In the figure,DE parallel BC. If DE = 3 text{ | vedclass

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(D) Given that $DE \parallel BC$.In $\Delta ADE$ and $\Delta ABC$:$\angle ADE = \angle ABC$ (Corresponding angles)$\angle AED = \angle ACB$ (Correspon

Vedclass · Accepted Answer

$60$

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(D) Given that $DE \parallel BC$.In $\Delta ADE$ and $\Delta ABC$:$\angle ADE = \angle ABC$ (Corresponding angles)$\angle AED = \angle ACB$ (Corresponding angles)By $AA$ similarity criterion,$\Delta ADE \sim \Delta ABC$.We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Therefore,$\frac{	ext{Area}(\Delta ADE)}{	ext{Area}(\Delta ABC)} = \frac{DE^2}{BC^2}$.Substituting the given values:$\frac{15}{	ext{Area}(\Delta ABC)} = \frac{3^2}{6^2}$$\frac{15}{	ext{Area}(\Delta ABC)} = \frac{9}{36}$$\frac{15}{	ext{Area}(\Delta ABC)} = \frac{1}{4}$$	ext{Area}(\Delta ABC) = 15 	imes 4 = 60 	ext{ cm}^2$.

In the figure,$DE \parallel BC$. If $DE = 3 \text{ cm}$,$BC = 6 \text{ cm}$,and the area of $\Delta ADE = 15 \text{ cm}^2$,then the area of $\Delta ABC$ is ............ $\text{cm}^2$.

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