Question

class 10 chapter triangles...pls solve this and help me

Answer 1

Hey there!

Solution :

Given : ABCD is a trapezium in which AB || DC, 2AB = 3DC.

            $\frac{AB}{DC} = \frac{3}{2}$

To find : area Δ AOB : area Δ COD.

Proof : In Δs AOB and COD, we have

       ∠AOB = ∠COD            [Vertcally opp. ∠s]

and, ∠OAB = ∠OCD            [Alternate ∠s]

∴ By AA - criterion of similarity, we have

    Δ AOB ~ Δ COD

⇒ $\frac{Area \ (Δ \ AOB)}{Area \ (Δ \ COD)} = \frac{AB^{2}}{DC^{2}}$

⇒ $\frac{Area \ ( Δ \ AOB)}{Area \ (Δ \ COD)} = \frac{9}{4}$

⇒ Hence, area (Δ AOB) : area (Δ COD) = 9 : 4