Answer:
140
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,
![AB^2=BE^2 +AE^2,\\ \\CD^2=CF^2+DF^2.](https://img.qammunity.org/2020/formulas/mathematics/high-school/zq33ln2tsqaw0c1wqmcuob7ht93q183078.png)
Thus,
![13^2=y^2+x^2,\\ \\14^2=(15-x)^2+y^2.](https://img.qammunity.org/2020/formulas/mathematics/high-school/mdotng0w4hbyen5mnzpq5ep7ap1u9r6saw.png)
Subtract from the second equation the first one:
![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.](https://img.qammunity.org/2020/formulas/mathematics/high-school/vfhld7vxl5mmg7yafg6c9ewtzn1sxg8jkg.png)
Therefore,
![169=6.6^2+y^2,\\ \\y^2=169-43.56,\\ \\y^2=125.44,\\ \\y=11.2.](https://img.qammunity.org/2020/formulas/mathematics/high-school/k3uboboohl361ehppgb08siie8tg098wbq.png)
The area of the trapezoid is
![A=(5+20)/(2)\cdot 11.2=140](https://img.qammunity.org/2020/formulas/mathematics/high-school/8vncp5nvnr86lrlsmzpwqxa7dgw99thdu8.png)