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Question Number 77164 by Maclaurin Stickker last updated on 03/Jan/20

Commented by Maclaurin Stickker last updated on 03/Jan/20

I got x=5. Is my answer correct?

$${I}\:{got}\:{x}=\mathrm{5}.\:{Is}\:{my}\:{answer}\:{correct}? \\ $$

Commented by mr W last updated on 03/Jan/20

correct!

$${correct}! \\ $$

Answered by mr W last updated on 03/Jan/20

b=width of rectangle (√(10^2 −x^2 ))+((bx)/(10))=10 ⇒b=((10)/x)(10−(√(10^2 −x^2 ))) b+x+(b/(10))(√(10^2 −x^2 ))=10 x+((10)/x)(10−(√(10^2 −x^2 )))(1+((√(10^2 −x^2 ))/(10)))=10 10x^2 +10(10−(√(10^2 −x^2 )))(10+(√(10^2 −x^2 )))=100x 20x^2 =100x ⇒x=5

$${b}={width}\:{of}\:{rectangle} \\ $$$$\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{{bx}}{\mathrm{10}}=\mathrm{10} \\ $$$$\Rightarrow{b}=\frac{\mathrm{10}}{{x}}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right) \\ $$$${b}+{x}+\frac{{b}}{\mathrm{10}}\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\mathrm{10} \\ $$$${x}+\frac{\mathrm{10}}{{x}}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{10}}\right)=\mathrm{10} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} +\mathrm{10}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)\left(\mathrm{10}+\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)=\mathrm{100}{x} \\ $$$$\mathrm{20}{x}^{\mathrm{2}} =\mathrm{100}{x} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$

Commented by Maclaurin Stickker last updated on 03/Jan/20

oh yeah!

$${oh}\:{yeah}! \\ $$

Commented by Tawa11 last updated on 29/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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