Key Concept: Properties of a 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 square are equal in length and bisect each other at 90°, which is true because it is one of the properties of a square. The reason states that a square is a quadrilateral with all sides equal and all angles equal to 90°, which is also true and explains why the diagonals have these properties. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Definition of Square

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

Key Concept: 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: Square as a Rectangle

b) Every square is a rectangle, but not every rectangle is a square
[Solution Description]
Every square is a rectangle because it meets all the conditions of a rectangle (all angles $90^\circ$, opposite sides equal), but not every rectangle is a square since rectangles may have unequal adjacent sides.

Key Concept: Diagonals of a Square

c) The diagonals are not perpendicular to each other.
[Solution Description]
The properties of the diagonals of a square are: (1) They are equal in length, (2) They bisect each other at $90^{\circ}$, and (3) They bisect the angles of the square. The statement "The diagonals are not perpendicular to each other" is false because, in a square, the diagonals are always perpendicular.

Key Concept: Square diagonals and angle bisection

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

Key Concept: Diagonals of a Rectangle

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

Key Concept: Properties of a rectangle

d) Opposite sides are equal and parallel.
[Solution Description]
A rectangle has four right angles (each 90°), opposite sides equal and parallel, and diagonals that are equal in length and bisect each other.

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: Diagonals of a Square

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

Key Concept: Diagonals of a Rhombus

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

Key Concept: Diagonals of a Rectangle

d) The point $O$ divides both diagonals into two equal parts.
[Solution Description]
The diagonals of a rectangle bisect each other, so the point of intersection $O$ divides both diagonals into two equal parts. Each diagonal will be split into two segments of equal length.

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: Quadrilaterals, Diagonals of a Rectangle and Square

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

Key Concept: Diagonals, Angles in Quadrilaterals

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

Key Concept: 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 of Square

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

Key Concept: 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 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: Quadrilateral Properties, Congruence

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

Key Concept: Definition of a Rectangle

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

Key Concept: Diagonals of Rectangle

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

Key Concept: 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 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: 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 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: Square Properties

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

Key Concept: Properties of 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: Properties of a Rectangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a quadrilateral with all angles equal to $90^\circ$ is always a rectangle, which is true by definition. The reason states that a quadrilateral with diagonals that bisect each other and are equal in length must have all angles equal to $90^\circ$. While this is also true, the reason does not correctly explain the assertion because the assertion is about angles, whereas the reason is about diagonals. Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of 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: Special Case: Square

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

Key Concept: Properties of 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: Quadrilateral Angle Sum, Algebraic Equation

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

Key Concept: Sum of angles in a quadrilateral

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: Properties of a rectangle

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

Key Concept: 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: 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: Impossible quadrilateral configuration

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

Key Concept: 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 quadrilateral, properties of triangles

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

Key Concept: 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: Properties of quadrilaterals

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

Key Concept: Quadrilateral Angle Sum Property

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

Key Concept: 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: Angle Sum in Quadrilaterals

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

Key Concept: 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: Definition of Parallelogram

c) Both pairs of opposite sides are parallel.
[Solution Description]
A parallelogram is a quadrilateral with both pairs of opposite sides parallel and equal, opposite angles equal, adjacent angles supplementary, and diagonals bisecting each other.

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

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: Congruent Triangles in Parallelogram

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

Key Concept: Diagonals of a parallelogram

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

Key Concept: 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: Opposite Sides, Diagonals

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

Key Concept: 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

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

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

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

Key Concept: Venn Diagram: Relationship between quadrilaterals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that every square is a rhombus. This is true because a square has all sides equal, which is the defining property of a rhombus. The Reason states that a square has all sides equal and all angles equal to $90^\circ$, satisfying the definition of a rhombus. This is also correct because having all sides equal is sufficient to be a rhombus, regardless of angle measures. Additionally, the Reason correctly explains why the Assertion is true. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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

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

Key Concept: 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: Comparison with Square

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

Key Concept: Properties of Rhombus

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

Key Concept: Angle Properties of Rhombus

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

Key Concept: Properties of Rhombus, Angles in Rhombus

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

Key Concept: Adjacent Angles in a Rhombus

d) $110^\circ$
[Solution Description]
In a rhombus, adjacent angles are supplementary, meaning they add up to $180^\circ$. Given $\angle A = 70^\circ$, we can find $\angle B$ using the property of adjacent angles:
$\angle A + \angle B = 180^\circ$
$70^\circ + \angle B = 180^\circ$
$\angle B = 180^\circ - 70^\circ = 110^\circ$
Therefore, the correct answer is option d).

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: Rhombus Properties

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

Key Concept: 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: 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: 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: Adjacent Angles of Rhombus

b) $110^\circ$
[Solution Description]
Property 3 mentions that adjacent angles in a rhombus add up to $180^\circ$. If one angle is $70^\circ$, then its adjacent angle will be $180^\circ - 70^\circ = 110^\circ$.

Key Concept: Folding Paper Activity

c) Kite

[Solution Description]

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

Key Concept: Extending Diagonals of Quadrilaterals

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

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

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

Key Concept: 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: Geoboard Activity, Quadrilateral Properties

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that such a quadrilateral must be a square, but the reason only guarantees it is a rectangle (as per syllabus observations). For it to be a square, additional conditions (all sides equal) are needed. Thus, the assertion is false, but the reason is true as it correctly describes properties leading to a rectangle.

Key Concept: Diagonals of Quadrilaterals, Rectangle Formation

d) Square
[Solution Description]
When the diagonals of a quadrilateral are equal in length and bisect each other at $90^\circ$, all angles formed are $90^\circ$, and opposite sides are equal. This satisfies the properties of a square. However, since the sides are not necessarily equal, it could also be a rectangle. But given the diagonals intersect at $90^\circ$, it must be a square.

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: 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: 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: Joining Triangles, Kite Properties

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

Key Concept: Constructing quadrilaterals using equilateral triangles

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

Key Concept: Kite Properties

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

Key Concept: Kite Properties

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: Kite properties, Diagonal bisection

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

Key Concept: Kite and Trapezium

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

Key Concept: 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: 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 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, angle bisector

b) $60^\circ$
[Solution Description]
First, since $BD$ bisects $\angle ABC$ and $\angle ABD = 30^\circ$, then $\angle ABC = 60^\circ$ because the angle bisector splits $\angle ABC$ into two equal angles.
Now, in a kite, one pair of opposite angles are equal. Therefore, $\angle ABC = \angle ADC = 60^\circ$.

Key Concept: Trapezium Angle Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the property of trapezium, when one pair of opposite sides is parallel (here $AB \parallel CD$), the sum of the angles on the same side of the transversal is $180^\circ$. Therefore, $\angle A + \angle D = 180^\circ$ and $\angle B + \angle C = 180^\circ$. The reason given is also true as it states the general property of angles formed by a transversal intersecting two parallel lines. Further, the reason correctly explains why the assertion is true.

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: Isosceles Trapezium: Property Verification

c) The non-parallel sides are equal and the base angles are equal.
[Solution Description]
In an isosceles trapezium, the non-parallel sides are equal in length, and the angles adjacent to each base are equal. This means that if $UVWX$ is an isosceles trapezium with $UV \parallel XW$ and $UX = VW$, then $\angle U = \angle V$.

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: Diagonal Properties of a Kite

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

Key Concept: Kite 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: Properties of Isosceles Trapezium - Angles

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