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Question Number 200377 by mr W last updated on 18/Nov/23

Commented by mr W last updated on 18/Nov/23

$t h e s i d e l e n g t h s o f t h e h e x a g o n a r e$ $g i v e n . f i n d x + y + z = ?$
	Answered by mr W last updated on 18/Nov/23

	Commented by mr W last updated on 18/Nov/23

	Commented by mr W last updated on 18/Nov/23

	$r e d d i a g o n a l = z$ $b l u e d i a g o n a l = y$ $g r e e n d i a g o n a l = x$
	Commented by mr W last updated on 18/Nov/23

	$a p p l y i n g {P t o l e m y}^{'} s t h e o r e m$ $t o ◻ A D E F :$ $z^{2} = 81^{2} + 81 y . . . (i)$ $t o ◻ A C D F :$ $y^{2} = 81^{2} + x z . . . (i i)$ $t o ◻ A B C D :$ $x z = 31 \times 81 + 81 y . . . (i i i)$ $f r o m (i i) & (i i i) :$ $y^{2} - 81 y - 81 \times 112 = 0$ $\Rightarrow y = \frac{81 + 9 \times 23}{2} = 144 ✓$ $f r o m (i) :$ $z = \sqrt{81^{2} + 81 \times 144} = 135 ✓$ $f r o m (i i) :$ $x = \frac{y^{2} - 81^{2}}{z} = \frac{144^{2} - 81^{2}}{135} = 105 ✓$ $\Rightarrow x + y + z = 105 + 144 + 135 = 384 ✓$
	Commented by ajfour last updated on 18/Nov/23

	$i s e e, t h a n k y o u s i r .$
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