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.


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Question Number 50818 by ajfour last updated on 20/Dec/18

$I f a r i g h t a n g l e d t r i a n g l e h a s$ $s a m e a r e a a n d d o u b l e p e r i m e t e r$ $a s t h a t o f a c i r c l e o f u n i t r a d i u s,$ $f i n d t h e m u t u a l l y p e r p e n d i c u l a r$ $s i d e s o f t h e t r i a n g l e .$
	Answered by mr W last updated on 20/Dec/18

	$\frac{1}{2} a b = π \times 1^{2}$ $\Rightarrow a b = 2 π$ $a + b + \sqrt{a^{2} + b^{2}} = 2 \times 2 π \times 1$ $\Rightarrow a + b + \sqrt{a^{2} + b^{2}} = 4 π$ $\Rightarrow a^{2} + b^{2} = 16 π^{2} - 8 π (a + b) + a^{2} + b^{2} + 2 a b$ $\Rightarrow 0 = 16 π^{2} - 8 π (a + b) + 4 π$ $\Rightarrow a + b = 2 π + \frac{1}{2}$ $\Rightarrow x^{2} - (2 π + \frac{1}{2}) x + 2 π = 0$ $\Rightarrow x = \frac{4 π + 1 \pm \sqrt{{(4 π + 1)}^{2} - 32 π}}{4}$ $\Rightarrow a, b = \frac{4 π + 1 \pm \sqrt{{(4 π + 1)}^{2} - 32 π}}{4}$
	Commented by ajfour last updated on 21/Dec/18

	$T h a n k y o u S i r . Q u i t e e l e g a n t!$
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