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Question Number 60812 by peter frank last updated on 26/May/19

Commented by tanmay last updated on 26/May/19

$a r e a o f B C A D B = 2 \times \frac{1}{2} \times 5 \times 12 = 60$ $60 = 2 [\frac{1}{2} \times 13 \times h]$ $h = \frac{60}{13}$ $s i n θ_{1} = \frac{h}{12} a n d s i n θ_{2} = \frac{h}{5}$ $s i n θ_{1} = \frac{60}{13 \times 12} = \frac{5}{13} s i n θ_{2} = \frac{12}{13}$ $a r e a o f t r i a n g l e △ B C D = \frac{1}{2} \times C D \times B O$ $= \frac{1}{2} \times 2 h \times 12 c o s θ_{1} [c o s θ_{1} = \frac{B O}{12}]$ $= \frac{60}{13} \times 12 \times \frac{12}{13} = 60 \times {(\frac{12}{13})}^{2}$ $a r e a o f t r i s n g l e △ C A D = \frac{1}{2} \times C D \times A O$ $= \frac{1}{2} \times 2 h \times 5 c o s θ_{2} [c o s θ_{2} = \frac{A O}{5}]$ $= \frac{60}{13} \times 5 \times \frac{5}{13} = 60 \times {(\frac{5}{13})}^{2}$ $a r e a o f s e c t o r = \frac{π r^{2}}{2 π} \times θ = \frac{1}{2} r^{2} θ$ $h e r e o n e s e c t o r a r e a = \frac{1}{2} \times 12^{2} \times 2 θ_{1} =$ $a n o t h e r s e c t o r a r e a = \frac{1}{2} \times 5^{2} \times 2 θ_{2}$ $s h a d e d a r e a = [12^{2} θ_{1} + 5^{2} θ_{2}] - [60 \times {(\frac{12}{13})}^{2} + 60 \times {(\frac{5}{13})}^{2}]$ $= 144 {s i n}^{- 1} (\frac{5}{13}) + 25 {s i n}^{- 1} (\frac{12}{13}) - 60$
Commented by peter frank last updated on 26/May/19

$f i n d a r e a o f s h a d e d r e g i o n$
Commented by tanmay last updated on 26/May/19

Commented by peter frank last updated on 26/May/19

$t h a n k y o u s i r$
Commented by tanmay last updated on 26/May/19

$m o s t w e l c m e . . .$
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