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Question 9
If $$\sqrt{86 - 60\sqrt{2}} = a - b\sqrt{2}$$, then what will be the value of $$\sqrt{a^2 + b^2}$$, correct to one decimal place?
Solution
$$\sqrt{86 - 60\sqrt{2}} = a - b\sqrt{2}$$
$$\sqrt{36 + 50 - 2 \times 30\sqrt{2}} = a - b\sqrt{2}$$
$$\sqrt{6^2 + (5\sqrt{2})^2 - 2 \times 6 \times 5\sqrt{2}} = a - b\sqrt{2}$$
$$\sqrt{(6 - 5\sqrt{2})^2 } = a - b\sqrt{2}$$
$$6 - 5\sqrt{2} = a - b\sqrt{2}$$
On compare,
a = 6,
b = 5
Now,
$$\sqrt{a^2 + b^2}$$
=Â $$\sqrt{6^2 + 5^2}$$
= $$\sqrt{61}$$ = 7.8
$$\therefore$$ The correct answer is option C.
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