The bases of an isosceles trapezoid are 1 cm and 17 cm, and the diagonal divides its obtuse

The bases of an isosceles trapezoid are 1 cm and 17 cm, and the diagonal divides its obtuse angle in half. Find the area of the trapezoid.

Since the bases of the trapezoid are parallel, the diagonal is a secant, and when intersecting parallel straight secant lines in a cross, the lying angles are equal, then the angle between the diagonal of the trapezoid and the smaller base is equal to the angle between the diagonal and the larger base. In this trapezoid, the diagonal divides the obtuse angle in half, which means the angles between the diagonal and the large base and between the diagonal and the side are also equal. Consequently, the triangle formed by the side of the trapezoid, its diagonal and the large base is isosceles, where the diagonal of the trapezoid is the base of this triangle, the side of the trapezoid and the larger base of the trapezoid are the sides of the isosceles triangle, equal to each other. This means that the side of the trapezoid is equal to the larger base and is equal to 17 cm.
The larger base of an isosceles trapezoid is equal to the sum of the smaller base and two equal projections of the lateral sides. Let’s find the projection of the lateral side onto the larger base: (17-1) / 2 = 16/2 = 8 cm.
Consider a right-angled triangle, in which the lateral side of the trapezium is the hypotenuse, the legs are the height of the trapezium and the projection of the lateral side to the larger base. The sum of the squares of the legs is equal to the square of the hypotenuse, we find the height of the trapezoid: h ^ 2 = 17 ^ 2-8 ^ 2 = 289-64 = 225, h = √225 = 15 cm.
The area of ​​the trapezoid is equal to the product of the height and the midline: S = h * m = 15 * (17 + 1) / 2 = 15 * 9 = 135 cm2.