Question

The hypotenuse of a right triangle is √18 units. what are the lengths of the legs of the triangle? How many different answers can you find? Sketch a triangle for each answer. Explain your strategies.
Please help me... Can i Have the whole thing how you do it?

Answer 1

Since a right triangle has special rules to it such that the hypotenuse squared equals the sum of the squares of the other 2 sides. In other words, if the hypotenuse = c, and the 2 smaller sides are a and b, then:
${c}^{2} = {a}^{2} + {b}^{2}$
Solving for c (hypotenuse), we get:
$\sqrt{({c}^{2})} = \sqrt{({a}^{2}+{b}^{2} )} \\ c = \sqrt{({a}^{2}+{b}^{2} )}$
Therefore c is the root of the square of the other sides. So by having root18, it's like saying:
$\sqrt{18} = \sqrt{({a}^{2} + {b}^{2} )} \\ 18 = {a}^{2} + {b}^{2}$
Getting rid of the square root, so sides a and b must have their squares total 18:
the only squares < 18 are: 16 (4×4), 9 (3×3), 4 (2×2), and 1 (1×1)
of those above added in any order of two of them, only 16+4
$18 = 17 + 1 - - > \sqrt{17} \: and \: 1 \\ 18 = 16 + 2 - - > \sqrt{16} \: and \: \sqrt{2} \\ ... \: 4 \: and \: \sqrt{2} \\ 18 = 15 + 3 - - > \sqrt{15} \: and \: \sqrt{3}$
$18 = 14 + 4 - - > \sqrt{14} \: and \: \sqrt{4} \\ ... \sqrt{14} \: and \: 2 \\ 18 = 13 + 5 - - > \sqrt{13} \: and \: \sqrt{5} \\ 18 = 12+ 6 - - > \sqrt{12} \: and \: \sqrt{6}$
$18 = 11 + 7 - - > \sqrt{11} \: and \: \sqrt{7} \\ 18 = 10 + 8 - - > \sqrt{10} \: and \: \sqrt{8} \\ ... \sqrt{10} \: and \: 2 \sqrt{2} \\ 18 = 9 + 9 - - > \sqrt{9} = 3 \: and \: 3$
I HOPE THAT IS ALONG THE LINE THAT WAS ASKED FOR!!! :-D