Q. The diagonal of a square A is (a+b). The diagonal of a square whose are is twice the area of square A, is
Answer: $latex \sqrt{2}(a+b)$
Notes: 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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