A rectangle PQRS is inscribed in a circle. One side of the rectangle, SR is 12cm and another side QR is \(4\sqrt{3}\) cm. What is the length of the minor arc QR?

Explanation: The diagonal of the inscribed rectangle (\(SQ\)) is the diameter (\(D\)) of the circle. The triangle SQR is a right-angled triangle. By Pythagoras theorem, \(D^2 = SR^2 + QR^2\). \(D^2 = 12^2 + (4\sqrt{3})^2 = 144 + 16 \times 3 = 144 + 48 = 192\). \(D = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}\) cm. Radius (\(r\)) = \(\frac{D}{2} = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\) cm. We need the length of the minor arc QR. The arc length formula is \(s = r \cdot \theta\) (\(\theta\) in radians) or \(s = 2\pi r \cdot \frac{\theta}{360}\) (\(\theta\) in degrees). Consider the triangle QOR, where O is the center. \(OQ = OR = r = 4\sqrt{3}\) cm. \(QR = 4\sqrt{3}\) cm. Since all three sides of \(\triangle QOR\) are equal, it is an equilateral triangle. The central angle \(\angle QOR = \theta\) is \(60^\circ\). Arc Length \(s = 2\pi r \cdot \frac{\theta}{360}\). \(s = 2\pi (4\sqrt{3}) \cdot \frac{60}{360}\). \(s = 8\pi\sqrt{3} \cdot \frac{1}{6}\). \(s = \frac{8\pi\sqrt{3}}{6} = \frac{4\pi\sqrt{3}}{3}\) cm.