Question Number 187549 by Rupesh123 last updated on 18/Feb/23
Answered by HeferH last updated on 18/Feb/23
$${The}\:{height}\:{of}\:\bigtriangleup{BIC}\:{is}\:{the}\:{inradius}. \\ $$$${Area}\:=\:{s}\centerdot{I} \\ $$$$\:\frac{\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}}{{s}}\:=\:{I} \\ $$$$\:{s}\:=\:\frac{\mathrm{7}+\mathrm{9}+\mathrm{12}}{\mathrm{2}}\:=\:\mathrm{14} \\ $$$$\:{I}\:=\:\sqrt{\frac{\left(\mathrm{14}−\mathrm{9}\right)\left(\mathrm{14}−\mathrm{7}\right)\left(\mathrm{14}−\mathrm{12}\right)}{\mathrm{14}}}\:=\:\sqrt{\mathrm{5}} \\ $$$$\:{Area}\:{of}\:{BIC}\:=\:\frac{\mathrm{12}\centerdot\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\mathrm{6}\sqrt{\mathrm{5}}\:{u}^{\mathrm{2}} \\ $$
Commented by Rupesh123 last updated on 18/Feb/23
Great, bro!