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 square bisect its angles, which is true as per the properties of a square. The reason states that a square has all sides equal and diagonals perpendicular to each other, which is also true. However, the reason does not explain why the diagonals of a square bisect its angles. The correct explanation would involve the fact that the diagonals of a square are angle bisectors due to the congruence of triangles formed by the diagonals and sides of the square. Thus, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

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 a Rectangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that all angles in a rectangle are $90°$, which is a fundamental property of rectangles as per the syllabus. The reason provided is that a rectangle is a quadrilateral with opposite sides equal and parallel. While this is also a true statement, it does not explain why the angles are $90°$. Therefore, both statements are true, but the reason does not correctly explain the assertion.

Key Concept: Square properties

c) Square
[Solution Description]
A quadrilateral with all sides equal and one angle as 90° must have all angles as 90° (since opposite angles are equal and adjacent angles are supplementary). Hence, it is a square.

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: 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: Square Properties, Diagonal

c) 48 cm
[Solution Description]
For a square with side length $s$, the diagonal is given by $d = s\sqrt{2}$. Given $PR = 12\sqrt{2}$ cm:
$s\sqrt{2} = 12\sqrt{2} \implies s = 12 \text{ cm}$
The perimeter of the square is $4s$.
$4 \times 12 = 48 \text{ cm}$

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 the diagonals of rectangle $ABCD$, $AC$ and $BD$, are equal in length, which is a known property of rectangles. Therefore, the assertion is true.
The reason states that the diagonals of a rectangle bisect each other, which is also a known property of rectangles. However, this property does not explain why the diagonals are equal in length—it is an independent property. Thus, the reason is true, but it is not the correct explanation for the assertion.

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: Diagonals of Rectangle and Square, Multi-step Solution

c) 26
[Solution Description]
Step 1: In a rectangle, diagonals are equal and bisect each other. So, AO = OC and BO = OD. Since AO = BO, we set $3x + 1 = 5x - 7$.
Step 2: Solve for x: subtract 3x from both sides to get $1 = 2x - 7$. Add 7 to both sides: $8 = 2x$. Divide by 2: $x = 4$.
Step 3: Substitute x back into AO: $AO = 3(4) + 1 = 13$.
Step 4: Diagonal AC is twice AO (since AO = OC): $AC = 2 \times 13 = 26$.

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: Diagonals of Squares and Rectangles

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that if the diagonals of a quadrilateral bisect each other at right angles and are equal in length, then the quadrilateral must be a square. This is not entirely accurate because a rhombus also has diagonals that bisect each other at right angles, but its diagonals are not necessarily equal in length. However, the reason given is that such a quadrilateral satisfies all properties of a square, which is true only for squares. Therefore, while both statements seem plausible, the assertion overgeneralizes by not considering other quadrilaterals (like rhombuses) that partially satisfy these conditions.
Thus:
- Assertion (A) is false because other quadrilaterals (e.g., rhombus) can have diagonals bisecting at right angles without being squares.
- Reason (R) is true because, for a square, the diagonals do indeed satisfy these properties, but this does not mean the quadrilateral must necessarily be a square.

Key Concept: Diagonals of a square

c) $10\ \text{cm}$
[Solution Description]
For a square with diagonal $d$, the side length $s$ is related by $d = s\sqrt{2}$. Given $d = 10\sqrt{2}\ \text{cm}$, solving for $s$ gives:
$s = \frac{d}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10\ \text{cm}$

Key Concept: Angles between diagonals of a rectangle

a) $50^\circ, 50^\circ, 130^\circ$
[Solution Description]
In a rectangle, the diagonals bisect each other and form two pairs of equal angles. Given one angle as $x = 50^\circ$, the other angles will be $50^\circ$, $180^\circ - 50^\circ = 130^\circ$, and $130^\circ$ as they must add up to $360^\circ$ in total.

Key Concept: Diagonal Intersection, Angle Calculation

c) $100^\circ$
[Solution Description]
In a rectangle, the diagonals bisect each other and are equal in length, forming congruent isosceles triangles. For the given quadrilateral formed by the diagonals, the angles are $80^\circ$, $100^\circ$, $80^\circ$, and $100^\circ$.
Since the sum of angles around a point is $360^\circ$, the angle between the diagonals is:
$x + 180 - x = 180^\circ.$
Here, the given angles suggest two pairs of supplementary angles ($80^\circ + 100^\circ = 180^\circ$), which implies the angle between the diagonals alternates as $80^\circ$ and $100^\circ$. Thus, $x$ is either $80^\circ$ or $100^\circ$.
However, for a rectangle, the answer must align with the property that diagonals form congruent isosceles triangles, so $x = 100^\circ$ is correct (as the acute angle would be $80^\circ$).

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 diagonals

c) Square
[Solution Description]
A quadrilateral with equal diagonals that bisect each other at $90^\circ$ is a square because all angles will be $90^\circ$ and all sides will be equal.

Key Concept: Properties of Rectangle

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

Key Concept: Properties of Rectangles and Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that if a quadrilateral has equal-length diagonals that bisect each other at 90°, it must be a square. This is true because such properties are unique to squares among all quadrilaterals. The Reason correctly explains the Assertion as it lists the defining properties of the diagonals of a square. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Square Construction, Diagonals

c) $90°$
[Solution Description]
In a square, the diagonals are equal in length, bisect each other, and intersect at right angles (90°). Thus, to construct a square with a diagonal of 6 cm, the angle between the diagonals must be 90°.

Key Concept: Diagonals of Rectangle

d) 8 cm, joined at their midpoints
[Solution Description]
For a quadrilateral to be a rectangle, its diagonals must be equal in length and bisect each other. Given one diagonal is 8 cm, the other must also be 8 cm. The strips should be joined at their midpoints to ensure the diagonals bisect each other.

Key Concept: Quadrilateral Angles, Side Lengths

b) Opposite sides are equal.
[Solution Description]
A quadrilateral with all angles equal to $90^\circ$ is a rectangle, where opposite sides are equal in length. This can be proved by dividing the quadrilateral into two congruent right triangles using a diagonal and applying the properties of congruent triangles. Thus, opposite sides must be equal.

Key Concept: Diagonals of Square

a) Diagonals must be perpendicular
[Solution Description]
For a quadrilateral to be a square, its diagonals must not only be equal in length and bisect each other but also intersect at 90\degree. This ensures all sides are equal and all angles are 90\degree.

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: Rectangle Properties

a) Diagonals bisect each other at $90^\circ$
[Solution Description]
A rectangle has the following properties: all angles are $90^\circ$, opposite sides are equal and parallel, and diagonals are equal and bisect each other. The option that does not match any of these properties is the correct answer.

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: 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: 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: 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: 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: 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: Square and Rectangle Relationship

b) Every square is a rectangle, but not every rectangle is a square
[Solution Description]
A square is a special type of rectangle where all four sides are of equal length. Therefore, every square is a rectangle, but not every rectangle is a square because rectangles do not necessarily have all sides equal.

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: 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: Quadrilateral Angle Sum, Triangle Angle Property

b) $90^\circ$
[Solution Description]
We know that the sum of all angles in a quadrilateral is $360^\circ$. Let angles R and S be $x$ each. So, we can write the equation:
$60^\circ + 120^\circ + x + x = 360^\circ$.
Simplifying this: $180^\circ + 2x = 360^\circ$.
Subtract $180^\circ$ from both sides: $2x = 180^\circ$.
Divide by 2: $x = 90^\circ$.
Therefore, angle R 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: Sum of angles in a quadrilateral, diagonals creating triangles

b) 40°
[Solution Description]
Step 1: In triangle EFG, angles are 50°, 60°, and 70° (sum = 180°).
Step 2: Diagonal EG divides the quadrilateral into two triangles, EFG and EHG.
Step 3: Given one angle of triangle EHG is 40°. Let the other two angles be x and y ⇒ $40° + x + y = 180°$ ⇒ $x + y = 140°$.
Step 4: The quadrilateral's angles are the combination of both triangles' angles: 50°, 60°, 70°, 40°, x, y.
But since the quadrilateral has only four angles, they must be 50°, 60°, 70°, and 40° (the unique angles from both triangles). However, this leads to a sum of $50° + 60° + 70° + 40° = 220° ≠ 360°$, which is incorrect.
Upon reevaluating, the quadrilateral's angles are formed by combining adjacent angles from the two triangles. For example, $∠FEG + ∠HEG$ forms angle E, and similarly for others. Therefore, the quadrilateral's angles are:
$∠E = ∠FEG + ∠HEG$, $∠F = ∠EFG$, $∠G = ∠EGF + ∠EGH$, $∠H = ∠EHG$.
Given $∠F = 60°$ (from triangle EFG), $∠H = 40°$ (given in EHG), $∠FEG = 50°$, $∠EGF = 70°$, and let $∠HEG = z$, $∠EGH = w$.
Since $∠FEG + ∠HEG = ∠E ⇒ 50° + z = ∠E$.
And $∠EGF + ∠EGH = ∠G ⇒ 70° + w = ∠G$.
Also, in triangle EHG: $40° + z + w = 180° ⇒ z + w = 140°$.
Now, sum of quadrilateral angles: $∠E + ∠F + ∠G + ∠H = 360° ⇒ (50° + z) + 60° + (70° + w) + 40° = 360° ⇒ 220° + z + w = 360° ⇒ z + w = 140°$.
This confirms consistency but doesn't provide new information. To find the smallest angle, note that the quadrilateral's angles include $∠F = 60°$, $∠H = 40°$, and $∠E$ and $∠G$ are combinations of angles greater than or equal to their components. Thus, the smallest angle is $∠H = 40°$.

Key Concept: Rectangle properties

d) Rectangle
[Solution Description]
A rectangle is defined as a quadrilateral with all four angles equal to $90^\circ$, opposite sides equal, and opposite sides parallel.

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]
The assertion states that the sum of angles in any quadrilateral is 360°, which is true based on the syllabus. The reason explains that when a quadrilateral is split into two triangles via a diagonal, the sum of angles in each triangle is 180°, leading to a total of 360° for the quadrilateral. Therefore, both the assertion and reason are correct, and the reason justifies the assertion.

Key Concept: Sum of angles in quadrilateral, properties of triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
1. Assertion (A): Any quadrilateral can indeed be divided into two triangles by drawing one diagonal. This is true as a quadrilateral has four sides and connecting two non-adjacent vertices with a diagonal splits it into two distinct triangles.
2. Reason (R): The sum of the interior angles of any quadrilateral is $360^\circ$. This is true because when you divide the quadrilateral into two triangles, the sum of angles in each triangle is $180^\circ$. Adding them gives $180^\circ + 180^\circ = 360^\circ$. Thus, the reason correctly explains why the sum of angles in a quadrilateral is $360^\circ$.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Sum of angles in a quadrilateral

c) $360^\circ$
[Solution Description]
The sum of all angles in any quadrilateral is always $360^\circ$. This can be derived by dividing the quadrilateral into two triangles and adding their angle sums ($180^\circ + 180^\circ = 360^\circ$).

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: 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 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 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: 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: Properties of Parallelograms and Rhombus

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 rhombus bisect each other at $90^\circ$, which is a known property of a rhombus. The reason states that a rhombus is a special type of parallelogram with all sides equal, which is also true. However, the reason does not directly explain why the diagonals bisect at $90^\circ$; it merely defines a rhombus. Therefore, both statements are true, but the reason is not the correct explanation of the 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: 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: Properties of Parallelogram, Diagonal Bisection

a) $\frac{20}{3}$
[Solution Description]
The diagonals of a parallelogram bisect each other. Therefore, in parallelogram $PQRS$, $PO = OR$.
Given:
$PO = 4x - 5$
$OR = x + 15$
Since $PO = OR$, we set them equal:
$4x - 5 = x + 15$
Subtract $x$ from both sides:
$3x - 5 = 15$
Add $5$ to both sides:
$3x = 20$
Divide by $3$ on both sides:
$x = \frac{20}{3}$
The value of $x$ is $\frac{20}{3}$.

Key Concept: Congruent Triangles in Parallelogram

c) The opposite sides are equal and parallel.
[Solution Description]
The congruence of triangles $WXY$ and $YZW$ implies certain properties of the parallelogram $WXYZ$.
For triangles $WXY$ and $YZW$ to be congruent, their corresponding sides and angles must be equal. Specifically:
1. Opposite sides of a parallelogram are equal and parallel, so $WX = YZ$ and $XY = WZ$.
2. Diagonally opposite angles in a parallelogram are equal, so $\angle WXY = \angle YZW$ and $\angle XYW = \angle ZWY$.
This means the diagonal $WY$ divides the parallelogram into two congruent triangles. Additionally, the diagonals of a parallelogram bisect each other but do not necessarily have to be equal in length.
Therefore, the correct property deduced from the given information is that the opposite sides of $WXYZ$ are equal and parallel.

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: 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: Diagonal Properties

c) Diagonals are equal and bisect each other.
[Solution Description]
The diagonals of a rectangle are equal in length and bisect each other. This is a key property that distinguishes rectangles from general parallelograms.

Key Concept: Opposite Sides, Diagonals

b) Yes, because the sides satisfy Pythagoras' theorem
[Solution Description]
Step 1: Check if the given sides satisfy the Pythagorean theorem for a right angle. Here, $6^2 + 8^2 = 36 + 64 = 100 = 10^2$.
Step 2: This means the angle between sides $6$ cm and $8$ cm is $90°$.
Step 3: For a parallelogram to be a rectangle, all its angles must be $90°$. Since one angle is already $90°$ and opposite angles in a parallelogram are equal, all angles will be $90°$.
Step 4: Hence, the given conditions make it a rectangle.

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: 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 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 of Quadrilaterals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a square is both a rhombus and a rectangle because it meets the criteria of all three types of quadrilaterals. This is true since a square has all sides equal (rhombus property), all angles equal to $90^\circ$ (rectangle property), and opposite sides parallel (parallelogram property). The Reason explains that in a Venn diagram, squares lie in the intersection of rhombuses and rectangles, which correctly represents their hierarchical relationship. Since a square inherits properties from both rhombuses and rectangles, the Venn diagram must show this overlap, making the Reason the correct explanation of the Assertion.

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

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: Properties of Rhombus

a) 50°, 50°, 130°, 130°
[Solution Description]
In a rhombus, opposite angles are equal, and adjacent angles add up to 180°. Given one angle is 50°, its opposite angle is also 50°. The adjacent angle will be $180° - 50° = 130°$. Therefore, the angles are 50°, 130°, 50°, and 130°.

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: 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: Diagonals of Rhombus, Right Triangles

b) 40 cm
[Solution Description]
All sides of a rhombus are equal, so the perimeter is simply 4 times the side length. Given the side length is 10 cm:
$\text{Perimeter} = 4 \times 10\,\text{cm} = 40\,\text{cm}$
Note that the information about the diagonal is irrelevant for calculating the perimeter.

Key Concept: Adjacent Angles of a Rhombus

c) $180^\circ$
[Solution Description]
Adjacent angles in a rhombus add up to $180^\circ$.

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: Diagonals of a Rhombus

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

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: 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 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: Construction of Parallelograms

b) At the midpoints of both strips
[Solution Description]
To form a parallelogram, the strips must be joined at their midpoints. The lengths of the strips determine the sides of the parallelogram, but the key property is that opposite sides must be parallel and equal in length, which is ensured when the strips intersect at their midpoints.

Key Concept: Folding Paper Activity

c) Kite

[Solution Description]

Folding the paper into quarters and making a triangular crease results in a kite shape due to the symmetry of the folds and the crease.

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

d) Square
[Solution Description]
When the diagonals of a quadrilateral are perpendicular and equal in length, the quadrilateral must be a square. In a square, the diagonals bisect each other at $90^{\circ}$ and are equal in length.

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: Properties of Quadrilaterals

c) Square
[Solution Description]
The diagonals of a quadrilateral are equal in length and bisect each other at 90\textdegree. This property is satisfied by a square, where all sides and angles are equal.

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: 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: 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: 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: Constructing quadrilaterals by joining triangles of equal sides

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion (A) is true because two equilateral triangles of side length 8 cm can indeed be joined along one of their sides to form a quadrilateral (specifically, a rhombus). However, the Reason (R) is false because joining two congruent triangles along their equal sides does not necessarily result in a rectangle; it could also form other types of quadrilaterals such as a rhombus or parallelogram.

Key Concept: Constructing quadrilaterals using equilateral triangles

c) Rhombus
[Solution Description]
When two identical equilateral triangles are joined along one common side, the resulting figure has four sides. Since all sides of the original triangles are equal and they are joined along a common side, the new quadrilateral will have opposite sides equal in length. This makes it a rhombus.

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: 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 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 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: Property of Diagonals in Kite

b) Bisects $\angle ABC$ and $\angle ADC$
[Solution Description]
According to the properties of a kite, diagonal BD bisects $\angle ABC$ and $\angle ADC$. Additionally, it bisects the other diagonal AC, making $AO = OC$, and is perpendicular to AC.

Key Concept: Properties of Kite

c) The other diagonal can be of any length.
[Solution Description]
Given that in a kite, one diagonal bisects the other perpendicularly. Thus, if one diagonal (BD) is given as 6 cm, it bisects the other diagonal (AC) but does not specify its length. The other diagonal can vary based on the side lengths and angles of the kite.

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: 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: 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: Trapezium, Isosceles Trapezium, Angle Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Given an isosceles trapezium $UVWX$ with $UV \parallel XW$ and $UX = VW$.
Step 1: According to the property of a trapezium, since $UV \parallel XW$, the consecutive angles on the same side of the legs are supplementary. Therefore, $\angle U + \angle W = 180^\circ$ and $\angle V + \angle X = 180^\circ$. This makes the Assertion (A) true.
Step 2: For an isosceles trapezium, the non-parallel sides are equal by definition, i.e., $UX = VW$. This makes the Reason (R) true.
Step 3: However, the Reason (R) states a defining property of an isosceles trapezium but does not explain why $\angle U$ and $\angle W$ are supplementary. The supplementary nature of these angles arises from the parallel sides and the properties of a trapezium, not directly from the equality of the non-parallel sides.
Conclusion: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).

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: Diagonal Properties of a Kite

b) $4\,cm$
[Solution Description]
According to the properties of a kite, the diagonal $BD$ bisects the diagonal $AC$ at right angles. Hence, $AO = OC$. Given $AO = 4\,cm$, it follows that $OC = 4\,cm$.

Key Concept: Kite Diagonal Properties, Trapezium Angle Relationships

c) Assertion is true, but Reason is false.
[Solution Description]
For assertion (A), in a kite, one diagonal (main diagonal) is known to be perpendicular to the other and bisects it. This property makes $\angle AOB = 90^\circ$ true because diagonals of a kite are perpendicular by definition. Therefore, Assertion (A) is true.
For reason (R), the diagonals of a trapezium do not necessarily bisect each other unless it is an isosceles trapezium. Even in an isosceles trapezium, diagonals are equal but they only bisect each other if additional conditions are met. Hence, Reason (R) is false.
Thus, Assertion (A) is true, but Reason (R) is false.

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: Properties of Trapezium - Angles

c) $110^\circ$
[Solution Description]
In a trapezium with one pair of sides parallel (PQ$\parallel$SR), the sum of consecutive angles on the same side of the legs is $180^\circ$. Therefore:
$\angle P + \angle S = 180^\circ$
Substituting the given value:
$70^\circ + \angle S = 180^\circ \implies \angle S = 110^\circ$
Hence, the measure of $\angle S$ is $110^\circ$.