The incircle of radius 4 of a triangle ABC touches the side BC at D. If BD=6, DC=10, then what is the area of triangle ABC?
+526 Okay, let's imagine the scenario as in this diagram
Length of 2 tangents drawn from same point are equal
BE = BD = 6 cm
CF = CD = 10 cm
AF = AE = x cm
Semi-perimeter of circle
\(s={AB+BC+CA \over 2}\)
\(= {6+10+10+6+x+x \over 2}\)
\(={32+2x \over 2}\)
\(=16+x\)
Area of △ABC \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{(16+x)(16+x-16)(16+x-x-6)(16+x-x-10)}\)
\(=\sqrt{(16+x)60x}\) ...(1)
Also, area of △ABC \(=2 ×\) area of\((△AOE+ △OBD+△OCD)\)
\(=2×[({1\over 2}×4×x)+({1\over 2}×4×6)+({1\over 2}×4×10)]\)
\(=4x+24+40\)
\(=4x+64\) ...(2)
Equating eq (1) and (2)
\(\sqrt{(16+x)60x}=4x+64\)
\(960x+60x^2=16x^2+512x+4096\)
\(44x^2-448x-4096=0\)
⇒\(x=16\)
Thus, area of △ABC \(=64+64\)
\(=128\) sq. units
~ Hope you got it. Thanks.
