In triangle JKL, we have JK = JL = 25 and KL = 30. Find the inradius of triangle JKL and the circumradius of triangle JKL.

The circumradius is 57/4.

Draw altitude $JM$. Because the triangle is isosceles, $KM = ML = 15$. 

Then, by the Pythagorean Theorem, $JM = \sqrt{25^2 - 15^2} = 20$.

This means that the area of the triangle is $2 \times( 20 \times 15 \div 2) = 300$.

Now, recall the formula for inradius, which is ${\text{Area} \over \text{semiperimeter} } = {300 \over 40} = \color{brown}\boxed{7.5}$

Now, recall the formula for circumradius, which is ${\text{product of the side lengths} \over 4\times \text{Area}} = {18,750 \over 1,200} = \color{brown}\boxed{15.625}$