Question

if the diagonal of a square is 'a' units,what is the diagonal of square,whose area is double that of the first square

Answer 1

Refer to given image. It contains the solution.

Answer 2

Answer:

$\sqrt{2}a$

Step-by-step explanation:

Diagonal of square = $\sqrt{2} \times Side$

We are given that the diagonal of a square is 'a' units

So,  $\sqrt{2} \times Side=a$

$Side=\frac{a}{\sqrt{2}}$

$Side=\frac{\sqrt{2}a}{2}$

Area of square = $Side^2 =( \frac{\sqrt{2}a}{2})^2= \frac{2a^2}{4}=\frac{a^2}{2}$

Now we are given that Area of another square is double the area of first square = $\frac{a^2}{2}\times 2 = a^2$

Area of Square = $Side^2 = a^2$

So, Side of another square =a

So, Diagonal of square = $\sqrt{2} \times Side =\sqrt{2}a$

Hence  the diagonal of square,whose area is double that of the first square is $\sqrt{2}a$