In triangle JKL, we have JK = JL = 25 and KL = \(10\). Find the circumradius
+9674 We calculate the area of the triangle.
Area of triangle JKL = \(\dfrac12 \cdot 10 \cdot \sqrt{25^2 - \left(\dfrac{10}2\right)^2} = 50\sqrt 6\)
Then, we use the formula \(\text{circumradius} = \dfrac{\text{product of 3 sides}}{4(\text{area})}\)
\(\text{circumradius} = \dfrac{25(25)(10)}{4(50\sqrt 6)} = \dfrac{125}{24}\sqrt6\)
+9674 We calculate the area of the triangle.
Area of triangle JKL = \(\dfrac12 \cdot 10 \cdot \sqrt{25^2 - \left(\dfrac{10}2\right)^2} = 50\sqrt 6\)
Then, we use the formula \(\text{circumradius} = \dfrac{\text{product of 3 sides}}{4(\text{area})}\)
\(\text{circumradius} = \dfrac{25(25)(10)}{4(50\sqrt 6)} = \dfrac{125}{24}\sqrt6\)
