A trapezium with a chord has two parallel sides $a$ and $b(a<b)$. The longer base can be seen at an angle of $45^{\circ}$ from one end of the shorter base. What could be the area of the trapezium?
Let's use the notations of the figures. We assume that the circular trapezoid is convex. The conditions can be satisfied by two different types of trapezoids if $b$ is not a diameter (see Figure 1 and Figure 2). We answer the question of the problem by expressing the area of the trapezoid in terms of $a$ and $b$.

The $45^{\circ}$ inscribed angle implies that $A_{1} O_{1} B_{1} \varangle = A_{2} O_{2} B_{2} \varangle = 90^{\circ}$. Therefore, $O_{1} T_{1} = O_{2} T_{2} = \frac{a}{2}$, and the radius of the circle is $r = \frac{a}{\sqrt{2}}$. By the Pythagorean theorem,

$$
O_{1} P_{1} = \sqrt{\frac{a^{2}}{2} - \frac{b^{2}}{4}} = \frac{\sqrt{2 a^{2} - b^{2}}}{2}
$$

and thus, for example, the height of the trapezoid in Figure 1 is:

$$
T_{1} P_{1} = \frac{a}{2} - \frac{\sqrt{2 a^{2} - b^{2}}}{2} = \frac{a - \sqrt{2 a^{2} - b^{2}}}{2}
$$

Similarly, we get $T_{2} P_{2} = \frac{a + \sqrt{2 a^{2} - b^{2}}}{2}$. The area of the trapezoid is

$$
t_{1} = \frac{(a + b) \left(a - \sqrt{2 a^{2} - b^{2}}\right)}{4} \quad \text{or} \quad t_{2} = \frac{(a + b) \left(a + \sqrt{2 a^{2} - b^{2}}\right)}{4}
$$

If $b$ is the diameter of the circle, then $b = \frac{2 a}{\sqrt{2}}$, which means $\sqrt{2 a^{2} - b^{2}} = 0$. In this case, $t_{1} = t_{2} = \frac{a(a + b)}{4}$.

Zsuzsa Erdei (Hajdúszoboszló, Hőgyes. E. Gymnasium, 9th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_eaceaeb717a2100bd845g-1.jpg?height=1052&width=616&top_left_y=1022&top_left_x=749)