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Question Number 180813 by mr W last updated on 17/Nov/22

Commented by mr W last updated on 17/Nov/22

$t h e a r e a s o f t h r e e s q u a r e s a r e g i v e n .$ $f i n d t h e a r e a s o f t h e t h r e e b l u e$ $s q u a r e s .$
Commented by a.lgnaoui last updated on 17/Nov/22

Commented by mr W last updated on 18/Nov/22

Commented by mr W last updated on 18/Nov/22

$a^{2} = \frac{2 (q^{2} + r^{2}) - p^{2}}{9} = \frac{2 (49 + 36) - 16}{9} = \frac{154}{9}$ $b^{2} = \frac{2 (r^{2} + p^{2}) - q^{2}}{9} = \frac{2 (36 + 16) - 49}{9} = \frac{55}{9}$ $c^{2} = \frac{2 (p^{2} + q^{2}) - r^{2}}{9} = \frac{2 (16 + 49) - 36}{9} = \frac{94}{9}$ $a r e a s o f b l u e s q u a r e s :$ $16 a^{2} = \frac{2464}{9}$ $16 b^{2} = \frac{880}{9}$ $16 c^{2} = \frac{1504}{9}$
Commented by mr W last updated on 19/Nov/22

$p r o o f s e e Q 180882$
	Answered by a.lgnaoui last updated on 18/Nov/22

	$A r e a 1 (R i g h t) = {(5 \sqrt{2} + \frac{\sqrt{94}}{3})}^{2} ≅ 106$ $A r e a 2 (L e f t) = {(\frac{11 \sqrt{2}}{2} + \frac{\sqrt{154}}{3})}^{2} =≅ 142$ $A r e a (d o w n) = 4^{2} = 16$
	Commented by a.lgnaoui last updated on 18/Nov/22

	$c o t e = \frac{6 \sqrt{2}}{3} + \frac{4 \sqrt{2}}{2} + b = 5 \sqrt{2} + b$ $A r e a (R i g h t) = {(5 \sqrt{2} + b)}^{2} = 106$ $A r e a (l e f t)$ $c o t e = \frac{4 \sqrt{2} + 7 \sqrt{2}}{2} + c$ $A r e a = 142$ $D o w n A r e a ?$
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