Question 400983
Let {{{a}}} = the smaller leg
Let {{{b}}} = longer leg
Let {{{c}}} = hypotenuse
given:
{{{b = a + 7}}} in
{{{c = a + 8}}} in
It's a right tringle, so
{{{a^2 + b^2  = c^2}}}
By substitution:
{{{a^2 + (a + 7)^2  = (a + 8)^2}}}
{{{a^2 + a^2 + 14a + 49 = a^2 + 16a + 64}}}
{{{a^2 - 2a  = 15}}}
I can complete the square to find {{{a}}}
{{{a^2 - 2a + (-2/2)^2 = 15 + (-2/2)^2}}}
{{{a^2 - 2a + 1 = 15 + 1}}}
Both sides are now a perfect square
{{{(a - 1)^2 = 4^2}}}
Take the square root of both sides
{{{a - 1 = 4}}}
{{{a = 5}}}
It is also true that
{{{a - 1 = -4}}}, but I can't use a negative result
Since
{{{b = a + 7}}}
{{{b = 12}}}
and, since
{{{c = a + 8}}}
{{{c = 13}}}
The sides are 5, 12, and 13
check answer:
{{{5^2 + 12^2 = 13^2}}}
{{{25 + 144 = 169}}}
{{{169 = 169}}}