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Question Number 53962 by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

$F i n d m a x i m u m p e r i m e t e r o f$ $t r i a n g l e A P Q . (h a v e i p o s t e d t h i s$ $q u e s t i o n, b e f o r e, i^{'} m n o t s u r e! ?)$
Commented by ajfour last updated on 27/Jan/19

$i s i t j u s t = a + b + \sqrt{a^{2} + b^{2}} ?$
Commented by mr W last updated on 27/Jan/19

$y e s s i r .$
	Answered by mr W last updated on 27/Jan/19

	Commented by mr W last updated on 27/Jan/19

	$A^{'} D = b$ $A^{″} B = a$ $A Q = A^{'} Q$ $P A = {P A}^{″}$ $p e r i m e t e r o f Δ A P Q = A^{'} Q + Q P + {P A}^{″}$ $m i n . p e r i m e t e r = A^{'} A^{″} = 2 \sqrt{a^{2} + b^{2}}$ $m a x . p e r i m e t e r = A^{'} D + D B + {B A}^{″}$ $= b + \sqrt{a^{2} + b^{2}} + a$
	Commented by ajfour last updated on 28/Jan/19

	$g r e a t t e c h n i q u e t o p r o v e i t S i r, t h a n k y o u!$
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