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Question Number 49536 by behi83417@gmail.com last updated on 07/Dec/18

Commented by behi83417@gmail.com last updated on 07/Dec/18

$△ a n d △ a r e e q u i l a t e r a l t r i a n g l e s w i t h$ $p a r a l l e l s i d e s a n d PM, i s t h e s a m e f o r$ $a l l s i d e s . i f : a r e a (△) = 32 a n d a r e a (△) = 6,$ $t h e n : v a l v e o f : PM = ?$
Commented by Kunal12588 last updated on 07/Dec/18
is answer 1.40704302?
	Answered by mr W last updated on 07/Dec/18

	$A = \frac{\sqrt{3} a^{2}}{4}$ $\Rightarrow a = \sqrt{\frac{4 A}{\sqrt{3}}} = \frac{2 \sqrt{3 \sqrt{3} A}}{3}$ $a_{r e d} = \frac{2 \sqrt{3 \sqrt{3} \times 32}}{3} = \frac{8 \sqrt{6 \sqrt{3}}}{3}$ $a_{b l u e} = \frac{2 \sqrt{3 \sqrt{3} \times 6}}{3} = 2 \sqrt{2 \sqrt{3}}$ $l e t δ = P M$ $a_{r e d} - a_{b l u e} = 2 \sqrt{3} δ$ $\frac{8 \sqrt{6 \sqrt{3}}}{3} - 2 \sqrt{2 \sqrt{3}} = 2 \sqrt{3} δ$ $\Rightarrow δ = \frac{4 \sqrt{2 \sqrt{3}} - \sqrt{6 \sqrt{3}}}{3} \approx 1.407$
	Commented by behi83417@gmail.com last updated on 07/Dec/18

	$n i c e w o r k s i r . t h a n k s i n a d v a n c e .$
	Commented by mr W last updated on 07/Dec/18

	$t h a n k y o u s i r f o r c h e c k i n g i t! i t$ $i s f i x e d n o w .$
	Answered by behi83417@gmail.com last updated on 07/Dec/18

	$a_{r e d}^{2} = \frac{4}{\sqrt{3}} . 32 = \frac{128}{\sqrt{3}} \Rightarrow a_{r e d} = \frac{8 \sqrt{2}}{\sqrt[4]{3}}$ $a_{b l u e}^{2} = \frac{4}{\sqrt{3}} . 6 = \frac{24}{\sqrt{3}} \Rightarrow a_{b l u e} = \frac{2 \sqrt{6}}{\sqrt[4]{3}}$ $\frac{P M}{s i n 30^{∙}} + h_{b l u e} + P M = h_{r e d}$ $P M = \frac{h_{r e d} - h_{b l u e}}{\frac{1 + s i n 30}{s i n 30}} = \frac{\frac{\sqrt{3}}{2} (a_{r e d} - a_{b l u e})}{3} =$ $You can't use 'macro parameter character #' in math mode$
	Commented by Kunal12588 last updated on 07/Dec/18
	sir what is the meaning of #
	Commented by behi83417@gmail.com last updated on 07/Dec/18

	$i u s e t h i s s i g n t o s a y : a p p r o x i m a t e l y$ $o r a l m o s t o r n e a r l y . . . .$
	Answered by Kunal12588 last updated on 08/Dec/18
	$a_(blue) =(√(((24)/(√3)) )), a_(red) =(√((128)/(√3))) join T & A and S & B , a_(red) ∥a_(blue) so quad.ABST is a trapezium ar(ABST)=(1/3){ar(△)−ar(△)}=(1/3)(32−6) (1/2)(a_(blue) +a_(red) )PM= ((26)/3) PM{(√(((24)/(√3)) ))+(√((128)/(√3)))}=((52)/3) PM=((52)/(3((√(((24)/(√3)) ))+(√((128)/(√3))))))=1.40704302...$
	$a_{b l u e} = \sqrt{\frac{24}{\sqrt{3}}}, a_{r e d} = \sqrt{\frac{128}{\sqrt{3}}}$ $j o i n T & A a n d S & B, a_{r e d} ∥ a_{b l u e}$ $s o q u a d . A B S T i s a t r a p e z i u m$ $a r (A B S T) = \frac{1}{3} {a r (△) - a r (△)} = \frac{1}{3} (32 - 6)$ $\frac{1}{2} (a_{b l u e} + a_{r e d}) P M = \frac{26}{3}$ $P M {\sqrt{\frac{24}{\sqrt{3}}} + \sqrt{\frac{128}{\sqrt{3}}}} = \frac{52}{3}$ $P M = \frac{52}{3 (\sqrt{\frac{24}{\sqrt{3}}} + \sqrt{\frac{128}{\sqrt{3}}})} = 1.40704302 . . .$
	Commented by Kunal12588 last updated on 08/Dec/18

	$t h a n k s f o r c h e c k i n g s i r$
	Commented by behi83417@gmail.com last updated on 07/Dec/18

	$t h a n k y o u s i r . b e a u t i f u l m e t h o d!$ $\frac{54}{3} \Rightarrow \frac{52}{3} (r e p l a c e p l e a s e)$
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