In parallelogram ABCD, the bisector of an acute angle A intersects the side BC at point M.

univerkov | June 19, 2021 | education

In parallelogram ABCD, the bisector of an acute angle A intersects the side BC at point M. It is known that AD = 10cm, BM = 4cm. a) Determine the type of quadrangle AMCD. b) Find the perimeter ABCD and the center line of the trapezoid AMCD.

1. The bisector AM of the parallelogram ABCD cuts off the triangle ABM, which is isosceles. Therefore, AB = BM = 4 cm.

2. The total length of all sides of the parallelogram (perimeter) = 2AB + 2 AD = 2 x 4 + 2 x 10 = 8 + 20 = 28 centimeters.

3. CM = BC – BM = 10 – 4 = 6 centimeters.

4. Quadrilateral AMCD is a trapezoid, since the CM side is parallel to the AD side, and the other two sides AM and CD are not parallel.

5. The middle line of the trapezoid AMCD = (CM + AD): 2 = (6 + 10): 2 = 8 centimeters.

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