D and E are points on sides AB and AC of Delt | vedclass

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(A) Given that $DE \parallel BC$,by the Basic Proportionality Theorem,$\Delta ADE \sim \Delta ABC$.Since the triangles are similar,the ratio of their

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$4:21$

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(A) Given that $DE \parallel BC$,by the Basic Proportionality Theorem,$\Delta ADE \sim \Delta ABC$.Since the triangles are similar,the ratio of their areas is equal to the square of the ratio of their corresponding sides.Given $\frac{AD}{DB} = \frac{2}{3}$,we have $AD = 2k$ and $DB = 3k$.Thus,$AB = AD + DB = 2k + 3k = 5k$.The ratio of the areas is $\frac{	ext{Area}(\Delta ADE)}{	ext{Area}(\Delta ABC)} = \left(\frac{AD}{AB}ight)^2 = \left(\frac{2k}{5k}ight)^2 = \frac{4}{25}$.Let $	ext{Area}(\Delta ADE) = 4x$ and $	ext{Area}(\Delta ABC) = 25x$.Then,$	ext{Area}(BDEC) = 	ext{Area}(\Delta ABC) - 	ext{Area}(\Delta ADE) = 25x - 4x = 21x$.The ratio of the area of $\Delta ADE$ to the area of quadrilateral $BDEC$ is $\frac{4x}{21x} = 4:21$.

$D$ and $E$ are points on sides $AB$ and $AC$ of $\Delta ABC$. $DE$ is parallel to $BC$. If $AD:DB = 2:3$,what is the ratio of the area of $\Delta ADE$ to the area of quadrilateral $BDEC$?

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