In an isosceles trapezoid, the height drawn from the top of an obtuse angle divides the larger

In an isosceles trapezoid, the height drawn from the top of an obtuse angle divides the larger base into 2 segments, the larger of which is 18 find the area if the height is 12.

Draw a straight line through vertex C parallel to AB. A quadrilateral ABCK is a parallelogram, since its opposite sides are parallel in pairs. Then BC = AK.

The SDK triangle is isosceles, since СD = СK, then DН = KН.

The middle line of the trapezoid is: RM = (BC + AD) / 2 = (AK + AK + KН + DН) / 2 = (2 * AK + 2 * KН) / 2 = (AK + KN) = AH = 18 cm.

Then the area of the trapezoid is equal to: Savsd = AH * CH = 18 * 12 = 216 cm2.

Answer: The area of the trapezoid is 216 cm2.