# In the parallelogram ABCD, the sides AB and BC are 4 and 7, respectively. The bisectors AK and BM

**In the parallelogram ABCD, the sides AB and BC are 4 and 7, respectively. The bisectors AK and BM of the angles of the parallelogram intersect at point O (points K and M lie on the sides BC and AD, respectively). How many times is the area of the pentagon OKСDM greater than the area of triangle OAB?**

Since BM and AK are bisectors of angles, they cut off the isosceles triangles ABM and ABK, and which AB = BK = AM, therefore, the quadrilateral ABKM is a rhombus.

Let’s draw the height BH. The area of the rhombus ABKM will be equal to: Sawkm = AM * BH = 4 * BH cm2.

The diagonals of the rhombus divide it into four equal triangles, then Soav = Scom = Savkm / 4 = BH cm2.

Let us determine the area of the parallelogram СDMK, in which DM = СK = BC = BK = 7 – 4 = 3 cm.

Then Ssdmk = DM * BH = 3 * BH cm2.

Then the area of the OKСDM pentagon will be equal to: Soksdm = Ssdmk + Scom = 3 * BH + BH = 4 * BH cm2.

Then Soksdm / Soav = 4 * BH / BH = 4.

Answer: The area of the pentagon is 4 times larger.