In the figure,if DE parallel BC, find the rat | vedclass

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(D) Given,$DE \parallel BC,$ $DE = 6 \, cm$ and $BC = 12 \, cm.$In $	riangle ABC$ and $	riangle ADE,$$\angle ABC = \angle ADE$ [Corresponding angles

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$1: 3$

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(D) Given,$DE \parallel BC,$ $DE = 6 \, cm$ and $BC = 12 \, cm.$In $	riangle ABC$ and $	riangle ADE,$$\angle ABC = \angle ADE$ [Corresponding angles]$\angle ACB = \angle AED$ [Corresponding angles]$\angle A = \angle A$ [Common angle]Therefore,$	riangle ADE \sim 	riangle ABC$ [By $AAA$ similarity criterion].We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.$\frac{\operatorname{ar}(ADE)}{\operatorname{ar}(ABC)} = \frac{(DE)^2}{(BC)^2}$$= \frac{(6)^2}{(12)^2} = \left(\frac{6}{12}ight)^2 = \left(\frac{1}{2}ight)^2 = \frac{1}{4}.$Let $\operatorname{ar}(ADE) = k,$ then $\operatorname{ar}(ABC) = 4k.$Now,$\operatorname{ar}(DECB) = \operatorname{ar}(ABC) - \operatorname{ar}(ADE) = 4k - k = 3k.$Therefore,the required ratio is $\operatorname{ar}(ADE) : \operatorname{ar}(DECB) = k : 3k = 1 : 3.$

In the figure,if $DE \parallel BC,$ find the ratio of $\operatorname{ar}(ADE)$ and $\operatorname{ar}(DECB).$

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