The diagonal of a square A is (a+b). The diagonal of a square whose are is twice the area of square A, is

The diagonal of a square A is (a+b). The diagonal of a square whose are is twice the area of square A, is
[A]$latex \sqrt{2}(a-b)$
[B]$latex \sqrt{2}(a+b)$
[C]$latex 2(a+b)^{2}$
[D]$latex 2(a+b)$

$latex \mathbf{\sqrt{2}(a+b)}$
Area of the square A = $latex \frac{\left ( diagonal \right )^{2}}{2}= \frac{\left ( a+b \right )^{2}}{2}&s=1$
Area of the new square
$latex = \frac{\left ( a+b \right )^{2}}{2}\times 2 = \left ( a+b \right )^{2}&s=1$
$latex => side = (a+b)$
$latex \therefore Diagonal = \sqrt{2}\times side$
$latex = \sqrt{2}(a+b)$
Hence option [B] is the right answer.


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