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Difficulty: HardCircle Geometry: Arc Length and Sector Area

A region ABCDABCD is bounded by two concentric circular arcs of radii RR and rr (where R>rR > r) and two radial line segments, all sharing a central angle of 72∘72^\circ. The area of region ABCDABCD is 60π cm260\pi\text{ cm}^2, and the total perimeter of region ABCDABCD is (12π+20) cm(12\pi + 20)\text{ cm}. What is the value of the outer radius RR, in centimeters?

  1. A
    10
  2. B
    15
  3. 20Answer
  4. D
    25
  5. E
    30

Answer

The outer radius RR is 20 centimeters.
The central angle of 72∘72^\circ represents 72360=15\frac{72}{360} = \frac{1}{5} of a circle. The area of the region is π5(R2−r2)=60π\frac{\pi}{5}(R^2 - r^2) = 60\pi, which simplifies to R2−r2=300R^2 - r^2 = 300, or (R+r)(R−r)=300(R+r)(R-r) = 300. The total perimeter is the sum of the outer arc 2πR5\frac{2\pi R}{5}, the inner arc 2πr5\frac{2\pi r}{5}, and the two straight side segments 2(R−r)2(R-r). Setting 2π5(R+r)+2(R−r)=12π+20\frac{2\pi}{5}(R+r) + 2(R-r) = 12\pi + 20 yields 25(R+r)=12  ⟹  R+r=30\frac{2}{5}(R+r) = 12 \implies R+r = 30 and 2(R−r)=20  ⟹  R−r=102(R-r) = 20 \implies R-r = 10. Solving the system R+r=30R+r = 30 and R−r=10R-r = 10 by adding the equations gives 2R=402R = 40, so R=20 cmR = 20\text{ cm}.

Step-by-Step Solution

1
Determine the fraction of the full circle represented by the 72∘72^\circ central angle.
The fraction is 72∘360∘=15\frac{72^\circ}{360^\circ} = \frac{1}{5}.
Arc lengths and sector areas are proportional to the ratio of the central angle to 360∘360^\circ.
2
Set up and simplify the equation for the area of the region ABCDABCD.
15πR2−15πr2=60π  ⟹  R2−r2=300  ⟹  (R+r)(R−r)=300\frac{1}{5}\pi R^2 - \frac{1}{5}\pi r^2 = 60\pi \implies R^2 - r^2 = 300 \implies (R+r)(R-r) = 300.
The region's area is the difference between the outer sector area and inner sector area.
3
Set up and simplify the equation for the perimeter of region ABCDABCD.
15(2πR)+15(2πr)+2(R−r)=12π+20  ⟹  2π5(R+r)+2(R−r)=12π+20\frac{1}{5}(2\pi R) + \frac{1}{5}(2\pi r) + 2(R - r) = 12\pi + 20 \implies \frac{2\pi}{5}(R + r) + 2(R - r) = 12\pi + 20.
The perimeter consists of the outer arc, the inner arc, and two radial segments each of length R−rR - r.
4
Equate corresponding rational and π\pi-coefficient terms to solve for (R+r)(R+r) and (R−r)(R-r).
25(R+r)=12  ⟹  R+r=30\frac{2}{5}(R + r) = 12 \implies R + r = 30, and 2(R−r)=20  ⟹  R−r=102(R - r) = 20 \implies R - r = 10.
Equating the algebraic components yields a system of linear equations.
5
Solve the system of equations for the outer radius RR.
Adding the equations gives (R+r)+(R−r)=30+10  ⟹  2R=40  ⟹  R=20 cm(R + r) + (R - r) = 30 + 10 \implies 2R = 40 \implies R = 20\text{ cm}.
Eliminating rr isolates the required variable RR.

Key Concept

Arc length and sector area of concentric circular regions
Estimated Time:2m 0s
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Topics

Circle Geometry: Arc Length and Sector Area

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