Key Concept: Artwork Construction, Support Lines

c) Reference guides for positioning and symmetry
[Solution Description]
Horizontal support lines serve as reference guides for positioning both the compass tip and ensuring symmetrical curve placement. They help maintain consistent height for all elements and allow precise alignment of the upper and lower curves that form the eye shape.

Key Concept: Artwork Construction

a) On the same vertical line, one above and one below the horizontal center line
[Solution Description]
To draw symmetrical eyes, points A and B (where the compass tip is placed) must be equidistant from the center line of the eye. The upper and lower curves are drawn with the same radius but mirrored positions. This ensures symmetry as both curves will have equal curvature and alignment.

Key Concept: Circle Construction

c) Distance between the pencil and the needle tip
[Solution Description]
The radius of the circle is equal to the distance between the tip of the compass and the pencil point. To draw a circle of radius 4 cm, you must set the compass such that the distance from the needle to the pencil is exactly 4 cm.

Key Concept: Freehand vs. instrument drawing, precision

b) Compass enforces uniform radius throughout
[Solution Description]
A compass ensures that every point on the drawn circle is exactly the same distance (radius) from the center, which is hard to achieve by hand due to natural human error in maintaining consistent curvature and radius measurement without tools.

Key Concept: Wave Construction, Compass Use

c) 0.75 cm
[Solution Description]
Given AB = 6 cm. AX is one-fourth of AB, so $AX = \frac{1}{4} \times 6\,\text{cm} = 1.5\,\text{cm}$. The first half-circle wave uses AX as its diameter. The radius of this half-circle would be half of AX, so $\frac{1.5\,\text{cm}}{2} = 0.75\,\text{cm}$.

Key Concept: Compass and Radius

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, a compass can be used to draw a circle by setting the distance between its tip (center point) and pencil to the desired radius. When this is done, moving the pencil around the fixed tip draws a circle. The reason correctly states that the distance from the center to any point on the circle is the radius, which explains why setting the compass to 4 cm results in a circle with radius 4 cm.

Key Concept: Circle Properties

b) $6$ cm
[Solution Description]
For any point lying on the circumference of the circle, its distance from the center $P$ is equal to the radius of the circle. Here, the radius is given as $6$ cm, so the distance between $P$ and $Q$ is $6$ cm.

Key Concept: Circle Construction

b) $5$ cm
[Solution Description]
To draw a circle with radius $5$ cm, the compass must be opened such that the distance between its tip and pencil is exactly equal to the radius. Therefore, the required distance is $5$ cm.

Key Concept: Centre, Radius, Compass Construction

b) 5 cm
[Solution Description]
The radius of a circle is defined as the distance between the centre (P) and any point on the circumference (Q). Here, it is given that this distance is 5 cm. Thus, the radius must be equal to 5 cm.

Key Concept: Radius and Centre of a Circle

c) Exactly 5 cm
[Solution Description]
The distance between the centre of the circle and any point on the circle is called the radius. Given that the radius is 5 cm, the distance is therefore 5 cm.

Key Concept: Symmetrical Figure Construction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The syllabus states that the upper and lower curves should form a symmetrical figure, which requires placing the compass tips (Points A and B) at equal distances from the center. This ensures both curves are identical, leading to symmetry. The reason correctly explains the assertion as it ties the placement of points A and B to achieving symmetry through equal radii and distance.

Key Concept: Drawing Curved Figures

c) Points A and B should be at equal distances from the center line
[Solution Description]
To draw symmetrical curves like those in the neck portion, points A and B must be placed at equal distances from the center line. This ensures both curves will mirror each other perfectly, creating identical waves smaller than a half circle.

Key Concept: Radius of semi-circle, Construction of identical waves

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given that $AB = 8$ cm is the central line, the first semi-circular wave is constructed with its diameter coinciding with $AB$. The radius ($r$) of a semi-circle is half its diameter ($d$). Therefore, $r = d/2$. Since the diameter of the semi-circle equals the length of $AB$, we have $d = 8$ cm. Substituting, $r = 8/2 = 4$ cm. Hence, the assertion is true. The reason correctly explains the assertion because the property of a semi-circle states that its diameter is indeed the straight-line segment it is constructed on, validating the explanation.

Key Concept: Locating the Center

b) At the midpoint of AB
[Solution Description]
The first semi-circle is drawn with diameter AB. To draw a semi-circle, the compass tip must be placed at the midpoint of AB. The midpoint divides AB into two equal parts: $AX = XB = \frac{8}{2} = 4$ cm.

Key Concept: Symmetry in Eye Construction

b) At equal distances above and below the centerline
[Solution Description]
To achieve symmetry, points A and B must be equidistant from the centerline of the eye. The upper curve is drawn by placing the compass tip at point A, and the lower curve is drawn by placing the compass tip at point B. Both points should be at the same vertical distance above and below the centerline, respectively.

Key Concept: Compass Placement for Curves

c) The same radius must be used for both curves
[Solution Description]
The radius must be consistent for both the upper and lower curves to create a symmetrical look. If different radii are used, the eye will appear lopsided. Trial and adjustment may be needed to find the right radius that fits the intended design of the eye.

Key Concept: Circle Construction

c) A circle
[Solution Description]
To solve this, fix the compass tip at point P and set the radius to 4 cm. Rotate the pencil around point P while keeping the distance constant. All points at 4 cm from P will lie on a circle because a circle is defined as the set of all points equidistant from its center.

Key Concept: constructing points using arcs, fixed distance

c) 2
[Solution Description]
The condition for two circles (or arcs) to intersect is that the distance between centers ($d$) must satisfy $|r_1 - r_2| < d < r_1 + r_2$. Here, $d = 6\,\text{cm}$, $r_1 = r_2 = 4\,\text{cm}$. $|4-4| = 0 < 6 < 4+4 = 8$ is false because $6 < 8$ but $0 r_1 + r_2$ is not true and $d = 6 \not< r_1 + r_2 = 8$, they intersect at two distinct points.

Key Concept: Properties of Rotated Squares, Diagonals of Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a rotated square retains the properties of its diagonals being equal in length and bisecting each other at $90^\circ$. This is true because rotation does not change the intrinsic properties of a square, including the equality and perpendicularity of its diagonals.
The reason states that the diagonals of a square are congruent and perpendicular to each other regardless of rotation. This is also true as it is a fundamental property of squares.
Moreover, the reason correctly explains the assertion because the invariance of diagonal properties under rotation is directly due to the inherent properties mentioned in the reason.

Key Concept: Naming Rectangles

c) 8
[Solution Description]
A rectangle can be named starting from any corner and moving in a clockwise or anti-clockwise direction. The possible sequences are ABCD, BCDA, CDAB, DABC, ADCB, DCBA, CBAD, BADC. Hence, there are 8 valid names.

Key Concept: Naming rectangles

c) ABDC
[Solution Description]
Valid names for a rectangle must follow the order of travel around the rectangle. The name ABDC does not follow this order because it skips from B to D instead of moving sequentially to C.

Key Concept: Naming Rectangles

d) PRSQ
[Solution Description]
The valid names for a rectangle must follow the order of travel around it. Therefore, names like PQSR (going from P to Q to S to R) are invalid because they skip a vertex when going around the rectangle. All other options follow the correct sequence.

Key Concept: Diagonals of a Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in rectangle PQRS, diagonal PR divides angle P into two angles of $45°$ each. For this to happen, the triangle formed by the sides and the diagonal must be an isosceles right-angled triangle. This implies that the adjacent sides of the rectangle are equal, which means the rectangle must actually be a square.
The reason states that a rectangle becomes a square when its diagonals divide opposite angles into equal parts. This is true because only in a square are all sides equal, leading to diagonals that bisect the angles equally ($45°$ each).
Therefore, both the assertion and reason are true, and the reason correctly explains why the assertion holds.

Key Concept: Angle equality, Square properties

c) $45°$ and $45°$
[Solution Description]
In a square, all angles are $90°$, and the diagonals bisect the angles, meaning they divide them into two equal parts. Therefore, each angle formed by the diagonal will be:
$\frac{90°}{2} = 45°$

Key Concept: Rotated Square

b) Yes, because all sides and angles remain unchanged
[Solution Description]
Rotating a square does not change the lengths of its sides or the measures of its angles. Therefore, the rotated figure still satisfies both properties of a square (equal sides and right angles).

Key Concept: Construction Activity, Diagonals of a Square

a) AB = AD and BD = AF
[Solution Description]
For the figure AFGD to be a perfect square, all sides must be equal, and all angles must be $90°$. Since AF and AG are already perpendicular and equal in length, points B and D must be marked such that AB = AD = AF = AG = GD = GF, ensuring all sides are equal. Moreover, the diagonal FG must divide the opposite angles equally, preserving the square's properties. Hence, the distance from A to B must be equal to the distance from A to D.

Key Concept: Rotated squares, Properties validation

a) The side lengths change.
[Solution Description]
When a square is rotated by $90°$, its properties remain unchanged. All sides stay equal, and all angles remain $90°$. The labeling of vertices changes based on rotation direction.
Analyze each option:
1) "The side lengths change." This is false because rotation preserves side lengths.
2) "The angles remain $90°$." This is true as rotation does not alter angles.
3) "The new vertex order is Q, R, S, P when rotated clockwise from P, Q, R, S." This is true as each vertex shifts to the next position clockwise.
4) "The diagonals remain equal in length." This is true since rotation preserves diagonal lengths.
Thus, the false statement is the one claiming side lengths change.

Key Concept: Naming Conventions of Squares and Rectangles

d) Assertion is false, but Reason is true.
[Solution Description]
A square's naming convention requires that the vertices must appear in an order of travel either clockwise or anti-clockwise. Here, PQSR does not follow this rule because after P to Q, it skips R and goes to S, then comes back to R, which is not a valid sequence of travel around the square. The correct sequences would be PQRS, QRSP, RSPQ, or SPQR (clockwise), or any anti-clockwise variant like SRQP. Thus, Assertion (A) is false because PQSR is not a valid name. However, Reason (R) is true because the rule states that names must follow the order of travel around the square.

Key Concept: Rotated Squares and Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a rotated square remains a square. This is true because all the properties of a square remain unchanged after rotation, i.e., all sides stay equal in length and all angles stay $90^\circ$. The reason correctly explains the assertion as it states that rotating a square does not alter the lengths of the sides or the measures of the angles, ensuring it still satisfies the properties of a square.

Key Concept: Rotated Squares and Rectangles, Side Lengths

c) It remains a square because all sides and angles are unchanged by rotation.
[Solution Description]
The rotated figure still has all sides equal and all angles $90°$. Therefore, it remains a square. Rotation does not alter the intrinsic properties of a square or rectangle.

Key Concept: Properties of Squares and Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when a square is rotated, it remains a square because its sides and angles are unchanged. The reason explains why: rotation preserves the lengths of sides and the measures of angles.
Step 1: A square has four equal sides and four right angles ($90^\circ$).
Step 2: When a square is rotated, its sides do not stretch or shrink—they remain equal in length.
Step 3: Similarly, the angles between the sides remain $90^\circ$ after rotation.
Step 4: Therefore, the rotated figure still satisfies the properties of a square.
Step 5: Hence, the assertion is true, and the reason correctly explains why the assertion holds.

Key Concept: Rectangle construction, Property verification

b) It has all angles 90° but opposite sides unequal
[Solution Description]
To verify if it's a rectangle:
1) Check opposite sides: EF = 4-0=4 units, GH = $\sqrt{(0-4)^2+(3-2)^2}$=$\sqrt{16+1}$=$\sqrt{17}$ units (not equal). EH = 3-0=3 units, FG = 2-0=2 units (not equal).
2) Check angles: Slope of EF = 0, EH = undefined (vertical), so ∠E is 90°. Slope of FG = (2-0)/(4-4)=undefined (vertical), GH = (3-2)/(0-4)=-1/4. Product of perpendicular slopes = 0 × -1/4 = 0 ≠ -1, so ∠G is not 90°.
Thus, some angles are 90° but opposite sides aren't equal - violating rectangle properties.

Key Concept: Constructing Squares, Properties of Squares

b) 4 cm
[Solution Description]
First, we need to visualize the square PQRS with side 6 cm. Since T is inside the square at 4 cm from P along a 45° line, we can calculate its coordinates relative to P (origin). The x and y components would both be $4\cos(45°) = 4\times\frac{\sqrt{2}}{2} \approx 2.828$ cm each. The nearest corner would be P (0,0) or Q (6,0). Distance to P is already 4 cm. Distance to Q is $\sqrt{(6-2.828)^2 + (0-2.828)^2} = \sqrt{10.03 + 8} = \sqrt{18.03} \approx 4.246$ cm. So the minimum distance is 4 cm.

Key Concept: Properties of Squares and Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a rectangle are always equal in length, which is true as per the properties of rectangles. The reason given is that a rectangle has all angles equal to $90^\circ$ and opposite sides equal, which is also true. However, while both statements are correct, the equality of diagonals in a rectangle is directly derived from the Pythagorean theorem applied to the congruent right triangles formed by the diagonals, not just from the angles being $90^\circ$ or the opposite sides being equal. Hence, the reason does not correctly explain the assertion.

Key Concept: Rectangle construction, Angle bisector, Diagonal properties

b) Bisect the right angle at A into $60°$ and $30°$
[Solution Description]
First draw AB = 5 cm. At point A, construct a $90°$ angle using perpendicular bisector method since all rectangle angles are $90°$. Now divide this right angle into $60°$ and $30°$ using angle bisector method from point A. The $60°$ portion will be towards side AB and $30°$ portion towards AD. This gives the direction for side AD.

Key Concept: Construction of Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a rectangle can be constructed if the lengths of its two adjacent sides are known. This is true because, with two adjacent sides, we can determine all four sides of the rectangle since opposite sides are equal. The reason given is that the opposite sides of a rectangle are equal and parallel, which is also a true property of rectangles. Moreover, the reason correctly explains why knowing two adjacent sides is sufficient for constructing a rectangle, as it implies the lengths of all sides.

Key Concept: Constructing a Perpendicular Line for a Square

c) Using a protractor to measure 90 degrees from PQ at P
[Solution Description]
After drawing PQ, we need to draw a perpendicular to PQ through point P to construct the adjacent side PS. This can be done using a protractor or by constructing a right angle geometrically.

Key Concept: Constructing squares with constraints, center alignment

b) 6 cm
[Solution Description]
First find the center of the rectangle, which is at (5 cm, 3 cm) from any corner. The maximum square inside must fit within the shorter dimension (6 cm). Thus, its side cannot exceed 6 cm. However, centering it requires equal distances on all sides. So, the maximum square has side length equal to the smaller dimension of the rectangle, which is 6 cm.

Key Concept: Rectangle Construction

c) 7.62 cm
[Solution Description]
To find the diagonal of the rectangle, we use the Pythagorean theorem: $d = \sqrt{s_1^2 + s_2^2}$. Here, $s_1 = 3$ cm and $s_2 = 7$ cm.
Step 1: Square both sides: $3^2 = 9$, $7^2 = 49$.
Step 2: Add them: $9 + 49 = 58$.
Step 3: Take the square root: $\sqrt{58} \approx 7.62$ cm.

Key Concept: Constructing Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a quadrilateral with sides 4 cm, 5 cm, 4 cm, and 5 cm and all right angles must be a rectangle. By definition, a rectangle has four right angles and opposite sides equal in length. Here, the given sides (4 cm, 5 cm, 4 cm, 5 cm) show equality of opposite sides, and all angles are right angles. Therefore, it satisfies the properties of a rectangle. The reason correctly explains this property of rectangles. Hence, both the assertion and reason are true, and the reason is the correct explanation.

Key Concept: Constructing rectangles using compass, verifying properties

b) Draw an arc of $6\,\text{cm}$ from $\text{A}$ and an arc of $10\,\text{cm}$ from $\text{B}$
[Solution Description]
To construct a right angle at $\text{B}$ using a compass:
1. Draw side $\text{AB} = 8\,\text{cm}$.
2. At point $\text{B}$, draw an arc of radius $6\,\text{cm}$ from $\text{A}$.
3. From $\text{B}$, draw another arc of radius $10\,\text{cm}$. The intersection point of these two arcs will form the right angle.
4. Join $\text{B}$ to the intersection point to complete the perpendicular line for $\text{BC}$.
Verification: Using the Pythagorean theorem, $6^2 + 8^2 = 10^2$ ensures the angle at $\text{B}$ is $90^\circ$.

Key Concept: Constructing rectangles using ruler and compass, Verification of properties

d) Assertion is false, but Reason is true.
[Solution Description]
To verify whether the constructed figure is a rectangle, we need to ensure two things: (1) Opposite sides must be equal, and (2) All angles must be $90^\circ$. The assertion states that verifying $\angle$DAB as $90^\circ$ is sufficient, but this alone does not confirm the other three angles or opposite side equality. The reason claims that the compass method guarantees all angles are $90^\circ$, which is true because the compass ensures perpendicularity when constructing arcs and intersecting them accurately. However, the assertion oversimplifies the verification process and does not consider the full set of conditions required for a rectangle. Therefore, the assertion is false, but the reason is true.

Key Concept: Side Lengths and Angles

c) It is a rectangle because opposite sides are equal and all angles are 90 degrees.
[Solution Description]
Given two pairs of equal opposite sides (3 cm and 6 cm) and all angles as 90 degrees, the figure satisfies the properties of a rectangle.

Key Concept: Constructing Squares and Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we first need to check if the given sides satisfy the properties of a rectangle. The sides are given as 4 cm, 6 cm, 4 cm, and 6 cm, which means the opposite sides are equal (4 cm = 4 cm and 6 cm = 6 cm). Next, we need to ensure that all angles are 90$^\circ$. The reason states correctly that in a rectangle, opposite sides are equal and all angles are 90$^\circ$. Since both conditions are satisfied, the assertion is true, and the reason explains why the assertion is true.

Key Concept: Naming Rectangles

d) PRQS
[Solution Description]
In a rectangle, the corners must be named in an order of travel around the rectangle (clockwise or counter-clockwise). Any deviation breaks this rule.

Key Concept: Rotation of Shapes

c) It remains a square with unchanged properties.
[Solution Description]
Rotating a square by any angle does not change its properties. All sides remain equal, and all angles remain 90°.

Key Concept: Diagonal Properties

c) They must be equal
[Solution Description]
In any rectangle, a diagonal divides each vertex angle into two equal parts because diagonals bisect the angles in rectangles (making them congruent triangles).

Key Concept: Rectangle Construction

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that any quadrilateral with all angles equal to 90\textdegree must be a square. This is false because a rectangle also has all angles equal to 90\textdegree, but its sides are not necessarily equal. Only when all sides are equal in addition to the angles being 90\textdegree does it become a square. The reason states that in a rectangle, opposite sides are always equal, which is true by definition of a rectangle. Since the assertion is false but the reason is true, the correct answer is d).

Key Concept: Constructing squares within rectangles, Center alignment

c) 4 cm
[Solution Description]
For the centers of the rectangle and the square to coincide, the square must fit symmetrically within the rectangle. The maximum possible side length of the square is constrained by the shorter side of the rectangle, which is 4 cm. Hence, the side length of the square cannot exceed 4 cm.

Key Concept: Dividing rectangle into identical squares, Compass construction

b) 3 cm
[Solution Description]
Since the rectangle is divided into two identical squares, the longer side must be twice the shorter side. Let the shorter side be $x$ cm. Therefore, the longer side is $2x$ cm. Given that the longer side is 6 cm, we have:
$2x = 6 \implies x = \frac{6}{2} = 3 \text{ cm}$
So, the shorter side is 3 cm.

Key Concept: Dimension constraints, multiple concepts

c) 8 cm
[Solution Description]
Step 1: Let square side be $s$. Then rectangle dimensions are $3s \times s$ (since squares fit along length).
Step 2: Area of rectangle = $3s \times s = 3s^2$.
Step 3: Given area is 192: $3s^2 = 192$.
Step 4: Solve: $s^2 = 64$ → $s = 8$ cm.
Step 5: Verify: Rectangle becomes 24cm × 8cm, allowing 3 squares of 8cm side.

Key Concept: Centered square construction, diagonal properties

d) $3\sqrt{2}$ cm
[Solution Description]
Step 1: The center of both shapes must match. For the maximum square, its corners will touch the rectangle's sides.
Step 2: Let $s$ be the side of the square. The distance from center to any corner of the square is $\frac{s\sqrt{2}}{2}$.
Step 3: The rectangle's half-diagonal is $\frac{\sqrt{12^2 + 6^2}}{2} = \frac{\sqrt{180}}{2} = \frac{6\sqrt{5}}{2} = 3\sqrt{5}$ cm.
Step 4: However, the limiting factor comes from the shorter dimension. The maximum distance from center to square's corner cannot exceed half the rectangle's height: $\frac{s\sqrt{2}}{2} \leq 3$.
Step 5: Solve for s: $s\leq \frac{6}{\sqrt{2}} = 3\sqrt{2} \approx 4.2426$ cm.

Key Concept: Rectangle Division into Identical Squares

a) 3s cm by s cm
[Solution Description]
To divide a rectangle into three identical squares:
1. The squares must be arranged in a row, so the rectangle's length will be 3s and height will be s.
2. Alternatively, they could be arranged in an L-shape or other configurations, but the simplest case is a single row.
3. Thus, the rectangle’s dimensions are length = 3s cm and height = s cm.

Key Concept: Rectangle Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us analyze the scenario step-by-step.
1. The rectangle ABCD is given with AB = 7 cm and BC = 4 cm. Points X and Y lie on sides AD and BC respectively.
2. If the distance of X from A is equal to the distance of Y from B, let’s say both distances are 'd' cm.
3. Therefore, the figure formed by points A, B, Y, X is a quadrilateral where AX = BY = d cm, and AB = 7 cm, XY becomes parallel to AB when distances are equal.
4. Since ABCD is a rectangle, angle at A and angle at B are right angles (90°).
5. Using Pythagoras' theorem in triangle AXY:
$XY = \sqrt{AB^2 + (AX - BY)^2}$, but since AX = BY, $XY = AB = 7$ cm.
6. This proves the Assertion is true.
7. The Reason states that in a rectangle opposite sides are equal and all angles are 90°. This is a basic property of a rectangle, which is also correct.
8. The Reason explains why the Assertion holds true because it's due to the properties mentioned in the Reason that the figure behaves as described.
Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Distance comparison

c) XY is less than AB
[Solution Description]
Since X is 1 cm away from A and Y is 1 cm away from B, the horizontal distance between them is |AB - (AX + BY)| = |7 cm - (1 cm + 1 cm)| = 5 cm. The vertical distance is BC = 4 cm. Thus, $XY = \sqrt{(5)^2 + (4)^2} = \sqrt{25 + 16} = \sqrt{41}$ cm ≈ 6.40 cm, which is less than AB (7 cm).

Key Concept: Exploring distances between two moving points in a rectangle

d) Assertion is false, but Reason is true.
[Solution Description]
In rectangle $ABCD$, let $AB = 7$ cm and $BC = 4$ cm. Let point $X$ be on side $AD$ and point $Y$ be on side $BC$. When both $X$ and $Y$ are placed at the same distance from $A$ and $B$ respectively, their horizontal separation is zero (directly opposite each other), and the vertical separation is equal to the length $BC = 4$ cm. Hence, the distance $XY$ is simply the length of side $BC$, which is 4 cm, not equal to the length of $AB$ (7 cm). Therefore, the assertion is false, but the reason is true because points $X$ and $Y$ are indeed closest when they are directly opposite each other.

Key Concept: Shortest and farthest distances in a rectangle

b) 4 cm
[Solution Description]
To find the minimum distance between X and Y, we can consider them moving along AB and CD. The shortest distance occurs when both X and Y are at the same distance from A and C respectively. Let's assume X is x cm from A, then Y must be x cm from C to minimize XY. The distance XY is given by the Pythagorean theorem: $XY = \sqrt{(6 - 2x)^2 + 4^2}$. For minimization, set $6 - 2x = 0$ → $x = 3$. Thus, $XY = \sqrt{0 + 16} = 4$ cm.

Key Concept: Distance between points on rectangle sides

a) $2\sqrt{10}$ cm
[Solution Description]
Draw rectangle ABCD with AB = 8 cm and AD = 6 cm. Point X is 3 cm from A on AB. Since AB is 8 cm, point X is 5 cm from B (8 cm - 3 cm). Point Y is 3 cm from C on CD.
The vertical distance between X and Y is the height of the rectangle, which is AD = 6 cm.
The horizontal distance between X and Y is the difference in their positions: (8 cm - 3 cm) - 3 cm = 2 cm.
Now, use the Pythagorean theorem to find XY:
$XY = \sqrt{(6 \text{ cm})^2 + (2 \text{ cm})^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ cm}$

Key Concept: Rectangle construction, diagonal and side comparison

c) 12 cm
[Solution Description]
Let the given side be $a = 5$ cm and the diagonal be $d = 13$ cm. We need to find the other side $b$.
Using the Pythagorean theorem for rectangles:
$d = \sqrt{a^2 + b^2}$. Squaring both sides gives $d^2 = a^2 + b^2$.
Substituting the known values:
$13^2 = 5^2 + b^2$ leads to $169 = 25 + b^2$.
Solving for $b^2$, we get $b^2 = 144$, so $b = \sqrt{144} = 12$ cm.

Key Concept: Rectangle diagonal properties, angle bisection

c) $53.13^\circ$ and $36.87^\circ$
[Solution Description]
First, calculate the length of the diagonal PR using the Pythagorean theorem:
$PR = \sqrt{PQ^2 + QR^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
Next, since PR is a diagonal of the rectangle, it bisects the right angle at P in a rectangle only if the rectangle is a square. Here, the sides are unequal (6 cm and 8 cm), so the diagonal does not bisect the angle P equally. Instead, we use trigonometry to find angles $c$ and $d$.
Angle $c$ is the angle between PQ and PR. We can find it using the tangent function:
$\tan(c) = \frac{QR}{PQ} = \frac{8}{6} \approx 1.333$, so $c \approx \arctan(1.333) \approx 53.13^\circ$.
Angle $d$ is the remaining part of the right angle at P, so $d = 90^\circ - c \approx 36.87^\circ$.

Key Concept: Distance between points in a rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the distance between points X and Y will be minimum when they are directly opposite each other on the rectangle's sides. This is true because the shortest distance between two parallel lines (as sides AB and CD are parallel in rectangle ABCD) is indeed the perpendicular distance between them. Therefore, when X is placed such that it is directly opposite Y, the line XY becomes perpendicular to both AB and CD, minimizing the distance between X and Y. Hence, the reason correctly explains the assertion.

Key Concept: Tracking distances tabularly

a) 7.21 cm
[Solution Description]
Distance of X from A = 2 cm.
Distance of Y from B = 4 cm.
Distance of X from B along AB = 8 cm - 2 cm = 6 cm.
This forms a right-angled triangle with:
Horizontal leg (along AB): 6 cm.
Vertical leg (along BC where Y is placed): 4 cm.
Using Pythagoras' theorem:
$XY = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$ cm.

Key Concept: Constructing Rectangles with Given Diagonal Angles

c) $60^{\circ}$ and $120^{\circ}$
[Solution Description]
In a rectangle, the diagonals bisect each other and are equal in length. When one diagonal divides the opposite angles into $60^{\circ}$ and $30^{\circ}$, it means that the angles at the vertices are split accordingly. Since the sum of angles around a point is $360^{\circ}$, the other angles formed by the intersection of the diagonals can be calculated as follows:
Let the angles at one vertex be $60^{\circ}$ and $30^{\circ}$. The remaining angles at the intersection point must also total $180^{\circ}$ on either side of the diagonal. Therefore, the other two angles must be $90^{\circ}$ each because $60^{\circ} + 30^{\circ} + 90^{\circ} + 90^{\circ} = 270^{\circ}$ (which is incorrect; the correct calculation should consider linear pairs). Actually, if one angle is $60^{\circ}$, its adjacent angle must be $120^{\circ}$ (linear pair), and similarly for the $30^{\circ}$ angle ($150^{\circ}$). But since diagonals intersect, they form vertical angles, so the opposite angles are equal. Thus, the four angles at the intersection are $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.
However, to match standard rectangle properties where diagonals create isosceles triangles, the angles should balance out to two pairs of equal angles, summing to $360^{\circ}$. The correct angles at the intersection are $60^{\circ}$ and $120^{\circ}$.

Key Concept: Diagonals of Rectangles, Construction Planning

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, let's analyze the assertion: Constructing a rectangle divided into 3 identical squares means the rectangle's sides will have a ratio of 1:3 or 3:1 (depending on orientation). For example, if each square has side length $x$, the rectangle could have dimensions $x \times 3x$. The diagonals ($d$) of such a rectangle can be calculated using Pythagoras' theorem:
$d = \sqrt{x^2 + (3x)^2} = \sqrt{x^2 + 9x^2} = \sqrt{10x^2} = x\sqrt{10}$
Since both diagonals are calculated the same way, they are equal.
Now, the reason states that diagonals of any rectangle are equal and bisect each other. This is a general property of rectangles, which holds true for all cases, including this specific construction.
However, while the assertion is correct, the reason does not explain why the diagonals are equal in this specific case (division into 3 squares). It only provides a general property. Hence, the reason is not the correct explanation of the assertion.

Key Concept: Farthest Distance in Divided Rectangle

c) XY is less than AB.
[Solution Description]
Given AB = 10 cm and AX = BY = 3 cm.
The remaining part XB = AB - AX = 10 - 3 = 7 cm.
Since Y is placed 3 cm from B, XY = XB - BY = 7 - 3 = 4 cm.
So XY is less than AB ($4 < 10$).

Key Concept: Breaking Rectangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To divide a rectangle into two identical squares, the rectangle's length must be exactly twice its width. This ensures that when the rectangle is split vertically or horizontally, each resulting part is a square with sides equal to the width of the original rectangle. The reason correctly explains why this dimensional ratio (2:1) is necessary for achieving identical squares.

Key Concept: Constructing a square within a rectangle

b) 4 cm
[Solution Description]
To find the maximum side length of the square, we need to ensure the square fits entirely within the rectangle while sharing the same center. The diagonal of the square should not exceed the dimensions of the rectangle.
The rectangle's dimensions are 8 cm (length) and 4 cm (height). The center of the rectangle is at (4 cm, 2 cm). For the square to fit, the distance from its edges to the center must be equal in both x and y directions.
Let the side length of the square be $s$. The distance from the center to any edge of the square will be $\frac{s}{2}$.
For the square to fit vertically: $\frac{s}{2} \leq 2$ cm $\Rightarrow s \leq 4$ cm.
For the square to fit horizontally: $\frac{s}{2} \leq 4$ cm $\Rightarrow s \leq 8$ cm.
The limiting dimension here is the height (4 cm), so the maximum possible side length is 4 cm.

Key Concept: Rectangle Division into Identical Squares

c) 5 cm x 10 cm
[Solution Description]
To divide a rectangle into three identical squares, the length must be exactly three times the width. Alternatively, the width could be three times the length if reversed. However, if the ratio of length to width is not 3:1 or 1:3, it's impossible to divide it into three identical squares. Let’s evaluate each option:
a) 6 cm x 2 cm: This can be divided horizontally or vertically into three 2 cm squares.
b) 9 cm x 3 cm: This can be divided into three 3 cm squares.
c) 5 cm x 10 cm: Here, neither dimension is a multiple of three of the other (5/10 ≈ 0.5), so division is impossible.
d) 12 cm x 4 cm: This can be divided into three 4 cm squares.
Thus, option c) is the correct choice.

Key Concept: Rotated Squares and Rectangles

c) It remains a square after rotation.
[Solution Description]
According to the syllabus, rotating a square does not change its lengths or angles. Hence, all sides remain equal, and all angles remain $90^\circ$, meaning it still satisfies the properties of a square.

Key Concept: Exploring Diagonals of Rectangles and Squares

c) PR and QS are equal in length.
[Solution Description]
The syllabus states that in a rectangle, the diagonals are equal in length. Therefore, PR and QS must have the same length.

Key Concept: Shadings - Area Calculation

a) 48 cm$^2$
[Solution Description]
First, calculate the area of one square. The area of a square is given by $side^2$. For one square with side 4 cm:
$Area = 4 \times 4 = 16 \text{ cm}^2$
Since there are three such squares, the total area is:
$Total\,Area = 3 \times 16 = 48 \text{ cm}^2$

Key Concept: Four-Sided Figures

b) Rhombus
[Solution Description]
While squares have all sides equal and all angles as 90 degrees, a rhombus also has all sides equal but its angles are not necessarily 90 degrees. Therefore, a rhombus fits this description.

Key Concept: Square with Curves, Constructing a Square within a Rectangle

c) $2\sqrt{5}$ cm
[Solution Description]
The rectangle has length 8 cm and width 4 cm, so its center is at (4 cm, 2 cm). The square's corners touch the midpoints of the rectangle's sides. The distance from the center to any midpoint of the shorter sides (4 cm) is 2 cm (half of the width), and to any midpoint of the longer sides (8 cm) is 4 cm (half of the length). The side length of the square is the hypotenuse of a right triangle formed by these distances. Using the Pythagorean theorem: $l = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ cm.

Key Concept: Rotated Squares, Properties of a Square

c) $1:2$
[Solution Description]
Let the side length of the original square be $s$. After rotating the square by $45^\circ$, the vertices of the rotated square are now aligned with the diagonals. The smallest enclosing square will have side lengths equal to the diagonal of the original square, which is $s\sqrt{2}$. The area of the original square is $s^2$, and the area of the enclosing square is $(s\sqrt{2})^2 = 2s^2$. The ratio is $\frac{s^2}{2s^2} = \frac{1}{2}$.

Key Concept: Properties of Squares and Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that in a rectangle, the diagonals are always equal in length. This is true because rectangles have opposite sides equal, and by the properties of quadrilaterals, this ensures their diagonals are equal.
The reason provided is that a rectangle has all angles equal to $90^{\circ}$. While this is true, it does not directly explain why the diagonals are equal in length. The correct explanation would involve the congruent triangles formed by the diagonals due to the equal opposite sides and right angles.
Thus, both assertion and reason are true, but the reason does not correctly explain the assertion.

Key Concept: Hole Placement in Squares

c) At equal distances from each other
[Solution Description]
For uniform distribution without overlapping, holes must be placed at equal distances from each other and from the square's edges. Common placements include the corners or midpoints of the sides, ensuring symmetry and spacing.

Key Concept: Square with Curves, Arc Construction

c) $64 + 16\pi$
[Solution Description]
First, find the area of the square: $8 \times 8 = 64$ cm$^2$. Next, since there are four quarter-circles, together they form a full circle. The radius of each quarter-circle is half the side length, so $r = 4$ cm. The area of one full circle is $\pi r^2 = 16\pi$ cm$^2$. Adding the square's area and the circle's area gives $64 + 16\pi$ cm$^2$.

Key Concept: Square with Curves - Measurement

b) 1 cm away
[Solution Description]
For uniform bulging curves, no point on the arc should extend further than 1 cm from the original straight side, maintaining symmetry. This keeps the overall curved shape balanced while still clearly showing the bulge effect.

Key Concept: Diagonals of Rectangles and Squares

d) Assertion is false, but Reason is true.
[Solution Description]
In a rectangle, the diagonals are equal in length and bisect each other. This means that when the diagonals intersect, they divide each other into two equal parts.
Consider a rectangle PQRS with diagonals PR and QS intersecting at point O. Since the diagonals are equal and bisect each other, PO = OR = QO = OS. Additionally, all angles in a rectangle are $90^\circ$.
Now, let's analyze the triangles formed by the diagonals:
- Triangle POQ: It has sides PO and QO (equal in length as diagonals bisect each other), angle POQ is some angle $\theta$ (not necessarily $90^\circ$).
- Triangle QOR: It has sides QO and OR (equal in length), angle QOR is $(180^\circ - \theta)$ because POQ and QOR lie on a straight line.
However, for the triangles to be congruent, their corresponding sides and angles must be equal. Here, only the sides are equal, but the included angle between them differs unless $\theta = 90^\circ$, which would imply the diagonals are perpendicular. But in a general rectangle, the diagonals are not perpendicular (except in squares). Thus, the Assertion that the four triangles are congruent is not generally true for rectangles.
The Reason states that a diagonal divides the rectangle into two congruent right-angled triangles, which is correct. For example, triangle PQR is congruent to triangle PSR by the Side-Angle-Side (SAS) criterion (since PQ = SR, PR is common, and angle PQR = angle SRP = $90^\circ$).
But the Reason does not correctly explain why the diagonals would form four congruent triangles, as this requires specific conditions (like diagonals being perpendicular, as in squares).
Therefore, the Assertion is false (as it overgeneralizes beyond squares), but the Reason is true (as it correctly describes how a single diagonal divides the rectangle into two congruent triangles).

Key Concept: Special Case: Square

c) The diagonals bisect each other at 90 degrees and are equal in length.
[Solution Description]
In a square, both diagonals are equal in length and bisect each other at 90 degrees. They also divide the opposite angles equally (each into 45 degrees).
Length of diagonal calculation:
Diagonal = $\sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$ cm.

Key Concept: Diagonal Length, Angle Division

c) 90°
[Solution Description]
Step 1: The diagonals of a rectangle are equal in length and bisect each other. So, PO = OR = QO = OS.
Step 2: Calculate the diagonal PR using the Pythagorean theorem: $PR = \sqrt{PQ^2 + QR^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
Step 3: Triangles POQ and QOR are congruent because PO = OR, PQ = SR, and $\angle PQR = \angle SRQ = 90°$. Thus, $\angle POQ = \angle QOR = 90°$.
Step 4: Therefore, angle POS is supplementary to angle POQ, so $\angle POS = 180° - 90° = 90°$.

Key Concept: Angle Division by Diagonal

d) Cannot determine without measurements
[Solution Description]
A diagonal of a rectangle bisects the opposite angles when it is a square. However, in a general rectangle, the angles formed may not be equal unless specified otherwise.

Key Concept: Diagonal properties, Pythagorean theorem

a) 5 cm
[Solution Description]
A rectangle's diagonal divides it into two right-angled triangles. Using the Pythagorean theorem for a right triangle with sides 3 cm and 4 cm, the diagonal (hypotenuse) $d$ is calculated as $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ cm.

Key Concept: Diagonals divide opposite angles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a square, all sides are equal, and all angles are $90^\circ$. The diagonals are equal in length and intersect at $90^\circ$. Each diagonal divides the opposite angles into two equal angles of $45^\circ$ because the diagonals are angle bisectors in a square. Therefore, both Assertion and Reason are true, and Reason correctly explains why the diagonals divide the opposite angles equally.

Key Concept: Square properties, Angle equality

c) $x = y$
[Solution Description]
Step 1: In a square, diagonals bisect the angles. Thus, angle $x = 45^\circ$ because it is half of angle A ($90^\circ$).
Step 2: Similarly, angle $y$ is also $45^\circ$ as it is half of angle B ($90^\circ$).
Step 3: Both angles $x$ and $y$ are equal because diagonals in a square bisect the angles equally.

Key Concept: Rectangle diagonals, Side lengths

a) $\frac{3}{4}$
[Solution Description]
Step 1: The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side.
Step 2: For angle $m$, if $\tan(m) = \frac{3}{4}$, this suggests that the sides opposite and adjacent to angle $m$ are in ratio 3:4.
Step 3: The total sides are LM = 6 (adjacent to angle L) and MN = 8 (opposite to angle L when considering triangle LMN).
Step 4: This means angle $m$ corresponds to the angle whose tangent is $\frac{MN}{LM} = \frac{8}{6} = \frac{4}{3}$ in triangle LMN, leading to confusion. Instead, refer to the angle between diagonal LN and side LM.
Step 5: The correct interpretation is $\tan(m) = \frac{ON}{LO}$ in triangle LON, where ON = 8 and LO = 6, so $\tan(m) = \frac{8}{6} = \frac{4}{3}$. But the given $\tan(m) = \frac{3}{4}$ contradicts this, implying a different configuration.
Step 6: Given the contradiction, reassess using the entire rectangle: $\tan(\text{angle between LN and LM}) = \frac{MN}{LM} = \frac{8}{6} = \frac{4}{3}$.
Step 7: Therefore, angle $n$ would be such that $\tan(n) = \frac{LM}{MN} = \frac{6}{8} = \frac{3}{4}$.

Key Concept: Square Diagonal Angles

b) $45^\circ$
[Solution Description]
In a square, all angles are $90^\circ$. A diagonal divides any vertex angle into two equal angles. Therefore, each smaller angle will be:
$\frac{90^\circ}{2} = 45^\circ$

Key Concept: Diagonal Angles in a Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a square, all sides are equal and all angles are $90^\circ$. The diagonals of a square bisect each other at $90^\circ$ and also bisect the vertex angles. This means that each diagonal divides the $90^\circ$ angle at each vertex into two angles of $45^\circ$ each. Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion because the property of diagonals being angle bisectors directly leads to the division of opposite angles into equal parts.

Key Concept: Angles formed by diagonals in a rectangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In a rectangle, the diagonals are equal in length and bisect each other. This property ensures that the diagonals divide the rectangle into four triangles. Each pair of opposite angles in the rectangle is split by a diagonal into two smaller angles. Due to the symmetry and properties of rectangles, these smaller angles are equal. For example, if one angle is divided into $c$ and $d$, then $c = d$. Similarly, for the other angles divided by the diagonals. Therefore, the assertion is true. The reason states that the diagonals are equal in length and bisect each other, which is also true. However, while this property contributes to the overall symmetry, it does not directly explain why the diagonals divide the opposite angles into equal parts. Thus, the reason is true but not the correct explanation of the assertion.

Key Concept: Diagonals in Rectangles

c) The angles $c$ and $d$ formed by the diagonal PR are equal.
[Solution Description]
In a rectangle, the diagonals are equal in length and bisect each other. This means they divide each other into two equal parts. The angles created where the diagonals intersect will form pairs of vertically opposite angles, which are always equal. Additionally, the diagonal divides the opposite angles of the rectangle into two equal parts. So, for example, if PR divides $\angle P$ into $c$ and $d$, then $c = d$.

Key Concept: Constructing a Rectangle with Given Diagonal Angles

a) The other angles are divided into $50^\circ$ and $40^\circ$
[Solution Description]
Since diagonals in a rectangle divide opposite angles equally, if one angle is divided into $50^\circ$ and $40^\circ$, the opposite angle must also be divided into $50^\circ$ and $40^\circ$. This ensures symmetry in the rectangle.

Key Concept: Rectangle construction with given side and diagonal

c) 12 cm
[Solution Description]
Here's how to approach this problem:
1. Let the rectangle be ABCD with AB = 5 cm and diagonal AC = 13 cm.
2. We need to find the length of side BC.
3. Using Pythagoras' theorem for triangle ABC: $AB^2 + BC^2 = AC^2$
4. Plugging in known values: $5^2 + BC^2 = 13^2$
5. Calculating squares: $25 + BC^2 = 169$
6. Subtracting: $BC^2 = 169 - 25 = 144$
7. Taking square root: $BC = \sqrt{144} = 12$ cm
Hence, the other side measures 12 cm.

Key Concept: Equidistant Points Geometry

c) No such point exists
[Solution Description]
Let the treasure's coordinates be $(a, a)$ since it lies on $y = x$. The distance from the treasure to $T_1$ is $\sqrt{(a-2)^2 + (a-7)^2}$, and the distance to $T_2$ is $\sqrt{(a-6)^2 + (a-3)^2}$. Setting them equal and squaring both sides gives $(a-2)^2 + (a-7)^2 = (a-6)^2 + (a-3)^2$. Expanding both sides: $a^2 -4a +4 + a^2 -14a +49 = a^2 -12a +36 + a^2 -6a +9$. Simplifying, $-18a +53 = -18a +45$. This leads to $53 = 45$, which is not possible. Therefore, there is no solution, meaning the treasure does not exist on the line $y = x$.

Key Concept: Circle Construction, Points Equidistant from Two Given Points

b) Two arcs
[Solution Description]
To locate point $A$ equidistant from points $B$ and $C$, we need to draw two arcs: one centered at $B$ with radius 5 cm and another centered at $C$ with radius 5 cm. The intersection of these two arcs will give the position of $A$. Drawing only part of the circles (arcs) is sufficient since we are interested in their intersection point.

Key Concept: Points Equidistant from Two Given Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that point A can be found by drawing arcs (not full circles) of radius 5 cm from points B and C. This is true because the intersection point of these arcs will be at a distance of 5 cm from both B and C. The reason explains why this method works: the intersection point is equidistant from both B and C. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Equidistant points, Arc construction

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that there is exactly one point equidistant from B and C when constructed using equal-radius arcs. However, this depends on the configuration:
1) If the distance between B and C is less than twice the radius used for the arcs (i.e., $d 2r$, the circles do not intersect, so no equidistant point exists under the given radius constraint.
The reason claims that equal-radius circles centered at B and C always intersect at one point. This is false because they can intersect at 0, 1, or 2 points depending on the distance between B and C relative to the radius.
Thus, the assertion is false (as it disregards multiple possible cases), while the reason is also false (as it oversimplifies the intersection scenario).

Key Concept: Equidistant Points Construction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The construction involves drawing an arc that is part of the house-shaped figure. For the arc to correctly connect points B and C, point A must be equidistant from both B and C because the arc is drawn using the compass set to this equal distance (5 cm). Thus, the assertion is true, and the reason correctly explains why point A must be equidistant from B and C.

Key Concept: Points Equidistant Using Arcs

c) To locate the point equidistant from both points
[Solution Description]
The arcs help find the intersection point which is at equal distances from both given points. This method uses the concept that any point on the arc from one center is at a fixed distance (radius) from that center.

Key Concept: Wave Construction

b) 4 cm
[Solution Description]
To draw a half-circle wave on AB = 8 cm:
1. For a half-circle, the diameter equals the length of the wave segment.
2. Since the first half-circle spans the entire 8 cm line, its diameter is 8 cm.
3. The radius is half the diameter: $8/2 = 4$ cm.
4. Thus, AX (distance from A to center X) must equal the radius: 4 cm.

Key Concept: Intersection of arcs, Equidistant points

c) 2
[Solution Description]
The points equidistant from P and Q lie on the perpendicular bisector of PQ.
Step 1: The midpoint M divides PQ into two parts of 4 cm each.
Step 2: The perpendicular distance from M to either intersection point can be found using the Pythagorean theorem:
$6^2 = x^2 + 4^2 \implies 36 = x^2 + 16 \implies x^2 = 20 \implies x = \sqrt{20} \approx 4.47 \text{ cm}$
There are two points of intersection, one on each side of the line PQ, each at $\sqrt{20}$ cm from M.