The drawing below shows two equilateral triangles with side length a. The triangle are horizontally shifted by (a/2). Find the intersection area A of the two triangles (grey area).


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Question Number 174591 by Mastermind last updated on 05/Aug/22

$The drawing below shows two$ $equilateral triangles with side length$ $a . The triangle are horizontally shifted$ $by \frac{a}{2} . Find the intersection area A of$ $the two triangles (grey area) .$
	Answered by behi834171 last updated on 05/Aug/22

	$a s l o h a v e a n e q u i l a t e r a l t r i a n g l e$ $w i t h s i d e l e n g h t : \frac{a}{2}$ $s o, w a n t e d a r e a = \frac{{(\frac{a}{2})}^{2}}{4} \sqrt{3} = \frac{a^{2} \sqrt{3}}{16} .$
	Commented by Mastermind last updated on 05/Aug/22

	$I did not understand what {you}^{'} re trying$ $to say here, explain it well$
	Commented by Mastermind last updated on 05/Aug/22

	Commented by behi834171 last updated on 05/Aug/22

	$4 A = \frac{a^{2} \sqrt{3}}{4} \Rightarrow A = \frac{a^{2} \sqrt{3}}{16} .$ $4 o f s u c h s h a d e d a r e a s = w h o l e o f a$ $e q u i l a t e r a l t r i a n g l e w i t h s i d e : a$ $t h a t {i t}^{'} s a r e a = \frac{a^{2} \sqrt{3}}{4} .$ $i g o t w h a t y o u m e a n t s i r .$ $i t i s t o o e a s y t o s a y m o r e . . . .$
	Commented by Mastermind last updated on 05/Aug/22

	$Tank you so much man$
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