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Problem 229
229
In the diagram, \(OAXB\) is a sector of a circle with centre \(O\) and radius 10 cm. The length of the chord \(AB\) is 12 cm. The line \(OX\) passes through \(M\), the mid-point of \(AB\), and \(OX\) is perpendicular to \(AB\). The shaded region is bounded by the chord \(AB\) and by the arc of a circle with centre \(X\) and radius \(XA\).
(i) Show that angle \(AXB\) is 2.498 radians, correct to 3 decimal places.
(ii) Find the perimeter of the shaded region.
(iii) Find the area of the shaded region.
Solution
(i) Letting \(M\) be the midpoint of \(AB\), we have \(OM = 8\) cm using the Pythagorean theorem. Therefore, \(XM = 2\) cm. Using trigonometry, \(\tan AXM = \frac{6}{2}\), so \(AXB = 2 \times \tan^{-1}(3) = 2.498\) radians.
(ii) The length \(AX = \sqrt{6^2 + 2^2} = \sqrt{40}\). The arc \(AYB = r\theta = \sqrt{40} \times 2.498\). The perimeter is \(12 + \text{arc} = 27.8\) cm.
(iii) The area of sector \(AXBY = \frac{1}{2} \times (\sqrt{40})^2 \times 2.498\). The area of triangle \(AXB = \frac{1}{2} \times 12 \times 2\). Subtract these to find the shaded area: 38.0 cm\(^2\).