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-

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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