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
Triangle and semi-circles
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cans
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Cans!!
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Cans!!
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shankar.ashwin
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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)
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)
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tpr-becky
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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.
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.
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
Master GMAT Instructor
The Princeton Review
Irvine, CA
