# The bases of the rectangular trapezoid are 9 cm and 17 cm, and the diagonal is the bisector

**The bases of the rectangular trapezoid are 9 cm and 17 cm, and the diagonal is the bisector of its obtuse angle. Calculate the area of the trapezoid.**

Since ABCD is a trapezoid, then BC is parallel to AD, and then the angle ACB = CAD as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AC.

Angle ABC = ACD since AC is bisector, then ACD triangle is isosceles with the base of AC, which means CD = BP = 17 cm.

Let’s build the height of the CH. ABCH is a rectangle, then AH = BC = 9 cm, then DH = AD – AH = 17 – 9 = 8 cm.

In a right-angled triangle SDN, according to the Pythagorean theorem, CH ^ 2 = CD ^ 2 – DH ^ 2 = 289 – 64 = 225.

CH = 15 cm.

Determine the area of the trapezoid.

Sabsd = (BC + AD) * CH / 2 = (9 + 17) * 15/2 = 195 cm2.

Answer: The area of the trapezoid is 195 cm2.