Vertex D of parallelogram ABCD is connected to point O on the BC side. The segment DO intersects

| September 18, 2021 | education

Vertex D of parallelogram ABCD is connected to point O on the BC side. The segment DO intersects the dioganal AC at point K. S of the triangle KOC = 8 S of the triangle CDK = 20 Find the area of the parallelogram.

Consider triangles CKO and CKD, the areas of which are known. Both triangles have a common height CH, then the ratio of the areas of these triangles is equal to the ratio of their bases.

Sco / Sskd = OK / DK.

8/20 = OK / DK = 2/5.

Triangles OKS and AKD are similar, since the angle O is common for them, and the angle OCS = DAK is as criss-cross. The coefficient of similarity of triangles is OK / DK = 2/5.

Then the ratio of the areas of these triangles is equal to the square of the similarity coefficient.

Soks / Sacd = 4/25.

8 / Sackd = 4/25.

Sacd = 8 * 25/4 = 50 cm2.

The area of ​​the triangle ACD will be equal to: Sacd = Sacd + Sscd = 50 + 20 = 70 cm2.

Since the diagonal of the parallelogram divides it into two equal triangles, then Savsd = 2 * Sasd = 2 * 70 = 140 cm2.

Answer: The area of ​​the parallelogram is 140 cm2.



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