Question

Difficulty: Very hardCircles, Arc Lengths, and Sector Areas

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?

Answer: 18

Answer

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.

Step-by-Step Solution

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.

Key Concept

Relationship between arc length, radius, central angle, and sector area
Estimated Time:2m 30s
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