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I am asked to find the shaded region. The markscheme states that it is found by taking $$POQ + POR + QOR - PQR$$which would be $$\frac{r^{2}sin(π-θ)}{2}+\frac{r^{2}sin(π-θ)}{2}+\frac{r^{2}2θ}{2}+\frac{l^{2}θ}{2}$$However, isn't $$POQ + POR + QOR = PQR$$Why does $PQR$ not include the shaded region?

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user373534
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  • By $QOR$ they definitely meant the sector of smaller circle. So $QOR$ includes the shaded region. – jonsno Sep 28 '17 at 10:27
  • Yes indeed, but why is it that the calculation for PQR, using the formula for the area of a sector, specifically pertains to the smaller circle? What stops it from being used for the bigger circle? – user373534 Sep 28 '17 at 10:29
  • You can, but then you can't calculate its area easily. It would not be a sector. Its a sector only of the lower, smaller circle. – jonsno Sep 28 '17 at 10:30
  • See, $PQR$ does not contain shaded portion, and is part of bigger circle. But here consider $QOR$ as part of smaller circle, and it includes shaded area. – jonsno Sep 28 '17 at 10:39

1 Answers1

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It should have been written as $[\triangle POQ] + [\triangle POR] + [sector(QOR)] - [sector(QPR)]$; where sector QOR is centered at O with radius = r, while sector QPR is centered at P with radius = l.

Mick
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