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Question Number 181432 by mr W last updated on 25/Nov/22

Commented by mr W last updated on 25/Nov/22

$f i n d t h e a r e a o f c i r c l e .$
Commented by universe last updated on 25/Nov/22

Commented by universe last updated on 25/Nov/22

${A E}^{2} + {B E}^{2} = 4$ ${C E}^{2} + {E D}^{2} = 16$ ${A E}^{2} + {B E}^{2} + {C E}^{2} + {E D}^{2} = 20$ $4 r^{2} = 20 \Rightarrow r^{2} = 5$ $a r e a o f c r i c l e = π r^{2} = 5 π$
Commented by mr W last updated on 25/Nov/22

$t h a n k s s i r!$ $p l e a s e p r o v e t h a t$ ${A E}^{2} + {B E}^{2} + {C E}^{2} + {E D}^{2} = 4 r^{2} .$ $b e c a u s e {i t}^{'} s n o t o b v i o u s .$ $b t w, p l e a s e p o s t y o u r a n s w e r a s$ $‘ ‘ {a n s w e r}^{″}, n o t a s ‘ ‘ {c o m m e n t}^{″}!$ $t h a n k y o u!$
Commented by universe last updated on 25/Nov/22
https://youtu.be/1EkoaQJyrfk
	Answered by HeferH last updated on 25/Nov/22

	Commented by HeferH last updated on 25/Nov/22

	${(\frac{a + 2 b}{2})}^{2} + {(\frac{2 a + b}{2} - b)}^{2} = r^{2}$ $\frac{a^{2} + 4 b^{2} + 4 a b + {(2 a - b)}^{2}}{4} = r^{2}$ $\frac{a^{2} + 4 b^{2} + 4 a b + 4 a^{2} + b^{2} - 4 a b}{4} = r^{2}$ $a^{2} + b^{2} + 4 (a^{2} + b^{2}) = 4 r^{2}$ $b u t a^{2} + b^{2} = 4 \Rightarrow$ $4 + 4 (4) = 4 r^{2}$ $20 = 4 r^{2}$ $5 = r^{2}$ $A = π r^{2} = 5 π u^{2}$
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