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.

Correct result:

r =  3.33
R =  7.04

Solution:

h2=b2(c/2)2 h=13252=12 (bc/2)2+r2=(hr)2 82+r2=h22hr+r2 82=h22hr 82=122212r r=(12282)/(212)=3.33 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(2Rh)  R=c2/4+h22h R=52+122212 R=7.04(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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