The bases of an isosceles trapezoid are 62 and 92, the side is 39. Find the length of the diagonal of the trapezoid.

Given:
lateral trapezoid ABCE,
BC = 62,
AE = 92.
Find the AC diagonal -?
Solution:
1. Let’s draw the heights of AO and CH.
2. Triangle ABO = triangle CHE by hypotenuse and acute angle. Then HE = AO = (92 – 60): 2 = 30: 2 = 15 (centimeters);
3. Consider a right-angled triangle CHE. By the Pythagorean theorem (the square of the hypotenuse is equal to the sum of the squares of the legs):
CH ^ 2 + HE ^ 2 = CE ^ 2;
CH ^ 2 = CE ^ 2 – HE ^ 2;
CH ^ 2 = 39 ^ 2 – 15 ^ 2;
CH ^ 2 = 1521-225;
CH ^ 2 = 1,296;
CH = 36;
4. Consider a right-angled triangle ACH. By the Pythagorean theorem:
CH ^ 2 + AH ^ 2 = AC ^ 2;
36 ^ 2 + 77 ^ 2 = AC ^ 2;
5 929 + 1 296 = AC ^ 2;
AC ^ 2 = 7,225;
AC = 85.
Answer: 85.