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Question Number 183227 by mr W last updated on 23/Dec/22

Commented by mr W last updated on 23/Dec/22

$f i n d t h e l e n g t h o f o t h e r d i a g o n a l .$
	Answered by mr W last updated on 24/Dec/22

	Commented by mr W last updated on 24/Dec/22

	$B D = \sqrt{{(6 + 6)}^{2} - 7^{2}} = \sqrt{95}$
	Commented by manxsol last updated on 24/Dec/22

	$W e d i d n o t h a v e t h e$ $v i s i o n b u t i t i s a$ $s p e c i a l c a s e . I s e n d a n o t h e r$ $c a s e$
	Commented by manxsol last updated on 24/Dec/22

	Commented by mr W last updated on 24/Dec/22

	$y e s {i t}^{'} s a s p e c i a l c a s e . b u t t h e q u e s t i o n$ $o n l y r e q u e s t s y o u t o s o l v e t h i s s p e c i a l$ $c a s e, n o t h i n g e l s e .$
	Commented by mr W last updated on 24/Dec/22

	$a s f o r t h e c a s e y o u s e n t, s e e Q 183281$
	Commented by manxsol last updated on 24/Dec/22

	$O h, S i r W . b u t m y c a s e h a s$ $a s u r p r i s e$
	Answered by Frix last updated on 23/Dec/22

	$19 \times 5 \times 7^{2} = (12 + x) (12 - x) x^{2}$ $x \neq 7 \Rightarrow x = \sqrt{95}$
	Answered by a.lgnaoui last updated on 23/Dec/22
	${ ((x^2 +y^2 =36 (1))),((x^2 +z^2 =49 (2))) :} (2)−(1)⇒z^2 −y^2 =13 (z−y)(z+y)=13 6(z−y)=13 ⇒z−y=((13)/6) { ((z+y=6 z=((49)/(12)))),((z−y=((13)/6) y=((23)/(12)))) :} x^2 =6^2 −y^2 =36−(((23)/(12)))^2 BD^2 =DE^2 +BE^2 =x^2 +(2y+z)^2 BD=(√((36−y^2 )+(2y+z)^2 )) BD =(√(95 )) =9,746794$
	${\begin{cases} x^{2} + y^{2} = 36 (1) \\ x^{2} + z^{2} = 49 (2) \end{cases}$ $(2) - (1) \Rightarrow z^{2} - y^{2} = 13$ $(z - y) (z + y) = 13$ $6 (z - y) = 13 \Rightarrow z - y = \frac{13}{6}$ ${\begin{cases} z + y = 6 z = \frac{49}{12} \\ z - y = \frac{13}{6} y = \frac{23}{12} \end{cases}$ $x^{2} = 6^{2} - y^{2} = 36 - {(\frac{23}{12})}^{2}$ ${BD}^{2} = {DE}^{2} + {BE}^{2}$ $= x^{2} + {(2 y + z)}^{2}$ $BD = \sqrt{(36 - y^{2}) + {(2 y + z)}^{2}}$ $BD = \sqrt{95} = 9, 746794$
	Commented by a.lgnaoui last updated on 23/Dec/22

	Commented by a.lgnaoui last updated on 23/Dec/22

	Answered by manxsol last updated on 23/Dec/22

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