Question 788833
Legs x and y. Hypotenuse h.
x=120
y=?
h=32+y


Pythagorean Theorem:  {{{h^2=x^2+y^2}}}
{{{(32+y)^2=120^2+y^2}}}
{{{32^2+2*32y+y^2=120^2+y^2}}}
See y^2 as term on both sides of the equation.  Subtract y^2 from both sides.
{{{32^2+2*32y=120^2}}}
Simplify and solve for y.
{{{64y+32^2=120^2}}}
{{{64y=120^2-32^2}}}
{{{y=(120+32)(120-32)/64}}}   ------because expression shows difference of two squares.
{{{y=(152)(88)/64}}}
Keep simplifying,
{{{y=76*2*8*11/(8*8)=2*38*2*8*11/(8*8)=2*2*19*2*8*11/(8*8)=19*11}}}
{{{highlight(y=19*11)}}}
That is the leg, y, in factored form.  Just use it in computed form to help compute the value of h.


{{{highlight(h=32+19*11)}}}