In a rectangle, the point of intersection of the diagonals is 4 cm farther from the smaller

| September 6, 2021 | education

In a rectangle, the point of intersection of the diagonals is 4 cm farther from the smaller side than from the larger side. The square corner is 56cm. Find its sides

Let the length of the segment OM = X cm, then, by condition, the length of the segment OK = (X + 4) cm.

In a rectangle, the diagonals are equal and are halved at the intersection. ОВ = DO = ВD / 2.

The OAD triangle is isosceles, then OM is the height and median of the triangle. AM = DM, then OM is the middle line of the triangle ABD, and AB = 2 * OM.

Similarly, AD = 2 * OK.

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

8 * X = 56 – 16 = 40.

X = OM = 40/8 = 5 cm.

OK = 4 + 5 = 9 cm.

Then AB = 2 * OM = 2 * 5 = 10 cm, AD = 2 * OK = 2 * 9 = 18 cm.

Answer: The sides of the rectangle are 10 cm and 18 cm.



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