On the figure given show that $¯ ¯¯¯¯ ¯ O C$ $bar(OC)$ is $\sqrt{2}$ $sqrt(2)$ ?

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On the figure given show that $¯ ¯¯¯¯ ¯ O C$ $bar(OC)$ is $\sqrt{2}$ $sqrt(2)$ ?

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Answer 1 · 2016-09-14T16:21:37

WOW...I finally got it...although it seems too easy...and probably it is not the way you wanted it!

Explanation:

I considered the two small circles as equal and having radius $1$ $1$ , each of them (or $u$ $u$ as unity in distance $¯ ¯¯¯¯ ¯ P O$ $bar(PO)$ ...I think). So the entire base of the triangle (diameter of big circle) should be $3$ $3$ .
According to this, the distance $¯ ¯¯¯¯¯ ¯ O M$ $bar(OM)$ should be $0.5$ $0.5$ and the distance $¯ ¯¯¯¯¯ ¯ M C$ $bar(MC)$ should be one big cirlce radius or $\frac{3}{2} = 1.5$ $3/2=1.5$ .
Now, I applied Pythagoras to the triangle $O M C$ $OMC$ with:
$¯ ¯¯¯¯ ¯ O C = x$ $bar(OC)=x$
$¯ ¯¯¯¯¯ ¯ O M = 0.5$ $bar(OM)=0.5$
$¯ ¯¯¯¯¯ ¯ M C = 1.5$ $bar(MC)=1.5$ and I got:
${1.5}^{2} = x^{2} + {0.5}^{2}$ $1.5^2=x^2+0.5^2$
or:
$x^{2} = {1.5}^{2} - {0.5}^{2} = {(\frac{3}{2})}^{2} - {(\frac{1}{2})}^{2} = \frac{8}{4} = 2$ $x^2=1.5^2-0.5^2=(3/2)^2-(1/2)^2=8/4=2$
so:
$x = \sqrt{2}$ $x=sqrt(2)$

Does it make sense...?

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On the figure given show that $¯ ¯¯¯¯ ¯ O C$ $bar(OC)$ is $\sqrt{2}$ $sqrt(2)$ ?

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On the figure given show that $¯ ¯¯¯¯ ¯ O C$ $bar(OC)$ is $\sqrt{2}$ $sqrt(2)$ ?