Question

O is a point inside a parallelogram ABCD. Find the area of (∆AOB + ∆COD), when area of parallelogram ABCD is 40 sq.cm.

Answer 1

Let the height of the parallelogram be h and the height of COD be x So the height of AOB will be h-x. Opposite sides are equal So bases if both the triangles is equal. Ar(aob)+ar(cod)=1/2*(h-x)*base + 1/2 * x*base =1/2*base[h-x+x] =1/2*(base* h) Base* height= 40 Area of both triangles is 1/2*40 That is 20

Answer 2

Answer:

If point O present inside parallelogram

Triangle AOB + triangle COD is half of the parallelogram

Then;

Triangle BOC + triangle DOA is another half of parallelogram

If triangle AOB + triangle COD = 16 sq. cm

Then area of parallelogram = 2×[ triangle AOB + triangle COD]

= 2× 16 sq. cm

= 32 sq. cm

Step-by-step explanation:

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