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.

Correct result:

S =  157.5 cm²

Solution:

$$\begin{gathered} r=5 \\ y=14 \\ s=(x+y+z)\mathrm{/}2 \\ {x}^2=y^2+z^2 \\ S=sr \\ S=z_y\mathrm{/}2 \\ {z}_y\mathrm{/}2=(z+y+\sqrt{z^2+y^2})\mathrm{/}(2r) \\ z=2\cdot (y\cdot r-r^2)\mathrm{/}(y-2\cdot r)=2\cdot (14\cdot 5-5^2)\mathrm{/}(14-2\cdot 5)={\displaystyle \frac{45}{2}}=22\frac{1}{2}=22.5 \\ x=\sqrt{y^2+z^2}=\sqrt{14^2+22.5^2}={\displaystyle \frac{53}{2}}=26\frac{1}{2}=26.5 \\ S=z\cdot y\mathrm{/}2=22.5\cdot 14\mathrm{/}2={\displaystyle \frac{315}{2}}=157\frac{1}{2}={\displaystyle \frac{315}{2}}{\text{cm}}^2=157.5{\text{cm}}^2 \end{gathered}$$

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