Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 214427 by ajfour last updated on 08/Dec/24

Commented by ajfour last updated on 08/Dec/24
https://youtu.be/pnMSZ77i2P8?si=OaLd0p_DlkwNYv8d
	Answered by a.lgnaoui last updated on 08/Dec/24

	$Resolution de l equation donne$ $x = 0, 60048518$ $alors Aire = x + 1 = 1, 60048518$ $autre Methode$ $Aire = Aire [Care (OAED)] = 1$ $2 Aire (OEC) = 2 \times \frac{x}{2} = x$ $A = A (Care) + 2 A (OEC) = x + 1$
	Commented by a.lgnaoui last updated on 08/Dec/24

	Commented by a.lgnaoui last updated on 08/Dec/24

	Commented by mr W last updated on 08/Dec/24

	$c l e a r l y w r o n g! e v e n f r o m t h e$ $d i a g r a m o n e c a n s e e t h a t y o u r x$ $s h o u l d b e \approx 1 a n d > 1 . s o y o u r a n s w e r$ $x = 0, 60048518 i s w r o n g .$
	Answered by mr W last updated on 08/Dec/24

	Commented by mr W last updated on 08/Dec/24

	$y^{2} = x^{2} + 2^{2}$ ${(\sqrt{x^{2} - 1^{2}} - 1)}^{2} + {(\sqrt{y^{2} - 1^{2}} - 1)}^{2} = 2^{2}$ ${(\sqrt{x^{2} - 1} - 1)}^{2} + {(\sqrt{x^{2} + 3} - 1)}^{2} = 4$ $\Rightarrow x \approx 2.1673$ $A r e a = \frac{(1 + 1) x}{2} = x \approx 2.1673$ $f o r e x a c t s o l u t i o n :$ $x^{2} - \sqrt{x^{2} - 1} = \sqrt{x^{2} + 3}$ $x^{4} - 4 = 2 x^{2} \sqrt{x^{2} - 1}$ $x^{8} - 4 x^{6} - 4 x^{4} + 16 = 0$ $l e t t = x^{2}$ $t^{4} - 4 t^{3} - 4 t^{2} + 16 = 0$ $. . . . . .$
	Commented by ajfour last updated on 08/Dec/24

	$\sin θ = \frac{1}{x} = \frac{\sqrt{x^{2} - 1} - 1}{2}$ ${(1 + \frac{2}{x})}^{2} = x^{2} - 1$ $A r e a = x \approx 2.1673$
	Answered by Spillover last updated on 11/Dec/24

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum