Question

Given ABCD is rectangle and E is the midpoint of BC and a line drawn from E meets the AD at G when extended. If DG = 2 and area of DFG = 40 and area of EFC = 10, find the area of the rectangle ABCD.

• A. 40
• B. 80
• C. 140
• D. 120

Answer 1

The correct option is D

120

In $\Delta DFG$ and $\Delta EFC$

$\angle EFC=\angle DFG$ (vertically opposite angles)

$\angle GDF=\angle ECF$ (Perpendicular)

$\angle DGF=\angle EFC$ (alternate interior angles)

$\Delta DFG\approx \Delta CFE$ (AAA similarity)

Ratio of areas of similar triangles is equalto the ratio of the squares of their corresponding sides

$\frac{areaof\Delta DFG}{areaof\Delta EFC}=\frac{D{G}^2}{E{C}^2}$

EC=1

Similarly DF : FC = 2 : 1

BC = 2 EC

=2

DC = DF + FC

=FC + 2FC

=3FC

$\frac{areaofABCD}{areaof\Delta EFC}=\frac{2\times AD\times DC}{EC\times FC}$

=12

Area of ABCD = 120