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Question Number 6056 by Rasheed Soomro last updated on 11/Jun/16

Commented by Rasheed Soomro last updated on 11/Jun/16

$I a n d H a r e m i d - p o i n t s o f l i n e s e g m e n t s .$ $E i s t h e c e n t r e o f t h e s q u a r e .$ $c e n t e r o f b o t h a r c s i s B .$
Commented by Yozzii last updated on 11/Jun/16

$m A B = x m e a n s t h a t ∣ A B ∣= x ?$
Commented by Rasheed Soomro last updated on 11/Jun/16

$Y e s .$
	Answered by Yozzii last updated on 11/Jun/16

	$◻ A B C D, H l i e s o n D C, F l i e s o n B C,$ $G l i e s o n A B, I l i e s o n A D, E l i e s$ $o n a r c G F, E i s t h e c e n t r e o f ◻ A B C D,$ $H I i s a n a r c, c e n t r e o f b o t h a r c s i s B,$ $∣ H D ∣= 0.5 ∣ D C ∣, ∣ A I ∣= 0.5 ∣ A D ∣,$ $∣ A B ∣= x > 0, a r e a o f A G E F C H I = ?$ $- - - - - - - - - - - - - - - - - - - - - - - - - -$ $N o t e t h a t ∣ A B ∣=∣ B C ∣=∣ C D ∣=∣ D A ∣= x > 0 .$ $B y t h e P h y t h a g o r e a n t h e o r e m,$ $∣ E B ∣= \sqrt{{(\frac{∣ A B ∣}{2})}^{2} + {(\frac{∣ B C ∣}{2})}^{2}} = \sqrt{\frac{x^{2}}{4} \times 2} = x \frac{\sqrt{2}}{2}$ $∠ A B C = \frac{π}{2} . L e t t h e a r e a o f t h e s e c t o r F E G B$ $b e d e n o t e d b y A . ∴ A = \frac{1}{2} r^{2} θ$ $A = \frac{1}{2} ∣ E B ∣^{2} (∠ A B C) = \frac{1}{2} (\frac{x^{2}}{2}) \frac{π}{2} = \frac{π x^{2}}{8} {u n i t s}^{2} .$ $F o r △ H C B, i t s a r e a i s g i v e n b y$ $A_{1} = \frac{1}{2} ∣ B C ∣∣ C H ∣= \frac{1}{2} x \times \frac{1}{2} x = \frac{x^{2}}{4} {u n i t s}^{2} .$ $B y s y m m e t r y o f t h e f i g u r e, f o r △ I A B$ $i t s a r e a i s a l s o A_{1} = \frac{x^{2}}{4} {u n i t s}^{2} .$ $N o w, i n △ B C H, t a n ∠ H B C = \frac{∣ H C ∣}{∣ B C ∣} = \frac{x / 2}{x} = \frac{1}{2}$ $S i n c e 0 < ∠ H B C < \frac{π}{2} \Rightarrow ∠ H B C = {t a n}^{- 1} 0.5 .$ $A l s o, ∠ I B A = {t a n}^{- 1} 0.5 .$ $T h u s, ∠ I B H = \frac{π}{2} - (∠ I B A + ∠ H B C)$ $∠ I B H = \frac{π}{2} - 2 {t a n}^{- 1} 0.5 .$ $N o w, ∣ H B ∣^{2} =∣ I B ∣^{2} =∣ H C ∣^{2} + ∣ B C ∣^{2} b y$ $t h e P h y t h a g o r e a n t h e o r e m .$ $∴∣ H B ∣^{2} = \frac{x^{2}}{4} + x^{2} = \frac{5 x^{2}}{4} .$ $T h u s, t h e a r e a o f t h e s e c t o r I H B i s$ $g i v e n b y A_{2} = \frac{1}{2} ∣ H B ∣^{2} (∠ I B H)$ $o r A_{2} = \frac{1}{2} (\frac{5 x^{2}}{4}) (\frac{π}{2} - 2 {t a n}^{- 1} 0.5)$ $A_{2} = \frac{5 x^{2} (π - 4 {t a n}^{- 1} 0.5)}{16} {u n i t s}^{2} .$ $T h e a r e a o f t h e r e g i o n A G B F C H I i s$ $g i v e n b y A_{3} = A_{2} + 2 A_{1}$ $A_{3} = \frac{5 x^{2} (π - 4 {t a n}^{- 1} 0.5)}{16} + 2 \times \frac{x^{2}}{4}$ $A_{3} = \frac{x^{2}}{2} (\frac{5 π - 20 {t a n}^{- 1} 0.5 + 16}{16})$ $A_{3} = \frac{x^{2} (5 π + 16 - 20 {t a n}^{- 1} 0.5)}{32}$ $T h e f i n a l a n s w e r i s g i v e n b y$ $A_{3} - A = x^{2} (\frac{5 π + 16 - 20 {t a n}^{- 1} 0.5}{32} - \frac{π}{8})$ $A_{3} - A = x^{2} (\frac{π + 16 - 20 {t a n}^{- 1} 0.5}{32}) {u n i t s}^{2}$
	Commented by Rasheed Soomro last updated on 11/Jun/16

	$\overset{V}{G_{OO} D} S t r a t e g y!$ $A S t r a t e g y I {d i d n}^{'} t t h i n k o f!$
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