Posted by **Sanjna** on Friday, September 30, 2011 at 4:44am.

In parallelogram ABCD, P is mid-point of AB.CP and BD intersect each other at point O. If area of triangle POB = 40 cm^2,

find : (i) OP :OC

(ii) areas of triangle BOC and PBC

(iii) areas of /\ ABC and parallelogram ABCD ? ?

- Geometry -
**Reiny**, Friday, September 30, 2011 at 9:37am
Did you make your sketch?

Notice that ∆PBO is similar to ∆CDO

let PB = 1, then CD = 2 (remember AB was bisected)

then OP : OC = 1 : 2

let's fill in areas:

∆POB = 40 (given)

∆PBO = 40 , (same base, same height)

∆COD = 160 ( areas proportional to square of sides)

let ∆BCO = x

let ∆AOD = y

x + 160 = y + 80

y = x+80

now look at ∆PBC and ABD , they have the same height, but bases of 1 and 2

so x+40 = (1/2)(y+80)

2x + 80 = y+80

2x = y , ahhhh!

then 2x = x+80

x = 80 and y = 160

Now we know it all, take it from here.

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