Now suppose we are given a square, and we want to construct a square whose area is half that of the original square. How would you do it?

Question

Homi · Accepted Answer

Understand the Question

The area of a square with side length $s$ is $s^2$ .
A square with half the area must have an area of $\dfrac{s^2}{2}$ , which requires a side length of $\dfrac{s}{\sqrt{2}}$ .
Geometrically, connecting the consecutive midpoints of the sides of the original square forms a new square whose side length is precisely $\dfrac{s}{\sqrt{2}}$ and whose area is half the original.

Step 1 · Calculate the Required Side Length

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

\begin{aligned} \text{Area of original square} &= s \times s \\ &= s^2 \end{aligned}

$\text{Desired area of new square} = \dfrac{s^2}{2}$

Let the side length of the new square be $s'$ . $s'^2 = \dfrac{s^2}{2}$

$s' = \sqrt{\dfrac{s^2}{2}} = \dfrac{s}{\sqrt{2}}$

Step 2 · Geometric Verification via Pythagoras Theorem

Join the midpoints of the four sides of the original square. Each corner forms a right-angled triangle with legs of length $\dfrac{s}{2}$ .

By Pythagoras theorem:

\begin{aligned} (\text{side of new square})^2 &= \left(\dfrac{s}{2}\right)^2 + \left(\dfrac{s}{2}\right)^2 \\[0.6em] &= \dfrac{s^2}{4} + \dfrac{s^2}{4} \\[0.6em] &= \dfrac{2s^2}{4} \\[0.6em] &= \dfrac{s^2}{2} \end{aligned}

$\text{Side of new square} = \sqrt{\dfrac{s^2}{2}} = \dfrac{s}{\sqrt{2}}$

All four sides are equal and all internal angles are $90^\circ$ , so the inner shape is a square with area $\left(\dfrac{s}{\sqrt{2}}\right)^2 = \dfrac{s^2}{2}$ .

Step 3 · Steps for Construction

Diagram 1

Draw the given square of side length $s$ .
Mark the midpoint of each of its four sides.
Connect these four midpoints consecutively with straight lines.
The inner figure formed is the required square with half the area of the original square.

Answer

Connect the midpoints of the four sides of the original square consecutively to construct a square with half the area.

Common Mistakes

Halving the Side Length: Halving the side length to $\dfrac{s}{2}$ yields an area of $\left(\dfrac{s}{2}\right)^2 = \dfrac{s^2}{4}$ (one-fourth of the area), not half.
Incorrect Side Scaling: The side length must scale by a factor of $\dfrac{1}{\sqrt{2}}$ , not $\dfrac{1}{2}$ , because $\text{Area} \propto \text{side}^2$ .

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