Squares and Square Roots | A

Question 3

Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?

Fold the square paper inward, as shown, such that the crease lines pass through the midpoints of the sides. $PQRS$ is the required square with half the area.

Solution

Understand the Question

The area of a square is given by $\text{Area} = \text{side}^2$ .
If the side length is halved, both the length and width are halved, reducing the area by a factor of $2 \times 2 = 4$ (one-fourth, not half).
Therefore, it takes $4$ smaller squares of half the sidelength to fill the original square.
To construct a square with exactly half the area, the corners of the original square are folded inward such that the crease lines connect the midpoints of the sides to form square $PQRS$ .

Step 1 · Area of Square with Half the Sidelength

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

$\text{Area}_{\text{original}} = L^2$

For a new square with side length $\dfrac{L}{2}$

\begin{aligned} \text{Area}_{\text{new}} &= \left(\frac{L}{2}\right)^2 \\[0.6em] &= \frac{L^2}{4} \end{aligned}

Since $\dfrac{L^2}{4}$ is one-fourth of the original area, it is not half.

Step 2 · Reason Why Area is One-Fourth

The area of a square depends on the square of its side length:

$\text{Area} = \text{side} \times \text{side}$

Halving the side length reduces both dimensions by a factor of $2$ , so the total area is reduced by a factor of $2^2 = 4$ :

$\left(\frac{L}{2}\right)^2 = \frac{L^2}{4}$

Therefore, the area becomes one-fourth rather than one-half.

Step 3 · Calculate Number of Smaller Squares

Diagram 2

To find how many smaller squares fill the original square, divide the original area by the new area

\begin{aligned} \text{Number of squares} &= \frac{\text{Area}_{\text{original}}}{\text{Area}_{\text{new}}} \\[0.6em] &= \frac{L^2}{\frac{L^2}{4}} \\[1.1em] &= 4 \end{aligned}

Step 4 · Area of Folded Square PQRS

Let the side length of the original square paper be $s$ , so $\text{Area}_{\text{original}} = s^2$ . Diagram 3

Points $P, Q, R, S$ are midpoints of the four sides. Each folded corner is a right-angled triangle with base and height equal to $\dfrac{s}{2}$ .

\begin{aligned} \text{Area}_{\text{one corner triangle}} &= \frac{1}{2} \times \frac{s}{2} \times \frac{s}{2} \\[0.6em] &= \frac{s^2}{8} \end{aligned}

Total area of the 4 corner triangles

\begin{aligned} \text{Area}_{\text{four corner triangles}} &= 4 \times \frac{s^2}{8} \\[0.6em] &= \frac{s^2}{2} \end{aligned}

Subtracting the four corners from the original square gives the area of $PQRS$

\begin{aligned} \text{Area}_{PQRS} &= s^2 - \frac{s^2}{2} \\[0.6em] &= \frac{s^2}{2} \end{aligned}

Answer

(i) No, the square having half the sidelength will not have half the area.

(ii) Area depends on the square of the side length: $\left(\dfrac{L}{2}\right)^2 = \dfrac{L^2}{4}$ , so the area becomes one-fourth.
(iii) $4$ such squares will fill the original square.

Common Mistakes

Linear Scaling Fallacy: Assuming that halving the side length halves the area. Since $\text{Area} = \text{side}^2$ , scaling the side by $\frac{1}{2}$ scales the area by $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ .
Confusing Halved Sidelength with Half Area: A square with half the side length has $\frac{1}{4}$ the area, whereas folding the midpoints creates a square $PQRS$ with $\frac{1}{2}$ the area.

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