In a rectangle ABCD. AB = 6 cm AD = 10 cm k – bisector of angle A. (K – belongs to BC)

In a rectangle ABCD. AB = 6 cm AD = 10 cm k – bisector of angle A. (K – belongs to BC). Determine the midline of the AKCD trapezoid.

1. The angle BAK of the triangle ABK is equal to 45 °, since the bisector AK divides the angle 90 ° in half.

3. Angle AKB is also equal to 180 ° – 90 ° – 45 = 45 °, that is, an isosceles triangle.

Therefore, BK = AB = 6 cm.

4. Considering that the opposite sides of the rectangle are equal, that is, AD = BC, we calculate the length of the CS segment:

KС = ВС – ВK = 10 – 6 = 4 cm.

5. Considering that the middle line of the trapezoid АКСD is equal to the half-sum of the bases, we calculate its length: (10 + 4) / 2 = 7 cm.

Answer: the length of the midline of the trapezoid AKCD is 7 cm.