Question : The value of $\frac{1}{\sqrt{17+12 \sqrt{2}}}$ is closest to ______.
Option 1: 1.4
Option 2: 1.2
Option 3: 0.14
Option 4: 0.17
Correct Answer: 0.17
Solution :
According to the question
$\frac{1}{\sqrt{17+12 \sqrt{2}}}$
= $\frac{1}{\sqrt{3^{2}+(2 \sqrt{2})^{2} + 2 × 3 × 2\sqrt{2}}}$
= $\frac{1}{\sqrt(3 + 2\sqrt{2})^{2}}$
= $\frac{1}{(3 + 2\sqrt{2})}$
= $\frac{3 – 2\sqrt{2}}{9 – 8}$ (After rationalization)
= 3 – 2 × 1.414
= 3 – 2.828 = 0.17
Hence, the correct answer 0.17.
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