1.

By rationalising the denominator of \(\frac{1}{(6-\sqrt{3})}\), we get __________(a) \(\frac{6-\sqrt{3}}{3}\)(b) \(\frac{6-\sqrt{3}}{33}\)(c) \(\frac{6+\sqrt{3}}{3}\)(d) \(\frac{6+\sqrt{3}}{33}\)This question was addressed to me in quiz.I would like to ask this question from Representing Real Numbers on the Number Line & Real Numbers Operations in division Number Systems of Mathematics – Class 9

Answer»

Right answer is (d) \(\frac{6+\sqrt{3}}{33}\)

BEST explanation: When the denominator of an expression contains a term with a square root, the process of converting it to an EQUIVALENT expression whose denominator is a rational number is called rationalising the denominator.

By MULTIPLYING \(\frac{1}{(6-\sqrt{3})}\) by \(6+\sqrt{3}\), we will GET same expression since \(\frac{6+\sqrt{3}}{6+\sqrt{3}}\) = 1.

Therefore, \(\frac{1}{(6-\sqrt{3})} = \frac{1}{(6-\sqrt{3})} * \frac{6+\sqrt{3}}{6+\sqrt{3}} = \frac{6+\sqrt{3}}{(6*6) – (\sqrt{3}*\sqrt{3})}\)

= \(\frac{6+\sqrt{3}}{(36-3)}\)

= \(\frac{6+\sqrt{3}}{33}\).


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