The bases of an isosceles trapezoid are 7 and 51, the tangent of the angle is 5/11, find the height of the trapezoid?

ABCD – trapezoid, AB = CD, AD = 54, BC = 7, tgA = tgC = 5/11.
1. From the top B draw the height BH to the base AD. Point H divides AD into segments, one of which is equal to the half-sum of the bases (DH), and the second is the half-difference of the bases (AH).
Thus, AH = (AD – BC) / 2 = (51 – 7) / 2 = 44/2 = 22.
2. The tangent of an acute angle A in a right-angled triangle AHB is the ratio of the opposite leg BH to the adjacent AH, then:
tgA = BH / AH.
Substitute the known values:
BH / 22 = 5/11;
BH = 22 * 5/11 (in proportion);
BH = 10.
Answer: BH = 10.