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Question Number 72194 by mr W last updated on 26/Oct/19

Commented by mr W last updated on 26/Oct/19

$f i n d t h e s u m o f a r e a s o f t w o s q u a r e s$ $i n s i d e t h e s e m i c i r c l e w i t h r a d i u s 10 .$
	Answered by MJS last updated on 27/Oct/19

	$I turned it around$ $if the larger square has side length 1, which$ $side lengths are possible for the smaller square ?$ $put the right upper vertice (\begin{matrix} 1 \\ 1 \end{matrix}), the left upper$ $one (\begin{matrix} - s \\ s \end{matrix})$ $both must be on a circle with center on the$ $x - axis$ ${(x - p)}^{2} + y^{2} = r^{2}$ $\Rightarrow$ $upper semi circle : y = \sqrt{2 s - 2 (s - 1) x - x^{2}}$ $the upper vertices of the larger square must$ $fit into the circle \Rightarrow y ⩾ 1 at x = 0$ $\Rightarrow \frac{1}{2} ⩽ s < 1$ $now choose s$ $\Rightarrow area = s^{2} + 1; r = \sqrt{s^{2} + 1}$ $area within circle with radius 10 is$ $\frac{s^{2} + 1}{{(\sqrt{s^{2} + 1})}^{2}} \times 100 = 100$
	Commented by mr W last updated on 27/Oct/19

	$t h a n k s s i r!$
	Answered by mr W last updated on 27/Oct/19

	Commented by mr W last updated on 27/Oct/19

	$∠ A O B = \frac{2 π}{4} = \frac{π}{2}$ ${A B}^{2} = {(a + b)}^{2} + {(b - a)}^{2} = 2 (a^{2} + b^{2})$ ${A B}^{2} = r^{2} + r^{2} = 2 r^{2}$ $\Rightarrow a^{2} + b^{2} = r^{2} = 10^{2} = 100$ $i . e . t h e s u m o f b o t h s q u a r e s i s c o n t a n t .$
	Commented by ajfour last updated on 27/Oct/19

	$G r e a t w a y S i r, i m y s e l f h a d$ $t h o u g h t t o a t t e m p t i t s o m e w h a t$ $t h i s w a y . .$
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