Question

In a trapezium ABCD with AB || CD and AB = 2CD. The two
diagonals AC and BD intersect at X. Prove that AC = 3/2
AX​

Answer 1

Answer:

ABCD is a trapezium in which, AB ∥ CD and diagonals AC and BD intersect each other at O.

In ΔAOB and ΔCOD we have

∠1=∠2 [Alt ∠s]

∠3=∠4 [vert opp. \angle s\therefor \Delta AOB \sim \Delta COD$$

[By AA criterion of similarity]

⇒

ar(ΔCOD)

ar(ΔAOB)

=

(CD)

2

(2CD)

2

[AB = 2CD (Given)]

=

CD

2

4CD

2

=4

=

ar(ΔCOD)

84

=4;

∴ar(ΔCOD)=84+4=21cm

2

solution