Find the area of a trapezoid whose diagonals are 7 cm and 15 cm, the middle line is 10 cm.

Through the vertex C we draw a segment of the CК parallel to the ВD diagonal.

The quadrangle of the ВСKD is a parallelogram, then DK = ВС, СK = ВD = 15 cm.

AK = AD + DK = AD + BC.

The middle line of the trapezoid is equal to: MН = (BP + BC) / 2 = 10 cm.

(BP + ВС) = 2 * 10 = 20 cm.

Then AK = 20 cm.

In triangle ACK, according to Heron’s theorem, we define its area.

The half-perimeter of the triangle is: p = (7 + 15 + 20) / 2 = 21 cm.

Then Sask = √21 * (21 – 7) * (21 – 15) * (21 – 20) = √1764 = 42 cm.

Since the triangle ACK and the trapezium have a total height СР, and the length of the base of the triangle is equal to the sum of the bases of the trapezium, then Sask = Savsd = 42 cm2.

Answer: The area of ​​the trapezoid is 42 cm2.