Why is a square? Why is its area half that of the original paper? Explain by connecting and , finding the different angles formed, and then using tringle congruence.

Understand the Question A square sheet of paper with side length has midpoints on sides respectively. Connecting these midpoints creates quadrilateral inside the paper. To prove is a square, we: 1. Use SAS triangle congruence on the four corner right-angled triangles to show all sides are equal (). 2. Use the angle sum on a straight line to show each interior angle is . 3. Verify with diagonals and that they are equal in length and perpendicular to each other. To find its area, we apply the Pythagoras theorem on one of the corner triangles to find , which is exactly half of the original area (). Part (i) Why is a square? Step 1 · Define the Squares and Side Lengths Let the original square be with side length and area . Since are the midpoints of respectively: Step 2 · Prove is a Square Using Triangle Congruence and Angles Consider the four corner right-angled triangles , , , and : By the congruence rule: Therefore, their corresponding hypotenuses are equal: Thus, is a rhombus. In isosceles right-angled triangle , since and : Similarly, in , . Since is a straight line, the sum of angles at point is : Since all four sides are equal and each interior angle is , is a square. Step 3 · Confirm Using Diagonals and Connect diagonals and . connects midpoints of and , so and . connects midpoints of and , so and . Therefore: Since , it follows that: The diagonals of are equal and perpendicular bisectors of each other, confirming is a square. Answer > (i) is a square because its four sides are equal () and all its interior angles are . Part (ii) Why is its area half that of the original paper? Step 1 · Calculate the Area of In right-angled triangle , by Pythagoras theorem: The area of square is: Comparing with the area of the original square (): Answer > (ii) Common Mistakes Rhombus vs. Square: Proving that all four sides are equal () only shows that is a rhombus. You must also prove that the interior angles are (or that diagonals are equal and perpendicular) to establish that it is a square. Side Length vs. Area: Halving the area does not mean the side length is halved. The side length of is (not ), which yields an area of .

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Question 7

Why is $PQRS$ a square? Why is its area half that of the original paper?

Explain by connecting $QS$ and $PR$ , finding the different angles formed, and then using tringle congruence.

Solution

Understand the Question

A square sheet of paper $ABCD$ with side length $a$ has midpoints $P, Q, R, S$ on sides $CD, BC, AB, DA$ respectively.
Connecting these midpoints creates quadrilateral $PQRS$ inside the paper.
To prove $PQRS$ $P QR S$ is a square, we:
1. Use SAS triangle congruence on the four corner right-angled triangles to show all sides are equal ( $PQ = QR = RS = SP$ ).
2. Use the angle sum on a straight line to show each interior angle is $90^\circ$ .
3. Verify with diagonals $QS$ and $PR$ that they are equal in length and perpendicular to each other.
To find its area, we apply the Pythagoras theorem on one of the corner triangles to find $\text{Side}^2 = \dfrac{a^2}{2}$ , which is exactly half of the original area ( $a^2$ ).

(i) Why is $PQRS$ a square?

Step 1 · Define the Squares and Side Lengths

Let the original square be $ABCD$ with side length $a$ and area $a^2$ . Diagram 1

Since $P, Q, R, S$ are the midpoints of $CD, BC, AB, DA$ respectively: $AR = RB = BQ = QC = CP = PD = DS = SA = \dfrac{a}{2}$

Step 2 · Prove $PQRS$ is a Square Using Triangle Congruence and Angles

Consider the four corner right-angled triangles $\triangle ARS$ , $\triangle BQR$ , $\triangle CPQ$ , and $\triangle DSP$ :

$AR = BR = CP = DP = \dfrac{a}{2}$
$AS = BQ = CQ = DS = \dfrac{a}{2}$
$\angle A = \angle B = \angle C = \angle D = 90^\circ$

By the $\text{SAS}$ congruence rule: $\triangle ARS \cong \triangle BQR \cong \triangle CPQ \cong \triangle DSP$

Therefore, their corresponding hypotenuses are equal: $RS = QR = PQ = SP$ Thus, $PQRS$ is a rhombus.

In isosceles right-angled triangle $\triangle ARS$ , since $AR = AS$ and $\angle A = 90^\circ$ : $\angle ARS = \angle ASR = \dfrac{180^\circ - 90^\circ}{2} = 45^\circ$

Similarly, in $\triangle BQR$ , $\angle BRQ = 45^\circ$ .

Since $ARB$ is a straight line, the sum of angles at point $R$ is $180^\circ$ :

\begin{aligned} \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 = 90^\circ \end{aligned}

Since all four sides are equal and each interior angle is $90^\circ$ , $PQRS$ is a square.

Step 3 · Confirm Using Diagonals $QS$ and $PR$

Connect diagonals $QS$ and $PR$ . Diagram 2

$QS$ connects midpoints of $AD$ and $BC$ , so $QS \parallel AB$ and $QS = a$ .
$PR$ connects midpoints of $CD$ and $AB$ , so $PR \parallel AD$ and $PR = a$ .

Therefore: $QS = PR = a$

Since $AB \perp AD$ , it follows that: $QS \perp PR$

The diagonals of $PQRS$ are equal and perpendicular bisectors of each other, confirming $PQRS$ is a square.

Answer

(i) $PQRS$ is a square because its four sides are equal ( $PQ = QR = RS = SP$ ) and all its interior angles are $90^\circ$ .

(ii) Why is its area half that of the original paper?

Step 1 · Calculate the Area of $PQRS$

In right-angled triangle $\triangle ARS$ , by Pythagoras theorem:

\begin{aligned} RS^2 &= AR^2 + AS^2 \\[0.6em] &= \left(\dfrac{a}{2}\right)^2 + \left(\dfrac{a}{2}\right)^2 \\[0.6em] &= \dfrac{a^2}{4} + \dfrac{a^2}{4} \\[0.6em] &= \dfrac{2a^2}{4} \\[0.6em] &= \dfrac{a^2}{2} \end{aligned}

The area of square $PQRS$ is: $\text{Area}(PQRS) = RS^2 = \dfrac{a^2}{2}$

Comparing with the area of the original square ( $a^2$ ):

\begin{aligned} \text{Area}(PQRS) &= \dfrac{1}{2} \times a^2 \\[0.6em] &= \dfrac{1}{2} \times \text{Area}(\text{Original Paper}) \end{aligned}

Answer

(ii) $\text{Area}(PQRS) = \dfrac{1}{2} \times \text{Area}(\text{Original Paper}) = \dfrac{a^2}{2}$

Common Mistakes

Rhombus vs. Square: Proving that all four sides are equal ( $PQ = QR = RS = SP$ ) only shows that $PQRS$ is a rhombus. You must also prove that the interior angles are $90^\circ$ (or that diagonals are equal and perpendicular) to establish that it is a square.
Side Length vs. Area: Halving the area does not mean the side length is halved. The side length of $PQRS$ is $\dfrac{a}{\sqrt{2}}$ (not $\dfrac{a}{2}$ ), which yields an area of $\left(\dfrac{a}{\sqrt{2}}\right)^2 = \dfrac{a^2}{2}$ .

Question 7

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