[ Area of a trapezoid ] [Pythagorean theorem (direct and inverse).]

Find the area of a trapezoid if its diagonals are 17 and 113, and the height is 15.
Through the vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$ ($AC=17, BD=113$), draw a line parallel to the diagonal $BD$. Let $K$ be the point of intersection of this line with the line $AD$. Then $S_{ABCD} = S_{ACK}$. The sides $AC, CK=113$, and the height $CM$ of this triangle are known. From the right triangles $ACM$ and $KCM$, we find that $AM^2 = 17^2 - 15^2 = (17-15)(17+15) = 64$, $KM^2 = 113^2 - 15^2 = (113-15)(113+15) = 98 \cdot 128 = 49 \cdot 256 = (7 \cdot 16)^2 = 112^2$.

If point $M$ lies between points $A$ and $K$, then $S_{ACK} = \frac{1}{2}(AM + KM) \cdot CM = 60 \cdot 15 = 900$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-30.jpg?height=228&width=1195&top_left_y=2513&top_left_x=434)

If point $A$ lies between points $M$ and $K$, then $S_{AKC} = \frac{1}{2}(KM - AK) \cdot CM = 52 \cdot 15 = 780$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-31.jpg?height=231&width=1189&top_left_y=-1&top_left_x=432)

## Answer

900 or 780.