Key Concept: Square Area, Rectangle Perimeter

b) 100 cm$^2$
[Solution Description]
1. First, find the perimeter of the rectangle: $P_{rect} = 2(l + w) = 2(12 + 8) = 40$ cm.
2. Since the perimeter of the square is equal to that of the rectangle, $P_{square} = 40$ cm.
3. For a square, the perimeter formula is $P = 4s$, where $s$ is the side length. Solving for $s$ gives $s = \frac{40}{4} = 10$ cm.
4. The area of the square is then $A = s^2 = 10^2 = 100$ cm$^2$.

Key Concept: Properties of Rectangle and Square

b) Both have diagonals that are equal and bisect each other.
[Solution Description]
The properties common to both rectangles and squares are:
- All angles are $90^\circ$.
- Opposite sides are equal and parallel.
- Diagonals are equal in length and bisect each other.
Only option b) satisfies all these conditions.

Key Concept: Properties of a Rectangle

b) The diagonals are equal and bisect each other
[Solution Description]
The diagonals of a rectangle are equal and bisect each other, as stated in the properties of a rectangle.

Key Concept: Rectangle properties, Diagonal congruence

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that $\triangle AOB \cong \triangle COD$ by SAS congruence in rectangle $ABCD$.
Step 1: Properties of rectangle diagonals - Diagonals are equal in length ($AC = BD$) and bisect each other at point $O$. So, $AO = OC$ and $BO = OD$.
Step 2: Since opposite sides of rectangle are equal, $AB = CD$.
Step 3: Included angle between sides $AO$ and $BO$ is $\angle AOB$, which is equal to $\angle COD$ (vertically opposite angles).
Thus, by SAS criteria ($AO = OC$, $\angle AOB = \angle COD$, $BO = OD$), $\triangle AOB \cong \triangle COD$.
The reason correctly states that diagonals are equal and bisect each other, but this property alone does not directly explain why SAS congruence holds because it doesn't mention the vertically opposite angles or the equality of opposite sides. Hence, both assertion and reason are true, but reason is not the correct explanation.

Key Concept: Square vs Rectangle

a) A square has all sides equal, while a rectangle does not.
[Solution Description]
A square is a special type of rectangle where all sides are equal. While all squares are rectangles, not all rectangles are squares. Rectangles only require opposite sides to be equal and all angles to be $90^\circ$.

Key Concept: Properties of Squares

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 square bisect each other at $90^\circ$, which is true because the diagonals of a square are perpendicular bisectors. The reason states that the diagonals are equal in length and bisect the angles, which is also true as it's a property of squares. However, the reason does not correctly explain why the diagonals bisect each other at $90^\circ$ (it only mentions their equality and angle bisection). Thus, both statements are true, but the reason is not the correct explanation for the assertion.

Key Concept: Square properties

c) All its sides are equal.
[Solution Description]
A square has all the properties of a rectangle (right angles, equal opposite sides, and equal diagonals that bisect each other), but additionally, all its sides are equal, and its diagonals are perpendicular and bisect the angles.

Key Concept: Diagonals of a Rectangle

b) 5 cm
[Solution Description]
The diagonals of a rectangle are equal in length and bisect each other. Given $PR = 10$ cm, which means each half-diagonal $PO$ and $OR$ will be half of $PR$.
$PO = \frac{PR}{2} = \frac{10}{2} = 5 \text{ cm}$
Therefore, the correct answer is 5 cm.

Key Concept: Diagonals of a Square

c) 90°
[Solution Description]
In a square, the diagonals bisect each other at right angles. Therefore, the angle between the diagonals is 90°.

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 is true because the diagonals of a rectangle are always equal in length, as proven by the properties of rectangles. The reason is also true because the diagonals of a rectangle bisect each other at their midpoints. Furthermore, the reason correctly explains why the assertion is true, as both properties (equality and bisecting each other) are fundamental to the definition and properties of a rectangle.

Key Concept: Diagonal Properties of Square and Rhombus, Conceptual Complexity

c) 32
[Solution Description]
Step 1: In a square, diagonals are equal and bisect each other at right angles. So, $PR = QS$ and $PT = TR = QT = TS = 4\sqrt{2}$. Thus, full diagonal $PR = 8\sqrt{2}$.
Step 2: Let side of square be s. By Pythagoras' theorem in triangle PQR: $PR^2 = PQ^2 + QR^2$. But since all sides are equal, $PQ = QR = s$. So, $(8\sqrt{2})^2 = s^2 + s^2 = 2s^2$.
Step 3: Simplify: $128 = 2s^2$ → $s^2 = 64$ → $s = 8$.
Step 4: Perimeter = $4 \times 8 = 32$.

Key Concept: Angles Between Diagonals

c) $90^\circ$
[Solution Description]
In a square, the diagonals bisect each other at right angles ($90^\circ$). This is a unique property of squares compared to other rectangles.

Key Concept: Quadrilaterals, Diagonals of a Rectangle and Square

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that if diagonals of a quadrilateral bisect each other at $90^\circ$, the quadrilateral must be a rhombus or square. This is true because a rhombus and square have diagonals intersecting at right angles, while rectangles without equal adjacent sides (non-squares) do not necessarily meet this condition. However, the reason states that a quadrilateral with diagonals bisecting each other at $90^\circ$ has all sides equal, which is only true for rhombuses and squares but not for general quadrilaterals. Therefore, the assertion is true, but the reason is false because it incorrectly generalizes the property to all quadrilaterals instead of specific ones like rhombuses and squares.

Key Concept: Diagonals, Angles in Quadrilaterals

c) $90^\circ$
[Solution Description]
Since $AO = BO = CO = DO$, the triangles $AOB$, $BOC$, $COD$, and $DOA$ are all isosceles. Given that the angle between the diagonals ($\angle AOB$) is $60^\circ$, we can find the base angles of $\triangle AOB$ as follows:
Let the base angles be $a$.
$2a + 60 = 180 \implies 2a = 120 \implies a = 60.$
So, $\angle OAB = \angle OBA = 60^\circ$.
Similarly, in $\triangle BOC$, $\angle BOC = 120^\circ$ (since angles around point $O$ add to $360^\circ$). The base angles ($b$) of $\triangle BOC$ are:
$2b + 120 = 180 \implies 2b = 60 \implies b = 30.$
Thus, $\angle ABC = \angle ABO + \angle OBC = 60^\circ + 30^\circ = 90^\circ$.

Key Concept: Assertion-Reason

d) Assertion is false, but Reason is true.
[Solution Description]
In a rectangle, the diagonals are equal and bisect each other, but they do not necessarily intersect at $90^\circ$ unless it is a square. Thus, Assertion (A) is false. However, Reason (R) is true because diagonals of a rectangle are equal and bisect each other.

Key Concept: Angles between diagonals of a rectangle

d) Assertion is false, but Reason is true.
[Solution Description]
In a rectangle, diagonals are equal in length and bisect each other, forming two pairs of isosceles triangles. For any angle $x$ between the diagonals, the base angles of these isosceles triangles can be calculated as $a = 90^\circ - \frac{x}{2}$ and $b = \frac{x}{2}$. The sum of the angles around the point of intersection must satisfy $a + b = 90^\circ$, leading to all four angles being $90^\circ$. However, while the diagonals of a rectangle are equal and bisect each other (Reason), this property alone does not necessarily imply that they intersect at $90^\circ$ (Assertion is false). Only in special cases like squares or rhombuses do diagonals intersect at $90^\circ$.

Key Concept: Quadrilateral Properties, Congruence

a) Yes, because all sides and angles are equal.
[Solution Description]
A quadrilateral with four equal sides is a rhombus. If one angle is $90°$, then by properties of a rhombus, all angles must be $90°$ (since opposite angles are equal and consecutive angles are supplementary). Hence, it is a square.

Key Concept: Construction of Square

c) $3\sqrt{2}$ cm
[Solution Description]
For a square, the relationship between the diagonal ($d$) and the side length ($s$) is given by $d = s\sqrt{2}$. Rearranging this, $s = \frac{d}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ cm.

Key Concept: Properties of Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In a rectangle, the diagonals are always equal in length. This is a proven property of rectangles. The diagonals also bisect each other, but this property alone does not explain why they are equal in length. Both statements are true, but the Reason does not correctly explain the Assertion.

Key Concept: Properties of square diagonals

c) $90^\circ$
[Solution Description]
In a square, the diagonals bisect each other at $90^\circ$. Therefore, the angle between the diagonals is $90^\circ$.

Key Concept: Rectangle Diagonals, Angle between Diagonals

c) $90^\circ$
[Solution Description]
In a quadrilateral with all angles equal to $90^\circ$ (rectangle), the diagonals are equal in length and bisect each other. The angle between the diagonals can be determined by considering two congruent right triangles formed by the diagonals. Let the point of intersection be O. Using properties of rectangles, $\angle AOB$ and $\angle BOC$ are equal and supplementary ($\angle AOB + \angle BOC = 180^\circ$). Hence, each angle must be $90^\circ$.

Key Concept: Properties of Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
From the syllabus, we know that if the diagonals of a quadrilateral are equal and bisect each other, then it must be a rectangle. So, the assertion (A) is true. However, the reason (R) states that the diagonals bisect at $90^\circ$, which is only true for squares, not all rectangles. Therefore, while both statements are individually correct, the reason does not correctly explain the assertion.

Key Concept: Definition of a Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a quadrilateral with diagonals equal in length and bisecting each other is a rectangle, which aligns with the alternative definition of a rectangle mentioned in the syllabus. The reason correctly states that in a rectangle, the diagonals are equal and bisect each other. Moreover, the reason provides the justification for why such a quadrilateral must be a rectangle, making it the correct explanation of the assertion.

Key Concept: Rectangle Definition

c) 8 cm
[Solution Description]
In a rectangle, both diagonals are equal in length. Since one diagonal is given as 8 cm, the other diagonal must also be 8 cm.

Key Concept: Properties of Squares and Rectangles, Geometric Reasoning

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A square has diagonals that bisect each other at $90^\circ$ and are equal in length (Property 4 of squares). Since all sides of a square are equal (Property 1), the four triangles formed by the intersection of diagonals have two sides equal (half the diagonal) and the included angle as $90^\circ$. By SAS congruency, these triangles are congruent. Hence, both Assertion and Reason are true, and the Reason explains why the Assertion is true.

Key Concept: Quadrilateral Construction

c) Square
[Solution Description]
When the diagonals of a quadrilateral are equal in length and bisect each other at $90^\circ$, the quadrilateral must be a square. This is because such properties are specific to squares and not to other quadrilaterals like rectangles or rhombuses unless all sides are also equal.

Key Concept: Rectangle Construction

d) Connect sticks at midpoints and join ends with thread
[Solution Description]
To construct a right angle using two equal-length sticks and thread:
1) Join the sticks at their midpoints (5 cm from either end) forming an X-shape.
2) Use the thread to connect all four ends of the sticks sequentially - this forms a quadrilateral.
3) Since diagonals are equal and bisect each other, the quadrilateral is a rectangle (Property 4).
4) All angles in a rectangle are $90°$, so the angle formed by adjacent sides is a right angle.
Key points:
- Equal-length sticks act as diagonals.
- Bisecting them ensures rectangle formation.
- Rectangle guarantees right angles at vertices.

Key Concept: Properties of Squares

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 square bisect each other at $90^\circ$, which is true as per Property 4 of squares. The reason states that a square is a special type of rectangle with all sides equal, which is also true (Property 1 of squares). However, the reason does not explain why the diagonals bisect at $90^\circ$; it only describes another property of squares. Therefore, both statements are true, but the reason is not the correct explanation of the assertion.

Key Concept: Construction of Parallelograms and Relation to Rectangles

a) 6.24 cm
[Solution Description]
To find the shorter diagonal of the parallelogram, we use the Law of Cosines.
Step 1: Let the sides be $a = 5$ cm and $b = 7$ cm, with the included angle $\theta = 60^\circ$.
Step 2: The shorter diagonal $d_1$ can be calculated using:
$d_1 = \sqrt{a^2 + b^2 - 2ab \cos(\theta)}$
Step 3: Substituting the values:
$d_1 = \sqrt{5^2 + 7^2 - 2 \times 5 \times 7 \cos(60^\circ)} = \sqrt{25 + 49 - 70 \times 0.5} = \sqrt{74 - 35} = \sqrt{39} \approx 6.24 \text{ cm}$
Therefore, the length of the shorter diagonal is approximately $6.24$ cm.

Key Concept: Properties of a Rectangle

c) They are equal in length and bisect each other
[Solution Description]
The diagonals of a rectangle have two key properties: they are equal in length and bisect each other. This means they divide each other into two equal parts at the point of intersection.

Key Concept: Special Case: Square

d) Square
[Solution Description]
A quadrilateral with all sides equal and all angles equal to $90^\circ$ satisfies the definition of a square. Hence, it is a square.

Key Concept: Properties of Rectangles

c) The diagonals are equal in length.
[Solution Description]
Since all angles are $90^\circ$, the quadrilateral is a rectangle. By Property 2, opposite sides must be equal and parallel (Property 3). Therefore, the correct statement is that its diagonals are equal in length.

Key Concept: Properties of a rectangle

c) Rectangle
[Solution Description]
A quadrilateral with all four angles equal must have each angle equal to $90^\circ$ because the total sum of angles is $360^\circ$. Therefore, such a quadrilateral is a rectangle.

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let's verify both statements:
1. Assertion: In a parallelogram, adjacent angles are supplementary (sum to $180^\circ$) because they are consecutive interior angles formed by a transversal intersecting parallel sides. This is true.
2. Reason: Opposite angles in a parallelogram are equal due to its properties. This is also true.
However, the equality of opposite angles does not explain why adjacent angles are supplementary. They are separate properties derived from different geometric principles.

Key Concept: Quadrilateral Angle Sum, Algebraic Equation

c) $76^\circ$
[Solution Description]
Let angle B be $x$. Then:
- Angle A = $2x$,
- Angle C = $x + 20^\circ$,
- Angle D = $x - 40^\circ$.
The sum of all angles is $360^\circ$, so:
$2x + x + (x + 20^\circ) + (x - 40^\circ) = 360^\circ$.
Combine like terms: $5x - 20^\circ = 360^\circ$.
Add $20^\circ$ to both sides: $5x = 380^\circ$.
Divide by 5: $x = 76^\circ$.
Therefore, angle B is $76^\circ$.

Key Concept: Diagonals of a quadrilateral

b) Opposite sides are equal
[Solution Description]
If the diagonals of a quadrilateral bisect each other, then the opposite sides of the quadrilateral are equal in length. This property is characteristic of parallelograms, rectangles, rhombuses, and squares.

Key Concept: Angles formed by diagonals in a quadrilateral

c) 360°
[Solution Description]
The sum of angles in the first triangle is $40° + 60° + 80° = 180°$.
The sum of angles in the second triangle is $50° + 70° + 60° = 180°$.
Adding these gives the total sum of angles in the quadrilateral: $180° + 180° = 360°$.

Key Concept: Sum of angles in a quadrilateral

b) $360^\circ$
[Solution Description]
The sum of the interior angles of any quadrilateral is always $360^\circ$. This is because splitting the quadrilateral into two triangles (by drawing a diagonal) gives two triangles, each with angles summing to $180^\circ$.

Key Concept: Sum of angles in a quadrilateral, properties of diagonals

c) 110°
[Solution Description]
Step 1: The sum of all angles in a quadrilateral is 360°.
Step 2: Given angles A, B, and C are 80°, 100°, and 70° respectively.
Step 3: Calculate the sum of known angles: $80° + 100° + 70° = 250°$.
Step 4: Subtract the sum from 360° to find angle D: $360° - 250° = 110°$.

Key Concept: Sum of angles in a quadrilateral

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, let's calculate the fourth angle:
Sum of given angles = $80° + 100° + 90° = 270°$.
Since the total sum of angles in a quadrilateral is $360°$, the fourth angle must be $360° - 270° = 90°$. This matches the assertion.
The reason states that the sum of interior angles in any quadrilateral is $360°$, which is true and correctly explains why the fourth angle must be $90°$ in this case.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Construction of quadrilaterals

c) The fourth angle must be a right angle
[Solution Description]
The sum of angles in any quadrilateral is $360^\circ$. If a quadrilateral has three right angles ($90^\circ$ each), their sum would be $270^\circ$. Therefore, the fourth angle must be:
$360^\circ - 270^\circ = 90^\circ$
Hence, if a quadrilateral has three right angles, the fourth angle must also be a right angle.

Key Concept: Quadrilateral angles, diagonal properties

c) The fourth angle must be $90^\circ$
[Solution Description]
Let's analyze the given quadrilateral PQRS with three right angles (let's say $\angle P = \angle Q = \angle R = 90^\circ$) and diagonals intersecting at O forming angle POQ = $60^\circ$.
Since sum of angles in any quadrilateral is $360^\circ$, we have:
$\angle P + \angle Q + \angle R + \angle S = 360^\circ$
Substituting known values:
$90^\circ + 90^\circ + 90^\circ + \angle S = 360^\circ$
$270^\circ + \angle S = 360^\circ$
Therefore, $\angle S = 90^\circ$.
The information about diagonal intersection angle doesn't affect the angle sum property. Hence, the fourth angle must also be $90^\circ$.

Key Concept: Properties of quadrilaterals

c) $\angle S$ is a right angle
[Solution Description]
Since the sum of angles in a quadrilateral is $360^\circ$, we can calculate $\angle S$ as follows:
$\angle P + \angle Q + \angle R + \angle S = 360^\circ$
Substituting the given angles:
$90^\circ + 90^\circ + 90^\circ + \angle S = 360^\circ$
$270^\circ + \angle S = 360^\circ$
$\angle S = 90^\circ$
Thus, $\angle S$ must also be $90^\circ$.

Key Concept: Angles in a Quadrilateral, Sum of Angles

d) Assertion is false, but Reason is true.
[Solution Description]
To verify, assume a quadrilateral has three right angles ($90^{\circ}$ each). Let the fourth angle be $x$. According to the property of quadrilaterals, the sum of all four angles must be $360^{\circ}$. Thus:
$90^{\circ} + 90^{\circ} + 90^{\circ} + x = 360^{\circ}.$
Simplifying:
$270^{\circ} + x = 360^{\circ},$
which gives:
$x = 90^{\circ}.$
Therefore, if three angles are $90^{\circ}$, the fourth must also be $90^{\circ}$. Hence, it is impossible for a quadrilateral to have exactly three right angles and one non-right angle. The Assertion is false, but the Reason is true.

Key Concept: Properties of Parallelograms

c) $115^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal and adjacent angles are supplementary. Given $\angle A = 65^\circ$, $\angle B$ will be $180^\circ - 65^\circ = 115^\circ$. Since $\angle D$ is opposite to $\angle B$, $\angle D = 115^\circ$.

Key Concept: Diagonals of Rhombus

c) $10$ cm
[Solution Description]
The diagonals of a rhombus bisect each other at $90^\circ$. So, half-diagonals are $6$ cm and $8$ cm. Using the Pythagorean theorem:
Side length = $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.

Key Concept: Properties of a Parallelogram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a parallelogram, by definition, both pairs of opposite sides are parallel. The assertion states that the opposite sides are equal and parallel, which is true as per the properties of a parallelogram. The reason correctly defines a parallelogram as a quadrilateral with both pairs of opposite sides parallel. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Special Cases: Rhombus, Rectangle, Square

c) All sides are equal
[Solution Description] A rhombus has all sides equal, but its angles are not necessarily $90^\circ$. A rectangle has all angles equal to $90^\circ$, but its sides are not necessarily equal. Therefore, the property that applies to all rhombuses but not necessarily to all rectangles is "All sides are equal".

Key Concept: Congruent triangles in a parallelogram

c) $\Delta ABD \cong \Delta BDC$
[Solution Description]
In parallelogram ABCD, opposite sides are equal ($AB = CD$ and $AD = CB$). By the SSS (Side-Side-Side) congruence condition, $\Delta ABD \cong \Delta CDB$ because $AB = CD$, $AD = CB$, and the common side BD is equal to itself. Hence, these two triangles are congruent.

Key Concept: Opposite sides and angles of a parallelogram

b) $110^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal. Here, $\angle P$ and $\angle R$ are opposite angles, while $\angle Q$ and $\angle S$ are opposite angles. Therefore, $\angle S = \angle Q$. However, adjacent angles in a parallelogram add up to $180^\circ$, so $\angle Q = 180^\circ - \angle P = 180^\circ - 70^\circ = 110^\circ$. Thus, $\angle S = 110^\circ$.

Key Concept: Definition of Parallelogram

c) A quadrilateral in which opposite sides are parallel
[Solution Description]
A parallelogram is defined as a quadrilateral in which opposite sides are parallel.

Key Concept: Adjacent Angles in Parallelogram

b) $92^{\circ}$
[Solution Description]
In a parallelogram, adjacent angles are supplementary, meaning they add up to $180^{\circ}$. Therefore, $\angle A + \angle D = 180^{\circ}$.
Given:
$\angle A = (3x + 10)^{\circ}$
$\angle D = (2x + 40)^{\circ}$
Set up the equation:
$(3x + 10) + (2x + 40) = 180$
Combine like terms:
$5x + 50 = 180$
Subtract $50$ from both sides:
$5x = 130$
Divide by $5$ on both sides:
$x = 26$
Now find $\angle A$:
$\angle A = 3(26) + 10 = 88^{\circ}$
In a parallelogram, opposite angles are equal, so $\angle B = \angle D$.
Thus, $\angle B = 2(26) + 40 = 92^{\circ}$
The measure of $\angle B$ is $92^{\circ}$.

Key Concept: Special Cases of Parallelograms

d) Square
[Solution Description]
A quadrilateral with all sides equal and all angles equal to $90^\circ$ is called a square. This is because a square is a special type of rectangle where all sides are equal.

Key Concept: Properties of Rectangles

c) All angles are equal to $90^{\circ}$ and diagonals are equal.
[Solution Description]
A rectangle is a parallelogram with all angles equal to $90^{\circ}$. It satisfies all properties of a parallelogram, and additionally, its diagonals are equal in length.

Key Concept: Rectangle as a Parallelogram

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
A rectangle satisfies the definition of a parallelogram because it has two pairs of opposite sides that are parallel and equal in length (Property 3 of Rectangle). Additionally, the diagonals of a rectangle bisect each other (Property 4 of Rectangle), which is also true for any parallelogram. However, the Reason does not explain why a rectangle is a "special" type of parallelogram (i.e., due to its all angles being $90^\circ$). Thus, while both Assertion and Reason are true, the Reason does not fully explain the Assertion.

Key Concept: Rectangles as a type of parallelogram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A rectangle is a quadrilateral with all angles equal to $90^\circ$ and opposite sides parallel and equal, which satisfies the definition of a parallelogram. The reason correctly states that a rectangle has opposite sides parallel and equal. Since both statements are true, and the Reason correctly explains why the Assertion is true, the correct answer is option a).

Key Concept: Properties of Quadrilaterals, Venn Diagram

d) Square
[Solution Description]
A quadrilateral with diagonals that are equal in length is either a rectangle or a square. Additionally, if the diagonals bisect each other at $90^\circ$, it must be a square because rectangles have equal diagonals but they do not necessarily bisect each other at $90^\circ$. Only squares satisfy both conditions: equal diagonals and perpendicular bisectors.

Key Concept: Venn Diagram: Relationship between Rectangles, Rhombuses, Squares, and Parallelograms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that all squares are rectangles, which is true because a square is a special type of rectangle where all sides are equal and all angles are $90^\circ$. The reason given is that a square has all angles equal to $90^\circ$, which is also true and correctly explains why all squares are rectangles, as this property is essential for any quadrilateral to be classified as a rectangle.

Key Concept: Venn Diagram, Classification

c) Every square is a rhombus.
[Solution Description]
A square is a special type of rhombus where all angles are $90^\circ$. However, not all rhombuses are squares because a rhombus does not require all angles to be $90^\circ$. Therefore, every square is a rhombus, but not every rhombus is a square.

Key Concept: Venn Diagram: Relationship between squares, rectangles, and rhombuses

d) Nested circles: Square inside both Rectangle and Rhombus, which are inside Parallelogram.
[Solution Description]
A square is both a rectangle (all angles are $90^\circ$) and a rhombus (all sides equal). A rectangle has opposite sides equal and all angles $90^\circ$, while a rhombus has all sides equal but angles not necessarily $90^\circ$. Thus, the set of squares lies within both the sets of rectangles and rhombuses, which are subsets of parallelograms.

Key Concept: Properties of Rhombus and Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that every square is a rhombus, which is true because a square has all sides equal, satisfying the definition of a rhombus. The Reason states that a rhombus with all angles equal to $90^\circ$ is a square, which is also true as this defines a square. Moreover, the Reason correctly explains why the Assertion is true, since it describes the condition under which a rhombus becomes a square.

Key Concept: Diagonals and side length relationship

b) $2\sqrt{5}$ cm
[Solution Description]
Area of rhombus = $\frac{d_1 \times d_2}{2}$. Let $d_1 = x$ cm, then $d_2 = 2x$ cm.
So, $\frac{x \times 2x}{2} = 16 \implies x^2 = 16 \implies x = 4$ cm.
Thus, $d_1 = 4$ cm, $d_2 = 8$ cm.
Now, half-diagonals are $2$ cm and $4$ cm.
Side length = $\sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ cm.

Key Concept: Types of Quadrilaterals

c) Rhombus
[Solution Description]
A rhombus has all sides equal, but its angles are not necessarily equal (unless it is a square). Therefore, a rhombus fits this description.

Key Concept: Properties of Rhombus, Diagonal properties

b) 20 cm
[Solution Description]
A rhombus has diagonals that bisect each other at $90^\circ$. Given $AC = 8$ cm, so $AO = CO = 4$ cm. Given $BD = 6$ cm, so $BO = DO = 3$ cm. Using the Pythagorean theorem in $\triangle AOB$:
$AB = \sqrt{AO^2 + BO^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ cm.
Since all sides of a rhombus are equal, perimeter $= 4 \times AB = 4 \times 5 = 20$ cm.

Key Concept: Rhombus Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that all sides of a rhombus are equal, which is true because one of the defining properties of a rhombus is that all its sides have the same length. The reason correctly explains this property by stating that a quadrilateral with all sides equal is defined as a rhombus. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Properties of Rhombus, Angle Relationships

c) $180^\circ$
[Solution Description]
Opposite angles in a rhombus are equal, so $\angle PQR = \angle PSR = 110^\circ$. Adjacent angles in a rhombus are supplementary, so $\angle QPS = 180^\circ - \angle PQR = 70^\circ$. Therefore, the sum is:
$\angle QPS + \angle RSP = 70^\circ + 110^\circ = 180^\circ$

Key Concept: Properties of Rhombus, Diagonals

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, analyze the Assertion: A rhombus has diagonals that are perpendicular bisectors and divide the rhombus into four congruent right-angled triangles. However, unless it is a square, the diagonals are unequal. So, the Assertion is true.
Next, examine the Reason: It states that the diagonals bisect each other at right angles, which is a fundamental property of a rhombus. Hence, the Reason is also true.
Now, check if the Reason explains the Assertion: The fact that diagonals bisect at right angles does not imply they must be unequal (e.g., in a square, they are equal). Therefore, although both statements are true, the Reason does not explain why the diagonals are unequal.
Thus, the correct option is: Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Diagonals of a Rhombus

c) $90^\circ$
[Solution Description]
The diagonals of a rhombus bisect each other at right angles ($90^\circ$).

Key Concept: Construction of Rhombus

c) All sides must be equal.
[Solution Description]
A rhombus has all four sides of equal length. Therefore, for construction, we must ensure that all sides are equal.

Key Concept: Properties of Rhombus

c) They are equal.
[Solution Description]
According to Property 3 of a rhombus, the opposite angles are equal. For example, if one angle is $50^\circ$, its opposite angle will also be $50^\circ$.

Key Concept: Diagonals of Rhombus

c) $90^\circ$
[Solution Description]
Property 4 states that the diagonals of a rhombus bisect each other at right angles ($90^\circ$). Therefore, the angle between the diagonals is $90^\circ$.

Key Concept: Angle properties, External points

c) 60$^\circ$
[Solution Description]
Since $XWZB$ is a straight line, $\angle BXW = 180^\circ$. Triangle $AXB$ is equilateral (since $XA = XB$ and $\angle AXB = 60^\circ$). In rhombus $XYZW$, adjacent angles add up to $180^\circ$ and opposite angles are equal. Let $\angle YXW = y$. Then $\angle WXZ = 180^\circ - y$. Also, $\angle AXW = 60^\circ + y$ (exterior angle). But $\angle AXW$ must equal $\angle BXW - \angle BXA = 180^\circ - 60^\circ = 120^\circ$. Thus $60^\circ + y = 120^\circ$ → $y = 60^\circ$.

Key Concept: Properties of Quadrilaterals

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that a quadrilateral with diagonals intersecting at right angles and equal in length is a square. This is true because such properties are unique to squares among quadrilaterals. However, the reason claims that a quadrilateral with all sides equal and diagonals equal in length must be a rectangle. This is false because rhombuses also have all sides equal, but their diagonals are not necessarily equal unless they are squares.
Therefore, the assertion is true, but the reason is false.

Key Concept: Quadrilateral Identification

d) Square
[Solution Description]
When two diagonals are perpendicular and of equal length, the quadrilateral formed is a square because all sides and angles are equal.

Key Concept: Extending Diagonals of Quadrilaterals

b) Square
[Solution Description]
Initially, the square has diagonals that are equal in length and perpendicular bisectors. Extending both diagonals equally by 2 cm on both sides means:
- The new endpoints are symmetrically placed around the original square.
- The distance between any two adjacent new vertices will be equal due to symmetry.
- The angles at the vertices remain 90 degrees because the extensions preserve the perpendicularity of the diagonals.
Thus, the resulting quadrilateral formed by joining the new endpoints is another square, as all sides and angles remain equal and perpendicular.

Key Concept: Diagonals of Quadrilaterals, Properties of Squares and Rhombuses

c) Rhombus
[Solution Description]
The properties described here are:
- Diagonals are perpendicular to each other.
- Diagonals bisect each other.
- Diagonals are not equal in length.
From the given conditions, the quadrilateral cannot be a square because squares have equal-length diagonals. It also cannot be a rectangle since rectangles do not necessarily have perpendicular diagonals. However, a rhombus satisfies all the given conditions: its diagonals are perpendicular bisectors but are not equal in length. Therefore, the quadrilateral must be a rhombus.

Key Concept: Quadrilateral Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When two identical equilateral triangles are joined side-by-side, the resulting figure will have four sides of equal length since all sides of an equilateral triangle are equal. Additionally, opposite sides will be parallel because the angles at each vertex will be 60°, making the sum of adjacent angles 180°, which ensures parallelism. Therefore, the quadrilateral meets all the properties of a rhombus: all sides equal and opposite sides parallel. Hence, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Properties of Diagonals in Quadrilaterals

c) Square
[Solution Description]
If the diagonals of a quadrilateral are equal in length, bisect each other, and intersect at $90^\circ$, the quadrilateral must be a square. This is because these conditions ensure all sides are equal and all angles are $90^\circ$.

Key Concept: Geoboard Activity: Properties of Quadrilaterals

c) Square
[Solution Description]
When two diagonals are equal in length and perpendicular to each other, joining their endpoints forms a square. This is because all sides become equal due to the properties of symmetry and the right angles between the sides.

Key Concept: Diagonal Properties

c) It is a square
[Solution Description]
A quadrilateral with diagonals that are equal and perpendicular must be a square because it satisfies all properties of a square: equal sides, equal angles, and diagonals that bisect each other at 90\textdegree.

Key Concept: Constructing quadrilaterals using isosceles triangles

b) Kite
[Solution Description]
When two identical isosceles triangles (with sides 8 cm, 8 cm, 6 cm) are joined along their unequal sides (6 cm), the resulting quadrilateral will have two pairs of adjacent sides equal. This forms a kite because $AB = BC$ (8 cm each) and $CD = DA$ (8 cm each).

Key Concept: Constructing quadrilaterals using scalene triangles

d) Parallelogram
[Solution Description]
When two identical scalene triangles (with sides 6 cm, 9 cm, and 12 cm) are joined along their 6 cm sides, the resulting quadrilateral will have sides measuring 9 cm, 12 cm, 12 cm, and 9 cm. Since both pairs of opposite sides are equal, it forms a parallelogram.

Key Concept: Trapezium Angles

d) $110^\circ$
[Solution Description]
In an isosceles trapezium, the base angles on the same side are supplementary ($\angle U + \angle W = 180^\circ$). Given $\angle U = 70^\circ$, we can find $\angle W$ as follows:
$\angle W = 180^\circ - \angle U = 180^\circ - 70^\circ = 110^\circ$

Key Concept: Joining Triangles, Kite Properties

d) Assertion is false, but Reason is true.
[Solution Description]
Consider two scalene triangles ABC and CDA with sides 6 cm, 9 cm, and 12 cm. When joined along the 9 cm side (AC), the quadrilateral ABCD will have AB = CD = 6 cm and BC = DA = 12 cm. This satisfies the definition of a kite ($AB = BC$ is false here; correct condition for kite is $AB = AD$ and $CB = CD$). The assertion incorrectly states the condition for a kite, while the reason provides the correct property. Thus, Assertion is false but Reason is true.

Key Concept: Kite Properties

c) $120^\circ$
[Solution Description]
The properties of a kite state that the angles opposite to the equal sides are equal. Given $\angle ABC = 120^\circ$, its opposite angle $\angle ADC$ will also be $120^\circ$.

Key Concept: Kite properties, Diagonal bisection

b) 60 cm$^2$
[Solution Description]
The diagonal $BD$ bisects $AC$ perpendicularly, so $AO = OC = 6$ cm. Since $\Delta AOB \cong \Delta COB$, we can calculate the area of one triangle and multiply by 2.
Area of $\Delta AOB = \frac{1}{2} \times AO \times BO$. To find $BO$, use the property that $BD = 10$ cm and $BD$ is split into $BO$ and $OD$. However, without additional information, we recognize that the area of the kite is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 12 = 60$ cm$^2$.

Key Concept: Kite and Trapezium

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In a kite $ABCD$, by definition, $AB = BC$ and $CD = DA$. The diagonal $BD$ divides the kite into two congruent triangles $\triangle ABD$ and $\triangle CBD$ by SSS congruence because $AB = CB$, $AD = CD$, and $BD$ is common. Since these triangles are congruent, $\angle ABD = \angle CBD$ and $\angle ADB = \angle CDB$, which means $BD$ bisects $\angle ABC$ and $\angle ADC$. However, the Reason states that $\triangle AOB$ and $\triangle COB$ are congruent by SSS, but it does not directly explain why $BD$ bisects the angles. Instead, $\triangle AOB$ and $\triangle COB$ are congruent by SAS (since $AB = CB$, $\angle ABO = \angle CBO$, and $BO$ is common), which shows that $AO = OC$ and that $BD$ is perpendicular to $AC$.
Hence, while both Assertion and Reason are true, the Reason does not correctly explain the Assertion since it incorrectly uses SSS instead of SAS.

Key Concept: Kite and Trapezium combined, Angle calculation

c) 120$^\circ$
[Solution Description]
The quadrilateral described is a combination of kite and trapezium properties. As it has two pairs of adjacent sides equal, it must be a kite. But it also has one pair of parallel sides, making it a trapezium. Hence, it is an isosceles trapezium where non-parallel sides are equal. Given one angle as $120^\circ$, the adjacent angle on the same side of the transversal will be $60^\circ$ (since they add up to $180^\circ$). The opposite angles in an isosceles trapezium are supplementary, so the largest angle will be $120^\circ$.

Key Concept: Congruency in Kite

c) Both sides and angles are equal.
[Solution Description]
In kite ABCD, diagonal BD bisects $\angle ABC$ and $\angle ADC$. Hence, $\angle ABO = \angle CBO$. Also, $AB = BC$ by definition. Since BD is common to both $\Delta AOB$ and $\Delta COB$, they are congruent by SAS (Side-Angle-Side) rule. Therefore, corresponding parts like $AO = OC$ and $\angle AOB = \angle COB$ (both right angles).

Key Concept: Kite construction, side lengths

c) $30$ cm
[Solution Description]
A kite must satisfy the triangle inequality for each pair of triangles formed by the diagonals. For sides 6 cm and 9 cm, let the shorter sides be repeated.
Then, the sides are 6, 6, 9, 9, giving a perimeter of $6 + 6 + 9 + 9 = 30$ cm. This satisfies all geometric constraints.

Key Concept: Classification of Quadrilaterals

a) Every rhombus is a kite, but not every kite is a rhombus.
[Solution Description]
A rhombus is a special case of a kite where all four sides are equal. However, not every kite is a rhombus because a kite only requires two distinct pairs of adjacent sides to be equal.

Key Concept: Kite diagonals, congruence

b) $10$ cm[Solution Description]The area of a kite is given by half the product of its diagonals. Here, $PR = PO + RO = 5 + 5 = 10$ cm and the area is 50 cm\textsuperscript{2}.
Using the formula: $\text{Area} = \frac{1}{2} \times PR \times QS$.
So, $50 = \frac{1}{2} \times 10 \times QS \implies 100 = 10 \times QS \implies QS = 10$ cm.

Key Concept: Trapezium angles, Isosceles trapezium properties

b) $105^\circ$
[Solution Description]
Property of isosceles trapezium states that angles adjacent to each base are equal. Here, $\angle A = \angle B = 75^\circ$. Sum of angles on same side of leg in trapezium is $180^\circ$, so $\angle D = 180^\circ - 75^\circ = 105^\circ$. Since it's isosceles, $\angle C = \angle D = 105^\circ$.

Key Concept: Sum of angles in a quadrilateral

c) $20^\circ$
[Solution Description]
The sum of the interior angles of any quadrilateral is $360^\circ$. Let the fourth angle be $x$. Then:
$105^\circ + 135^\circ + 100^\circ + x = 360^\circ$
Simplifying:
$105 + 135 = 240^\circ$
$240 + 100 = 340^\circ$
$x = 360^\circ - 340^\circ = 20^\circ$
So, the fourth angle is $20^\circ$.

Key Concept: Trapezium angle sum, Supplementary angles

b) $110^\circ$ and $70^\circ$
[Solution Description]
Let trapezium be ABCD with AB $\parallel$ CD. Given $\angle A = 110^\circ$ and $\angle B = 70^\circ$. Angles on same side of leg are supplementary: $\angle A + \angle D = 180^\circ \Rightarrow \angle D = 70^\circ$; $\angle B + \angle C = 180^\circ \Rightarrow \angle C = 110^\circ$. Remaining angles are $110^\circ$ and $70^\circ$.

Key Concept: Kite diagonals, Congruency

b) 6 cm
[Solution Description]
Diagonals of a kite intersect at right angles, bisecting one diagonal. Here, PR is bisected by QS, so PO = OR = 4 cm. Using Pythagoras' theorem in $\triangle$PQO: $5^2 = 4^2 + QO^2 \Rightarrow QO = 3$ cm. Thus, full diagonal QS = 2 $\times$ QO = 6 cm.

Key Concept: Properties of Kite: Diagonals

c) Assertion is true, but Reason is false.
[Solution Description]
According to the property of a kite, one of the diagonals (the one connecting the vertices with equal adjacent sides) bisects the other diagonal at $90^\circ$ and also bisects the angles at those vertices. However, the diagonals of a kite do not necessarily bisect each other completely; only one diagonal is bisected by the other. Therefore, the Assertion is true, but the Reason is false.

Key Concept: Properties of Kite - Diagonal Bisection

b) $40^\circ$
[Solution Description]
Given that diagonal BD bisects $\angle ABC$, this means it splits $\angle ABC$ into two equal angles. Since $\angle ABC = 80^\circ$, $\angle ABD$ will be half of $\angle ABC$.
Calculation:
$\angle ABD = \frac{\angle ABC}{2} = \frac{80^\circ}{2} = 40^\circ$
Thus, the measure of $\angle ABD$ is $40^\circ$.

Key Concept: Angle Relationships in Trapezium

b) $110^\circ$
[Solution Description]
A property of a trapezium states that the sum of adjacent angles on the same side of the non-parallel sides is $180^\circ$. Thus, $\angle P + \angle S = 180^\circ$. Given $\angle P = 70^\circ$, we can calculate $\angle S$ as follows:
$\angle S = 180^\circ - 70^\circ = 110^\circ$

Key Concept: Kite properties, Angle relationships

b) $100^\circ$
[Solution Description]
First, recall that in a kite, one diagonal bisects the angles at the vertices it connects. So, diagonal $BD$ bisects $\angle ABC$ and $\angle ADC$.
Given $\angle ABC = 120^\circ$, so $\angle ABO = \angle CBO = \frac{120^\circ}{2} = 60^\circ$.
Similarly, $\angle ADC = 80^\circ$, so $\angle ADO = \angle CDO = \frac{80^\circ}{2} = 40^\circ$.
Now, consider triangle $AOD$. The sum of angles in a triangle is $180^\circ$.
We know $\angle OAD$ can be found using the property that diagonals of a kite are perpendicular. Thus, $\angle AOD = 90^\circ$.
However, let's verify this by calculating $\angle OAD$:
In quadrilateral $ABCD$, the sum of all angles is $360^\circ$.
So, $\angle BAD + \angle ABC + \angle BCD + \angle CDA = 360^\circ$.
Since $AB = BC$ and $CD = DA$, we can find $\angle BAD = \angle BCD$ as follows:
Let $\angle BAD = x$, then $\angle BCD = x$ (because $AB = BC$ and $CD = DA$).
So, $x + 120^\circ + x + 80^\circ = 360^\circ$.
Which gives $2x + 200^\circ = 360^\circ$.
So, $2x = 160^\circ$ and $x = 80^\circ$.
Now, $\angle OAD = \frac{\angle BAD}{2} = \frac{80^\circ}{2} = 40^\circ$.
In triangle $AOD$, $\angle AOD + \angle OAD + \angle ODA = 180^\circ$.
Substituting known values: $\angle AOD + 40^\circ + 40^\circ = 180^\circ$.
So, $\angle AOD = 100^\circ$.