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Question Number 211016 by mr W last updated on 26/Aug/24

Commented by mr W last updated on 26/Aug/24

$$findx=?$$$$$$$$theoriginalquestionwasdeletedby$$$$thequestionerforsomereasons.$$

Answered by mr W last updated on 26/Aug/24

Commented by mr W last updated on 26/Aug/24

$$MethodII$$$$FB=AD=x=FE$$$$\Rightarrow BG=GE=\frac{5+5+2}{2}=6$$$$GC=6-2=4$$$$FG=\sqrt{5^2-4^2}=3$$$$x=\sqrt{6^2+3^2}=3\sqrt{5}$$

Answered by A5T last updated on 26/Aug/24

$$LetAE=y$$$$\Rightarrow 10\times 5^2+10x^2=2^2\times 5+y^2\times 5$$$$\Rightarrow 50+2x^2=4+y^2\Rightarrow y^2-2x^2=46...\left(i\right)$$$$5\times 2\times 7+{10}^2\times 7=5y^2+2x^2\Rightarrow 5y^2+2x^2=770...\left(ii\right)$$$$\left(ii\right)+\left(i\right)\Rightarrow 6y^2=816\Rightarrow y^2=136\Rightarrow x=3\sqrt{5}$$

Commented by mr W last updated on 26/Aug/24

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Answered by mr W last updated on 26/Aug/24

Commented by mr W last updated on 26/Aug/24

$$METHODI$$$$2y^2+5x^2=7(5^2+5\times 2)...\left(i\right)$$$${10}^2+4y^2=2(x^2+5\times 5)...\left(ii\right)$$$$\left(ii\right)-2\times \left(i\right):$$$${10}^2-10x^2=2(x^2+25)-14\times 35$$$$\Rightarrow x=3\sqrt{5}$$

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