# 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