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March 27, 2017

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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 - ,

    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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