Question

ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that :
(i) $\Delta AOB\sim \Delta COD$
(ii) If OA = 6 cm, OC = 8 cm,
Find:

(a) $\frac{Area\left(\Delta AOB\right)}{Area\left(\Delta COD\right)}$
(b) $\frac{Area\left(\Delta AOD\right)}{Area\left(\Delta COD\right)}$

Answer 1

Given: ABCD is a trapezium in which AB II CD. The diagonals AC and BD intersect at O.OA = 6 cm and OC = 8 cm.

To prove: (i) ΔAOB ~ ΔCOD (ii) ar(ΔAOB)ar(ΔCOD)

Proof:
In ΔAOB and ΔCOD,
∠AOB = ΔCOD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate angles)
∴ ΔAOB ~ ΔCOD (AA similarity)
⇒ ar( ∆ AOB) ar( ∆ COD) = OA^ 2/ OC^ 2 (Area Theorem)
⇒ ar( ∆ AOB) ar( ∆ COD)
= ( 6 cm /8 cm )^ 2
= ( 3 /4 )^2 = 9 16