Question : The diagonal of the square is $8 \sqrt{2}$ cm. Find the diagonal of another square whose area is triple that of the first square.
Option 1: $8 \sqrt{5}$ cm
Option 2: $8 \sqrt{3}$ cm
Option 3: $8 \sqrt{2}$ cm
Option 4: $8 \sqrt{6}$ cm
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Correct Answer: $8 \sqrt{6}$ cm
Solution :
Given, Diagonal = $8\sqrt2$
$\therefore$ Side = $\frac{8\sqrt{2}}{\sqrt 2}=8$ cm
⇒ Area of this square = $(8)^2 = 64$ cm
2
The Area of the Second square is triple the first one.
⇒ Area of the second one $= 64×3$
$\therefore$ Side of the 2nd square $= \sqrt{64×3} = 8\sqrt3$
⇒ The diagonal 2nd square $= \sqrt2×8\sqrt3 = 8\sqrt{6}$ cm
Hence, the correct answer is $8\sqrt6$ cm.
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