CAT 2002 QA Question 38 · Geometry - Mensuration

Area of trapezium ABCD = $\frac{1}{2}$ BC (CD + AB)
Also BC = 2CD, AB = BC = 2CD

Area of trapezium ABCD = $\frac{1}{2}$ × CD (CD + 2DC) = 3CD²

Area of trapezium DEFG = $\frac{1}{2}$ EO(EF + DG)

EO = CD

EF = CD

DG = CH + HK + KL = CD + CD + $\frac{CD}{2}=\frac{5}{2CD}$

Area of trapezium DEFG = $\frac{1}{2}\times CD\left(CD+\frac{5}{2}CD\right)=\frac{1}{2}\times CD\left(\frac{7CD}{2}\right)=\frac{7C{D}^2}{4}...\left(ii\right)$ 

Ratio of ABCD : DEFG = 12 : 7

Hence, option (c).