Question 2

We will find the shape left over after the cuts and then show how to reassemble the cut-off corners into a new square.

Step 1 — The shape left over

Let us imagine a square named ABCD. Let the side length of this square be $2x$ units. We find the midpoints of each side. Let P be the midpoint of side AB. Let Q be the midpoint of side BC. Let R be the midpoint of side CD. Let S be the midpoint of side DA. This means that AP = PB = BQ = QC = CR = RD = DS = SA = $x$ units. The problem says we cut between midpoints of adjacent edges. So, we cut along the lines PQ, QR, RS, and SP. The shape left in the middle is PQRS. Let us look at the triangle PBQ. Angle B in the square ABCD is a right angle (90 degrees). So, triangle PBQ is a right-angled triangle. Its sides PB and BQ are both $x$ units long. We can find the length of the cut line PQ using the Pythagoras theorem. The Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

$PQ^2 = PB^2 + BQ^2$

$PQ^2 = x^2 + x^2$

$PQ^2 = 2x^2$

$PQ = \sqrt{2x^2}$

$\boxed{PQ = x\sqrt{2} \text{ units}}$

All four triangles at the corners (AP S, PBQ, QCR, RDS) are identical. So, all four cut lines (PQ, QR, RS, SP) will have the same length. Therefore, PQ = QR = RS = SP = $x\sqrt{2}$ units. This means the shape PQRS has four equal sides. It is a rhombus. Now, let us check the angles of PQRS. Consider triangle AP S. Angle A is 90 degrees. AP = AS = $x$. This means triangle AP S is an isosceles right-angled triangle. The angles opposite the equal sides must be equal. So, Angle APS = Angle ASP = $(180 - 90) / 2 = 45$ degrees. Similarly, in triangle PBQ, Angle BPQ = 45 degrees. The line segment AB is a straight line. Angle SPQ is formed by parts of this line. Angle SPQ = 180 degrees - Angle APS - Angle BPQ

$= 180^\circ - 45^\circ - 45^\circ$

$= 90^\circ$

All four angles of PQRS are 90 degrees. Since PQRS has four equal sides and four right angles, it is a square.

Step 2 — The cut-off corners

The four cut-off corner pieces are the triangles AP S, PBQ, QCR, and RDS. Each of these is a right-angled isosceles triangle. Each triangle has two shorter sides (legs) of length $x$ units. Each triangle has one longer side (hypotenuse) of length $x\sqrt{2}$ units. All four triangles are identical in shape and size. The area of one such triangle is half times base times height.

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

$\boxed{\text{Area of one triangle} = \frac{x^2}{2} \text{ square units}}$

The total area of the four corner triangles is $4 \times (x^2/2) = 2x^2$ square units. The area of the central square PQRS is $(\text{side length})^2 = (x\sqrt{2})^2 = 2x^2$ square units. This shows that the total area of the cut-off corners is exactly the same as the area of the central square.

Step 3 — Reassembling the corners

We can reassemble the four identical right-angled isosceles triangles to form another square. To do this, we arrange the four triangles such that their right-angle vertices (A, B, C, D from the original square) all meet at a single central point. Let's imagine this central point is O. Each triangle has legs of length $x$ units. When the four right-angle vertices meet at O, the legs of the triangles will extend outwards from O. For example, one leg of triangle AP S will point along the positive x-axis for $x$ units. Its other leg will point along the positive y-axis for $x$ units. The hypotenuses of the four triangles will then form the outer boundary of the new square. The corners of this new square will be the endpoints of these legs. These endpoints are at a distance of $x$ units from the center O along four different directions (like the points $(x,0)$, $(0,x)$, $(-x,0)$, $(0,-x)$ on a graph). The side length of this new square is the distance between two adjacent corners, for example, between the point $(x,0)$ and $(0,x)$. Using the distance formula (which is like Pythagoras theorem on a coordinate plane):

$\text{Side length of new square} = \sqrt{(x-0)^2 + (0-x)^2}$

$= \sqrt{x^2 + (-x)^2}$

$= \sqrt{x^2 + x^2}$

$= \sqrt{2x^2}$

$\boxed{\text{Side length of new square} = x\sqrt{2} \text{ units}}$

This new square has a side length of $x\sqrt{2}$ units. In Step 1, we found that the central square PQRS also has a side length of $x\sqrt{2}$ units. Since both squares have the same side length, they are of the same size.

Answer

(i) The shape left over is a square. (ii) The four corners are identical right-angled isosceles triangles. (iii) We can reassemble the four corner triangles to make another square by placing their right-angle vertices together at a central point. This new square will be of the same size as the central square.