SSC CGL Model Paper 9

Consider ΔAOE and ΔDOC
∠AOE = ∠DOC (vertically opposite angle)
∠EAO = ∠OCD; ∠AEO = ∠ODC (Line AE and line DC are parallel)
⇒ ΔAEO ~ ΔCDO
Given, = AE $=\frac{\mathrm{AB}}{2}$
⇒ $\mathrm{AE}=\frac{\mathrm{DC}}{2}$ [AB = DC side of square]
∴ Height of ΔODC = 2 height of ΔAOE
Let the height of ΔAOE = h
Height of ΔDOC = 2h
Side of the square = h + 2h = 3h
Area of ΔDOC =$\frac{1}{2}$ DC × 2h
DC = 3h[side of square] = $\frac{1}{2}\times 3\mathit{h}\times 2\mathit{h}=3{\mathit{h}}^2$
ATQ,
Area of ΔDOC $=20{\mathrm{cm}}^2$
⇒ $3{\mathit{h}}^2=20{\mathrm{cm}}^2$
Area of square $=3\mathit{h}\times 3\mathit{h}=9{\mathit{h}}^2$, Which is equal to 3 times area of ΔDOC $=3\times 20{\mathrm{cm}}^2=60{\mathrm{cm}}^2$