Trapezium bases 5 and 12, diagonals 12.16. Find the height.

Draw a straight line through point C parallel to the diagonal BD.

Then the quadrangle ВСКD is a parallelogram, and then DК = ВС = 4 cm, CK = ВD = 12 cm.

The length of the segment AK = (AD + DK) = (AD + BC) = 15 + 5 = 20 cm.

The area of ​​the triangle ACK is equal to: Sask = (AD + BC) * CH / 2.

The area of ​​the trapezium ABCD is equal to: Savsd = (AD + BC) * CH) / 2.

Sask = Savsd.

Let us determine the area of ​​the triangle ACK by Heron’s theorem.

The semi-perimeter of the triangle AСK is equal to: p = (AS + СK + AK) / 2 = (16 + 12 + 20) / 2 = 24 cm.

Then Sask = √24 * (24 – 16) * (24 – 12) * (24 – 20) = √24 * 8 * 12 * 4 = √9216 = 96 cm2.

CH = 2 * Sask / (AD + BC) = 2 * 96 / (15 + 5) = 9.6 cm.

Answer: The height of the trapezoid is 9.6 cm.