Question

the diagonal of a square a is(a+b) units. find the diagonal of a square b whose area is twice the area of a
pls give me correct answer with full explaination it's very very urgent​

Answer 1

Answer:

Area of the square A = \frac{\left ( diagonal \right )^{2}}{2}= \frac{\left ( a+b \right )^{2}}{2}

Area of the new square

= \frac{\left ( a+b \right )^{2}}{2}\times 2 = \left ( a+b \right )^{2}

=> side = (a+b)

\therefore Diagonal = \sqrt{2}\times side

= \sqrt{2}(a+b)