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Question Number 6157 by Rasheed Soomro last updated on 16/Jun/16

	Answered by Yozzii last updated on 16/Jun/16

	$∣ B C ∣=∣ C D ∣=∣ D A ∣=∣ A B ∣= x u n i t s > 0$ $F o r m t h e t r i a n g l e △ G B H w h e r e ∠ G B H = 90 °$ $a n d ∠ G H B = θ (0 < θ < 90 °) . L e t ∣ G H ∣= r > 0 . S i n c e$ $H i s t h e c e n t r e o f a r c G E, t h e n ∣ E H ∣=∣ G H ∣= r .$ $∣ A E ∣= \frac{1}{2} ∣ A B ∣= \frac{x}{2} \Rightarrow∣ E B ∣= \frac{x}{2} .$ $∣ G C ∣= \frac{1}{4} ∣ B C ∣= \frac{x}{4} \Rightarrow∣ G B ∣= \frac{3 x}{4} .$ $∣ B H ∣=∣ E H ∣ - ∣ E B ∣= r - \frac{x}{2} .$ $I n △ G B H, s i n θ = \frac{∣ B G ∣}{∣ G H ∣} = \frac{3 x}{4 r}$ $a n d c o s θ = \frac{∣ B H ∣}{∣ G H ∣} = \frac{r - \frac{x}{2}}{r} .$ $S i n c e {c o s}^{2} θ + {s i n}^{2} θ = 1$ $\Rightarrow {(\frac{3 x}{4 r})}^{2} + {(\frac{r - \frac{x}{2}}{r})}^{2} = 1$ $\Rightarrow \frac{9 x^{2}}{16} + {(r - \frac{x}{2})}^{2} = r^{2}$ $\frac{9 x^{2}}{16} + r^{2} - r x + \frac{x^{2}}{4} = r^{2}$ $\frac{x^{2}}{4} (\frac{9}{4} + 1) = r x$ $r = \frac{13 x}{16} u n i t s$ $∴ s i n θ = \frac{3 x}{4 \times \frac{13 x}{16}} = \frac{12}{13} \Rightarrow θ = {s i n}^{- 1} \frac{12}{13} .$ $F o r △ G B H, i t s a r e a i s$ $A_{1} = \frac{1}{2} (r - \frac{x}{2}) (\frac{3 x}{4})$ $A_{1} = \frac{1}{2} (\frac{5}{16}) \frac{3 x^{2}}{4} = \frac{15 x^{2}}{128}$ $F o r t h e s e c t o r G H E, i t s a r e a i s$ $A_{2} = \frac{1}{2} r^{2} θ = \frac{1}{2} {(\frac{13 x}{16})}^{2} {s i n}^{- 1} \frac{12}{13}$ $A_{2} = \frac{169 x^{2}}{512} {s i n}^{- 1} \frac{12}{13}$ $T h e a r e a o f t h e r e g i o n G B E i s$ $A_{3} = A_{2} - A_{1} = x^{2} (\frac{169}{512} {s i n}^{- 1} \frac{12}{13} - \frac{15}{128}) .$ $T h e a r e a o f t h e r e g i o n A E G C D i s$ $A = x^{2} - A_{3}$ $A = x^{2} (\frac{143}{128} - \frac{169}{512} {s i n}^{- 1} \frac{12}{13}) = \frac{x^{2}}{128} (143 - \frac{169}{4} {s i n}^{- 1} \frac{12}{13}) s q u a r e u n i t s .$
	Commented by Rasheed Soomro last updated on 16/Jun/16

	$NICE!$ $THANKS! A l t h o u g h I t h o u g h t a$ $d i f f e r e n t a p p r o a c h, l i k e y o u r$ $a p p r o a c h t o o . M y a p p r o a c h n e e d s$ $s o m e c o n s t r u c t i o n l i n e s .$ $S o I t h i n k y o u r a p p r o a c h i s b e t t e r!$
	Commented by Yozzii last updated on 16/Jun/16

	$I s e e . I t h i n k y o u s h o u l d s t i l l$ $s h a r e y o u r m e t h o d, j u s t t o$ $g i v e a n a l t e r n a t i v e a p p r o a c h .$
	Commented by Rasheed Soomro last updated on 17/Jun/16

	$N o t m y e n t i r e m e t h o d i s d i f f e r e n t . O n l y d i f f e r e n c e$ $i s i n c o m p u t i n g r a d i u s a n d a n g l e o f t h e s e c t o r . . . .$
	Commented by Yozzii last updated on 17/Jun/16

	$A h o k l o l .$
	Commented by Rasheed Soomro last updated on 17/Jun/16

	Commented by Rasheed Soomro last updated on 18/Jun/16

	$T h e r a d i u s a n d a n g l e o f t h e s e c t o r$ $c a n a l s o b e d e t e r m i n e d i n t h e f o l l o w i n g$ $w a y :$ $C o n s t r u c t i o n : J o i n E a n d G . D r a w$ $p e r p e n d i c u l a r - b i s e c t o r o f E G .$ $A l s o j o i n G a n d H$ $A c c o r d i n g t o {Y o z z i}^{'} s a n s w e r$ $∣ E B ∣= \frac{1}{2} x a n d ∣ B G ∣= \frac{3}{4} x$ $∙ S o i n △ E B G, ∣ E G ∣= \sqrt{{(x / 2)}^{2} + {(3 x / 4)}^{2}}$ $= \sqrt{\frac{x^{2}}{4} + \frac{9 x^{2}}{16}} = \frac{\sqrt{13}}{4} x$ $S o, ∣ E J ∣= \frac{1}{2} ∣ E G ∣= \frac{1}{2} (\frac{\sqrt{13}}{4} x) = \frac{\sqrt{13}}{8} x [\overset{\leftrightarrow}{H J} i s p e r p e n d i c u l a r b i s e c t o r]$ $L e t ∠ G E B = α, \cos α =∣ E B ∣ / ∣ E G ∣= (\frac{1}{2} x) / (\frac{\sqrt{13}}{4} x) = \frac{2}{\sqrt{13}}$ $α = \cos^{- 1} \frac{2}{\sqrt{13}}$ $∙ N o w i n △ E J H, ∣ E J ∣= \frac{\sqrt{13}}{8} x a n d α = \cos^{- 1} \frac{2}{\sqrt{13}}$ $\cos α =∣ E J ∣ / ∣ E H ∣$ $\cos (\cos^{- 1} \frac{2}{\sqrt{13}}) = \frac{(\sqrt{13} / 8) x}{r} [l e t ∣ E H ∣= r]$ $\frac{2}{\sqrt{13}} = \frac{\sqrt{13} x}{8 r}$ $16 r = 13 x$ $r = \frac{13}{16} x$ $L e t ∠ J H E = β a n d ∠ G H E = θ$ $β = 90 ° - α = 90 ° - \cos^{- 1} \frac{2}{\sqrt{13}}$ $θ = 2 β = 180 ° - {2 \cos}^{- 1} \frac{2}{\sqrt{13}}$ $F o r f u r t h e r s i m p l i f i c a t i o n s e e YOZZII^{'} s$ $You can't use 'macro parameter character #' in math mode$ $θ = \sin^{- 1} (\frac{12}{13})$
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