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 Rectangles

d) Diagonals intersect at $90^\circ$.
[Solution Description]
A rectangle has all angles equal to $90^\circ$, opposite sides equal and parallel, and diagonals that are equal and bisect each other. However, the diagonals do not necessarily bisect each other at $90^\circ$ in a rectangle (this is a property of a square).

Key Concept: Properties of Rectangle

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that in a rectangle, the diagonals are equal and bisect each other. This is true because it is one of the key properties of a rectangle as per the syllabus.
The reason claims that if the diagonals of a quadrilateral are equal and bisect each other, then the quadrilateral must be a rectangle. While this is true for parallelograms (since rectangles are a type of parallelogram), it is not necessarily true for all quadrilaterals. For example, an isosceles trapezoid has equal diagonals but is not a rectangle. Hence, the reason is false.
Therefore, Assertion is true, but Reason is false.

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: Area Calculation with Diagonals

c) $44\%$
[Solution Description]
Let the original diagonal be $d$ and the original side length be $s$. Then, $d = s\sqrt{2}$ and the area is $A = s^2 = \left(\frac{d}{\sqrt{2}}\right)^2 = \frac{d^2}{2}$. If the diagonal increases by $20\%$, the new diagonal is $1.2d$. The new area becomes:
$A' = \frac{(1.2d)^2}{2} = \frac{1.44d^2}{2} = 1.44A$
The percentage increase in area is $(1.44 - 1) \times 100\% = 44\%$.

Key Concept: Properties of Squares

d) Diagonals bisect each other at $90^\circ$
[Solution Description]
A square has all sides equal and diagonals bisecting at $90^\circ$, while a rectangle only requires opposite sides to be equal and diagonals to be equal and bisect each other. Thus, the property that is unique to squares is diagonals bisecting at $90^\circ$.

Key Concept: Opposite Sides of a Rectangle

b) 8 cm
[Solution Description]
By the property of rectangles, opposite sides are equal in length. Here, $AB$ is opposite to $CD$. Since $AB = 8$ cm, then $CD$ must also be 8 cm.

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 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: 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: 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 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: Angles between diagonals of a rhombus

d) $120^\circ$
[Solution Description]
The diagonals of a rhombus bisect the vertex angles and intersect at $90^\circ$ in a standard case. However, given an intersection angle of $60^\circ$, we calculate:
For triangle formed by half-diagonals, let the base angles be $a$.
Then, $a + a + 60^\circ = 180^\circ$ implies $2a = 120^\circ$, so $a = 60^\circ$.
Thus, each interior angle adjacent to the $60^\circ$ angle is $60^\circ + 60^\circ = 120^\circ$.

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 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: Construction and reasoning about squares

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that a quadrilateral with equal diagonals bisecting each other at 90° must be a square. The reason states that such conditions ensure all sides are equal and all angles are 90°.
First, consider the properties of a square:
- Diagonals are equal in length.
- Diagonals bisect each other at 90°.
- All sides are equal.
- All angles are 90°.
However, a rhombus also has diagonals that bisect each other at 90° and all sides equal, but its angles are not necessarily 90°. Hence, the assertion is partially correct but not entirely, as the given conditions could also describe a rhombus unless the angles are specified as 90°. The reason correctly explains the properties but fails to distinguish between a square and a rhombus uniquely. Thus, the assertion is false because the conditions alone do not guarantee a square, while the reason is true.

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: 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: 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: 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: 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: 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: Square Construction, Diagonal Properties

c) 8 cm
[Solution Description]
In a square, both diagonals are equal in length. Therefore, if one diagonal is 8 cm, the other diagonal must also be 8 cm. This is because squares have all properties of rectangles plus the additional property that all sides are equal, leading to equal diagonals.

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: Properties of Rectangle

c) The quadrilateral must be a rectangle but not necessarily a square
[Solution Description]
When diagonals of a quadrilateral are equal and bisect each other, it must be a rectangle. However, for rectangles, diagonals should intersect at $90°$ if it's a square, otherwise they form angles not equal to $90°$ unless specified. Here the angle is $60°$, which means it's a rectangle but not necessarily a square.
Step-by-step reasoning:
1) Diagonals being equal and bisecting each other confirms it's a rectangle (Property 4 of rectangles).
2) The angle between diagonals being $60°$ shows it's not a square (where diagonals intersect at $90°$).
3) Thus, the quadrilateral is a rectangle with unequal adjacent sides.

Key Concept: Properties of a Square

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 right angles. This is true as per Property 4 of a square mentioned in the syllabus. The reason given is that a square is a special type of rectangle where all sides are equal and all angles are $90^\circ$. This is also true as per the definition and properties of a square. However, the reason does not directly explain why the diagonals bisect each other at right angles; it only defines what a square is. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Diagonals of a Rectangle

b) 28 cm
[Solution Description]
A rectangle has equal diagonals that bisect each other. Given diagonal = 10 cm and one side = 6 cm. Using the Pythagorean theorem:
Let the sides be $x$ and $y$ ($x = 6$ cm).
$$ x^2 + y^2 = 10^2$$
$$ 36 + y^2 = 100$$
$$ y^2 = 64 \Rightarrow y = 8 \text{ cm}$$
Perimeter = $2(x + y) = 2(6 + 8) = 28$ cm.

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) 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: Square and Rectangle Properties

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 all sides equal and all angles equal to $90^{\circ}$, it must be a square. This is true because by definition, a square is a rectangle with all sides equal.
The reason states that every square is a rectangle (since it meets the rectangle's condition of having all angles equal to $90^{\circ}$ and opposite sides parallel and equal), but not every rectangle is a square (as rectangles can have unequal adjacent sides). This correctly explains why the assertion is true. Hence, both statements are true, and the reason correctly explains 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: Parallelogram angles

d) $120^\circ$
[Solution Description]
In a parallelogram, adjacent angles are supplementary, meaning they add up to $180^\circ$.
Given $\angle P = 60^\circ$, then $\angle Q = 180^\circ - 60^\circ = 120^\circ$.

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, 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: Quadrilateral Angle Sum, Variable Substitution

b) $50^\circ$
[Solution Description]
The sum of all angles in a quadrilateral is $360^\circ$. Let the fourth angle be $x$.
We can write the equation: $70^\circ + 110^\circ + 130^\circ + x = 360^\circ$.
Adding the known angles: $310^\circ + x = 360^\circ$.
Subtract $310^\circ$ from both sides: $x = 50^\circ$.
Therefore, the fourth angle is $50^\circ$.

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: 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: Parallelogram angles

c) $110^\circ$
[Solution Description]
In a parallelogram, adjacent angles are supplementary (add up to $180^\circ$). Given $∠A = 70^\circ$, then $∠B = 180^\circ - 70^\circ = 110^\circ$.

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: 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: 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: 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 it is impossible to have a quadrilateral with three right angles and one non-right angle. This is true because, as per the Reason, the sum of all angles in a quadrilateral must be $360^\circ$. If three angles are each $90^\circ$, their total would be $270^\circ$, leaving the fourth angle as $90^\circ$ ($360^\circ - 270^\circ = 90^\circ$). Therefore, the fourth angle cannot be anything other than $90^\circ$. Here, the Reason correctly explains why the Assertion 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 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: 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: Opposite Angles in Parallelogram

a) $70^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal. So, if $\angle A = 70^\circ$, then $\angle C$ will also be $70^\circ$ because they are opposite angles.

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: 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: 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: 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: 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: Properties of Rectangles

d) The diagonals bisect each other at $90^\circ$
[Solution Description]
A rectangle has all angles equal to $90^\circ$, opposite sides equal and parallel, and diagonals that bisect each other. However, its diagonals do not necessarily bisect at $90^\circ$ unless it is a square. Thus, the incorrect property among the options is that the diagonals bisect at $90^\circ$.

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: 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 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: Properties of Quadrilaterals

c) Either a Square or a Rhombus.
[Solution Description]
Diagonals of a rhombus bisect each other at right angles and also bisect its angles. A square also has this property since it is a special type of rhombus. However, not all quadrilaterals with perpendicular bisecting diagonals are squares—they could also be rhombuses. Rectangles have diagonals that bisect each other but not necessarily at right angles unless they are squares.

Key Concept: Venn Diagram Relationships

d) Square
[Solution Description]
A square has all the properties of both a rectangle and a rhombus. It has all angles equal to $90^\circ$ (like a rectangle) and all sides equal (like a rhombus). Therefore, the correct answer is Square.

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: Diagonals of Rhombus

b) They bisect each other at right angles.
[Solution Description]
The diagonals of a rhombus bisect each other at right angles and also bisect the angles of the rhombus. However, they are not necessarily equal in length.

Key Concept: Adjacent angles, Opposite angles

c) $130^\circ$
[Solution Description]
In a rhombus, adjacent angles add up to $180^\circ$. If one angle is $50^\circ$, then the adjacent angle will be $180^\circ - 50^\circ = 130^\circ$.

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: Basic Properties of a Rhombus

c) Opposite sides are parallel
[Solution Description]
A rhombus has all sides equal, opposite sides parallel, adjacent angles supplementary, and diagonals bisecting each other at right angles. Among the given options, only one correctly states a property that applies to all rhombuses.
Checking each option:
1) "All angles are 90 degrees" is incorrect because a rhombus does not necessarily have all angles as 90 degrees (that would make it a square).
2) "Diagonals are equal in length" is incorrect because diagonals of a rhombus are generally not equal unless it is a square.
3) "Opposite sides are parallel" is correct as this is a fundamental property of all rhombuses.
4) "Only two sides are equal" is incorrect because all four sides must be equal in a rhombus.
Thus, the correct answer is option c).

Key Concept: Relationship with Other Quadrilaterals

d) A quadrilateral with all sides equal
[Solution Description]
We need to identify the quadrilateral that is always a rhombus but not necessarily a rectangle. Let's evaluate the options:
1) "Square" is incorrect because a square is both a rhombus and a rectangle.
2) "Parallelogram" is incorrect because not all parallelograms have all sides equal (only rhombuses do).
3) "Rectangle" is incorrect because a rectangle does not necessarily have all sides equal unless it is a square.
4) "A quadrilateral with all sides equal" is correct because it defines a rhombus, which may or may not be a rectangle (unless it is a square).
The correct answer is option d).

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, 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

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: Diagonal Properties

c) They intersect at $90^\circ$.
[Solution Description]
The diagonals of a rhombus bisect each other at right angles ($90^\circ$) and also bisect the angles of the rhombus. Thus, the correct statement is that they intersect at $90^\circ$.

Key Concept: Properties of Rhombus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A rhombus has the property that its diagonals bisect each other at $90^{\circ}$. This means the diagonals intersect at right angles and divide each other into two equal parts. Additionally, the diagonals divide the rhombus into four congruent right-angled triangles because all sides of a rhombus are equal, and the diagonals bisect the angles. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Properties of 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 Property 4 from the syllabus, so it is true. The reason states that a rhombus is a quadrilateral with all sides equal in length, which is the definition of a rhombus (Property 1), so it is also true. However, the property of diagonals bisecting at $90^\circ$ does not depend on all sides being equal but is a fundamental geometric property, so the Reason does not explain the Assertion. Thus, both are true, but Reason is NOT the correct explanation of Assertion.

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: 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: 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: 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: 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: 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: 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: 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: Joining Triangles, Kite Properties

d) All of the above
[Solution Description]
When two equilateral triangles of side 8 cm are joined along one common side, the resulting quadrilateral is a rhombus. A rhombus has all sides equal and opposite angles equal. The diagonals bisect each other at 90 degrees and also bisect the vertex angles. Hence, all the given properties apply.

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 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: 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: 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: 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: Kite Properties

d) All of the above.
[Solution Description]
According to the properties of a kite, diagonal $BD$ bisects angles $\angle ABC$ and $\angle ADC$, bisects diagonal $AC$ at right angles, and makes $AO = OC$. Thus, the correct statement must reflect these properties.

Key Concept: Trapezium Angles

c) $110^\circ$
[Solution Description]
For a trapezium with parallel sides, the sum of adjacent angles is $180^\circ$. Given $\angle P = 70^\circ$, then $\angle S = 180^\circ - 70^\circ = 110^\circ$.

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: Kite Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a kite ABCD, where AB = BC and CD = DA, the diagonal BD divides the kite into two congruent triangles $\Delta ABD$ and $\Delta CBD$ by SSS congruency (since AB = BC, AD = DC, and BD is common). This congruence implies that $\angle ABD = \angle CBD$ and $\angle ADB = \angle CDB$, meaning BD bisects both $\angle ABC$ and $\angle ADC$. Additionally, since $\Delta AOB$ and $\Delta COB$ are congruent (by SAS congruency as AB = CB, $\angle ABO = \angle CBO$, and BO is common), AO = OC, proving BD bisects AC. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Kite Properties

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the diagonals of a kite bisect each other at right angles. According to the properties of a kite, one diagonal bisects the other and is perpendicular to it, but only one diagonal gets bisected (not both). Hence, the assertion is false.
The reason states that a kite has two pairs of adjacent sides equal, which is true as per the definition of a kite. Therefore, the assertion is false, but the reason is true.

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: Basic property of trapezium angles

c) $120^\circ$
[Solution Description]
Since $PQ || SR$, by the property of trapezium, the sum of angles on the same side of the non-parallel side is $180^\circ$. Therefore, $\angle P + \angle S = 180^\circ$. Given $\angle P = 60^\circ$, we can find $\angle S$ as follows:
$\angle S = 180^\circ - 60^\circ = 120^\circ$

Key Concept: Trapezium: Angle Calculation

b) $\angle B = 75^\circ$, $\angle C = 105^\circ$
[Solution Description]
Since $AB \parallel CD$ and $AD$ is the transversal, the sum of the angles on the same side of the transversal is $180^\circ$. Therefore, $\angle A + \angle D = 105^\circ + 75^\circ = 180^\circ$ which is correct. Now, the sum of all angles in a quadrilateral is $360^\circ$. So, $\angle B + \angle C = 360^\circ - (\angle A + \angle D) = 360^\circ - 180^\circ = 180^\circ$. Also, since $AB \parallel CD$ and $BC$ is the transversal, $\angle B + \angle C = 180^\circ$. Hence, both $\angle B$ and $\angle C$ must be supplementary but not necessarily equal unless it's an isosceles trapezium.

Key Concept: Isosceles trapezium properties

b) $70^\circ$
[Solution Description]
In an isosceles trapezium, the angles opposite to the equal sides are equal. Therefore, $\angle U = \angle V$. Given that $\angle U = 70^\circ$, it follows that $\angle V$ is also $70^\circ$.

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 - 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: Kite and Trapezium properties, Diagonals

a) $40$ cm
[Solution Description]
Let the kite be $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
Given, $AC = 16$ cm and $BD = 12$ cm.
Since diagonals of a kite intersect at right angles and one diagonal is bisected, assume $AO = OC = \frac{16}{2} = 8$ cm, and $BO = OD = \frac{12}{2} = 6$ cm.
Now, using Pythagoras' theorem in triangles $AOB$ and $BOC$:
For triangle $AOB$, $AB = \sqrt{AO^2 + BO^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm.
Similarly, for triangle $BOC$, $BC = \sqrt{BO^2 + CO^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
For a kite, $AB = BC$ and $CD = DA$.
So, the perimeter $P = AB + BC + CD + DA = 10 + 10 + 10 + 10 = 40$ cm.
However, looking back at the problem, it's given that only one diagonal is bisected (which is $AC$ in our assumption). But in a kite, both diagonals have specific properties: one is bisected ($AC$) and the other is perpendicularly bisected ($BD$).
But according to the problem statement, the perimeter calculation remains consistent as $AB = BC$ and $CD = DA$ due to the kite's properties.

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 $110^\circ$.

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.