What is the radius of the circle inscribed in triangle ABC if AB=22, AC=12, and BC=14?
triangle ABC
\(a=14\\ b=12\\ c=22\\ r=?\)
\(s=\frac{a+b+c}{2}=\frac{14+12+22}{2}\\ s=24\\ A_{\triangle}=\sqrt{s(s-a)(s-b)(s-c)}\\ A_{\triangle}=\sqrt{24(24-14)(24-12)(24-22)}\)
\(A_{\triangle}=\sqrt{2^3*3*2*5*2^2*3*2}=\sqrt{2^7*3^2*5}\\ \color{blue}A_{\triangle}=24\cdot \sqrt{10}\)
\(A_{\triangle}=\frac{r(a+b+c)}{2}\\ r=\frac{2A_{\triangle}}{a+b+c}=\frac{48\cdot\sqrt{10}}{48}\)
\(r=\sqrt{10}\approx 3.1627766..\)
Thanks Mathhemathh
\(The\ radius\ of\ the\ circle\\ inscribed\ in\ triangle\ ABC\ is =\sqrt{10}\approx 3.1627766..\)
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