In triangle ABC,AD and BE are medians drawn f | vedclass

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(C) Let $O$ be the centroid of $	riangle ABC$. The medians $AD$ and $BE$ intersect at $O$. We know that the centroid divides the median in the ratio

Vedclass · Accepted Answer

$\frac{49}{12}$

Vedclass · Answer

(C) Let $O$ be the centroid of $	riangle ABC$. The medians $AD$ and $BE$ intersect at $O$. We know that the centroid divides the median in the ratio $2:1$. Given $AD = \frac{7}{2}$,so $AO = \frac{2}{3} AD = \frac{2}{3} 	imes \frac{7}{2} = \frac{7}{3}$. In $	riangle AOB$,$\angle OAB = \frac{\pi}{8}$ and $\angle OBA = \frac{\pi}{4}$. Therefore,$\angle AOB = \pi - (\frac{\pi}{8} + \frac{\pi}{4}) = \pi - \frac{3\pi}{8} = \frac{5\pi}{8}$. Using the sine rule in $	riangle AOB$: $\frac{AO}{\sin(\frac{\pi}{4})} = \frac{BO}{\sin(\frac{\pi}{8})}$. $BO = \frac{AO \sin(\frac{\pi}{8})}{\sin(\frac{\pi}{4})} = \frac{7/3 	imes \sin(\frac{\pi}{8})}{1/\sqrt{2}} = \frac{7\sqrt{2}}{3} \sin(\frac{\pi}{8})$. The area of $	riangle AOB = \frac{1}{2} 	imes AO 	imes BO 	imes \sin(\angle AOB) = \frac{1}{2} 	imes \frac{7}{3} 	imes \frac{7\sqrt{2}}{3} \sin(\frac{\pi}{8}) 	imes \sin(\frac{5\pi}{8})$. Since $\sin(\frac{5\pi}{8}) = \cos(\frac{\pi}{8})$,the area of $	riangle AOB = \frac{49\sqrt{2}}{18} \sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}) = \frac{49\sqrt{2}}{18} 	imes \frac{1}{2} \sin(\frac{\pi}{4}) = \frac{49\sqrt{2}}{36} 	imes \frac{1}{\sqrt{2}} = \frac{49}{36}$. The area of $	riangle ABC = 3 	imes 	ext{Area of } 	riangle AOB = 3 	imes \frac{49}{36} = \frac{49}{12}$.

In $\triangle ABC$,$AD$ and $BE$ are medians drawn from $A$ and $B$. If $AD = \frac{7}{2}$,$\angle DAB = \frac{\pi}{8}$ and $\angle ABE = \frac{\pi}{4}$,then the area (in sq. units) of $\triangle ABC$ is

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