Question

in the adjoining figure, the diagonals of a parallelogram
intersect at O. OE is drawn parallel to CB to meet AB at
E, find
area of AAOE : area of |gm ABCD.​

Answer 1

Answer:

Step-by-step explanation:

In ∆ AOE & ∆ ACB

Angle ACB= Angle AOB. (corresponding)

Angle A = Angle A. (Common angles)

Therefore... ∆AOE ~ ∆ACB

As diagonals of a parallelogram bisect AO=OC

and as the 2 triangles are similar:

AE/AB = AO/AC

AE/AB =AO/AO+OC

AE/AB =AO/2AO( as AO=OC)

AE/AB=1/2

area of ∆AOE/ area of ∆ACB= AE^2/AB^2 = 1/4

Now parallelogram ABCD = 2* ∆ACB

= 2*4= 8

Area of ∆AOE /area of ABCD= 1/8 OR. 1:8

(DIAGRAM IS BELOW)