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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?

 Jun 10, 2021
 #1
avatar+524 
+2

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. 

 Jun 11, 2021
edited by amygdaleon305  Jun 11, 2021
 #2
avatar+121054 
+1

Very nice, amy  !!!!!

 

 

cool cool cool

 Jun 11, 2021
 #3
avatar+524 
0

Thanks a lot Phill! laugh

amygdaleon305  Jun 12, 2021

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