\right). Substituting r=9r = 9 and θ=40\theta = 40^\circ yields π(92)(40360)=81π(19)=9π\pi (9^2) \left(\frac{40^\circ}{360^\circ}\right) = 81\pi \left(\frac{1}{9}\right) = 9\pi square inches.

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1
Calculate the area of the entire circle using the formula Acircle=πr2A_{\text{circle}} = \pi r^2.
Acircle=π(92)=81πA_{\text{circle}} = \pi (9^2) = 81\pi square inches.
The full circle area provides the total measure from which the sector fraction is calculated.
2
Determine the fraction of the circle represented by the central angle of 4040^\circ.
40360=19\frac{40^\circ}{360^\circ} = \frac{1}{9}.
A circle contains 360360^\circ in total, so the central angle over 360360^\circ gives the proportion of the circle covered by the sector.
3
Multiply the total area of the circle by the fraction of the sector.
Sector Area=81π×19=9π\text{Sector Area} = 81\pi \times \frac{1}{9} = 9\pi square inches.
Applying the fraction to the total area yields the specific sector area.

Anahtar Kavram

Sector Area of a Circle
Soru 18Soru

Circle C1C_1 has radius r1r_1 and Circle C2C_2 has radius r2r_2. An arc on Circle C1C_1 subtended by a central angle of θ1\theta_1^\circ has the exact same length as an arc on Circle C2C_2 subtended by a central angle of θ2\theta_2^\circ. The sector formed by this arc in Circle C1C_1 has an area of 54π54\pi, and the sector formed by this arc in Circle C2C_2 has an area of 36π36\pi. If θ1+θ2=150\theta_1 + \theta_2 = 150, what is the value of r1r_1?

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Cevap: 18

Cevap

The radius r1r_1 of Circle C1C_1 is 18.
Using the relationship A=12LrA = \frac{1}{2} L r, the ratio of the two sector areas gives r1r2=54π36π=1.5\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = 1.5, so r2=23r1r_2 = \frac{2}{3} r_1. Expressing arc length as L=108πr1L = \frac{108\pi}{r_1}, the central angles are θ1=19440r12\theta_1 = \frac{19440}{r_1^2} and θ2=29160r12\theta_2 = \frac{29160}{r_1^2}. Setting their sum equal to 150150 leads to 48600r12=150\frac{48600}{r_1^2} = 150, giving r12=324r_1^2 = 324 and r1=18r_1 = 18.

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1
Relate sector area to arc length and radius
A1=12Lr1=54πA_1 = \frac{1}{2} L r_1 = 54\pi and A2=12Lr2=36πA_2 = \frac{1}{2} L r_2 = 36\pi
The area of a sector with arc length LL and radius rr is given by A=θ360πr2=12LrA = \frac{\theta}{360}\pi r^2 = \frac{1}{2} L r.
2
Find the ratio of r1r_1 to r2r_2
r1r2=54π36π=32    r2=23r1\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = \frac{3}{2} \implies r_2 = \frac{2}{3}r_1
Dividing the first area equation by the second cancels out 12L\frac{1}{2}L.
3
Express central angles in terms of r1r_1
θ1=19440r12\theta_1 = \frac{19440}{r_1^2} and θ2=29160r12\theta_2 = \frac{29160}{r_1^2}
Since L=108πr1L = \frac{108\pi}{r_1}, substituting into θ1=L2πr1×360\theta_1 = \frac{L}{2\pi r_1} \times 360 yields θ1=19440r12\theta_1 = \frac{19440}{r_1^2}, and substituting into θ2=L2π(23r1)×360\theta_2 = \frac{L}{2\pi (\frac{2}{3}r_1)} \times 360 yields θ2=29160r12\theta_2 = \frac{29160}{r_1^2}.
4
Solve for r1r_1 using the angle sum equation
r1=18r_1 = 18
Summing the angles gives 19440+29160r12=48600r12=150    r12=324    r1=18\frac{19440 + 29160}{r_1^2} = \frac{48600}{r_1^2} = 150 \implies r_1^2 = 324 \implies r_1 = 18.

Anahtar Kavram

Relationship between arc length, radius, central angle, and sector area
Tahmini Süre:2m 30s
Soru 19Soru

A square is inscribed in a circle with center OO. The perimeter of the square is 16216\sqrt{2} units. A sector of this circle has an area equal to the total area of the region inside the circle that lies outside the square. What is the arc length of this sector?

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Cevap: 8π168\pi - 16

Cevap

8π168\pi - 16
The correct answer is 8π168\pi - 16. The inscribed square has side length 424\sqrt{2} and diagonal 88. Since the diagonal of an inscribed square is the circle's diameter, the circle has radius r=4r = 4 and total area 16π16\pi. The square has area 3232, so the region outside the square has area 16π3216\pi - 32. For any sector of radius rr, the arc length LL and sector area AA satisfy L=2ArL = \frac{2A}{r}. Substituting A=16π32A = 16\pi - 32 and r=4r = 4 yields L=2(16π32)4=8π16L = \frac{2(16\pi - 32)}{4} = 8\pi - 16.

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1
Find the side length and diagonal of the inscribed square.
Side length s=1624=42s = \frac{16\sqrt{2}}{4} = 4\sqrt{2}. Diagonal d=s2=(42)2=8d = s\sqrt{2} = (4\sqrt{2})\sqrt{2} = 8.
The perimeter of a square is 4s4s, and the diagonal of a square with side ss is s2s\sqrt{2}.
2
Determine the radius and total area of the circle.
Diameter equals diagonal d=8d = 8, so radius r=4r = 4. Area of circle Acircle=πr2=16πA_{\text{circle}} = \pi r^2 = 16\pi.
A square inscribed in a circle has its diagonal aligned with the diameter of the circle.
3
Calculate the area of the region inside the circle but outside the square.
Area of square Asquare=(42)2=32A_{\text{square}} = (4\sqrt{2})^2 = 32. Area outside square Aoutside=16π32A_{\text{outside}} = 16\pi - 32.
Subtract the area of the inscribed square from the total area of the circle.
4
Relate sector area to arc length using L=2AsectorrL = \frac{2 A_{\text{sector}}}{r}.
L=2(16π32)4=32π644=8π16L = \frac{2(16\pi - 32)}{4} = \frac{32\pi - 64}{4} = 8\pi - 16.
Since sector area Asector=θ360πr2A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2 and arc length L=θ360(2πr)L = \frac{\theta}{360^\circ}(2\pi r), we have L=2AsectorrL = \frac{2A_{\text{sector}}}{r}.

Anahtar Kavram

Relationship between circle area, inscribed figures, sector area, and arc length
Tahmini Süre:1m 30s
Soru 20Soru

In circle OO, minor arc ABAB is subtended by a central angle of 120120^\circ, and the radius of circle OO is 66. A smaller circle, circle CC, is constructed such that it is tangent to chord ABAB at its midpoint and tangent to minor arc ABAB at its midpoint. What is the area of a sector of circle CC subtended by a central angle of 9090^\circ?

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Cevap: 9π16\frac{9\pi}{16}

Cevap

9π16\frac{9\pi}{16}
The distance from the center of circle OO to chord ABAB is 6cos(60)=36 \cos(60^\circ) = 3, leaving a height of 63=36 - 3 = 3 for the minor segment. Circle CC is tangent to the chord and arc midpoints, making its diameter equal to 33 and its radius 32\frac{3}{2}. The area of a 9090^\circ sector of circle CC is 90360π(32)2=9π16\frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{9\pi}{16}.

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1
Find the perpendicular distance from center OO to chord ABAB.
In isosceles triangle OABOAB with OA=OB=6OA = OB = 6 and central angle AOB=120\angle AOB = 120^\circ, the perpendicular bisector from OO to ABAB meets ABAB at midpoint MM. AOM=60\angle AOM = 60^\circ, so OM=OAcos(60)=6×12=3OM = OA \cos(60^\circ) = 6 \times \frac{1}{2} = 3.
Determining OMOM allows us to find the sagitta (height) of the minor segment bounded by chord ABAB and minor arc ABAB.
2
Calculate the height of the minor segment (diameter of circle CC).
The height of the minor segment along the radius passing through MM to arc midpoint NN is MN=ONOM=63=3MN = ON - OM = 6 - 3 = 3. Since circle CC is tangent to ABAB at MM and to the arc at NN, segment MNMN is a diameter of circle CC, giving a diameter of 33.
The space between the chord midpoint and arc midpoint bounds circle CC, defining its diameter.
3
Find the radius of circle CC.
Radius rC=diameter2=32r_C = \frac{\text{diameter}}{2} = \frac{3}{2}.
The radius of circle CC is needed to calculate its sector area.
4
Compute the area of the 9090^\circ sector of circle CC.
\text{Area} = \frac{\theta}{360^\circ} \pi r_C^2 = \frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{1}{4} \pi \left(\frac{9}{4}\right) = \frac{9\pi}{16}.
Multiply the fraction of the full circle represented by the central angle by the total area of circle CC.

Anahtar Kavram

Calculating sector area of an inscribed circle within a circle segment
Tahmini Süre:2m 0s
Sayfa 1 / 2Sonraki
Circles, Arc Lengths, and Sector Areas Alıştırma Soruları — GRE General Test | Examkin