A geometry problem by A Former Brilliant Member

Geometry Level 3

Incircle of radius 4 cm 4 \text{ cm} of a triangle A B C ABC touches the side B C BC at D D . If B D = 6 cm , D C = 8 cm BD=6 \text{ cm}, DC=8\text{ cm} , then what area of triangle A B C ABC ?

Give your answer in cm 2 \text{cm}^2 .


The answer is 84.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Let the incircle touch AC at E, AB at F. Tangents from vertex to in circle have equal lengths. ∴ AE=AF=X say. FB=BD=6. EC=DC=8. ⟹ AB=X+6. BC=6+8. AC=X+8. So semiperemeter S=X+14. ∴ A r e a 2 = ( X + 14 ) ∗ 8 ∗ X ∗ 6..... H e r o ′ s f o r m u l a . ∴ i n r a d i u s 2 = a r e a 2 S 2 , ⟹ 4 2 = ( X + 14 ) ∗ 8 ∗ X ∗ 6 ( X + 14 ) 2 S o 16 = 48 ∗ ( X + 14 ) , ⟹ X = 7. ∴ a r e a = ( 7 + 14 ) ∗ 8 ∗ 7 ∗ 6 = 84 c m 2 . \text{Let the incircle touch AC at E, AB at F. Tangents from vertex to in circle have equal lengths.}\\ \therefore\ \text{AE=AF=X say. FB=BD=6. EC=DC=8.}\\ \implies\ \text{AB=X+6. BC=6+8. AC=X+8. So semiperemeter S=X+14.}\\ \therefore\ Area^2=(X+14)*8*X*6.....Hero's formula.\\ \therefore\ inradius^2=\dfrac{area^2}{S^2},\ \implies\ 4^2=\dfrac{(X+14)*8*X*6}{(X+14)^2}\\ So\ 16=48*(X+14),\ \ \implies\ X=7.\\ \therefore\ area=\sqrt{(7+14)*8*7*6}=\ \ \ \ \color{#D61F06}{84}\ cm^2.

3 Helpful 1 Interesting 0 Brilliant 0 Confused
Ahmad Saad
Apr 19, 2016

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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