How can one construct a square having double the area of a given square?

Question

Homi · Accepted Answer

Understand the Question

If the side length of a square is $s$ , its area is $s^2$ .
A square with double the area has an area of $2s^2$ , which means its side length must be $\sqrt{2s^2} = s\sqrt{2}$ .
By the Pythagoras theorem, the diagonal of a square with side $s$ is $\sqrt{s^2 + s^2} = s\sqrt{2}$ .
Therefore, we can construct a square with double the area by using the diagonal of the original square as the side length of the new square.

Step 1 · Determine the Required Side Length

Let the side length of the original square be $s$ . Diagram 1

\text{Area of original square} = s^2

Let $S$ be the side length of the new square with double the area:

S^2 = 2s^2

Taking the square root on both sides:

\begin{aligned} S &= \sqrt{2s^2} \\[0.6em] &= s\sqrt{2} \end{aligned}

Thus, the new square must have a side length of $s\sqrt{2}$ .

Step 2 · Calculate the Diagonal of the Original Square

Consider the given square $\text{ABCD}$ with side length $s$ . Diagram 2

In right-angled triangle $\text{ABC}$ , by Pythagoras theorem:

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\[0.6em] &= s^2 + s^2 \\[0.6em] &= 2s^2 \end{aligned}

\begin{aligned} \text{AC} &= \sqrt{2s^2} \\[0.6em] &= s\sqrt{2} \end{aligned}

The diagonal $\text{AC}$ has length $s\sqrt{2}$ , which is exactly the required side length of the new square.

Step 3 · Steps of Construction

Diagram 3

Draw the given square $\text{ABCD}$ and draw its diagonal $\text{AC}$ .
At point $\text{A}$ , construct a line perpendicular to $\text{AC}$ .
At point $\text{C}$ , construct a line perpendicular to $\text{AC}$ .
Using a compass set to length $\text{AC}$ , draw an arc from $\text{A}$ intersecting the perpendicular line at $\text{P}$ (so $\text{AP} = \text{AC}$ ).
From $\text{C}$ , draw an arc with radius $\text{AC}$ intersecting the perpendicular line at $\text{Q}$ (so $\text{CQ} = \text{AC}$ ).
Join $\text{P}$ and $\text{Q}$ to complete the square $\text{APQC}$ .

The area of the constructed square is:

\text{Area} = (s\sqrt{2})^2 = 2s^2

Answer

Construct a square whose side length is equal to the diagonal of the given square.

Common Mistakes

Doubling the Side Instead of the Area: Doubling the side length ( $2s$ ) results in a square with $4$ times the area ( $(2s)^2 = 4s^2$ ), not $2$ times.
Linear vs Area Scaling: Area scales by the square of the side factor. To double the area, the side length must be multiplied by $\sqrt{2}$ , which is given by the diagonal of the original square.

Question 1