yogi02 wrote:
Bunuel, Cant we simply get it from diagonal alone. Any rectangle with diagonal 13 will have width and length = 12 and 5 in any order. As per Pythagoras triplet 13-12-5
What is even a need of given 2 statements
You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).
For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) ...
Although extending this discussion makes no sense but again I hope we can find the area using: d1xd2/2. d1=d2 (rectangle and hence congruent).
Just a question for concept clarity.
SACHIN SHARMA
努力の精神を養うこと (Fostering the spirit of Effort)