In a rectangle, the point of intersection of the diagonals is 4 cm farther from the smaller side than from the larger one

In a rectangle, the point of intersection of the diagonals is 4 cm farther from the smaller side than from the larger one. Perimeter = 56 cm find the sides of the rectangle

1. Let’s designate the vertices of the rectangle ABCD.

2. From the point of intersection of the diagonals O draw perpendiculars OK to the smaller side AB and OE to the larger side AD.

3. OK and OE – the middle lines of triangles ABD and ACD, since AD and AB are parallel, respectively, and drawn from the middle of the diagonals.

4. AD = 2OK, AB = 2 (OK – 4).

5. Let’s compose the expression:

2 x 2 OK + 2 x 2 (OK-4) = 56;

8OK – 16 = 56;

8OK = 72;

OK = 9 cm.

OE = 9 – 4 = 5 cm.

AB = 5 x 2 = 10 cm, AD = 2 x 9 = 18 cm.

Answer: AB = 10 cm, AD = 18 cm.