Question

Given a triangle whose sides are 42 cm, 50 cm, and 63 cm. Find the radius of a circle which is tangent to the 50 cm and 63 cm side of the triangle, and whose center lies on the third side.

Answer 1

$S_{ABC}=\sqrt{\smash[b]{p(p-a)(p-b)(p-c)}}$

$p={a+b+c \over 2}$

$AB=a=50, BC=b=63, AC=c=42$

$p={50+63+42 \over 2}={155 \over 2}=77.5$

$S_{ABC}=\sqrt{\smash[b]{77.5 \ast (77.5-50) \ast (77.5-63) \ast (77.5-42)}}$

$S_{ABC}=1.25 \ast \sqrt{702119}$

$S_{ABC}=S_{ABO}+S_{OBC}={R \ast AB \over 2}+{R \ast BC \over 2}$

$S_{ABC}={50 \ast R \over 2}+{63 \ast R \over 2}={113 \ast R \over 2}$

${113 \ast R \over 2}=1.25 \ast \sqrt{702119}$

$R={10 \ast \sqrt{702119} \over 452} \approx 18.5$