If a, b and c are sides of a triangle with c as the hypotenuse then,
a^2+b^2 = c^2, and we have the right triangle with say angle x = 90 (x is angle opposite to hypotenuse.
Now can we deduce the following also
if a^2+b^2< c^2, x > 90, and if a^2+b^2 > c^2, x < 90.
I came across this at https://www.urch.com/forums/gmat-data-su ... angle.html
Thanks
Amit
Pythagoras Theorem
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Yes, you can. You can see this visually: imagine a right angle triangle, with legs a and b, and hypotenuse c. We know that a^2 + b^2 = c^2. Imagine rotating just side a, so that the 90 degree angle gets larger, without changing the length of a and b. You'd need to make c longer to complete the triangle, and we haven't changed a^2 + b^2, so now a^2 + b^2 < c^2. If you rotate a in the other direction, so the angle gets smaller than 90 degrees, you'd need to make c smaller to draw the triangle.erjamit wrote:If a, b and c are sides of a triangle with c as the hypotenuse then,
a^2+b^2 = c^2, and we have the right triangle with say angle x = 90 (x is angle opposite to hypotenuse.
Now can we deduce the following also
if a^2+b^2< c^2, x > 90, and if a^2+b^2 > c^2, x < 90.
You can also see why this is true from the Cosine Law, from trigonometry (which you don't need to know for the GMAT!). In any triangle with sides a, b and c, if C is the angle opposite the side of length c, then
c^2 = a^2 + b^2 - 2ab*cos(C)
When C = 90, cos(90) = 0, and we get the Pythagorean Theorem.
When C < 90, cos(C) is positive, so c^2 < a^2 + b^2.
When C > 90, cos(C) is negative, so c^2 > a^2 + b^2.
I glanced at that thread, and there's some dodgy math there, so I wouldn't recommend reading it too closely. For example, someone says that in right angled triangles, the sides are always in a 3:4:5 ratio, which is absolutely untrue.erjamit wrote: I came across this at https://www.urch.com/forums/gmat-data-su ... angle.html
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