Circles, Arc Lengths, and Sector Areas

31 soru

Soru 21Soru

In a circle centered at point OO with a radius of 66 units, points PP and QQ lie on the circle such that central angle POQ\angle POQ measures 120120^\circ. Which of the following statements are true? Select all such statements.

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Cevap: The length of minor arc PQPQ is 4π4\pi.; The area of minor sector POQPOQ is 12π12\pi.; The length of chord PQPQ is 636\sqrt{3}.

Cevap

The true statements are that the length of minor arc PQPQ is 4π4\pi, the area of minor sector POQPOQ is 12π12\pi, and the length of chord PQPQ is 636\sqrt{3}.
The arc length of a 120120^\circ central angle is 13\frac{1}{3} of the total circumference 12π12\pi, which gives 4π4\pi. The sector area is 13\frac{1}{3} of the total area 36π36\pi, which gives 12π12\pi. Furthermore, by geometry of an isosceles triangle with vertex angle 120120^\circ and congruent sides 66, the base chord length is 2×6sin(60)=632 \times 6 \sin(60^\circ) = 6\sqrt{3}.

Adım Adım Çözüm

1
Calculate the arc length of minor arc PQPQ
Arc length = 4π4\pi
Arc length equals θ360×2πr=120360×2π(6)=4π\frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi(6) = 4\pi.
2
Calculate the area of sector POQPOQ
Sector area = 12π12\pi
Sector area equals θ360×πr2=120360×π(62)=12π\frac{\theta}{360^\circ} \times \pi r^2 = \frac{120^\circ}{360^\circ} \times \pi(6^2) = 12\pi.
3
Determine the perimeter of sector POQPOQ
Perimeter = 4π+124\pi + 12
The boundary of a sector consists of the curved arc length plus two straight radii: 4π+6+6=4π+124\pi + 6 + 6 = 4\pi + 12.
4
Calculate the length of chord PQPQ
Chord PQ=63PQ = 6\sqrt{3}
Using triangle OPQOPQ with sides OP=OQ=6OP = OQ = 6 and POQ=120\angle POQ = 120^\circ, law of cosines or bisecting POQ\angle POQ into two 3030^\circ-6060^\circ-9090^\circ triangles yields chord length 2×(6sin60)=2×33=632 \times (6 \sin 60^\circ) = 2 \times 3\sqrt{3} = 6\sqrt{3}.

Anahtar Kavram

Arc Length, Sector Area, and Chord Relationships in Circles
Tahmini Süre:1m 30s
Soru 22Soru

In a circle with center OO and radius 1212, radii OAOA and OBOB are perpendicular. Point CC lies on segment OAOA such that CC is the midpoint of OAOA. A line segment perpendicular to OAOA is drawn from point CC to intersect minor arc ABAB at point DD. What is the area of the region bounded by line segment CDCD, line segment CACA, and minor arc ADAD?

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Cevap: 24π18324\pi - 18\sqrt{3}

Cevap

24π18324\pi - 18\sqrt{3}
The area of the region bounded by line segment CDCD, segment CACA, and minor arc ADAD is obtained by subtracting the area of right triangle OCDOCD from the area of sector OADOAD. With OC=6OC = 6 and radius OD=12OD = 12, triangle OCDOCD is a 30609030^\circ-60^\circ-90^\circ right triangle, giving CD=63CD = 6\sqrt{3} and central angle AOD=60\angle AOD = 60^\circ. The sector area is 60360π(122)=24π\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = 24\pi, and the triangle area is 12663=183\frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}. Subtracting the triangle area from the sector area yields 24π18324\pi - 18\sqrt{3}.

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1
Determine the length of OCOC and the height CDCD
OC=6OC = 6 and CD=63CD = 6\sqrt{3}
Since CC is the midpoint of radius OA=12OA = 12, OC=6OC = 6. Triangle OCDOCD is a right triangle at CC with hypotenuse OD=12OD = 12 (radius of circle). By the Pythagorean theorem, CD=12262=108=63CD = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}.
2
Find the central angle AOD\angle AOD
AOD=60\angle AOD = 60^\circ
In right triangle OCDOCD, cos(AOD)=OCOD=612=12\cos(\angle AOD) = \frac{OC}{OD} = \frac{6}{12} = \frac{1}{2}, which implies AOD=60\angle AOD = 60^\circ.
3
Calculate the area of sector OADOAD
Area(Sector OAD)=24π\text{Area(Sector } OAD) = 24\pi$
The area of a sector with central angle 6060^\circ and radius 1212 is 60360π(122)=16144π=24π\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = \frac{1}{6} \cdot 144\pi = 24\pi.
4
Calculate the area of right triangle OCDOCD
Area(Triangle OCD)=183\text{Area(Triangle } OCD) = 18\sqrt{3}
The area of right triangle OCDOCD is 12baseheight=12663=183\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}.
5
Subtract the area of triangle OCDOCD from the area of sector OADOAD
Area of shaded region=24π183\text{Area of shaded region} = 24\pi - 18\sqrt{3}
The bounded region is formed by removing triangle OCDOCD from sector OADOAD.

Anahtar Kavram

Calculating the area of a region bounded by a circle arc and line segments by subtracting a right triangle area from a sector area.
Soru 23Soru

A sector of a circle with a radius of 1010 units has a total perimeter of 20+5π20 + 5\pi units. What is the area of this sector?

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Cevap: 25π25\pi

Cevap

The area of the sector is 25π25\pi.
The total perimeter of a sector with radius rr and arc length LL is given by P=2r+LP = 2r + L. Substituting r=10r = 10 gives 20+5π=20+L20 + 5\pi = 20 + L, so L=5πL = 5\pi. Using the sector area formula A=12rLA = \frac{1}{2} r L, the area is 12×10×5π=25π\frac{1}{2} \times 10 \times 5\pi = 25\pi.

Adım Adım Çözüm

1
Set up the formula for the perimeter of a sector.
P=2r+LP = 2r + L, where r=10r = 10 is the radius and LL is the arc length.
A sector's perimeter is bounded by two straight radii and one curved arc.
2
Solve for the arc length LL.
20+5π=2(10)+L    L=5π20 + 5\pi = 2(10) + L \implies L = 5\pi.
Subtracting the combined length of the two radii (2020) isolates the arc length.
3
Calculate the area of the sector.
Sector Area = 12rL=12(10)(5π)=25π\frac{1}{2} r L = \frac{1}{2} (10)(5\pi) = 25\pi.
The area of a sector can be directly evaluated using half the product of its radius and arc length.

Anahtar Kavram

Perimeter, Arc Length, and Area of a Circular Sector

Alternatif Yöntem

Find the central angle θ\theta first: Since L=5πL = 5\pi and circumference C=2π(10)=20πC = 2\pi(10) = 20\pi, the fraction of the circle is 5π20π=14\frac{5\pi}{20\pi} = \frac{1}{4}, which corresponds to θ=90\theta = 90^\circ. The area is then 14×π(102)=25π\frac{1}{4} \times \pi(10^2) = 25\pi.
Tahmini Süre:1m 30s
Soru 24Soru

Two concentric circles are centered at point OO. The inner circle has a radius of 66 units, and the outer circle has a radius of 636\sqrt{3} units. Radii OAOA and OBOB of the outer circle form a central angle AOB=60\angle AOB = 60^\circ and intersect the inner circle at points CC and DD, respectively. Which of the following statements must be true regarding the region and boundary lengths defined by these figures? Select all that apply.

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Cevap: The area of the region bounded by arc ABAB, arc CDCD, segment ACAC, and segment BDBD is equal to 12π12\pi square units.; The straight-line distance between points AA and BB is 636\sqrt{3} units.; The area of the circular segment bounded by chord ABAB and minor arc ABAB is 18π27318\pi - 27\sqrt{3} square units.

Cevap

The true statements are: the area of the region bounded by arc AB, arc CD, segment AC, and segment BD is 12π square units; the straight-line distance between points A and B is 6√3 units; and the area of the circular segment bounded by chord AB and minor arc AB is 18π - 27√3 square units.
The area of the region between the two concentric arcs is obtained by subtracting the inner sector area (6π) from the outer sector area (18π), giving 12π square units. The triangle OAB is equilateral because it has two sides of length 6√3 and an included angle of 60°, making chord AB equal to 6√3. Subtracting the area of this equilateral triangle (27√3) from the outer sector area (18π) yields the area of the circular segment bounded by chord AB and arc AB, which is 18π - 27√3.

Adım Adım Çözüm

1
Calculate the area of sector OAB and sector OCD to evaluate the area of the annular sector region.
Sector OAB area = (60/360) * π * (6√3)^2 = (1/6) * 108π = 18π. Sector OCD area = (60/360) * π * (6^2) = (1/6) * 36π = 6π. Region area = 18π - 6π = 12π.
The area between two concentric sector arcs bounded by the same radii is the difference in sector areas.
2
Determine the length of chord AB using triangle properties.
In triangle OAB, OA = OB = 6√3 and angle AOB = 60°. An isosceles triangle with a 60° vertex angle is equilateral, so AB = 6√3.
All internal angles of an isosceles triangle with a 60° angle must equal 60°.
3
Calculate the area of circular segment AB.
Area of equilateral triangle OAB = (√3 / 4) * (6√3)^2 = 27√3. Segment area = Sector OAB area - Triangle OAB area = 18π - 27√3.
A circular segment's area is found by subtracting the area of the subtended triangle from the area of the corresponding sector.
4
Evaluate arc length ratio and total perimeter of the annular sector.
Arc CD / Arc AB = 6 / (6√3) = 1 / √3 ≠ 1 / 3. Total perimeter = Arc AB + Arc CD + 2*(R - r) = 2√3π + 2π + 2*(6√3 - 6) = 2π(1 + √3) + 12√3 - 12.
Verifies that statements regarding ratio 1:3 and incomplete perimeter calculations are mathematically false.

Anahtar Kavram

Concentric circle geometry, arc length proportions, sector area calculations, and circular segment area formulation.
Soru 25Soru

In a circle centered at point OO, the ratio of the area of sector AOBAOB to the area of the entire circle is 3:83:8. If the total perimeter of sector AOBAOB is 24+9π24 + 9\pi, what is the radius of the circle?

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Cevap: 12

Cevap

The radius of the circle is 12.
Since the ratio of the sector area to the total circle area is 3:83:8, the sector accounts for 38\frac{3}{8} of the entire circumference 2πr2\pi r, making the arc length L=38(2πr)=34πrL = \frac{3}{8}(2\pi r) = \frac{3}{4}\pi r. The perimeter of the sector is 2r+L=2r+34πr2r + L = 2r + \frac{3}{4}\pi r. Equating this to 24+9π24 + 9\pi gives 2r=242r = 24, so r=12r = 12.

Adım Adım Çözüm

1
Relate sector area ratio to arc length
Arc length L=34πrL = \frac{3}{4}\pi r
The fraction of the circle occupied by the sector is 38\frac{3}{8}, so the arc length is 38\frac{3}{8} of the circle's circumference 2πr2\pi r.
2
Formulate the perimeter expression for sector AOBAOB
Perimeter = 2r+34πr2r + \frac{3}{4}\pi r
The boundary of a sector includes two straight radii of length rr plus the curved arc length LL.
3
Equate to given perimeter and solve for rr
r=12r = 12
Comparing 2r+34πr2r + \frac{3}{4}\pi r to 24+9π24 + 9\pi, setting 2r=242r = 24 yields r=12r = 12, which also satisfies 34π(12)=9π\frac{3}{4}\pi (12) = 9\pi.

Anahtar Kavram

Perimeter of a circle sector and fractional relationship between sector area, central angle, and arc length.
Soru 26Soru

In circle OO, points PP and QQ lie on the circumference such that the ratio of the minor arc length PQPQ to the radius rr of the circle is 5π6\frac{5\pi}{6}. If the area of sector POQPOQ is 30π30\pi, what is the total perimeter of sector POQPOQ?

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Cevap: 122+52π12\sqrt{2} + 5\sqrt{2}\pi

Cevap

122+52π12\sqrt{2} + 5\sqrt{2}\pi
The central angle θ\theta in radians is equal to the ratio of minor arc length to radius, so θ=5π6\theta = \frac{5\pi}{6}. Using the area of a sector formula A=12r2θA = \frac{1}{2}r^2\theta, we substitute A=30πA = 30\pi to get 12r2(5π6)=30π\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi, which simplifies to r2=72r^2 = 72, or r=62r = 6\sqrt{2}. The arc length is then s=rθ=(62)(5π6)=52πs = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi. The total perimeter of the sector includes both bounding radii and the arc length: 2r+s=122+52π2r + s = 12\sqrt{2} + 5\sqrt{2}\pi.

Adım Adım Çözüm

1
Relate the central angle in radians to the given ratio of arc length to radius.
The central angle θ\theta in radians is given by θ=arc lengthr=5π6\theta = \frac{\text{arc length}}{r} = \frac{5\pi}{6}.
By definition of radian measure, arc length s=rθs = r\theta, so sr=θ\frac{s}{r} = \theta.
2
Use the sector area formula to solve for the radius rr.
Setting 12r2(5π6)=30π\frac{1}{2}r^2\left(\frac{5\pi}{6}\right) = 30\pi yields 5π12r2=30π    r2=72    r=62\frac{5\pi}{12}r^2 = 30\pi \implies r^2 = 72 \implies r = 6\sqrt{2}.
The area of a sector with central angle θ\theta (in radians) is A=12r2θA = \frac{1}{2}r^2\theta.
3
Calculate the minor arc length ss.
Arc length s=rθ=(62)(5π6)=52πs = r\theta = (6\sqrt{2})\left(\frac{5\pi}{6}\right) = 5\sqrt{2}\pi.
Multiplying the radius by the central angle in radians gives the length of the subtended arc.
4
Calculate the total perimeter of sector POQPOQ.
Perimeter =2r+s=2(62)+52π=122+52π= 2r + s = 2(6\sqrt{2}) + 5\sqrt{2}\pi = 12\sqrt{2} + 5\sqrt{2}\pi.
The perimeter of a sector consists of the two bounding radii plus the arc length.

Anahtar Kavram

Relationship between radian measure, sector area, arc length, and sector perimeter
Tahmini Süre:2m 0s
Soru 27Soru

In circle OO, sector AOBAOB has an area of 45π45\pi square units and arc ABAB has a length of 6π6\pi units. What is the perimeter of sector AOBAOB?

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Cevap: 30+6π30 + 6\pi

Cevap

30+6π30 + 6\pi
The area of a sector is given by 12rL\frac{1}{2} r L, where rr is the radius and LL is the arc length. Substituting L=6πL = 6\pi and Area =45π= 45\pi yields 45π=12r(6π)45\pi = \frac{1}{2} r (6\pi), which simplifies to 45π=3πr45\pi = 3\pi r, so r=15r = 15. The perimeter of the sector is L+2r=6π+2(15)=30+6πL + 2r = 6\pi + 2(15) = 30 + 6\pi. Thus, the option equal to 30+6π30 + 6\pi is correct.

Adım Adım Çözüm

1
Express sector area and arc length in terms of radius rr and central angle θ\theta in degrees.
Sector Area = θ360×πr2=45π\frac{\theta}{360^\circ} \times \pi r^2 = 45\pi and Arc Length = θ360×2πr=6π\frac{\theta}{360^\circ} \times 2\pi r = 6\pi.
Relating both formulas allows solving for the radius directly using the ratio of area to arc length.
2
Divide the sector area formula by the arc length formula to find radius rr.
\frac{\text{Sector Area}}{\text{Arc Length}} = \frac{\frac{\theta}{360^\circ} \pi r^2}{\frac{\theta}{360^\circ} 2\pi r} = \frac{r}{2} = \frac{45\pi}{6\pi} = 7.5 \implies r = 15.
The central angle fraction θ360\frac{\theta}{360^\circ} and π\pi cancel out, leaving r2=7.5\frac{r}{2} = 7.5.
3
Calculate the total perimeter of sector AOBAOB.
\text{Perimeter} = \text{Arc Length} + 2r = 6\pi + 2(15) = 30 + 6\pi.
The perimeter of a sector consists of the outer arc length plus the two straight boundary radii (OAOA and OBOB).

Anahtar Kavram

The relationship between sector area, arc length, radius, and sector perimeter
Soru 28Soru

In circle OO, line segments ABAB and CDCD are perpendicular diameters, each of length 1212. An arc of a second circle, centered at point AA with radius ACAC, is drawn from point CC to point DD through the interior of circle OO. What is the area of the crescent-shaped region bounded by the semicircle CBDCBD of circle OO and arc CDCD of the second circle?

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Cevap: 36

Cevap

36
Circle OO has radius r=6r = 6, giving semicircle CBDCBD an area of 12π(62)=18π\frac{1}{2}\pi(6^2) = 18\pi. The distance AC=62+62=62AC = \sqrt{6^2 + 6^2} = 6\sqrt{2} is the radius of circle AA. Because CAD=90\angle CAD = 90^\circ, sector ACDACD has area 90360π(62)2=18π\frac{90^\circ}{360^\circ}\pi(6\sqrt{2})^2 = 18\pi. Subtracting the area of triangle ACDACD (12×12×6=36\frac{1}{2} \times 12 \times 6 = 36) yields a segment area of 18π3618\pi - 36. Subtracting this segment area from the semicircle area yields 18π(18π36)=3618\pi - (18\pi - 36) = 36.

Adım Adım Çözüm

1
Find the radii of circle OO and circle AA.
Radius of circle OO is r=6r = 6. In right triangle AOCAOC, OA=OC=6OA = OC = 6, so radius AC=62+62=62AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}.
Perpendicular diameters ABAB and CDCD intersect at center OO, dividing each diameter into radii of length 66.
2
Calculate the area of sector ACDACD of circle AA and triangle ACDACD.
Sector area =90360π(62)2=18π= \frac{90^\circ}{360^\circ} \pi (6\sqrt{2})^2 = 18\pi. Triangle area =12×12×6=36= \frac{1}{2} \times 12 \times 6 = 36.
Angle CAD=90\angle CAD = 90^\circ because ACD\triangle ACD is a right isosceles triangle with hypotenuse CD=12CD = 12.
3
Find the area of the circular segment bounded by chord CDCD and arc CDCD of circle AA.
Segment Area =18π36= 18\pi - 36.
The area of a circular segment is equal to the sector area minus the triangle area.
4
Subtract the segment area from the area of semicircle CBDCBD of circle OO.
Region Area =18π(18π36)=36= 18\pi - (18\pi - 36) = 36.
Semicircle CBDCBD has radius 66 and area 12π(62)=18π\frac{1}{2}\pi(6^2) = 18\pi. Subtracting the segment area leaves the crescent region.

Anahtar Kavram

Area of circular sectors, segments, and compound regions (Lune of Hippocrates)
Soru 29Soru

In a circle centered at point OO, the radius is 1212 units and central angle AOB\angle AOB measures 150150^\circ. Which of the following statements regarding sector AOBAOB and minor arc ABAB are correct? Select all such statements.

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Cevap: The length of minor arc ABAB is 10π10\pi units.; The area of sector AOBAOB is 60π60\pi square units.; The total perimeter of sector AOBAOB is 24+10π24 + 10\pi units.

Cevap

The correct statements are those indicating that the minor arc length is 10π10\pi units, the sector area is 60π60\pi square units, and the perimeter of the sector is 24+10π24 + 10\pi units.
The central angle fraction is 150360=512\frac{150^\circ}{360^\circ} = \frac{5}{12}. Multiplying the full circumference (24π24\pi) by 512\frac{5}{12} gives an arc length of 10π10\pi. Multiplying the full circle area (144π144\pi) by 512\frac{5}{12} gives a sector area of 60π60\pi. Adding the two radii (2×12=242 \times 12 = 24) to the arc length (10π10\pi) gives the total sector perimeter of 24+10π24 + 10\pi.

Adım Adım Çözüm

1
Determine the central angle fraction of the circle.
The fraction is 150360=512\frac{150^\circ}{360^\circ} = \frac{5}{12}.
Arc length and sector area are proportional to the ratio of the central angle to 360360^\circ.
2
Calculate the circumference of the circle and the arc length of minor arc ABAB.
Circumference = 2π(12)=24π2\pi(12) = 24\pi. Arc length = 512×24π=10π\frac{5}{12} \times 24\pi = 10\pi units.
Arc length equals full circumference multiplied by the central angle fraction.
3
Calculate the total area of the circle and the area of sector AOBAOB.
Circle Area = π(122)=144π\pi(12^2) = 144\pi. Sector Area = 512×144π=60π\frac{5}{12} \times 144\pi = 60\pi square units.
Sector area equals total circle area multiplied by the central angle fraction.
4
Calculate the total perimeter of sector AOBAOB.
Perimeter = arc length+2r=10π+2(12)=24+10π\text{arc length} + 2r = 10\pi + 2(12) = 24 + 10\pi units.
The perimeter of a sector consists of the bounding arc plus the two radii.

Anahtar Kavram

Circles, Arc Lengths, and Sector Areas
Soru 30Soru

In a circle centered at point OO, the radius is 1818 units. Points PP and QQ lie on the circle such that the area of sector POQPOQ is 54π54\pi square units. What is the ratio of the length of minor arc PQPQ to the total perimeter of sector POQPOQ?

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Cevap: ππ+6\frac{\pi}{\pi + 6}

Cevap

The ratio of the length of minor arc PQPQ to the total perimeter of sector POQPOQ is ππ+6\frac{\pi}{\pi + 6}.
The full circle area is 324π324\pi, making the sector 54π/324π=1/654\pi / 324\pi = 1/6 of the circle. The arc length is 1/6×36π=6π1/6 \times 36\pi = 6\pi. The sector perimeter is the arc length plus two radii (6π+366\pi + 36). Taking the ratio of arc length to sector perimeter yields 6π/(6π+36)=π/(π+6)6\pi / (6\pi + 36) = \pi / (\pi + 6).

Adım Adım Çözüm

1
Calculate the total area of the circle and determine the fractional size of sector POQPOQ.
Total area = π(18)2=324π\pi(18)^2 = 324\pi. Sector fraction = 54π324π=16\frac{54\pi}{324\pi} = \frac{1}{6}.
Determining the fraction of the circle represented by the sector is required to find the arc length.
2
Calculate the length of minor arc PQPQ and the total perimeter of sector POQPOQ.
Minor arc PQ=16×2π(18)=6πPQ = \frac{1}{6} \times 2\pi(18) = 6\pi. Sector perimeter = 6π+2(18)=6π+366\pi + 2(18) = 6\pi + 36.
The sector perimeter consists of the curved arc length plus the two straight radii OPOP and OQOQ.
3
Form and simplify the ratio of arc length to sector perimeter.
6π6π+36=6π6(π+6)=ππ+6\frac{6\pi}{6\pi + 36} = \frac{6\pi}{6(\pi + 6)} = \frac{\pi}{\pi + 6}.
Factoring out 6 from the numerator and denominator simplifies the expression to its lowest form.

Anahtar Kavram

Arc Length and Sector Perimeter Calculations
Soru 31Soru

In a circle centered at point OO, the radius is 66 units and central angle POQ\angle POQ measures 120120^\circ. Which of the following statements regarding sector POQPOQ and minor arc PQPQ must be true? Select all such statements.

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Cevap: The length of minor arc PQPQ is 4π4\pi units.; The area of sector POQPOQ is 12π12\pi square units.

Cevap

The correct statements are that the length of minor arc PQPQ is 4π4\pi units and the area of sector POQPOQ is 12π12\pi square units.
The correct statements accurately apply the arc length and sector area formulas by multiplying the full circumference (12π12\pi) and full area (36π36\pi) by the central angle fraction 120360=13\frac{120^\circ}{360^\circ} = \frac{1}{3}, yielding an arc length of 4π4\pi units and a sector area of 12π12\pi square units.

Adım Adım Çözüm

1
Calculate the central angle fraction of the circle.
The fraction of the circle represented by sector POQPOQ is 120360=13\frac{120^\circ}{360^\circ} = \frac{1}{3}.
Arc length and sector area are proportional to the ratio of the central angle to 360360^\circ.
2
Calculate the arc length of minor arc PQPQ.
\text{Arc length} = \frac{1}{3} \times 2\pi(6) = 4\pi \text{ units}.
Arc length equals the central angle fraction times the total circumference 2πr2\pi r.
3
Calculate the area of sector POQPOQ.
\text{Sector area} = \frac{1}{3} \times \pi(6^2) = 12\pi \text{ square units}.
Sector area equals the central angle fraction times the total circle area πr2\pi r^2.
4
Evaluate the given statements against the calculated values.
Statements asserting an arc length of 4π4\pi units and a sector area of 12π12\pi square units are true. Other statements miscalculate by omitting the fraction or adding incorrect boundary components.
Comparing calculated values confirms the valid choices.

Anahtar Kavram

Arc Length and Sector Area Formulas
ÖncekiSayfa 2 / 2
Circles, Arc Lengths, and Sector Areas Alıştırma Soruları — GRE General Test — Sayfa 2 | Examkin