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Problem 216
216
The diagram shows an arc BC of a circle with centre A and radius 5 cm. The length of the arc BC is 4 cm. The point D is such that the line BD is perpendicular to BA and DC is parallel to BA.
(i) Find angle BAC in radians.
(ii) Find the area of the shaded region BDC.
Solution
(i) The angle BAC in radians is given by the formula for arc length: \(\theta = \frac{\text{arc length}}{\text{radius}}\). Here, \(\theta = \frac{4}{5} = 0.8\) radians.
(ii) To find the area of the shaded region BDC, we calculate:
1. The length of BD: \(BD = 5 \sin(0.8) \approx 3.587\) cm.
2. The length of DC: \(DC = 5 - 5 \cos(0.8) \approx 1.516\) cm.
3. The area of sector BAC: \(\frac{1}{2} \times 5^2 \times 0.8 = 10\) cm².
4. The area of segment BC: \(\frac{1}{2} \times 5^2 \times (0.8 - \sin(0.8)) \approx 1.03\) cm².
5. The area of triangle BDC: \(\frac{1}{2} \times BD \times DC \approx 2.719\) cm².
6. The area of the trapezium: \(\frac{1}{2} \times (5 + DC) \times BD \approx 11.69\) cm².
7. The shaded area BDC: \(\text{Trapezium} - \text{Sector} + \text{Segment} = 11.69 - 10 + 1.03 = 1.69\) cm².
