SSC CGL Model Paper 15

Given,
AE = 7 cm
ED = 24 cm
AD $=\sqrt{{24}^2+7^2}=25\mathrm{cm}$
AD = BC = 25 cm (opposite side of rectangle)
BF = 15cm
FC $=\sqrt{{25}^2-{15}^2}=20\mathrm{cm}$
FG × BC = BF × FC
FG $=\frac{15\times 20}{25}=12\mathrm{cm}$
ΔAED ~ ΔFHD
(∠D = common)
(∠DHF = ∠AGD = 90°)
$\frac{\mathrm{AE}}{\mathrm{HF}}=\frac{\mathrm{AD}}{\mathrm{FD}}=\frac{\mathrm{ED}}{\mathrm{HD}}$
In ΔBFC and ΔFGC
∠C = common
∠BFC = ∠FGC = 90°
⇒ ΔBFC ~ ΔFGC
$\frac{\mathrm{BC}}{\mathrm{FC}}=\frac{\mathrm{FC}}{\mathrm{GC}}$
$\mathrm{GC}=\frac{{\mathrm{FC}}^2}{\mathrm{GC}}=\frac{400}{25}\mathrm{cm}=16\mathrm{cm}$
GC = HD [HDCG is a rectangle]
$\frac{\mathrm{AE}}{\mathrm{HF}}=\frac{\mathrm{ED}}{\mathrm{HD}}$
$\Rightarrow \mathrm{HF}=\frac{16\times 7}{24}=\frac{14}{3}\mathrm{cm}$
DC = HG = HF + FG $=\frac{14}{3}+12=\frac{50}{3}\mathrm{cm}$
Area $=\frac{50}{3}\times 25=\frac{1250}{3}{\mathrm{cm}}^2$