In the diagram above, angle \(C=90^o\) and \(AC=BC\). Point

BTGmoderatorLU wrote:Source: Magoosh



In the diagram above, angle \(C=90^o\) and \(AC=BC\). Point \(M\) is the midpoint of \(AB\). Arc \(AXB\) has its center at \(C\) and passes through \(A\) and \(B\). Arc \(AYB\) has its center at \(M\) and passes through \(A\) and \(B\). The shaded region between the two arcs is called a "lune". What is the ratio of the area of the lune to the area of triangle \(ABC\)?

A. \(1\)

B. \(\frac{\pi}{2}\)

C. \(\frac{\pi}{3}\)

D. \(\frac{3\pi}{8}\)

E. \(\frac{4\pi}{9}\)

The OA is A

Since the question wants the answer in ratio, we can assume a convenient value for the sides of the triangle.

Say for the rightangled triangle, AC = BC = 1; thus, hypotenuse, AB = √2

Given that Arc \(AXB\) has its center at \(C\) and passes through \(A\) and \(B\), area of segment AXBCA = (πr^2)/4; where r = AC = 1

Thus, the area of segment AXBCA = (πr^2)/4 = π*1^2/4 = π/4

Also, given that Arc \(AYB\) has its center at \(M\) and passes through \(A\) and \(B\), area of segment AYBA = (πR^2)/2; where R = AM = AB/2 = √2/2 = 1/√2

Thus, the area of segment AYBA = (πR^2)/2 = (π*(1/√2)^2)/2 = π/4

Thus, the area of "lune" = area of segment AYBA – area of segment AXBMA = area of segment AYBA – [area of segment AXBCA – ∆ABC] =

= area of segment AYBA – [area of segment AXBCA – ∆ABC]

= π/4 – [π/4 – 1/2*1*1] = 1/2

Thus, ratio of the area of the lune to the area of triangle \(ABC\) = (1/2) / (1/2) = 1

The correct answer: A

Hope this helps!

-Jay
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