14.1. (Czechoslovakia, 80). The length of the largest side of an isosceles trapezoid is 13, and the perimeter is 28.

a) Find the sides of the trapezoid if its area is 27.

b) Can the area of such a trapezoid be 27,001?
14.1. Let $AD$ be the larger base and $BH$ the height of the given trapezoid $ABCD$ (Fig. 92). Then

$$
AD=13
$$

(otherwise, $AB=CD=13, AD+BC=28-2 \cdot 13=2$ and $S_{ABCD}=$ $=BH \cdot(AD+BC) / 2 \leqslant 13 \cdot(2 / 2)=13<27)$,

$$
\begin{gathered}
AB=x, \\
BC=28-13-2 x=15-2 x, \quad AH=\frac{13-(15-2 x)}{2}=x-1_{i} \\
BH=\sqrt{AB^2-AH^2}=\sqrt{2 x-1}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-245.jpg?height=190&width=512&top_left_y=579&top_left_x=349)

Fig. 92

Applying the theorem of means, we get

$$
\begin{aligned}
S_{ABCD}=\sqrt{2 x-1} \cdot(28-2 x) / 2=\sqrt{(2 x-1)(14-x)^{2}} & \leqslant \sqrt{\left(\frac{(2 x-1)+(14-x)+(14-x)}{3}\right)^{3}}=\sqrt{\left(\frac{27}{3}\right)^{3}}=27
\end{aligned}
$$

with equality $S_{ABCD}=27$ possible only in the case

$$
2 x-1=14-x, \text { i.e., } x=5
$$

and $AB=BC=CD=5$, while the equality

$$
S_{ABCD}=27.001
$$

is impossible.