The bases of a rectangular trapezoid are 9 and 17 cm, and the diagonal bisects its obtuse angle. Find the area of the trapezoid.

Angle ACB = CAD as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AC.

Since AC is the bisector of the angle ВСD, then in the triangle АСD the angle АСD = CAD, and then the triangle АСD is isosceles, СD = АD = 17 cm.

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

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

CH = 15 cm.

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

Answer: The area of the trapezoid is 195 cm2.