IMC 2014 Q20
20. The diagram shows a regular pentagon and five circular arcs. The sides of the pentagon have length 4. The centre of each arc is a vertex of the pentagon, and the ends of the arc are the midpoints of the two adjacent edges.
What is the total shaded area?
- A \(8\pi\)
- B \(10\pi\)
- C \(12\pi\)
- D \(14\pi\)
- E \(16\pi\)
Intermediate Mathematical Challenge 2014, Question 20. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(14\pi\)
Each exterior angle of a regular pentagon is \(\tfrac{1}{5} \times 360^\circ = 72^\circ\). So each of the five circular arcs has radius 2 and so subtends an angle of (180 + 72)° at a vertex of the pentagon. Therefore the area of each of the five shaded major sectors is \(\tfrac{252}{360} \times \pi \times 2^2 = \tfrac{7}{10} \times \pi \times 4 = \tfrac{14\pi}{5}\). So the total shaded area is \(14\pi\).
Solution © UK Mathematics Trust, from the IMC 2014 solutions. Longer versions with investigations: Solutions and investigations (PDF).