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Question Number 119529 by I want to learn more last updated on 25/Oct/20

Commented by help last updated on 25/Oct/20
$Area of shaded portion {(4/7)×10^2 }=400/7cm^2$
$A r e a o f s h a d e d p o r t i o n$ ${\frac{4}{7} \times 10^{2}} = 400 / 7 {c m}^{2}$
Commented by benjo_mathlover last updated on 25/Oct/20

$\frac{1600}{7} = 228.571429$ $228.571429$ $i m p o s s i b l e$
Commented by help last updated on 25/Oct/20

$t h i s i s a s h o r t c u t a p p r o a c h . i c a n p r o v e a r e a o f o n e l e a f t o b e 4 / 7 x^{2} i f n e e d e d$
Commented by benjo_mathlover last updated on 25/Oct/20

$y o u r s h o r t c u t i s v a l i d i f r a d i u s = 7 k$ $w h e r e k \in Z$
Commented by help last updated on 25/Oct/20

$400 / 7 i s 100 % c o r r e c t . . . . {t h e r e}^{} w a s a t y p o b e f o r e w h i c h w a s c o r r e c t e d . . . s t i l l l e a r n i n g t h e a p p u s a g e t h o$
Commented by mr W last updated on 25/Oct/20

$p l e a s e {d o n}^{'} t w r i t e a l l i n o n e s i n g l e$ $l i n e!$
Commented by talminator2856791 last updated on 27/Oct/20

$can this diagram be drawn in the$ $drawing tool ? i dont know how to draw$ $it$
	Answered by mr W last updated on 25/Oct/20

	$s h a d e d a r e a = 8 s e c t o r s =$ $2 r e d c i r c l e s w i t h r a d i u s 5 -$ $2 g r e e n s q u a r e s w i t h s i d e l e n g t h 5 \sqrt{2}$ $= 2 \times π \times 5^{2} - 2 \times {(5 \sqrt{2})}^{2} = 50 (π - 2)$ $= 57.08 {c m}^{2}$
	Commented by mr W last updated on 25/Oct/20

	Commented by talminator2856791 last updated on 27/Oct/20

	$very nice drawing . did you draw it with$ $the drawing tool ?$
	Answered by ebi last updated on 25/Oct/20

	$l e t s t a k e t h e f i r s t q u a r t e r o f A B C D$ $a t p o i n t A$ $a r e a o f s e c t o r (c e n t e r a t m i d p o i n t$ $A B, r = 5 c m)$ $= \frac{1}{4} π r^{2}$ $= \frac{1}{4} (\frac{22}{7}) (5^{2}) = 19.64 {c m}^{2}$ $a r e a o f t r i a n g l e (r i g h t - a n g l e t r i a n g l e$ $a t m i d p o i n t A B)$ $= \frac{1}{2} (5) (5) = 12.5 {c m}^{2}$ $a r e a o f h a l f o f t h e s h a d e d r e g i o n a t$ $t h e f i r s t q u a r t e r$ $= 19.64 - 12.5 = 7.14 {c m}^{2}$ $a r e a o f s h a d e d r e g i o n a t t h e f i r s t$ $q u a r t e r$ $= 2 \times 7.14 = 14.28 {c m}^{2}$ $t h u s,$ $t o t a l a r e a o f s h a d e d r e g i o n$ $= 4 \times 14.28 = 57.12 {c m}^{2}$
	Answered by TANMAY PANACEA last updated on 25/Oct/20

	$a r e a o f e a c h l e a f = 2 x$ $a r e a o f e a c h t r i a n g u l a r c u r v e d a r e a y$ $4 (2 x) + 4 y = a^{2}$ $a g a i n a r e s o f t r i a n g l e$ $2 x + y = \frac{1}{2} \times a \times \frac{a}{2} = \frac{a^{2}}{4}$ $∡ D O C = \frac{π}{2}$ $s o r a d i u s o f c u r v e = \frac{a}{2}$ $2 (2 x) + y = \frac{π a^{2}}{2}$ $4 x + y = \frac{π a^{2}}{2} \times 4$ $8 x + 4 y = a^{2}$ $16 + 4 y = 2 π a^{2}$ $8 x + 4 y = a^{2}$ $8 x = a^{2} (2 π - 1)$ $a r e a s h a d d e d = 8 x$
	Commented by TANMAY PANACEA last updated on 25/Oct/20

	$P l s f i n d e r r o r i n t h i s c a l c u l a t i o n . . .$
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