In a trapezium ABCD, seg AB || seg DC seg BD seg AD, seg AC seg BC, If AD = 15, BC = 15 and AB = 25. Find

Construct DE ⊥ AB and CF ⊥ AB

In ΔADB, as BD ⊥ AD, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

(AB)² = (AD)² + (BD)²

⇒ 25² = 15² + BD²

⇒ BD² = 625 - 225 = 400

⇒ BD = 20 cm

Similarly,

AC = 20 cm

Now, In ΔAED and ΔABD

∠AED = ∠ADB [Both 90°]

∠DAE = ∠DAE [Common]

ΔAED ~ ΔABD [By Angle-Angle Criteria]

[Property of similar triangles]

As AD = 15 cm, BD = 20 cm and AB = 25 cm

⇒ DE = 12 cm

Also,

⇒ AE = 9 cm

Similarly, BF = 9 cm

Now,

DC = EF [By construction]

DC = AB - DE - AE

DC = 25 - 9 - 9 = 7 cm

Also, we know

Area of trapezium