SSC CHSL 20 Dec 2015 Shift 1 Paper

∆ODE and ∆OCB are similar. We know that sides of similar triangles are in the same ratio.
OE : OB = 2 : 1
If two triangles are similar then the ratio of their areas is equal to the ratio of squares of their corresponding sides.Equation can be written as,
$\frac{\mathit{O}{\mathit{E}}^2}{{\mathit{O}\mathit{B}}^2}=\frac{\mathrm{Area}\mathrm{of}\mathrm{triangle}\mathrm{ODE}}{\mathrm{Area}\mathrm{of}\mathrm{triangle}\mathrm{BCO}}$
Area of ∆BCO = 4(Area of ∆ODE)
$\frac{1}{3}$ (Area of ∆ABC) = 4 (Area of triangle ∆ODE) (∵ area of ∆BCO = $\frac{1}{3}$ area of ∆ ABC)
Area of ∆ ABC = 12 (Area of ∆ ODE)
∆ODE : ∆ABC = 1 : 12