Hussain15 wrote:
OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined.
Why not "B"?????
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5-12-13 is a Pythagorean triple. What is Pythagorean triple?
Triangles that have their sides in the ratio of
whole numbers (integers) are called Pythagorean Triples. The smallest one is 3:4:5. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.
A Pythagorean triple consists of three positive
integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\).
If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). There are 16 primitive Pythagorean triples with c ≤ 100:
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).
BUT: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 \(b=\sqrt{165}\) or \(a=4\) and \(b=\sqrt{153}\) or \(a=2.5\) and \(b=\sqrt{162.75}\)...
So knowing that the diagonal of a rectangle (hypotenuse) equals to one of the Pythagorean triple hypotenuse value is not sufficient to calculate the sides of this rectangle.
Hope it helps.