pafrompa wrote:
GMAT198917 wrote:
For statement (1) why can't we assume this is a common triangle?
We know the hypotenuse is 10 and that it is a right triangle. With this information the side lengths of the large triangle should be 6 & 8.
Could someone please elaborate on this a bit? I also put choice one, because I assumed that the triangles formed in the corners were 30,60,90 triangles. Therefore, I ended up picking A because I had just assumed that you can back into x,x\sqrt{3},2x where 2x=10.
You are saying yourself that you
assumed that it must be 30-60-90 triangle. Why? Do you have any reasons supporting this claim? No.
So, you assumed with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=10\).
Hope it's clear.
Clear as day, thank you. I guess sometimes I force myself to try to see things that aren't there with DS. Thanks Bunuel