Answer :

Here,

In ΔADE and ΔABC

∠ADE = ∠ABC by corresponding angles (DE∥BC)

∠DEA = ∠BCA by corresponding angles (DE∥BC)

∴ ΔAED ∼ ΔACB

Similarly,

ΔAGD ∼ ΔAFB (where AF ⊥ BC)

⇒ AF = 3AG (∵ AB = 3AD which is given) ----(1)

Similarly,

⇒ BC = 3×DE ----(2)

{by (1) and (2)}

⇒ Area of ΔABC = 9 × Area of ΔADE

(∵ Area of ΔABC = 72 cm^{2})

⇒ Area of ΔADE = 8 cm^{2}

⇒ Area of DCEB = Area of ΔABC - Area of ΔADE

⇒ Area of DCEB = 72 – 8 = 64cm^{2}

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