Let's calculate the area of an isosceles trapezoid if the two parallel sides are given as $a$ and $c$ ($a > c$), and we know that the height is equal to the difference between the leg and the smaller parallel side. What is the solvability condition? What is the area if $c = \frac{a}{6}$?

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Denoting the leg of the trapezoid by $b$, its area is

$$
t=\frac{a+c}{2} \cdot m=\frac{a+c}{2}(b-c)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_59ee505f8fa33e95b9c6g-1.jpg?height=271&width=567&top_left_y=317&top_left_x=779)

By the Pythagorean theorem,

$$
b^{2}=(b-c)^{2}+\left(\frac{a-c}{2}\right)^{2}
$$

from which

$$
2 b c=c^{2}+\frac{(a-c)^{2}}{4}
$$

thus

$$
b=\frac{c}{2}+\frac{(a-c)^{2}}{8 c}
$$

and therefore

$$
b-c=\frac{(a-c)^{2}}{8 c}-\frac{c}{2}=\frac{a^{2}-2 a c-3 c^{2}}{8 c}=\frac{(a+c)(a-3 c)}{8 c}
$$

Substituting the value of $b-c$ into (1),

$$
t=\frac{a+c}{2} \cdot \frac{(a+c)(a-3 c)}{8 c}=\frac{(a+c)^{2}(a-3 c)}{8 c}
$$

Since $t$ must be positive, the necessary condition for solvability is that $a-3 c>0$, or $a>3 c$. However, this condition is also sufficient, because if $a>3 c$, then according to (3), $b-c$ is also positive, and thus there exists a right triangle with legs $\frac{a-c}{2}$ and $b-c$, and hypotenuse $b$, which is unique and ensures the existence of the trapezoid.

If $c=\frac{a}{6}$, then the area of the trapezoid is

$$
T=\frac{\left(\frac{7 a}{6}\right)^{2} \frac{a}{2}}{\frac{8 a}{3}}=\frac{49 a^{3}}{72} \cdot \frac{3}{8 a}=\frac{49 a^{2}}{192}
$$

László Perniczky (Kaposvár, Táncsics g. I. o. t.)