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Difficulty: HardCircle Geometry: Arc Length and Sector Area

A decorative emblem is shaped as the region bounded by two concentric circular sectors sharing the same central angle of θ\theta radians. The outer sector has radius R cmR\text{ cm}, and the inner sector has radius r cmr\text{ cm}, where the difference between the two radii is R−r=4 cmR - r = 4\text{ cm}. If the area of the emblem is 15π cm215\pi\text{ cm}^2 and the length of the outer arc is 5π cm5\pi\text{ cm}, what is the value of θ\theta, in radians?

  1. A
    3π8\frac{3\pi}{8}
  2. B
    5π16\frac{5\pi}{16}
  3. 5π8\frac{5\pi}{8}Answer
  4. D
    3π4\frac{3\pi}{4}
  5. E
    5π4\frac{5\pi}{4}

Answer

The central angle θ\theta is 5π8\frac{5\pi}{8} radians.
Using the formulas for arc length (s=Rθs = R\theta) and sector area (A=12r2θA = \frac{1}{2}r^2\theta), the area of the emblem is A=12θ(R2−r2)=12θ(R−r)(R+r)A = \frac{1}{2}\theta(R^2 - r^2) = \frac{1}{2}\theta(R - r)(R + r). Substituting R−r=4 cmR - r = 4\text{ cm} gives 15π=2θ(R+r)15\pi = 2\theta(R + r). Since Rθ=5πR\theta = 5\pi, we have R+r=10πθ−4R + r = \frac{10\pi}{\theta} - 4. Substituting this into the area equation yields 2θ(10πθ−4)=15π2\theta\left(\frac{10\pi}{\theta} - 4\right) = 15\pi, which simplifies to 20π−8θ=15π20\pi - 8\theta = 15\pi, yielding θ=5π8\theta = \frac{5\pi}{8}.

Step-by-Step Solution

1
Express the outer arc length using the radian arc length formula.
souter=Rθ=5πs_{\text{outer}} = R\theta = 5\pi, which gives R=5πθR = \frac{5\pi}{\theta}.
Arc length in radians is given by s=rθs = r\theta.
2
Express the area of the emblem as the difference between the outer and inner sector areas.
A=12R2θ−12r2θ=12θ(R2−r2)=15πA = \frac{1}{2}R^2\theta - \frac{1}{2}r^2\theta = \frac{1}{2}\theta(R^2 - r^2) = 15\pi.
The emblem is formed by removing the inner sector from the outer sector.
3
Factor R2−r2R^2 - r^2 as (R−r)(R+r)(R - r)(R + r) and substitute R−r=4R - r = 4.
15π=12θ(4)(R+r)=2θ(R+r)15\pi = \frac{1}{2}\theta(4)(R + r) = 2\theta(R + r).
The difference of squares allows substituting the known difference between radii.
4
Substitute R=5πθR = \frac{5\pi}{\theta} and r=5πθ−4r = \frac{5\pi}{\theta} - 4 into the sum (R+r)(R + r).
R+r=10πθ−4R + r = \frac{10\pi}{\theta} - 4.
Expreing R+rR + r solely in terms of θ\theta allows solving a single variable equation.
5
Solve the resulting equation for θ\theta.
2θ(10πθ−4)=15π  ⟹  20π−8θ=15π  ⟹  8θ=5π  ⟹  θ=5π82\theta \left(\frac{10\pi}{\theta} - 4\right) = 15\pi \implies 20\pi - 8\theta = 15\pi \implies 8\theta = 5\pi \implies \theta = \frac{5\pi}{8}.
Distributing 2θ2\theta cancels θ\theta in the first term and leaves a linear equation in θ\theta.

Key Concept

Sector area and arc length formulas in radians applied to concentric regions
Estimated Time:2m 0s
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Topics

Circle Geometry: Arc Length and Sector Area

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