Question Number 77164 by Maclaurin Stickker last updated on 03/Jan/20
Commented by Maclaurin Stickker last updated on 03/Jan/20
$${I}\:{got}\:{x}=\mathrm{5}.\:{Is}\:{my}\:{answer}\:{correct}? \\ $$
Commented by mr W last updated on 03/Jan/20
$${correct}! \\ $$
Answered by mr W last updated on 03/Jan/20
$${b}={width}\:{of}\:{rectangle} \\ $$$$\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{{bx}}{\mathrm{10}}=\mathrm{10} \\ $$$$\Rightarrow{b}=\frac{\mathrm{10}}{{x}}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right) \\ $$$${b}+{x}+\frac{{b}}{\mathrm{10}}\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\mathrm{10} \\ $$$${x}+\frac{\mathrm{10}}{{x}}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{10}}\right)=\mathrm{10} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} +\mathrm{10}\left(\mathrm{10}−\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)\left(\mathrm{10}+\sqrt{\mathrm{10}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)=\mathrm{100}{x} \\ $$$$\mathrm{20}{x}^{\mathrm{2}} =\mathrm{100}{x} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$
Commented by Maclaurin Stickker last updated on 03/Jan/20
$${oh}\:{yeah}! \\ $$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$