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Once again. Can we solve this question without using trignometry?
1) these are NOT 30-60-90 triangles,
2) you don't care. just use properties of SIMILAR triangles. There are 3 of them in the diagram. _________________
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
Just thought it might help to explain how we choose the simillar triangles. Taking notation from wonder_gmats solution
Triangle PQR is simillar to triangle QZR. We use Angle side angle as the simillarity law. Size QZ is common. 90 degrees is common to both. now the Angle ZPQ = Angle ZQR. So prove this you have to know that angle PQR = 90 degrees.
From this you can deduce that angle ZPQ = angle ZQR, hence we have angle side angle as your simillarity criteria.
I hope this helps. Anyone has shorter way to prove the same ?
I think this is another way of solving this using Pythagore (it's also longer than the previous methods):
1) You have a=4
Thus PQ = v(4^2 + 2^2) = v20
(a + b)^2 = (v20)^2 + QR^2
But QR^2 is also equal to: (2^2 + b^2)
Substituting for QR we get: (4+b)^2 = (v20)^2 + (2^2 + b^2)
Now solving it gives us:
16+8b+b^2 = 20 + 4 + b^2
8b = 24 - 16 = 8
b = 1
The same logic would apply to statement 2 and the answer would be D, that you can solve it with either statement. Also, this is assuming that you know that a triangle inscribed within a semi-circle is a right triangle.