ABCD – parallelogram, O – intersection point of the diagonals of this parallelogram Saod = 7.5 find Sabcd =?

Diagonals ABCD – AC AНD BD. The area of the triangle is S aod = 1/2 * (AO * OD * sin <AOD = S boc = 7.5. Let us determine the areas of the other two triangles included in the area ABCD.

S aob = S cod = 1/2 * BO * AO * sin <AOB = 1/2 * BO * AO * sin <(180 – <AOD) = 1/2 * BO * AO * sin <AOD.

Now we need to prove that S boc = S aob. Since sin <AOB = sin <AOD, we prove that BO = OD (the diagonals are halved by the intersection point), similarly to AO = OC, that is, S boc = S aob.

Then S abcd = S aob + S boc + S cod + S aod = 4 * S boc = 4 * 7.5 = 30