Question

ABCD is a trapezium such that BC is parallel to AD and AB = 4cm.if the diagonals AC and BD intersect at O such that AO/ OC = DO/ OB = 1/ 2 , then BC = ?

de optons r...

a. 7cm

b.8cm

c.9cm

d.6cm

why?plz answer!!!

Answer 1

Ignore the pic, it's answer is wrong....

I think AD = 4cm

In ∆OAD and ∆OCD,
angle AOD = angle COB....(vertically opposite angles)

angle BCA = angle CAD ..(alternate angles)

By AA criterion OAD ~ OCD is a similarity.

AO/OC = AD/CB
1/2=4/BC
BC = 8cm

☺☺ HERE IS YOUR ANSWER ☺

Your answer is 8cm.

Answer 2

Given: ABCD is a trapezium such that BC is parallel to AD and AB = 4cm.if the diagonals AC and BD intersect at O such that AO/ OC = DO/ OB = 1/ 2.

To find: The value of BC.

Solution:

Let the angle AOD be equal to x. Since angles AOD and BOC are vertically opposite angles, the angle BOC is also equal to x. Let angle OAD be equal to y. Since the angles OAD and OCB are alternate angles, the angle OCB is also equal to y. The sides AD and BC are parallel to each other. Thus, it can be said that the triangles OAD and OBC are similar by the ASA (Angle-Side-Angle) postulate.

It is given in the question that AO/OC = DO/OB = 1/2. Now, the corresponding sides to the two triangles can be written as

$\frac{OD}{OB} = \frac{AD}{BC} = \frac{AO}{OC}$

$\frac{1}{2} = \frac{4}{BC} = \frac{1}{2}$

$BC = 8 cm$

Therefore, the value of BC is 8 cm.