Inscribed circle

XYZ is right triangle with right angle at the vertex X that has inscribed circle with a radius 5 cm. Determine area of the triangle XYZ if XZ = 14 cm.

Result

S =  157.5 cm2

Solution:

$r = 5 \ \\ y = 14 \ \\ s = (x+y+z)/2 \ \\ x^2 = y^2+z^2 \ \\ S = sr \ \\ S = zy/2 \ \\ zy/2 = (z+y+\sqrt{ z^2+y^2 })/(2r) \ \\ \ \\ z = 2 \cdot \ (y \cdot \ r-r^2)/(y-2 \cdot \ r) = 2 \cdot \ (14 \cdot \ 5-5^2)/(14-2 \cdot \ 5) = \dfrac{ 45 }{ 2 } = 22.5 \ \\ x = \sqrt{ y^2+z^2 } = \sqrt{ 14^2+22.5^2 } = \dfrac{ 53 }{ 2 } = 26.5 \ \\ S = z \cdot \ y / 2 = 22.5 \cdot \ 14 / 2 = \dfrac{ 315 }{ 2 } = 157.5 = 157.5 \ cm^2$

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