Consider ΔAOE and ΔDOC ∠AOE = ∠DOC (vertically opposite angle) ∠EAO = ∠OCD; ∠AEO = ∠ODC (Line AE and line DC are parallel) ⇒ ΔAEO ~ ΔCDO Given, = AE =
AB
2
⇒ AE=
DC
2
[AB = DC side of square] ∴ Height of ΔODC = 2 height of ΔAOE Let the height of ΔAOE = h Height of ΔDOC = 2h Side of the square = h + 2h = 3h Area of ΔDOC =
1
2
DC × 2h DC = 3h[side of square] =
1
2
×3h×2h=3h2 ATQ, Area of ΔDOC =20cm2 ⇒ 3h2=20cm2 Area of square =3h×3h=9h2, Which is equal to 3 times area of ΔDOC =3×20cm2=60cm2