# Isosceles IV

In an isosceles triangle ABC is |AC| = |BC| = 13 and |AB| = 10. Calculate the radius of the inscribed (r) and described (R) circle.

Result

r =  3.33
R =  7.04

#### Solution:

$h^2 = b^2 - (c/2)^2 \ \\ h = \sqrt{ 13^2 - 5^2 } = 12 \ \\ (b-c/2)^2 + r^2 = (h-r)^2 \ \\ 8^2 + r^2 = h^2 -2hr + r^2 \ \\ 8^2 = h^2 -2hr \ \\ 8^2 = 12^2 -2\cdot 12 \cdot r \ \\ r = (12^2 - 8^2)/(2\cdot 12) = 3.33 \ \\$
$(c/2)^2 = h(2R-h) \ \\ \ \\ R =\dfrac{ c^2/4 + h^2 }{2h} \ \\ R = \dfrac{ 5^2 + 12^2 }{2 \cdot 12 } \ \\ R = 7.04$

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