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help w this question ??
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yes he is correct?
is it clear that the diagonal OM is \(6\sqrt2\) ?
no, it is not? @satellite73
ok then lets make sure it is clear
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|dw:1416803629265:dw|
by pythagoras, \[6^2+6^2=d^2\\ 2\times 6^2=d^2\\ \sqrt{2\times 6^2}=d\\ 6\sqrt2=d\]
more generally the diagonal of a square with side \(a\) is \(a\sqrt2\)
now use pythagoras again to find the diagonal here |dw:1416803755472:dw|
13.42?
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\[12^2+6^2=d^2\\ \sqrt{12^2+6^2}=d\]
forget decimals
1342?
or just 13
just use radicals \[\sqrt{144+36}=\sqrt{180}=6\sqrt5\]
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in any case \[6\sqrt5\] is not 2 times \(6\sqrt2\) so the answer is NO
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