In a rectangle, the intersection of the diagonals is 6 cm farther from one side than from the other side.

| February 3, 2021 | education

In a rectangle, the intersection of the diagonals is 6 cm farther from one side than from the other side. The perimeter of the rectangle is 56 cm. Find the sides of the rectangle.

Let the length of the segment OM = X cm, then, by condition, the length of the segment OK = (X + 6) cm.

The diagonals of the rectangle are equal and are halved at the intersection. OВ = DO =ВD / 2.

Triangle OAB is isosceles, then OK is the height and median of the triangle. AK = ВK, then OK is the middle line of the AВD triangle, and AD = 2 * OK.

Similarly AB = 2 * OM.

The perimeter of the rectangle is: Ravsd = 2 * (AB + AD) = 2 * (2 * OM + 2 * OK) = 4 * (X + X + 6) = 56.

8 * X = 56 – 24 = 32.

X = OM = 32/8 = 4 cm.

OK = 4 + 6 = 10 cm.

Then AB = 2 * OM = 2 * 4 = 8 cm, AD = 2 * OK = 2 * 10 = 20 cm.

Answer: The sides of the rectangle are 8 cm and 20 cm.



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