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Question Number 51849 by ajfour last updated on 31/Dec/18

Commented by ajfour last updated on 31/Dec/18

$F i n d d i a m e t e r o f s e m i c i r c l e i n$ $g i v e n r e c t a n g l e .$
Commented by mr W last updated on 31/Dec/18

$i t h i n k o n l y i f \frac{a}{2} ⩽ b ⩽ 2 a t h e r e i s a$ $s o l u t i o n . a n d t h e r e i s o n l y o n e s u c h$ $s e m i c i r c l e i n a g i v e n r e c t a n g l e p o s s i b l e .$ $s o t h e q u e s t i o n i s t o f i n d t h e a r e a o f$ $t h e s e m i c i r c l e, n o t t h e m a x i m a l a r e a .$
Commented by ajfour last updated on 31/Dec/18

$y e s S i r, i^{'} v e e d i t e d; p l e a s e a t t e m p t .$
	Answered by MJS last updated on 31/Dec/18

	$M = (\begin{matrix} 0 \\ 0 \end{matrix}) S = (\begin{matrix} 0 \\ r \end{matrix}) T = (\begin{matrix} - r \\ 0 \end{matrix})$ $line P Q : y = a x$ $P = (\begin{matrix} - \frac{r}{\sqrt{a^{2} + 1}} \\ - \frac{a r}{\sqrt{a^{2} + 1}} \end{matrix}) Q = (\begin{matrix} \frac{r}{\sqrt{a^{2} + 1}} \\ \frac{a r}{\sqrt{a^{2} + 1}} \end{matrix})$ $(1) y_{S} - y_{P} = 3$ $(2) x_{Q} - x_{T} = 5$ $(1) r + \frac{a r}{\sqrt{a^{2} + 1}} = 3$ $(2) \frac{r}{\sqrt{a^{2} + 1}} + r = 5$ $(1) - a \times (2) \Rightarrow r = \frac{5 a - 3}{a - 1}$ $inserting in (1) or (2) leads to$ $r = 8 - \sqrt{30}$ $a = \frac{5}{7} - \frac{2 \sqrt{30}}{21}$
	Commented by ajfour last updated on 31/Dec/18

	$E x c e l l e n t S i r, T h a n k s a l o t!$ $HAPPY NEW YEAR!$
	Commented by MJS last updated on 31/Dec/18

	$thank you, and all the best for you too!$
	Answered by mr W last updated on 31/Dec/18

	Commented by mr W last updated on 31/Dec/18

	$R + R \sin α = b \Rightarrow \sin α = \frac{b - R}{R}$ $R + R \cos α = a \Rightarrow \cos α = \frac{a - R}{R}$ ${(\frac{b - R}{R})}^{2} + {(\frac{a - R}{R})}^{2} = 1$ $R^{2} - 2 (a + b) R + (a^{2} + b^{2}) = 0$ $\Rightarrow R = a + b - \sqrt{2 a b}$ $e x a m p l e s :$ $a = 2 b \Rightarrow R = 3 b - 2 b = b$ $b = 2 a \Rightarrow R = 3 a - 2 a = a$ $b = a \Rightarrow R = 2 a - \sqrt{2} a = (2 - \sqrt{2}) a$
	Commented by ajfour last updated on 31/Dec/18

	$b e s t w a y S i r!$
	Answered by ajfour last updated on 31/Dec/18

	$R = a + b - \sqrt{2 a b} (I f o u n d) .$
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