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# triangle

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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
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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
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Very nice, amy  !!!!!   Jun 11, 2021
#3
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Thanks a lot Phill! amygdaleon305  Jun 12, 2021