In triangle ABC, a line parallel to side BC i | vedclass

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(A) Given that $DE \parallel BC,$ therefore $	riangle ADE \sim 	riangle ABC$ by $AA$ similarity criterion.Since $D$ divides $AB$ in the ratio $1:4,$

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(A) Given that $DE \parallel BC,$ therefore $	riangle ADE \sim 	riangle ABC$ by $AA$ similarity criterion.Since $D$ divides $AB$ in the ratio $1:4,$ we have $AD:DB = 1:4.$ This implies $AD:AB = 1:(1+4) = 1:5.$The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Thus,$\frac{	ext{Area}(	riangle ADE)}{	ext{Area}(	riangle ABC)} = \left(\frac{AD}{AB}ight)^2 = \left(\frac{1}{5}ight)^2 = \frac{1}{25}.$Given $	ext{Area}(	riangle ABC) = 200 	ext{ cm}^2,$ we have $	ext{Area}(	riangle ADE) = \frac{1}{25} 	imes 200 = 8 	ext{ cm}^2.$The area of quadrilateral $DECB$ is $	ext{Area}(	riangle ABC) - 	ext{Area}(	riangle ADE) = 200 - 8 = 192 	ext{ cm}^2.$

In $\triangle ABC,$ a line parallel to side $BC$ intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $D$ divides $AB$ in the ratio $1:4.$ If the area of $\triangle ABC$ is $200 \text{ cm}^2,$ what is the area (in $\text{cm}^2$) of quadrilateral $DECB$?

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