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 \({\text{product of the side lengths} \over 4\times \text{Area}} = {18,750 \over 1,200} = \color{brown}\boxed{15.625}\)