Problem Solving

OA is not available.
Please help me in solving this problem

we know p^2 + b^2 = h^2 in a right angles triangle.
Thus let the side of A1 be p, A2 be b and A3=h.
a1 = pi * (p/2)^2 * 1/2 (=pi*r^2 / 2)
= pi/8 * p^2
similarly a2 = pi/8 * b^2
and a3 = pi/8 * h^2
Thus putting in (a1+a2)/a3 = p^2 + b^2 / h^2 = 1
IMO A

Pick any pythagorean triples say (3,4,5) or (1,1,Sqrt2)as the sides of the triangle.

You can find area of the semi circle corresponding to the sides. (Am ignoring Pi as its a ratio and it would get cancelled anyways)

Considering 3,4,5

3^2 + 4^2 / 5^2 = 25/25 = 1 (rad would be 3/2,4/2 and 5/2 for ease of calc i am doubling all of values)

A, just like Cans explained.

First, don't panic becuase there are no numbers - if it is on the GMAT it will work out. STart with what we know. It is a right triangle therefore a(a) + b(b) = c(c)

Also the diameter of each full circle is one side - the area of each will be 1/2(1/2d)^2(pi)

Set this up. 1/2(1/2a^2)(pi) + 1/2(1/2b^2(pi)/1/2(1/2c^2)(pi)

the first 1/2's and the pi's cancel out so you then have 1/4a^2 + 1/4b^2/(1/4 c^2) - the 1/4 cancel out of each term and you get a^2 + b^2/ c^2 - using the pythagorean we know that the numerator and demoninator are the same thus the fraction is equal to 1.