The point B is at the centre of the circle. The pints P and Q are in the circumference of the circle. Calculate the area of the shaded sector. Take pi to be 3.142 in your working. Give your final answer to 1 dp.
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Answer:
24.7cm^2
Step-by-step explanation:
Line PQ = 5.413590865cm
BQ = 9cm
Angle BPQ =
5.413590865² = 9² + 9² - 2 × 9 × 9 × cos(x)
29.30696605 = 81 + 81 - 162 × cos(x)
29.30696605 = 162 - 162 × cos(x)
-132.693034 = -162 × cos(x)
cos(x) = 0.8190928022
x = 35.00591739˚
Area of sector = 24.7cm^2
When point B is at the centre of the circle. So the area of the shaded sector is 24.7 [tex]cm^2[/tex].
Since
Line PQ = 5.413590865cm
BQ = 9cm
Now
[tex]5.413590865^2 = 9^2 + 9^2 - 2 \times 9 \times 9 \times cos(x)\\\\29.30696605 = 81 + 81 - 162 \times cos(x) \\\\29.30696605 = 162 - 162 \times cos(x)\\\\-132.693034 = -162 \times cos(x)\\\\[/tex]
cos(x) = 0.8190928022
x = 35.00591739˚
Now finally the
Area of sector = 24.7cm^2
Hence, When point B is at the centre of the circle. So the area of the shaded sector is 24.7 [tex]cm^2[/tex].
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