SSC CGL 12 Aug 2017 Shift 1 Paper

$\mathrm{\Delta }\mathrm{ABC}\sim \mathrm{\Delta }\mathrm{ADE}$
and since, $\frac{\mathit{D}\mathit{E}}{\mathit{B}\mathit{C}}=\frac{1}{3}$
The areas of $\mathrm{\Delta }\mathrm{ADE}$ and $\mathrm{\Delta }\mathrm{ABC}$ will be in the ratio $\frac{1^2}{3^2}=\frac{1}{9}.$
That means when the area of $\mathrm{\Delta }\mathit{A}\mathit{D}\mathit{E}$ is $20,$ the area of $\mathrm{\Delta }\mathit{A}\mathit{B}\mathit{C}$ will be $20\times 9=180$
Now, area of trapezium $\mathit{B}\mathit{C}\mathit{E}\mathit{D}$ is $180-20=160$
The line segment $\mathit{D}\mathit{C}$ will divide the trapezium $\mathit{B}\mathit{C}\mathit{E}\mathit{D}$ into two triangles viz. $\mathrm{\Delta }\mathit{C}\mathit{E}\mathit{D}$ and $\mathrm{\Delta }\mathit{C}\mathit{B}\mathit{D}$ in which $\mathrm{\Delta }\mathit{C}\mathit{E}\mathit{D}=\frac{1}{2}\times \mathit{D}\mathit{E}\times$ distance between the parallel lines $\mathit{D}\mathit{E}$ and $\mathit{B}\mathit{C}$.
Area of $\mathrm{\Delta }\mathit{C}\mathit{B}\mathit{D}=\frac{1}{2}\times \mathit{B}\mathit{C}\times$ distance between the parallel lines $\mathit{D}\mathit{E}$ and $\mathit{B}\mathit{C}$. But $\mathit{D}\mathit{E}=\frac{1}{2}\mathit{B}\mathit{C}$, hence $\mathit{D}\mathit{C}$ will divide the trapezium $\mathit{B}\mathit{C}\mathit{E}\mathit{D}$ is divided in the ratio 1: 3 and Area of $\mathrm{\Delta }\mathit{C}\mathit{E}\mathit{D}=\frac{1}{1+3}\times$ areof the trapezium BCED $=\frac{1}{4}\times 160=40{\mathrm{cm}}^2$