Question 7

The inner square PQRS is formed by connecting the midpoints of the sides of the original square.

Step 1 — Define the squares and their parts

Let the original square be $ABCD$. Let its side length be $a$. The area of the original paper is $a^2$. The points $P, Q, R, S$ are the midpoints of the sides $CD, BC, AB, DA$ respectively. This means that each segment from a vertex of the original square to a vertex of PQRS has length $a/2$. For example, $AR = RB = BQ = QC = CP = PD = DS = SA = a/2$.

Step 2 — Prove PQRS is a square using triangle congruence and angles

Consider the four triangles at the corners of the original square: $\triangle ARS$, $\triangle BQR$, $\triangle CPQ$, and $\triangle DSP$. In $\triangle ARS$: The side $AR = a/2$. The side $AS = a/2$. The angle $\angle A = 90^\circ$ (because it is an angle of the original square).

Similarly, for the other three triangles: In $\triangle BQR$: $BR = a/2$, $BQ = a/2$, and $\angle B = 90^\circ$. In $\triangle CPQ$: $CP = a/2$, $CQ = a/2$, and $\angle C = 90^\circ$. In $\triangle DSP$: $DS = a/2$, $DP = a/2$, and $\angle D = 90^\circ$.

By the SAS (Side-Angle-Side) congruence rule, all four triangles are congruent: $\triangle ARS \cong \triangle BQR \cong \triangle CPQ \cong \triangle DSP$

Since these triangles are congruent, their corresponding sides are equal. The hypotenuses of these right-angled triangles are $RS, QR, PQ, SP$. So, $RS = QR = PQ = SP$. This means that PQRS is a rhombus (a quadrilateral with all four sides equal).

Now, let us find the angles of PQRS. In $\triangle ARS$, since $AR = AS$ and $\angle A = 90^\circ$, it is an isosceles right-angled triangle. The base angles are equal: $\angle ARS = \angle ASR = (180^\circ - 90^\circ) / 2 = 45^\circ$ Similarly, in $\triangle BQR$, $\angle BRQ = 45^\circ$. The points $A, R, B$ lie on a straight line (the side $AB$ of the original square). The sum of angles on a straight line at point $R$ is $180^\circ$. $\angle ARS + \angle SRQ + \angle QRB = 180^\circ$ $45^\circ + \angle SRQ + 45^\circ = 180^\circ$ $90^\circ + \angle SRQ = 180^\circ$ $\angle SRQ = 180^\circ - 90^\circ$

$\boxed{\angle SRQ = 90^\circ}$ Since PQRS is a rhombus and has one angle equal to $90^\circ$, it must be a square.

Step 3 — Confirm with diagonals QS and PR

Let us connect the diagonals $QS$ and $PR$ of the inner square PQRS. The line segment $QS$ connects $S$ (midpoint of $AD$) and $Q$ (midpoint of $BC$). The line segment $PR$ connects $P$ (midpoint of $CD$) and $R$ (midpoint of $AB$). In a square $ABCD$, the line connecting the midpoints of opposite sides is parallel to the other pair of sides and equal in length to the side of the square. So, $QS$ is parallel to $AB$ and $CD$. The length of $QS$ is $a$. And $PR$ is parallel to $AD$ and $BC$. The length of $PR$ is $a$. Therefore, $QS = PR = a$. The diagonals of PQRS are equal. Since $QS$ is parallel to $AB$ and $PR$ is parallel to $AD$, and $AB$ is perpendicular to $AD$ (sides of a square), then $QS$ is perpendicular to $PR$. The diagonals of PQRS are equal and perpendicular. This further confirms that PQRS is a square.

Step 4 — Calculate the area of PQRS

Let the side length of the original square be $a$. The area of the original paper is $a^2$. We need to find the side length of the square PQRS. Let us call it $s'$. Consider the right-angled triangle $\triangle ARS$. By the Pythagorean theorem (which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides): $RS^2 = AR^2 + AS^2$ We know $AR = a/2$ and $AS = a/2$. $RS^2 = (a/2)^2 + (a/2)^2$ $RS^2 = a^2/4 + a^2/4$ $RS^2 = 2a^2/4$ $RS^2 = a^2/2$ The area of square PQRS is $RS^2$.

$\boxed{\text{Area(PQRS)} = a^2/2}$ The area of the original paper is $a^2$. So, the area of PQRS is half the area of the original paper. $\text{Area(PQRS)} = (1/2) \times a^2$ $\text{Area(PQRS)} = (1/2) \times \text{Area(Original Paper)}$

Answer

(i) PQRS is a square because its four sides (PQ, QR, RS, SP) are equal in length (proven by triangle congruence of the corner triangles), and its interior angles are all $90^\circ$ (proven by angles on a straight line). (ii) The area of PQRS is half that of the original paper. This is because if the original paper has side length $a$, its area is $a^2$. The side length of PQRS is $a/\sqrt{2}$ (found using the Pythagorean theorem), so its area is $(a/\sqrt{2})^2 = a^2/2$.