In rectangle ABCD, the bisector of angle D intersects side AB at point P. The segment AP is 6 times less than the segment BP

In rectangle ABCD, the bisector of angle D intersects side AB at point P. The segment AP is 6 times less than the segment BP. Find the sides of the rectangle if its perimeter is 80 cm.

1. The bisector of the rectangle divides the angle at the vertex D in half, that is, the angle ADP is equal to 45 °.

2. Angle APD = 180 ° – 90 ° – 45 ° = 45 °. That is, triangle ADP is isosceles.

Therefore, AP = AD.

3. For convenience of calculations, we take the length AD as x, the length AB as y.

4. AP = x; BP = 6x; AB = AP + BP = x + 6x = 7x.

5. Let’s compose two equations:

2x + 2y = 80;

y = 7x;

6. Substitute y = 7x in the first equation:

2x + 14x = 80;

16x = 80;

x = 5 cm;

y = 5 x 7 = 35 cm.

Answer: AD = BC = 5 cm, AB = CD = 35 cm.