Find the area of the trapezoid, the bases of which have lengths 7 and 9, and one of the lateral sides is 15

Find the area of the trapezoid, the bases of which have lengths 7 and 9, and one of the lateral sides is 15 and forms an angle of 150 ° with one of the bases.

The CO height cuts off the right-angled triangle COD in the trapezoid. <BCD = 150; <OCD = 150 – 90 = 60; hence, <D = 180 – 90 -60 = 30 (the sum of the angles of the triangle is 180 degrees).

CO is the height of the trapezoid. CO = 1/2 CD (in a right-angled triangle, the leg, which lies opposite an angle of 30 degrees, is equal to half the hypotenuse). CO = 15: 2 = 7.5

The area of the trapezoid is equal to the product of the half-sum of the bases and the height.

S tr. = (BC + AD) / 2 * CO;

S tr. = (7 + 9) / 2 * 7.5 = 16/2 * 7.5 = 8 * 7.5 = 60.

Answer. 60.