Suppose ABC is a triangle and D, E are points | vedclass

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(B) Let $\angle A = 	heta$.The area of $	riangle ABC$ is given by $	ext{ar}(	riangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta$.The a

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$\left(2, \frac{5}{2}\right]$

Vedclass · Answer

(B) Let $\angle A = 	heta$.The area of $	riangle ABC$ is given by $	ext{ar}(	riangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta$.The area of $	riangle ADE$ is given by $	ext{ar}(	riangle ADE) = \frac{1}{2} \cdot AD \cdot AE \cdot \sin 	heta$.Therefore,the ratio of the areas is:$\frac{	ext{ar}(	riangle ABC)}{	ext{ar}(	riangle ADE)} = \frac{\frac{1}{2} \cdot AB \cdot AC \cdot \sin 	heta}{\frac{1}{2} \cdot AD \cdot AE \cdot \sin 	heta} = \frac{AB}{AD} 	imes \frac{AC}{AE}$.Given $AD : AB = 3 : 5$,we have $\frac{AB}{AD} = \frac{5}{3}$.Given $AE : AC = 2 : 3$,we have $\frac{AC}{AE} = \frac{3}{2}$.Thus,the ratio is $\frac{5}{3} 	imes \frac{3}{2} = \frac{5}{2} = 2.5$.The value $2.5$ lies in the interval $\left(2, \frac{5}{2}ight]$.Hence,the correct option is $B$.

Suppose $ABC$ is a triangle and $D, E$ are points on the sides $AB$ and $AC$ respectively. If $AD : AB = 3 : 5$ and $AE : AC = 2 : 3$,then the ratio of the areas of the triangles $ABC$ and $ADE$ lies in the interval.

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