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.
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