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.
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 ? ?
1 answer