Question 128479
Let c = the hypotenuse and a and b the other two legs.
From the problem description, you can write:
{{{c = a+8}}} and {{{c = b+4}}} Rewrite these in terms of a and b to get:
{{{a = c-8}}} and {{{b = c-4}}} Now you can use the Pythagorean theorem to find c, the hypotenuse:
{{{c^2 = a^2 + b^2}}} Substitute the two equations above for a and b:
{{{c^2 = (c-8)^2+(c-4)^2}}} Simplify.
{{{c^2 = (c^2-16c+64)+(c^2-8c+16)}}} Combine like-terms.
{{{c^2 = 2c^2-24c+80}}} Subtract {{{c^2}}} from both sides.
{{{c^2-24c+80 = 0}}} Solve this quadratic by factoring:
{{{(c-4)(c-20) = 0}}}, so then...
{{{c = 4}}} or {{{c = 20}}} Discard the first solution {{{c = 4}}} as c must be greater than 8 or else you have one leg of the triangle equal to zero!
So, the hypotenuse is 20 and the other two legs are a=12 and b = 16.
Check:
{{{c^2 = a^2+b^2}}} Substitute c = 20, a = 12, and b = 16.
{{{20^2 = 12^2+16^2}}}
{{{400 = 144+256}}}
{{{400 = 400}}} OK!