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Question Number 50818 by ajfour last updated on 20/Dec/18
If a right angled triangle has  same area and double perimeter  as that of a circle of unit radius,  find the mutually perpendicular  sides of the triangle.
$${If}\:{a}\:{right}\:{angled}\:{triangle}\:{has} \\ $$$${same}\:{area}\:{and}\:{double}\:{perimeter} \\ $$$${as}\:{that}\:{of}\:{a}\:{circle}\:{of}\:{unit}\:{radius}, \\ $$$${find}\:{the}\:{mutually}\:{perpendicular} \\ $$$${sides}\:{of}\:{the}\:{triangle}. \\ $$
Answered by mr W last updated on 20/Dec/18
(1/2)ab=π×1^2   ⇒ab=2π  a+b+(√(a^2 +b^2 ))=2×2π×1  ⇒a+b+(√(a^2 +b^2 ))=4π  ⇒a^2 +b^2 =16π^2 −8π(a+b)+a^2 +b^2 +2ab  ⇒0=16π^2 −8π(a+b)+4π  ⇒a+b=2π+(1/2)  ⇒x^2 −(2π+(1/2))x+2π=0  ⇒x=((4π+1±(√((4π+1)^2 −32π)))/4)  ⇒a,b=((4π+1±(√((4π+1)^2 −32π)))/4)
$$\frac{\mathrm{1}}{\mathrm{2}}{ab}=\pi×\mathrm{1}^{\mathrm{2}} \\ $$$$\Rightarrow{ab}=\mathrm{2}\pi \\ $$$${a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{2}×\mathrm{2}\pi×\mathrm{1} \\ $$$$\Rightarrow{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{4}\pi \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16}\pi^{\mathrm{2}} −\mathrm{8}\pi\left({a}+{b}\right)+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$$\Rightarrow\mathrm{0}=\mathrm{16}\pi^{\mathrm{2}} −\mathrm{8}\pi\left({a}+{b}\right)+\mathrm{4}\pi \\ $$$$\Rightarrow{a}+{b}=\mathrm{2}\pi+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\left(\mathrm{2}\pi+\frac{\mathrm{1}}{\mathrm{2}}\right){x}+\mathrm{2}\pi=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{4}\pi+\mathrm{1}\pm\sqrt{\left(\mathrm{4}\pi+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{32}\pi}}{\mathrm{4}} \\ $$$$\Rightarrow{a},{b}=\frac{\mathrm{4}\pi+\mathrm{1}\pm\sqrt{\left(\mathrm{4}\pi+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{32}\pi}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 21/Dec/18
Thank you Sir. Quite elegant!
$${Thank}\:{you}\:{Sir}.\:{Quite}\:{elegant}! \\ $$

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