Find the area of the trapezoid whose diagonals are 13 and 11, and the middle line is 10

The area of ​​the trapezoid is equal to the product of the height by half the sum of the bases:
SABCD = h * (BC + AD) / 2 = h * l, where l is the midline of the trapezoid l = (BC + AD) / 2. Therefore, we need to find the height h.
Extend the base AD and draw a segment from the vertex C parallel to BD until it intersects with the extended base at point M (as shown in the figure).
In the BCMD quadrangle, the side CM || BD (we ourselves carried out the CM) and DM || BC (as defined by the trapezoid).
Therefore, the quadrangle BCMD is a parallelogram.
Then, by the parallelogram property, DM = BC.
AM = AD + DM = AD + BC = 2l = 2 * 10 = 20
Consider the triangle ACM.
We know the lengths of all its sides, therefore we can find the area through the half-perimeter:
Half-perimeter p = (AC + CM + AM) / 2 = (AC + BD + AM) / 2 = (13 + 11 + 20) / 2 = 22
SACM = √p (p-AC) (p-CM) (p-AM) = √22 (22-13) (22-11) (22-20) = √22 * 9 * 11 * 2 = √4356 = 66
According to another formula, SACM = h * AM / 2 = 66
h = 2 * 66 / AM = 2 * 66/20 = 6.6
Now we can calculate the area of ​​the trapezoid:
SABCD = h * l = 6.6 * 10 = 66
Answer: 66

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