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Question Number 36590 by ajfour last updated on 03/Jun/18

Commented by ajfour last updated on 03/Jun/18

$F i n d m a x i m u m a r e a o f △ A B C$ $i f o n e v e r t e x i s o n s m a l l e r c i r c l e$ $o f r a d i u s r a n d o t h e r t w o o n$ $l a r g e r c i r c l e o f r a d i u s R .$ $T h e d i s t a n c e b e t w e e n c e n t r e s o f$ $c i r c l e s b e i n g d .$
	Answered by ajfour last updated on 03/Jun/18

	$△ = \frac{1}{2} (x + d + r) \sqrt{R^{2} - x^{2}}$ $\frac{d △}{d x} = \frac{\sqrt{R^{2} - x^{2}}}{2} - \frac{x (x + d + r)}{2 \sqrt{R^{2} - x^{2}}} = 0$ $\Rightarrow R^{2} - x^{2} = x^{2} + x (d + r)$ $2 x^{2} + (d + r) x - R^{2} = 0$ $x = \frac{\sqrt{{(d + r)}^{2} + 8 R^{2}} - (d + r)}{4}$ $△_{m a x} = \frac{1}{2} [\frac{\sqrt{{(d + r)}^{2} + 8 R^{2}} + 3 (d + r)}{4}]$ $\times \sqrt{R^{2} - {(\frac{\sqrt{{(d + r)}^{2} + 8 R^{2}} - (d + r)}{4})}^{2}} .$
	Commented by MrW3 last updated on 11/Jun/18
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