## [Isosceles, Inscribed, and Circumscribed Trapezoids] Law of Cosines

A trapezoid with a smaller base of 4 is inscribed in a circle of radius $\sqrt{7}$. Through a point on this circle, a tangent parallel to one of the non-parallel sides of the trapezoid is drawn, and a chord of the circle parallel to the bases of the trapezoid and of length 5 is drawn. Find the length of the diagonal of the trapezoid and the area of the trapezoid.
Prove that the chord from the condition of the problem and the diagonal of the trapezoid subtend equal arcs and, therefore, are equal. Using the sine and cosine theorems, formulate a trigonometric equation with respect to the angle between the diagonal and the base of the trapezoid.

## Solution

Let \(ABCD\) be the given trapezoid with the smaller base \(BC = 4\), and \(M\) be the given point on the smaller arc \(AB\), such that the tangent to the circumscribed circle with radius \(R = \sqrt{7}\), drawn at this point, is parallel to the side \(AB\). Clearly, \(M\) is the midpoint of the arc \(AB\).

The inscribed trapezoid is isosceles. The line passing through point \(M\) parallel to the bases of the trapezoid intersects the smaller arc \(CD\) at its midpoint \(N\).

Arcs enclosed by parallel chords are equal, so the smaller arc \(CD\) is equal to the smaller arc \(AB\), and therefore, the halves of these arcs, i.e., the smaller arcs \(AM\) and \(CN\), are equal. Hence, the chords \(AC\) and \(MN\) are equal, as they subtend equal arcs \(AMBC\) and \(MBCN\). Thus, the diagonal \(AC\) is equal to 5.

Let \(\angle ACB = \angle CAD = \varphi\). Then \(AB = CD = 2R \sin \varphi = 2\sqrt{7} \sin \varphi\). By the cosine theorem,

\[
AB^2 = BC^2 + CA^2 - 2 \cdot BC \cdot CA \cdot \cos \angle ACB, \text{ or } 28 \sin^2 \varphi = 16 + 25 - 2 \cdot 4 \cdot 5 \cdot \cos \varphi
\]

From this equation, we find that \(\cos \varphi = \frac{13}{14}\) or \(\cos \varphi = \frac{1}{2}\).

Consider the first case.

Let \(CH\) be the height of the trapezoid. From the right triangle \(AHC\), we find that

\[
AH = AC \cdot \cos \varphi = 5 \cdot \frac{13}{14} = \frac{65}{14}, \quad CH = AC \cdot \sin \varphi = 5 \cdot \frac{3\sqrt{3}}{14} = \frac{15\sqrt{3}}{14}
\]

Since the trapezoid is isosceles, the segment \(AH\) (the projection of the diagonal of the trapezoid onto the larger base) is equal to the midline of the trapezoid. Therefore,

\[
S_{ABCD} = AH \cdot CH = \frac{65}{14} \cdot \frac{15\sqrt{3}}{14} = \frac{975\sqrt{3}}{196}
\]

In the second case, \(AH = 5 \cdot \frac{1}{2} = \frac{5}{2} < 4\), which is impossible, since the midline cannot be less than the smaller base of the trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-44.jpg?height=617&width=652&top_left_y=1641&top_left_x=703)

## Answer

\(5; \frac{975\sqrt{3}}{196}\)