An isosceles trapezoid ABCD has bases AD = 17cm, BC = 5cm, and leg AB = 10 cm. A line is drawn through vertex B so that it bisects diagonal AC, and intersect AD at point M. Find the area of ΔBDM. What is the area of ABCD?What to answer:Area of triangle BDM=?Area or triangle ABCD=?

1. Let E be the midpoint of AC. Then ΔBCE is congruent to ΔMAE, and MA = BC = 5 cm. Then MD = 17 cm -5 cm = 12 cm.

The height of the trapezoid is found from the Pythagorean theorem. If the center 5 cm of the trapezoid is removed, the remaining ends form an isosceles triangle with sides 10 cm and base 17 -5 = 12 cm. Its height in cm is
.. h = √(10² -(12/2)²) = √(100 -36) = 8

So ΔBDM has base 12 cm and height 8 cm. its area is
.. A = (1/2)(12 cm)(8 cm) = 48 cm²


2. The area of trapezoid ABCD is found from
.. A = (b₁ +b₂)/2*h
.. A = (17 cm +5 cm)/2*(8 cm)
.. A = 88 cm²


The area of ΔBDM is 48 cm².
The area of trapezoid ABCD is 88 cm².