Question 250077
Use the Pythagorean theorem,
{{{A^2+B^2=H^2}}}
{{{(x+3)^2+(4(x+2))^2=25^2}}}
{{{(x^2+6x+9)+16(x^2+4x+4)=625}}}
{{{x^2+16x^2+6x+64x+9+64=625}}}
{{{17x^2+70x+73=625}}}
{{{17x^2+70x-552=0}}}
{{{(17x+138)(x-4)=0}}}
Only the positive value for x is valid.
{{{x=4}}}
The lengths are then 7 and 24.