Key Concept: Square Area, Rectangle Perimeter

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

Key Concept: Properties of 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: Rectangles, Squares, Properties

c) Both I and II
[Solution Description]
Statement I: A quadrilateral with all angles $90^\circ$ must have opposite sides equal and parallel, making it a rectangle. This statement is correct because these are the defining properties of a rectangle.
Statement II: A quadrilateral with diagonals that are equal and bisect each other at $90^\circ$ could be either a square or a rhombus (if sides are not equal). However, since the diagonals are equal, it must be a square (as rhombuses do not necessarily have equal diagonals unless they are squares). Therefore, this statement is also correct.
Hence, both statements I and II are correct.

Key Concept: Properties of a Rectangle

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

Key Concept: Diagonals and Angles

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

Key Concept: Construction of a Square

b) 50 cm$^2$
[Solution Description]
Given the diagonal $d = 10$ cm. Using the relationship $d = s\sqrt{2}$, we find the side length $s$ as $s = \frac{d}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$ cm. The area of the square is $s^2 = (5\sqrt{2})^2 = 25 \times 2 = 50$ cm$^2$.

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: Diagonals of a rectangle

c) It must be a rectangle.
[Solution Description]
If the diagonals of a quadrilateral are equal and bisect each other, then it must be a rectangle because these properties ensure that the quadrilateral has all its angles as 90° and opposite sides equal.

Key Concept: Diagonals of Rectangle

c) Are equal in length and bisect each other
[Solution Description]
The diagonals of a rectangle are equal in length and bisect each other at their midpoints. They do not necessarily intersect at right angles unless the rectangle is a square.

Key Concept: Rectangle and Rhombus Properties, Non-Straightforward Path

c) Rhombus
[Solution Description]
Step 1: Diagonals bisecting each other imply it could be a rectangle, rhombus, or square.
Step 2: Diagonals bisecting at right angles eliminate rectangles (their diagonals are equal but don’t bisect at right angles unless it's a square).
Step 3: Unequal diagonals further eliminate squares (their diagonals are equal).
Conclusion: It must be a rhombus (diagonals bisect at right angles and are unequal).

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: Assertion-Reason

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

Key Concept: Diagonals, Congruence, Angle Properties

c) $36 \, \text{cm}^2$
[Solution Description]
The area of a rhombus is half the product of its diagonals. Given the shorter diagonal $d_1 = 6$ cm, the longer diagonal $d_2 = 2d_1 = 12$ cm.
The area is calculated as:
$\text{Area} = \frac{d_1 \times d_2}{2} = \frac{6 \times 12}{2} = 36 \, \text{cm}^2.$

Key Concept: Properties of diagonals in a square

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

Key Concept: Diagonals, Angles in Quadrilaterals

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

Key Concept: 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: Properties of Rectangles, Diagonals

b) 10 cm
[Solution Description]
In a rectangle, the diagonals are equal in length and bisect each other. Given AO = 5 cm, which is half of diagonal AC. Since diagonals are equal, BD = AC. Therefore, BD = 2 * AO = 10 cm.

Key Concept: Diagonals of a rectangle

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

Key Concept: 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: 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: 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: Properties of Rectangle

c) Opposite sides must be equal
[Solution Description]
In a quadrilateral with all angles as 90\degree, drawing a diagonal divides it into two congruent right-angled triangles. Using properties of congruent triangles, we deduce that opposite sides must be equal in length. Thus, such a quadrilateral must have opposite sides equal.

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: Quadrilateral Construction

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

Key Concept: Properties of 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: 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: Special Case: Square

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

Key Concept: 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: 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: Sum of angles in a quadrilateral

b) $110^\circ$
[Solution Description]
The sum of all angles in a quadrilateral is $360^\circ$. Given three angles: $85^\circ$, $95^\circ$, and $70^\circ$.
Let the fourth angle be $x$.
Therefore, $85^\circ + 95^\circ + 70^\circ + x = 360^\circ$.
Simplify: $250^\circ + x = 360^\circ$.
Subtract $250^\circ$ from both sides to find $x = 110^\circ$.

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: Sum of angles in a quadrilateral

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

Key Concept: Quadrilateral Angle Sum, 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

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 a quadrilateral

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

Key Concept: 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: Sum of angles in a quadrilateral

c) 110°
[Solution Description]
The sum of all angles in any quadrilateral is $360°$. Given three angles as $80°$, $100°$, and $70°$, let the fourth angle be $x$.
Therefore, $80° + 100° + 70° + x = 360°$.
Adding the known angles: $250° + x = 360°$.
Subtracting $250°$ from both sides: $x = 110°$.

Key Concept: Quadrilateral construction, angle sum

c) $100^\circ$
[Solution Description]
Using the angle sum property of quadrilaterals:
$\angle A + \angle B + \angle C + \angle D = 360^\circ$
Substituting known values:
$80^\circ + 90^\circ + 90^\circ + \angle D = 360^\circ$
$260^\circ + \angle D = 360^\circ$
Therefore, $\angle D = 100^\circ$
The side length AB = 5 cm is given but doesn't affect the angle calculation as angle sum is independent of side lengths in this case.

Key Concept: Quadrilateral angles, diagonal properties

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

Key Concept: Properties of quadrilaterals with three right angles

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

Key Concept: Sum of angles in a quadrilateral

b) $80^\circ$
[Solution Description]
The sum of all four angles in a quadrilateral is $360^\circ$. Given three angles as $90^\circ$, $90^\circ$, and $100^\circ$, let the fourth angle be $x$. Then:
$90^\circ + 90^\circ + 100^\circ + x = 360^\circ$
$280^\circ + x = 360^\circ$
$x = 80^\circ$
So, the fourth angle measures $80^\circ$.

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: Diagonals of Rhombus

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

Key Concept: Properties of 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: Properties of Parallelogram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a parallelogram, the adjacent angles add up to $180^\circ$ because they are consecutive interior angles formed by a transversal intersecting two parallel lines. This property allows us to deduce that the opposite angles must be equal: if one angle is $x$, then its adjacent angle is $180^\circ - x$. The angle opposite to $x$ will also be $x$ because it is adjacent to $180^\circ - x$ ($180^\circ - (180^\circ - x) = x$). Therefore, both Assertion and Reason are true, and Reason correctly explains why the opposite angles in a parallelogram are equal.

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: Opposite sides and angles of a parallelogram

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

Key Concept: Properties of Parallelogram

b) $5\,cm$
[Solution Description]
According to Property 1 of parallelograms, opposite sides are equal. Hence, if $AB = 5\,cm$, then $CD$ must also be equal to $5\,cm$.

Key Concept: Assertion-Reason Question

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the opposite sides of a parallelogram are equal, which is one of the fundamental properties of a parallelogram as given in the syllabus. The Reason states that a parallelogram has its opposite sides parallel, which is the definition of a parallelogram. While both statements are true, the Reason does not explain why the opposite sides are equal; it merely defines what a parallelogram is. Therefore, the correct option is b).

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: 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: Rectangles as a type of parallelogram, Diagonals in rectangles and parallelograms

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the diagonals of a rectangle are perpendicular to each other. However, this is not a property of rectangles. The correct property is that the diagonals of a rectangle are equal in length and bisect each other but are not necessarily perpendicular. They are perpendicular only in the case of a square, which is a special type of rectangle. Therefore, the Assertion is false. The reason states that the diagonals of a rectangle bisect each other at $90^{\circ}$, which is also false because while they do bisect each other, it is not necessary that they do so at $90^{\circ}$. Hence, both the Assertion and Reason are false. Thus, none of the given options is correct, but among the provided options, option d) best matches since the Reason is partially true (diagonals bisect each other, though not always at $90^{\circ}$). Note: There might be an issue with the question design as all options involve truth values that don't perfectly match.

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: Special Cases of Parallelograms

c) All its angles are $90^\circ$
[Solution Description]
A rectangle is a special parallelogram where all angles are $90^\circ$. While it has all properties of a parallelogram, the defining feature is its right angles. Thus, the correct answer is "All its angles are $90^\circ$."

Key Concept: Properties of Quadrilaterals, Venn Diagram

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

Key Concept: 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: Properties of Quadrilaterals

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

Key Concept: 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: Angle Properties of Rhombus

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

Key Concept: 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: 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: 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: Properties of a Rhombus

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, let's examine Assertion (A): "The diagonals of a rhombus bisect each other at right angles." This is true because one of the properties of a rhombus states that its diagonals bisect each other at $90^\circ$.
Next, let's check Reason (R): "In a rhombus, the adjacent angles add up to $180^\circ$ and the opposite angles are equal." This is also true as it is another property of a rhombus mentioned in the syllabus.
However, while both Assertion and Reason are true, they describe different properties of a rhombus. The fact that adjacent angles sum to $180^\circ$ and opposite angles are equal does not directly explain why the diagonals bisect at right angles. These are separate properties.
Therefore, both statements are correct, but the reason does not explain the assertion.

Key Concept: Adjacent Angles of a Rhombus

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

Key Concept: Diagonals of Rhombus

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

Key Concept: Angle properties, External points

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

Key Concept: Properties of 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: 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: Properties of Quadrilaterals

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

Key Concept: Quadrilateral Identification

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

Key Concept: Sum of Angles in a Quadrilateral

b) $100^{\circ}$
[Solution Description]
The sum of all interior angles of a quadrilateral is $360^{\circ}$. Given three angles:
$70^{\circ} + 110^{\circ} + 80^{\circ} = 260^{\circ}$
The fourth angle is calculated as:
$360^{\circ} - 260^{\circ} = 100^{\circ}$

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: Joining Triangles: Types of Quadrilaterals

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

Key Concept: Quadrilateral Properties

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

Key Concept: Properties of 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 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: Trapezium Angles

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

Key Concept: Construction of quadrilaterals using triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When two equilateral triangles are joined along a common side, the resulting quadrilateral will have all four sides equal because each side of the quadrilateral is either a side of the original equilateral triangle or the newly formed side from joining. Since an equilateral triangle has all three sides equal, this property ensures that all sides of the quadrilateral are equal. Moreover, the opposite angles of the quadrilateral will be equal due to the symmetry of the equilateral triangles. Therefore, the quadrilateral formed is a rhombus as it satisfies the conditions mentioned in the Reason. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Kite, Diagonals

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

Key Concept: Joining Triangles, Trapezium Properties

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

Key Concept: Kite and Trapezium combined, Angle calculation

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

Key Concept: 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: 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: Definition of Kite

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

Key Concept: Kite Properties, Congruent Triangles

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

Key Concept: Kite construction, side lengths

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

Key Concept: 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: 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: 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: 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: 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: Kite properties, Angle relationships

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

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: 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 $40^\circ$.