[ Midline of the trapezoid ] $[\quad$ Area of the trapezoid $\quad]$

The diagonals of the trapezoid are 6 and 8, and the midline is 5. Find the area of the trapezoid.
Let $h$ be the height of trapezoid $ABCD$ with bases $AD$ and $BC$ and diagonals $AC=6$ and $BD=8$, and let $l$ be the midline of the trapezoid. Draw a line through vertex $C$ parallel to diagonal $BD$ until it intersects the extension of base $AD$ at point $M$. Then quadrilateral $BCMD$ is a parallelogram, so

$$
CM = BD = 8, DM = BC, AM = AD + DM = AD + BC = 2l = 10.
$$

Thus, triangle $ACM$ is a right triangle ( $AM^2 = AC^2 + CM^2$ ). Its area is half the product of

$$
\text{ the legs, i.e., } \\
S_{\triangle ACM} = \frac{1}{2} AC \cdot CM = \frac{1}{2} \cdot 6 \cdot 8 = 24. \\
\text{ Therefore, } \\
S_{ABCD} = l h = \frac{1}{2}(AD + BC) h = \frac{1}{2}(AD + DM) \cdot h = \\
= \frac{1}{2}(AD + BC) \cdot h = \frac{1}{2} AM \cdot h = S_{\triangle ACM} = 24.
$$

## Answer

24.00

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