Fractals and Visualising Solids | A

Question 2

Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?

Solution

Understand the Question

Let the original square be $\text{ABCD}$ with side length $2x$ . The midpoints of the sides $\text{AB}$ , $\text{BC}$ , $\text{CD}$ , and $\text{DA}$ are $\text{P}$ , $\text{Q}$ , $\text{R}$ , and $\text{S}$ respectively.
Cutting between adjacent midpoints removes four identical right-angled isosceles triangles (the corners) and leaves behind a central quadrilateral $\text{PQRS}$ .
To determine the shape left over, we find the side lengths using the Pythagoras theorem and calculate the interior angles.
To reassemble the four corner triangles into a new square, we join their right-angled vertices together at a single central point.

Step 1 · Determine the Shape Left Over

Let the original square $\text{ABCD}$ have side length $2x$ , and let $\text{P}$ , $\text{Q}$ , $\text{R}$ , $\text{S}$ be the midpoints of sides $\text{AB}$ , $\text{BC}$ , $\text{CD}$ , $\text{DA}$ respectively, so: $\text{AP} = \text{PB} = \text{BQ} = \text{QC} = \text{CR} = \text{RD} = \text{DS} = \text{SA} = x$ Diagram 1

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

\begin{aligned} \text{PQ}^2 &= \text{PB}^2 + \text{BQ}^2 \\ &= x^2 + x^2 \\ &= 2x^2 \\ \text{PQ} &= \sqrt{2x^2} = x\sqrt{2} \end{aligned}

Since all four corner triangles are identical $\text{PQ} = \text{QR} = \text{RS} = \text{SP} = x\sqrt{2}$

In isosceles right triangle $\text{APS}$ $\angle \text{APS} = \angle \text{ASP} = \dfrac{180^\circ - 90^\circ}{2} = 45^\circ$

Similarly, $\angle \text{BPQ} = 45^\circ$ . Along the straight line $\text{AB}$

\begin{aligned} \angle \text{SPQ} &= 180^\circ - \angle \text{APS} - \angle \text{BPQ} \\ &= 180^\circ - 45^\circ - 45^\circ \\ &= 90^\circ \end{aligned}

Since $\text{PQRS}$ has four equal sides of length $x\sqrt{2}$ and four $90^\circ$ angles, the leftover shape is a square.

Step 2 · Find the Area of the Corner Pieces

Each of the four cut-off corners is a right-angled isosceles triangle with legs of length $x$ .

$\text{Area of one triangle} = \dfrac{1}{2} \times x \times x = \dfrac{x^2}{2}$

$\text{Total area of } 4 \text{ corner triangles} = 4 \times \left(\dfrac{x^2}{2}\right) = 2x^2$

$\text{Area of central square } \text{PQRS} = (x\sqrt{2})^2 = 2x^2$

The total area of the four corners equals the area of the leftover square.

Step 3 · Reassemble the Corners into a New Square

Place the right-angle vertices ( $\text{A}$ , $\text{B}$ , $\text{C}$ , $\text{D}$ ) of the four corner triangles together at a common center point $\text{O}$ . Diagram 2

The outer vertices lie at $(x, 0)$ , $(0, x)$ , $(-x, 0)$ , and $(0, -x)$ . The side length of the newly formed square is the distance between adjacent outer vertices

\begin{aligned} \text{Side length} &= \sqrt{(x-0)^2 + (0-x)^2} \\ &= \sqrt{x^2 + (-x)^2} \\ &= \sqrt{x^2 + x^2} \\ &= \sqrt{2x^2} = x\sqrt{2} \end{aligned}

This forms another square identical in size to the leftover central square.

Answer

The shape left over is a square.

The four corner right-angled triangles can be reassembled into another identical square by placing their right-angle vertices together at a single central point.

Common Mistakes

Incomplete Proof of Shape: Concluding that $\text{PQRS}$ is a square solely because its four sides are equal, without proving that each interior angle is $90^\circ$ (a rhombus also has four equal sides).
Incorrect Reassembly: Trying to join the triangles along their hypotenuses rather than bringing their $90^\circ$ corner vertices to a common center.

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