Let ABC be a triangle such that AB=4, BC=5 an | vedclass

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(C) Given $AB=c=4, BC=a=5, CA=b=6$. Points $D, E, F$ are on $AB, BC, CA$ such that $AD=2, BE=2, CF=2$. Note that $BD = AB - AD = 4 - 2 = 2$,$CE = BC -

Vedclass · Accepted Answer

$\frac{4}{15}$

Vedclass · Answer

(C) Given $AB=c=4, BC=a=5, CA=b=6$. Points $D, E, F$ are on $AB, BC, CA$ such that $AD=2, BE=2, CF=2$. Note that $BD = AB - AD = 4 - 2 = 2$,$CE = BC - BE = 5 - 2 = 3$,$AF = AC - CF = 6 - 2 = 4$. The area of $	riangle ABC = \Delta$. The area of $	riangle ADF = \frac{1}{2} \cdot AD \cdot AF \cdot \sin A = \frac{1}{2} \cdot 2 \cdot 4 \cdot \sin A = 4 \sin A$. Since $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 6 \cdot 4 \cdot \sin A = 12 \sin A$,we have $\sin A = \frac{\Delta}{12}$. Thus,Area$(	riangle ADF)$ = $4 \cdot \frac{\Delta}{12} = \frac{1}{3} \Delta$. Similarly,Area$(	riangle BED)$ = $\frac{1}{2} \cdot BE \cdot BD \cdot \sin B = \frac{1}{2} \cdot 2 \cdot 2 \cdot \sin B = 2 \sin B$. Since $\Delta = \frac{1}{2} ac \sin B = \frac{1}{2} \cdot 5 \cdot 4 \cdot \sin B = 10 \sin B$,we have $\sin B = \frac{\Delta}{10}$. Thus,Area$(	riangle BED)$ = $2 \cdot \frac{\Delta}{10} = \frac{1}{5} \Delta$. Similarly,Area$(	riangle CFE)$ = $\frac{1}{2} \cdot CF \cdot CE \cdot \sin C = \frac{1}{2} \cdot 2 \cdot 3 \cdot \sin C = 3 \sin C$. Since $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin C = 15 \sin C$,we have $\sin C = \frac{\Delta}{15}$. Thus,Area$(	riangle CFE)$ = $3 \cdot \frac{\Delta}{15} = \frac{1}{5} \Delta$. Area$(	riangle DEF)$ = $\Delta - (	ext{Area}(	riangle ADF) + 	ext{Area}(	riangle BED) + 	ext{Area}(	riangle CFE)) = \Delta - (\frac{1}{3} + \frac{1}{5} + \frac{1}{5}) \Delta = \Delta - \frac{11}{15} \Delta = \frac{4}{15} \Delta$. Therefore,the ratio is $\frac{4}{15}$.

Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.

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