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Problem 256
256
The diagram shows a square ABCD of side 10 cm. The mid-point of AD is O and BXC is an arc of a circle with centre O.
Show that angle BOC is 0.9273 radians, correct to 4 decimal places.
Find the perimeter of the shaded region.
Find the area of the shaded region.
Solution
(i) To find angle BOC, use the formula for the angle in radians: \(\theta = 2 \tan^{-1} \left( \frac{1}{2} \right)\). Calculating gives \(\theta = 0.9273\) radians.
(ii) First, find the radius OB using the Pythagorean theorem: \(OB = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\) cm.
Next, calculate the arc length BXC using \(s = r\theta\), where \(r = \sqrt{125}\) and \(\theta = 0.9273\). Thus, \(s = 11.2 \times 0.9273 = 10.4\) cm.
The perimeter of the shaded region is the arc length plus the side BC of the square: \(10.4 + 10 = 20.4\) cm.
(iii) The area of the sector BXC is given by \(\frac{1}{2} r^2 \theta\). Substituting the values, \(\frac{1}{2} \times 11.2^2 \times 0.9273 = 7.96\) cm².
