Question

27. Locate √ 3on the real number line.​

Answer 1

JUSTIFICATION :

In right triangle OBC, we have

$oc {}^{2} = ob { }^{2} + bc {}^{2} \\ by \: pythagoras \: theorm$

therefore,

$oc = \sqrt{ob {}^{2} } + \sqrt{bc {}^{2} } = \sqrt{2 {}^{2} } + \sqrt{1 {}^{2} } = \sqrt{2 + 1 } = \sqrt{3} \: units \:$