Answer:
140
Step-by-step explanation:
Consider trapezoid ABCD. Draw two heights BE and CF. Quadrilateral BEFC is a rectangle, then EF = BC = 5 and BE = CF = y.
Let AE = x, then FD = AD - AE - EF = 20 - x - 5 = 15 - x.
Triangles ABE and CDF are two right triangles. By the Pythagorean theorem,
[tex]AB^2=BE^2 +AE^2,\\ \\CD^2=CF^2+DF^2.[/tex]
Thus,
[tex]13^2=y^2+x^2,\\ \\14^2=(15-x)^2+y^2.[/tex]
Subtract from the second equation the first one:
[tex]14^2-13^2=(15-x)^2-x^2,\\ \\196-169=225-30x+x^2-x^2,\\ \\30x=225-196+169,\\ \\30x=198,\\ \\x=6.6.[/tex]
Therefore,
[tex]169=6.6^2+y^2,\\ \\y^2=169-43.56,\\ \\y^2=125.44,\\ \\y=11.2.[/tex]
The area of the trapezoid is
[tex]A=\dfrac{5+20}{2}\cdot 11.2=140[/tex]
