Key Concept: Definition of Square

c) A quadrilateral with all angles $90^\circ$ and all sides equal.
[Solution Description]
A square is a quadrilateral with all angles equal to $90^\circ$ and all sides of equal length. It also has diagonals that are equal, bisect each other at $90^\circ$, and bisect the angles of the square.

Key Concept: Properties of Squares and Rectangles

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the diagonals of a rectangle bisect each other at $90^\circ$. However, this is false because in a rectangle, the diagonals bisect each other but not necessarily at $90^\circ$ (they do so only in squares). The reason states that a rectangle has all sides equal and all angles equal to $90^\circ$, which is also false since rectangles have opposite sides equal but not necessarily all sides equal (only squares satisfy this condition).
Therefore, both Assertion and Reason are false. But according to the given options, the closest correct choice is:
Assertion is false, but Reason is true.

Key Concept: Rectangles, Diagonals

c) $10$ cm
[Solution Description]
To find the length of the diagonal of a rectangle, we can use the Pythagorean theorem. The diagonal divides the rectangle into two right-angled triangles. For a rectangle with sides $a$ and $b$, the diagonal $d$ is given by $d = \sqrt{a^2 + b^2}$. Here, $a = 8$ cm and $b = 6$ cm. So,
$d = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}.$
Thus, the diagonal is $10$ cm.

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 diagonals and angle bisection

d) Assertion is false, but Reason is true.
[Solution Description]
Let us analyze both statements:
**Assertion Analysis:**
For a quadrilateral to be classified as a square, all its sides must be equal, all its angles must be 90°, and its diagonals must be equal in length and bisect each other at right angles. However, while these conditions guarantee that the quadrilateral is a rectangle or rhombus, they do not exclusively confirm it is a square unless all sides are equal (which makes it a rhombus) and all angles are 90° (which makes it a rectangle). A square fulfills both these criteria (rhombus + rectangle).
**Reason Analysis:**
Property 5 states that the diagonals of a square bisect its angles, which is correct and unique to squares. This property does not hold for other types of rectangles or rhombuses.
Here, the Assertion is partially correct but incomplete—it describes properties of a rhombus and rectangle, not just a square. The Reason is true for squares but does not fully explain why the Assertion guarantees a square. Hence, the Assertion is false, but the Reason is true.

Key Concept: Diagonals and Angles

b) $45^\circ$
[Solution Description]
In a square, the diagonals bisect each other at right angles, forming four congruent right-angled triangles. In triangle $AOB$, $\angle AOB = 90^\circ$ and the diagonals also bisect the angles of the square. Thus, $\angle OAB$ is half of $\angle DAB$ since the diagonal $AC$ bisects $\angle DAB$. Since all angles in a square are $90^\circ$, $\angle DAB = 90^\circ$ and $\angle OAB = 45^\circ$.

Key Concept: Properties of Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that in rectangle $ABCD$, diagonals $AC$ and $BD$ are equal in length. This is true because one of the properties of rectangles is that their diagonals are equal in length.
The reason states that the diagonals of a rectangle bisect each other. This is also true as it is another property of rectangles.
However, the reason does not explain why the diagonals are equal in length—it only states a different property of the diagonals. Thus, both statements are true, but the reason is not the correct explanation for the assertion.

Key Concept: Diagonals, Opposite Sides

b) 5 cm
[Solution Description]
First, find the length of diagonal $AC$ using the Pythagorean theorem since all angles in a rectangle are $90^\circ$.
$AC = \sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}$
Since the diagonals of a rectangle bisect each other, $AO$ is half of $AC$.
$AO = \frac{AC}{2} = \frac{10}{2} = 5 \text{ cm}$

Key Concept: Diagonals of a Rhombus

c) The diagonals bisect each other.
[Solution Description]
Both squares and rhombuses have diagonals that bisect each other. However, while squares also have equal-length diagonals, rhombuses do not necessarily have this property unless they are squares themselves.

Key Concept: Congruency of Triangles

c) Rectangle
[Solution Description]
A quadrilateral whose diagonals are equal and bisect each other must be a rectangle. While squares also satisfy this condition, the general case is a rectangle because squares have additional properties like all sides being equal and diagonals intersecting at $90^\circ$.

Key Concept: Diagonals of Rectangle and Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a rectangle always intersect at their midpoints. This is true because, in any rectangle, the diagonals are equal in length and they bisect each other at their midpoints. The reason states that a rectangle is a type of parallelogram where the diagonals bisect each other. This is also true as all rectangles are parallelograms with congruent diagonals that bisect each other. Moreover, the reason correctly explains why the assertion is true since it provides the underlying property (parallelogram diagonals bisecting each other) that leads to the assertion being valid.
Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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: Angles between diagonals in a rectangle

c) $90^\circ$
[Solution Description]
The diagonals of a rectangle bisect each other and are equal in length. When they intersect, they form four angles. Since the rectangle has all angles equal to $90^\circ$, the diagonals intersect at right angles. Thus, each angle at the intersection is $90^\circ$.

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: Properties of diagonals in a square

b) $90^\circ$, $90^\circ$, $90^\circ$
[Solution Description]
In a square, the diagonals are equal, bisect each other, and intersect at $90^\circ$. Therefore, the angles at the intersection are all equal. If one angle is $x = 90^\circ$, the other three angles will also be $90^\circ$ due to vertically opposite angles and linear pairs.

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: 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: Diagonals of Quadrilaterals

d) Square
[Solution Description]
If the diagonals of a quadrilateral are equal in length and bisect each other at $90^\circ$, then the quadrilateral is a square. This is because such properties ensure all angles are $90^\circ$ and all sides are equal.

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: Diagonals of a rectangle

b) 10 cm
[Solution Description]
The diagonals of a rectangle are equal in length. If one diagonal is 10 cm, the other diagonal must also be 10 cm.

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: Rectangle and Square Properties, Diagonal Congruence

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a quadrilateral to be a square, its diagonals must satisfy three conditions: 1) They must be of equal length, 2) They must bisect each other, and 3) They must intersect at right angles ($90^\circ$). The Assertion states that joining equal-length strips at their midpoints with a $90^\circ$ angle forms a square, which is correct because this setup ensures all three conditions are met. The Reason correctly identifies these properties as sufficient for a quadrilateral to be a square. Therefore, both Assertion and Reason are true, and Reason explains Assertion.

Key Concept: Angles Between Diagonals

c) $90^\circ$
[Solution Description]
For a square, the diagonals not only bisect each other but also intersect at right angles ($90^\circ$). So, the carpenter must ensure the angle between the diagonals is $90^\circ$.

Key Concept: Properties of Rectangle Diagonals

b) At the midpoint of both strips
[Solution Description]
The diagonals of a rectangle bisect each other at their midpoints. Therefore, the carpenter should join the strips at their midpoints.

Key Concept: Square Properties

c) They bisect each other at $90^\circ$
[Solution Description]
Both squares and rectangles have diagonals that are equal in length and bisect each other. However, the diagonals of a square additionally bisect each other at right angles ($90^\circ$), which is not necessarily true for all rectangles.

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: Properties of a Square

c) The diagonals are not equal in length.
[Solution Description]
Properties of a square:
1. All sides are equal.
2. All angles are $90^\circ$.
3. Diagonals are equal, bisect each other at $90^\circ$, and also bisect the angles.
The statement “The diagonals are not equal in length” is false for a square.

Key Concept: Square Properties

a) $3\sqrt{2}\ \text{cm}$
[Solution Description]
For any square, diagonals are equal and bisect each other perpendicularly. Given AB = 6 cm (side length), we find diagonal AC first using Pythagorean theorem:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}\ \text{cm}$.
Since diagonals bisect each other, AO = half of AC = $\frac{6\sqrt{2}}{2} = 3\sqrt{2}\ \text{cm}$.
Calculation steps:
1) Calculate diagonal AC using Pythagoras: $AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}\ \text{cm}$.
2) Find AO as half of AC: $AO = 3\sqrt{2}\ \text{cm}$.

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: Venn Diagram Representation

c) Square $\subset$ Rectangle
[Solution Description]
The set of all squares is a subset of the set of all rectangles because every square is a rectangle (all angles $90^\circ$ and opposite sides equal), but not all rectangles are squares. Thus, the correct representation is Square $\subset$ Rectangle.

Key Concept: Special Rectangle and Square Properties

c) Opposite sides are equal
[Solution Description]
We need to identify a property that is common to both squares and rectangles.
Property 1: All angles are $90^\circ$. This is true for both squares and rectangles.
Property 2: Opposite sides are equal. This is also true for both.
Property 3: Diagonals bisect each other. This is true for both.
However, only in a square are all sides equal and diagonals perpendicular. So, the correct statement must be one that applies to both but not exclusively to either.
The most general correct statement common to both is that “Opposite sides are equal.”

Key Concept: Angles in a Rectangle

c) $90^\circ$
[Solution Description]
By definition, all angles in a rectangle are right angles. A right angle measures $90^\circ$. Therefore, each angle in a rectangle is $90^\circ$.

Key Concept: Opposite angles in a parallelogram

c) $20^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal. Hence, $\angle A = \angle C$.
So, $3x + 10^\circ = 5x – 30^\circ$.
Subtract $3x$ from both sides: $10^\circ = 2x – 30^\circ$.
Add $30^\circ$ to both sides: $40^\circ = 2x$.
Divide by 2: $x = 20^\circ$.

Key Concept: Quadrilateral Angles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A quadrilateral can indeed be divided into two triangles by drawing one diagonal. The sum of angles in each triangle is $180^\circ$, so the total sum of angles for both triangles is $360^\circ$. Since the angles of the triangles collectively form the angles of the quadrilateral, the sum of the quadrilateral’s angles must also be $360^\circ$. Thus, both Assertion (A) and Reason (R) are true, and R correctly explains why A is true.

Key Concept: Sum of angles in a quadrilateral

c) 360$^\circ$
[Solution Description]
The sum of all interior angles of any quadrilateral is always $360^\circ$, as per the angle sum property of quadrilaterals.

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: Sum of angles in a quadrilateral, relationship between angles

d) 90°
[Solution Description]
Step 1: Given that angles P and R are supplementary, so $P + R = 180°$.
Step 2: Given $P = 110°$, so $R = 180° – 110° = 70°$.
Step 3: Angles Q and S are supplementary, so $Q + S = 180°$.
Step 4: Sum of all angles in quadrilateral PQRS is 360°.
Step 5: So, $P + Q + R + S = 360°$ ⇒ $110° + Q + 70° + S = 360°$.
Step 6: From step 3, substitute Q as $(180° – S)$ ⇒ $110° + (180° – S) + 70° + S = 360°$.
Step 7: Solve the equation: $110° + 180° – S + 70° + S = 360°$ ⇒ $360° = 360°$.
This shows that angle S can be any value as long as $Q + S = 180°$. However, since the options provided are specific, we need to consider possible constraints. Typically, for a valid quadrilateral, $S$ must be such that none of the angles exceed practical limits.

Key Concept: Impossible quadrilateral configuration

c) 120°, 110°, 100°, 20°
[Solution Description]
For a valid quadrilateral, the sum of all four angles must be $360°$.
Checking each option:
Option a) $90° + 90° + 90° + 90° = 360°$ (valid).
Option b) $100° + 80° + 70° + 110° = 360°$ (valid).
Option c) $120° + 110° + 100° + 20° = 350° \neq 360°$ (invalid).
Option d) $95° + 85° + 90° + 90° = 360°$ (valid).
Thus, option c does not satisfy the sum condition.

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

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: 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: Quadrilateral Angle Sum Property

d) Assertion is false, but Reason is true.
[Solution Description]
We know from the properties of quadrilaterals that the sum of all four interior angles is always $360^\circ$.
If three angles are each $90^\circ$, their sum is $3 \times 90^\circ = 270^\circ$. If the fourth angle were not $90^\circ$, say $x$, then the total sum would be $270^\circ + x$. For this to equal $360^\circ$, $x$ must satisfy:
$270^\circ + x = 360^\circ$
$x = 360^\circ – 270^\circ$
$x = 90^\circ$
This means the fourth angle must also be $90^\circ$ to satisfy the angle sum property. Hence, it is impossible for a quadrilateral to have three right angles and a non-right fourth angle.
The assertion claims it is possible, which is false, but the reason correctly states the angle sum property of quadrilaterals.

Key Concept: Properties of quadrilaterals with three right angles

b) $90^\circ$
[Solution Description]
Since the sum of angles in a quadrilateral is $360^\circ$, if three angles are $90^\circ$ each, the fourth angle must be $360^\circ – (3 \times 90^\circ) = 90^\circ$.

Key Concept: Diagonal properties, angle calculation

c) $95^\circ$
[Solution Description]
First, calculate the sum of given angles: $95^\circ + 85^\circ + 85^\circ = 265^\circ$.
Using angle sum property:
$265^\circ + \angle (\text{fourth angle}) = 360^\circ$
Thus, the fourth angle = $95^\circ$.
The information about diagonals intersecting at right angles is irrelevant for calculating the fourth angle in this case, as we already know three angles and can use the angle sum property directly.

Key Concept: Angle Sum in Quadrilaterals

c) $115^\circ$
[Solution Description]
The sum of angles in a quadrilateral is $360^\circ$. Given $\angle P + \angle Q + \angle R + \angle S = 360^\circ$. Substituting the values:
$80^\circ + 95^\circ + 70^\circ + \angle S = 360^\circ$
$245^\circ + \angle S = 360^\circ$
$\angle S = 360^\circ – 245^\circ = 115^\circ$.

Key Concept: Adjacent Angles in Parallelogram

b) $130^\circ$
[Solution Description]
Adjacent angles in a parallelogram are supplementary, meaning they add up to $180^\circ$. If one angle is $50^\circ$, the adjacent angle will be $180^\circ – 50^\circ = 130^\circ$.

Key Concept: Properties of Parallelograms, Supplementary Angles

d) $120^\circ$
[Solution Description] In a parallelogram, adjacent angles are supplementary, which means they add up to $180^\circ$. Given $\angle A = 60^\circ$, we can find $\angle B$ using the property:
$\angle A + \angle B = 180^\circ \implies 60^\circ + \angle B = 180^\circ \implies \angle B = 120^\circ$

Key Concept: Diagonals of Parallelogram, Bisection Property

c) $10$ cm
[Solution Description] The diagonals of a parallelogram bisect each other. This means $PO = OR$ and $QO = OS$. Here, $PO = 5$ cm and $OR = 3$ cm implies the diagonals are not bisecting each other equally, which contradicts the property. Therefore, the correct values must be equal. Assuming $PO = OR = x$, then $PR = PO + OR = x + x = 2x$. However, the problem states $PO = 5$ cm and $OR = 3$ cm, which is impossible for a parallelogram. But interpreting it as a misprint, if $PO = 5$ cm and $OR = 5$ cm, then $PR = 10$ cm.

Key Concept: Adjacent Angles in Parallelogram

c) $120°$
[Solution Description]
Property 3 of parallelograms states that adjacent angles add up to $180°$. So, if one angle is $60°$, the adjacent angle will be $180° – 60° = 120°$.

Key Concept: Diagonals of a parallelogram

b) They bisect each other.
[Solution Description]
The diagonals of a parallelogram bisect each other but are not necessarily equal in length. Only in special cases like rectangles or squares, the diagonals are equal. Hence, the correct statement is that the diagonals bisect each other.

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: Parallelogram Properties

d) Assertion is false, but Reason is true.
[Solution Description]
Assertion (A) claims that the diagonals of a parallelogram are equal in length. However, according to the syllabus, while the diagonals of a parallelogram bisect each other (Property 4), they need not be equal in length (as stated under “Note”). Therefore, the assertion is false.
Reason (R) states that the diagonals of a parallelogram bisect each other, which is true as per Property 4 in the syllabus.
Thus, Assertion is false while Reason is true.

Key Concept: Diagonals, Angles

b) 35°
[Solution Description]
Step 1: Draw rectangle $PQRS$ with diagonals $PR$ and $QS$ intersecting at $O$.
Step 2: In rectangles, diagonals are equal and bisect each other. Thus, $OP = OR = OQ = OS$.
Step 3: Triangle $POS$ has sides $OP = OS$, making it an isosceles triangle.
Step 4: Given $\angle POS = 70°$, the remaining angles in $\triangle POS$ must sum to $110°$. Since $OP = OS$, $\angle PSO = \angle OPS = 55°$ each.
Step 5: Now, $\angle OSR = 90° – \angle PSO = 35°$ because $PSR$ is a right angle ($90°$) in the rectangle.

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: Diagonal Properties

b) 10 cm
[Solution Description]
In a rectangle, the diagonals are always equal in length. Therefore, if diagonal AC is 10 cm, then diagonal BD must also be 10 cm.

Key Concept: Properties, Angles

b) Rhombus
[Solution Description]
Step 1: Quadrilateral with diagonals bisecting each other is a parallelogram. So, $ABCD$ is initially a parallelogram.
Step 2: Since $\angle AOB = 90°$, the diagonals intersect at right angles. A parallelogram with perpendicular diagonals is either a rhombus or a square.
Step 3: However, without knowing whether all sides are equal or angles are $90°$, we cannot confirm if it is a square. But the given condition guarantees at least rhombus properties.
Step 4: Therefore, $ABCD$ must be a rhombus among the options provided.

Key Concept: Angle Properties in Rhombus

d) $120^\circ$
[Solution Description]
Adjacent angles in a rhombus are supplementary (add up to $180^\circ$). Given $\angle A = 60^\circ$, we can find $\angle B$ using the property:
$\angle A + \angle B = 180^\circ$
$60^\circ + \angle B = 180^\circ$
$\angle B = 180^\circ – 60^\circ = 120^\circ$.
The opposite angles in a rhombus are equal, so $\angle D = \angle B = 120^\circ$.

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, Properties of Quadrilaterals

d) Parallelogram
[Solution Description]
The quadrilateral has both pairs of opposite sides equal and parallel, which means it is a parallelogram. Since the diagonals are not perpendicular and do not bisect the angles, it cannot be a rhombus or square. It also does not have all angles equal to $90^\circ$, so it is not a rectangle. Therefore, it is a general parallelogram.

Key Concept: Properties of Quadrilaterals

c) Rhombus
[Solution Description]
A rhombus has diagonals that are perpendicular bisectors of each other. This is one of its defining properties. Hence, the correct answer is Rhombus.

Key Concept: Properties of Rhombus

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that every rhombus is a square, which is false because a rhombus only requires all sides to be equal but does not require all angles to be $90^\circ$. The Reason correctly describes a square as having all sides equal and all angles equal to $90^\circ$, which is true but does not explain the Assertion since the Assertion is incorrect.
Therefore, the Assertion is false, but the Reason is true.

Key Concept: Angle Properties of Rhombus

c) $120^\circ$
[Solution Description]
Adjacent angles in a rhombus sum to $180^\circ$ (Property 3). Given $\angle A = 60^\circ$, then $\angle B = 180^\circ – 60^\circ = 120^\circ$.

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: Comparison with Square

c) A square has all angles equal to $90^\circ$
[Solution Description]
A square is a special case of a rhombus where all angles are $90^\circ$. Therefore, the key difference lies in the angle measures.

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, 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: Properties of Rhombus, Angles in Rhombus

c) $60^\circ$
[Solution Description]
Since diagonals of a rhombus bisect each other at right angles, we know $\angle AOD = 90^\circ$. Given $\angle OAD = 30^\circ$, we can find $\angle ADO$ using the sum of angles in triangle $AOD$:
$\angle OAD + \angle ADO + \angle AOD = 180^\circ \implies 30^\circ + \angle ADO + 90^\circ = 180^\circ \implies \angle ADO = 60^\circ$
The diagonal also bisects the angle, so $\angle ADC = 2 \times \angle ADO = 120^\circ$. Then, since adjacent angles in a rhombus are supplementary:
$\angle ABC = 180^\circ – \angle ADC = 180^\circ – 120^\circ = 60^\circ$

Key Concept: Opposite Angles of a Rhombus

b) $60^\circ$
[Solution Description]
In a rhombus, opposite angles are equal. Given $\angle A = 60^\circ$, its opposite angle $\angle C$ will also be $60^\circ$.

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: Opposite angles, Diagonal properties

b) 60
[Solution Description]
In a rhombus, diagonals bisect each other at $90^\circ$ and also bisect the angles. So $\angle QPR = 2 \times \angle QPO = 60^\circ$. Since all sides are equal, triangle $PQR$ is isosceles with $PQ = QR$. Therefore, $\angle PQR = 180^\circ – 2(30^\circ) = 120^\circ$. The diagonals intersect at right angles, so in triangle $ROQ$, $\angle ROQ = 180^\circ – 90^\circ – 30^\circ = 60^\circ$.

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: Side length, Diagonal relationship

c) 6 cm
[Solution Description]
In a rhombus, the diagonals bisect each other at right angles and divide the rhombus into four right-angled triangles. Let’s denote half of diagonal 1 as $d_1/2 = 4$ cm and half of diagonal 2 as $d_2/2 = x$ cm. Using Pythagoras theorem: $(d_1/2)^2 + (d_2/2)^2 = \text{side}^2$ → $4^2 + x^2 = 5^2$ → $16 + x^2 = 25$ → $x^2 = 9$ → $x = 3$ cm. Thus the full diagonal length is $6$ cm.

Key Concept: Folding Paper to Form Quadrilaterals

d) Kite
[Solution Description]
Folding the sheet into half creates a straight line (crease). Folding it again into a quarter creates another perpendicular straight line. Making a triangular crease at the middle corner introduces diagonal folds. When these creases are opened, they form intersecting lines within the paper. Specifically:
– Two main creases from folding into quarters (vertical and horizontal).
– Diagonal creases from the triangular fold.
The resulting pattern forms a kite shape due to the intersection of the diagonal folds with the main creases, creating two pairs of adjacent sides of unequal length meeting at the central point.

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: Properties of Quadrilaterals, Constructing Quadrilaterals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in a quadrilateral with two pairs of parallel sides, the diagonals bisect each other. This is true because such a quadrilateral is a parallelogram, and one of the properties of a parallelogram is that its diagonals bisect each other. The reason correctly identifies the quadrilateral as a parallelogram and states the property of diagonals bisecting each other, which explains the assertion. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Joining Triangles to Form Quadrilaterals

c) Rhombus
[Solution Description]
When two equilateral triangles are joined along one side, the resulting figure will have four sides. Each angle of an equilateral triangle is $60^\circ$. When two such triangles are joined, the adjacent angles add up to $120^\circ$. The opposite angles will also be $120^\circ$ each, making the quadrilateral a rhombus but not a square, as the angles are not $90^\circ$.

Key Concept: Joining Triangles: Types of Quadrilaterals

b) Rhombus
[Solution Description]
Joining two equilateral triangles along one common side results in a rhombus. Since all sides of the equilateral triangle are equal (8 cm), the resulting quadrilateral has four sides of equal length, making it a rhombus. Additionally, the angles will be $60^\circ$ and $120^\circ$.

Key Concept: Properties of Quadrilaterals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a quadrilateral formed by two perpendicular diagonals of equal length is a rectangle. This is true because when diagonals are equal and intersect at $90^\circ$, they form a rectangle as all angles become right angles and opposite sides become equal. The reason states that diagonals of a rectangle are equal in length and bisect each other at $90^\circ$. This is also true as it is a property of rectangles. Further, the reason correctly explains the assertion because the properties mentioned in the reason lead to the conclusion in the assertion.

Key Concept: Quadrilateral Transformation

b) Kite
[Solution Description]
When one diagonal of a square is extended equally on both sides, the angles between the sides change, but the adjacent sides remain equal. This results in a kite, as two pairs of adjacent sides are equal, but not all four sides.

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: Kite, Diagonals

b) They bisect each other perpendicularly
[Solution Description]
In a kite ABCD with AB = BC and CD = DA, diagonal BD bisects the other diagonal AC perpendicularly. This means BD divides AC into two equal parts (AO = OC) and forms right angles (90 degrees) with AC. Additionally, BD also bisects the angles ABC and ADC.

Key Concept: Joining Triangles, Trapezium Properties

c) Isosceles trapezium
[Solution Description]
When two isosceles triangles with sides 8 cm, 8 cm, and 6 cm are joined along their unequal sides (6 cm), the resulting quadrilateral is an isosceles trapezium. In an isosceles trapezium, the non-parallel sides are equal, and the base angles are equal. The diagonals are also equal in length.

Key Concept: Kite Properties

b) Kite
[Solution Description]
When two isosceles triangles with sides 8 cm, 8 cm, and 6 cm are joined along their unequal sides (6 cm), the resulting quadrilateral has adjacent sides equal (8 cm each). This satisfies the definition of a kite, where two pairs of adjacent sides are equal. Thus, the quadrilateral formed is a kite.

Key Concept: Trapezium Angles

d) $120^\circ$
[Solution Description]
One property of a trapezium states that the sum of angles on the same side of a leg is $180^\circ$. Here, $\angle P + \angle S = 180^\circ$. Given $\angle P = 60^\circ$, we calculate $\angle S = 180^\circ – 60^\circ = 120^\circ$.

Key Concept: Isosceles Trapezium Properties

b) $\angle U = \angle V$
[Solution Description]
In an isosceles trapezium, the non-parallel sides are equal, and the angles opposite these sides are equal. Thus, $\angle U = \angle V$.

Key Concept: Kite and Trapezium

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that in a kite, the diagonals are perpendicular to each other. This is true because one of the properties of a kite is that its diagonals intersect at right angles. The Reason claims that the diagonals bisect each other at right angles. While it’s true that they intersect at right angles, the longer diagonal bisects the shorter one, but not necessarily vice versa. Hence, the Reason is not entirely correct as it implies the diagonals bisect each other equally, which is true only for specific cases like rhombuses. Therefore, Assertion is true, but Reason is false.

Key Concept: Trapezium angles, Isosceles trapezium properties

c) 80$^\circ$
[Solution Description]
In an isosceles trapezium, the angles adjacent to each base are equal. Thus, $\angle P = \angle Q = 100^\circ$. Since the sum of angles on the same side of a transversal in a trapezium is $180^\circ$, we have $\angle P + \angle S = 180^\circ$. Substituting $\angle P = 100^\circ$, we get $\angle S = 80^\circ$.

Key Concept: Definition of Kite

c) A quadrilateral with two pairs of adjacent sides equal
[Solution Description]
A kite is a quadrilateral with two distinct pairs of adjacent sides equal. By definition, if a quadrilateral ABCD has $AB = BC$ and $CD = DA$, then it is a kite.

Key Concept: Kite Properties, Congruent Triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given that $\Delta AOB \cong \Delta COB$ in the kite ABCD, this implies that the corresponding angles $\angle ABO = \angle CBO$ and $\angle ADO = \angle CDO$ due to congruency. Therefore, the diagonal BD indeed bisects $\angle ABC$ and $\angle ADC$. The Reason correctly explains why the Assertion is true because congruent triangles ensure equal corresponding angles, which leads to the bisection of the angles by the diagonal BD.

Key Concept: Relationship Between Kite and Rhombus

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

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, 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: Kite: Diagonal Properties

b) Only one diagonal bisects the other and they are perpendicular.
[Solution Description]
In a kite, one diagonal bisects the other. Specifically, the diagonal connecting the vertices where the adjacent sides are equal (e.g., $BD$) bisects the other diagonal ($AC$) perpendicularly. Thus, $AO = OC$ and $BD \perp AC$.

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: Isosceles Trapezium Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An isosceles trapezium has two parallel sides (UV$\parallel$XW) and two non-parallel sides (UX = VW). According to Property 2 of the syllabus, in an isosceles trapezium, angles opposite to equal sides are equal ($\angle$U = $\angle$V). This equality arises because the trapezium can be divided into congruent triangles (e.g., $\triangle UXY \cong \triangle VWZ$), ensuring the base angles ($\angle$U and $\angle$V) are equal. The Reason correctly states the defining property of an isosceles trapezium (equal non-parallel sides), which directly leads to the Assertion.

Key Concept: Properties of Isosceles Trapezium – Angles

b) $115^\circ$
[Solution Description]
In an isosceles trapezium, the angles opposite to the equal sides are equal. Let these angles be $65^\circ$ each. The sum of all angles in any quadrilateral is $360^\circ$. Let the remaining two angles be $x$ each since they must also be equal.
Calculation:
$65^\circ + 65^\circ + x + x = 360^\circ \implies 130^\circ + 2x = 360^\circ \implies 2x = 230^\circ \implies x = 115^\circ$
Thus, each of the other two angles measures $115^\circ$.

Key Concept: Trapezium properties, Angle relationships

b) $23.75$
[Solution Description]
Recall that in an isosceles trapezium, the angles adjacent to each non-parallel side are equal.
So, $\angle P = \angle Q$ and $\angle S = \angle R$.
Also, since $PQ \parallel SR$, the sum of consecutive angles on the same side of the legs is $180^\circ$.
So, $\angle P + \angle S = 180^\circ$.
Given $\angle P = (3x + 10)^\circ$ and $\angle S = (5x – 20)^\circ$.
Therefore, $(3x + 10) + (5x – 20) = 180$.
Simplifying, $8x – 10 = 180$.
Then, $8x = 190$.
So, $x = \frac{190}{8} = 23.75$.

Key Concept: Diagonals of an Isosceles Trapezium

d) Diagonals are equal in length
[Solution Description]
One of the key properties of an isosceles trapezium is that its diagonals are equal in length. Therefore, $UW = VX$.

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.