Question : If $x=\sqrt{64}+\sqrt{121}-\sqrt{169}$, then find the value of $x^2$.
Option 1: 16
Option 2: 25
Option 3: 36
Option 4: 48
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Correct Answer: 36
Solution :
$x=\sqrt{64}+\sqrt{121}-\sqrt{169}$
$⇒x= 8+11-13$
$⇒x= 6$
$\therefore x^2=6^2 = 36$
Hence, the correct answer is 36.
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