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Question Number 206643 by cortano21 last updated on 21/Apr/24

Answered by mr W last updated on 21/Apr/24

Commented by mr W last updated on 21/Apr/24

$$\cos \alpha =\frac{x^2+b^2-a^2}{2xb}$$$$\cos \beta =\frac{x^2+b^2-c^2}{2xb}$$$$\alpha +\beta =90°$$$$\cos \alpha =\sin \beta$$$$\frac{x^2+b^2-a^2}{2xb}=\sqrt{1-{\left(\frac{x^2+b^2-c^2}{2xb}\right)}^2}$$$${(x^2+b^2-a^2)}^2={\left(2xb\right)}^2-{(x^2+b^2-c^2)}^2$$$$2x^4-2(a^2+c^2)x^2+a^4+c^4-2(a^2+c^2-b^2)b^2=0$$$$x^2=\frac{a^2+c^2\pm \sqrt{4(a^2+c^2-b^2)b^2-{(a^2-c^2)}^2}}{2}$$$$\Rightarrow x=\sqrt{\frac{a^2+c^2\pm \sqrt{4(a^2+c^2-b^2)b^2-{(a^2-c^2)}^2}}{2}}$$$$witha=3,b=4,c=5$$$$x=\sqrt{\frac{3^2+5^2\pm \sqrt{4(3^2+5^2-4^2)4^2-{(3^2-5^2)}^2}}{2}}$$$$=\sqrt{17\pm 4\sqrt{14}}\approx 5.654or1.426$$

Commented by A5T last updated on 21/Apr/24

$$Ifx=\sqrt{17-4\sqrt{14}},pointPcannotbeinsidethe$$$$squaresuchthatPC=5.$$

Commented by mr W last updated on 21/Apr/24

$$onesolutionisforpointPinside$$$$thesquareandtheothersolutionis$$$$forpointPoutsidethesquare.$$

Commented by A5T last updated on 21/Apr/24

$$Yea,justnotingthat,sincethequestionis$$$$specificaboutPbeinginsidethesquare.$$

Answered by dimentri last updated on 21/Apr/24

$$\underset{―}{\underbrace{}}\cancel{}$$

Commented by A5T last updated on 21/Apr/24

$$\left(i\right)\Rightarrow cos\alpha =\frac{x^2-9}{8x}$$$$\left(ii\right)\Rightarrow sin\alpha =\frac{x^2+7}{8x}\Rightarrow cos\alpha =\sqrt{1-{\left(\frac{x^2+7}{8x}\right)}^2}$$$$\left(i\right)\mathrm{\&}\left(ii\right)\Rightarrow \frac{x^2-9}{8x}=\sqrt{1-{\left(\frac{x^2+7}{8x}\right)}^2}$$$$\Rightarrow x=\sqrt{17+4\sqrt{14}}\approx 5.654$$

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