Question

The diagonal of a square A is a + b the diagonal of a square B with twice the area of A is

• A. $2(a+b)$
• B. $\sqrt{2}(a+b)$
• C. a + 2b
• D. 2a + 4b

Answer 1

The correct option is B $\sqrt{2}(a+b)$

Give Length of diagonal is a+b

Then $(a+b{)}^2=2\times (side{)}^2\Rightarrow side=\frac{1}{\sqrt{2}}(a+b)$

So side of square (A)=$\frac{1}{\sqrt{2}}(a+b)$

Then area of square (A)=$\left(\frac{1}{\sqrt{2}}\right(a+b){)}^2=\frac{1}{2}(a+b{)}^2$

So area of square (B)=$2\times \frac{1}{2}(a+b{)}^2=(a+b{)}^2$ (Given area of square A is twice the area of square B)

Then side of square B =(a+b)

So diagonal of square B=$\sqrt{2\times (side{)}^2}=\sqrt{2\times (a+b)}=\sqrt{2}(a+b)$