$[$ Transfer of side, diagonal, etc.]

The bases of the trapezoid are 1 and 7, and the diagonals are 6 and 10. Find the area of the trapezoid.
Through the vertex $C$ of the smaller base $BC$ of the trapezoid $ABCD (BC=1, AD=7, AC=6, BD=10)$, we draw a line parallel to the diagonal $BD$ until it intersects the line $AD$ at point $K$. In triangle $ACK$, it is known that

$$
AC=9, CK=BD=10, AK=AD+DK=AD+BC=7+1=8.
$$

Since $CK^2 = AC^2 + AD^2$, triangle $ACK$ is a right triangle. Its area is half the product of the legs, i.e.,

$$
S_{\triangle ACK}=\frac{1}{2} AC \cdot AK=\frac{1}{2} \cdot 8 \cdot 6=24.
$$

The area of trapezoid $ABCD$ is equal to the area of this triangle, since triangles $ABC$ and $CDK$ are equal in area ($BC=DK$, and the heights dropped to these sides are equal to the height of the trapezoid).

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## Answer

24.00