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Question Number 155277 by amin96 last updated on 28/Sep/21

	Answered by Rasheed.Sindhi last updated on 28/Sep/21
	$(big-square-area)−(non-shadow-area) big-square-area=8^2 =64 sq-units Non-shadow-area_(−) A:two-triangles=2((1/2).1.2)=2 unit^2 B:4{2^2 −(1/4)π(2)^2 }=4×((16−4π)/4)=16−4π C:2{4^2 −(1/4)π(4)^2 }=32−8π D(empty small squares):3(2^2 )=12 64−(A+B+C+D) =64−(2+16−4π+32−8π+12) =64−62+12π=2+12π$
	$(b i g - s q u a r e - a r e a) - (n o n - s h a d o w - a r e a)$ $b i g - s q u a r e - a r e a = 8^{2} = 64 s q - u n i t s$ $\underline{N o n - s h a d o w - a r e a}$ $A : t w o - t r i a n g l e s = 2 (\frac{1}{2} . 1.2) = 2 {u n i t}^{2}$ $B : 4 {2^{2} - \frac{1}{4} π {(2)}^{2}} = 4 \times \frac{16 - 4 π}{4} = 16 - 4 π$ $C : 2 {4^{2} - \frac{1}{4} π {(4)}^{2}} = 32 - 8 π$ $D (e m p t y s m a l l s q u a r e s) : 3 (2^{2}) = 12$ $64 - (A + B + C + D)$ $= 64 - (2 + 16 - 4 π + 32 - 8 π + 12)$ $= 64 - 62 + 12 π = 2 + 12 π$
	Commented by amin96 last updated on 28/Sep/21
	$B:4{2^2 −(1/4)π2^2 }=4((16−4π)/4)=16−4π C:2{4^2 −(1/4)π4^2 }=2((64−16π)/4)=32−8π shadow area =64−(16−4π+32−8π+12+2)= =64−62+12π=2+12π Sorry sir it has to be like that right?$
	$B : 4 {2^{2} - \frac{1}{4} π 2^{2}} = 4 \frac{16 - 4 π}{4} = 16 - 4 π$ $C : 2 {4^{2} - \frac{1}{4} π 4^{2}} = 2 \frac{64 - 16 π}{4} = 32 - 8 π$ $s h a d o w a r e a = 64 - (16 - 4 π + 32 - 8 π + 12 + 2) =$ $= 64 - 62 + 12 π = 2 + 12 π$ $S o r r y s i r i t h a s t o b e l i k e t h a t r i g h t ?$
	Commented by Rasheed.Sindhi last updated on 28/Sep/21

	$S o r r y f o r m y m i s t a k e s s i r . I^{'} l l e d i t$ $m y a n s w e r .$
	Answered by qaz last updated on 28/Sep/21

	$A = (1 + 2) \cdot 2 = 6$ $B = π 4^{2} / 4 = 4 π$ $C = π 2^{2} \cdot \frac{3}{4} = 3 π$ $D = 2^{2} = 4$ $E = π 2^{2} / 4 - 2^{2} / 2 = π - 2$ $F = π 4^{2} / 4 - 4^{2} / 2 = 4 π - 8$ $G = 2 \cdot 2 / 2 = 2$ $S = A + B + . . . + G = 12 π + 2$
	Commented by amin96 last updated on 28/Sep/21

	$t h a n k s s i r$
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