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 ~ ΔCDOGiven, = AE $= \frac{AB}{2}$⇒ $AE=\frac{DC}{2}$ [AB = DC side of square]∴ Height of ΔODC = 2 height of ΔAOELet the height of ΔAOE = hHeight of ΔDOC = 2hSide of the square = h + 2h = 3hArea of ΔDOC =$\frac{1}{2}$ DC × 2h DC = 3h[side of square] = $\frac{1}{2} \times 3h \times 2h = 3h^{2}$ATQ,Area of ΔDOC $=20\,\mathrm{cm}^{2}$⇒ $3h^{2} = 20\,\mathrm{cm}^{2}$Area of square $= 3h \times 3h = 9h^{2}$, Which is equal to 3 times area of ΔDOC $= 3 \times 20\,\mathrm{cm}^{2} = 60\,\mathrm{cm}^{2}$