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Question Number 205461 by mr W last updated on 21/Mar/24

Commented by mr W last updated on 21/Mar/24

$f i n d t h e a r e a o f t h e t h i r d p a r t$ $b e t w e e n t w o s q u a r e s .$
	Answered by mr W last updated on 22/Mar/24

	Commented by mr W last updated on 22/Mar/24

	$g r e e n h a t c h e d a r e a = 250 + 150 = 400$ $b l u e h a t c h e d a r e a = u n k n o w n a r e a ?$ $d u e t o s y m m e t r y$ $b l u e h a t c h e d a r e a = g r e e n h a t c h e d a r e a$ $\Rightarrow ? = 250 + 150 = 400 ✓$
	Answered by A5T last updated on 21/Mar/24

	$\frac{(a + s) e}{2} = 250; \frac{(a + s) d}{2} = 150$ $\Rightarrow \frac{(a + s) (d + e)}{2} = 400 \Rightarrow s^{2} - a^{2} = 800$ $a^{2} + 400 + ? = s^{2} \Rightarrow s^{2} - a^{2} = 400 + ? = 800 \Rightarrow ? = 400 {c m}^{2}$
	Commented by mr W last updated on 22/Mar/24
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	Commented by A5T last updated on 21/Mar/24

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