a piece of wire 40cm long is cut into two parts and each part is then bent into a square.if the sum of these squares is 68cm^2 find the lengths of the two pieces of wire.


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Question Number 40764 by mondodotto@gmail.com last updated on 27/Jul/18

$a piece of wire 40 cm long$ $is cut into two parts and each part$ $is then bent into a square . if the sum$ $of these squares is 68 {cm}^{2}$ $find the lengths of the two$ $pieces of wire .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

	$x = 4 a_{1} 40 - x = 4 a_{2}$ $a_{1}^{2} + a_{2}^{2} = 68$ $4 a_{1} + 4 a_{2} = 40$ $a_{1}^{2} + {(10 - a_{1})}^{2} = 68$ $2 a_{1}^{2} - 20 a_{1} + 32 = 0$ $a_{1}^{2} - 10 a_{1} + 16 = 0$ $(a_{1} - 2) (a_{1} - 8) = 0$ $i f a_{1} = 2 a_{2} = 8$ $i f a_{1} = 8 a_{2} = 2$ $l e n t t h o f t w o p i e c e a r e 4 \times 2 a n d 4 \times 8$ $8 c m a n d 32 c m$
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