## Intermediate Algebra for College Students (7th Edition)

$a=8,b=6$
As per the given figure, we have $a^2+b^2=10^2$ and $a^2+(b+9)^2=17^2$ Let us subtract the above equations. We get $b^2-(b+9)^2=100-289$ or, $-18b=-108$ or, $b=6$ Now, plug the value of $b=6$ in the equation: $a^2+b^2=10^2$, we get $a^2=100-36$ or, $a=8$ Hence, $a=8,b=6$