Key Concept: Venn Diagram Representation

a) Squares are entirely within the set of rectangles.
[Solution Description]
Since every square is a rectangle but not every rectangle is a square, the set of squares is entirely contained within the set of rectangles. Therefore, option a) is correct.

Key Concept: Rectangle-to-Square Diagonal Comparison

a) $\frac{4\sqrt{17}}{17}$
[Solution Description]
1. Find the perimeter of the rectangle: $P_{rect} = 2(10 + 6) = 32$ cm.
2. Since the square has the same perimeter, $P_{square} = 32$ cm, giving the side length as $s = \frac{32}{4} = 8$ cm.
3. Diagonal of the square: $d_{square} = s\sqrt{2} = 8\sqrt{2}$ cm.
4. Diagonal of the rectangle: $d_{rect} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.66$ cm.
5. The ratio is $\frac{d_{square}}{d_{rect}} = \frac{8\sqrt{2}}{\sqrt{136}} = \frac{8\sqrt{2}}{2\sqrt{34}} = \frac{4\sqrt{2}}{\sqrt{34}} = \frac{4\sqrt{68}}{34} \approx \frac{4 \times 8.246}{34} \approx 0.97$.
However, simplifying further algebraically:
$\frac{4\sqrt{2}}{\sqrt{34}} \times \frac{\sqrt{34}}{\sqrt{34}} = \frac{4\sqrt{68}}{34} = \frac{4 \times 2\sqrt{17}}{34} = \frac{8\sqrt{17}}{34} = \frac{4\sqrt{17}}{17}$.

Key Concept: Properties of a Rectangle

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

Key Concept: Rectangle properties

b) 10 cm
[Solution Description]
According to the properties of a rectangle, the diagonals are equal in length. Therefore, if diagonal $AC$ is 10 cm, diagonal $BD$ must also be 10 cm.

Key Concept: Side Length from Diagonal

c) $40$ cm
[Solution Description]
Let the side of the square be $s$. The relationship between the side and the diagonal of a square is given by $d = s\sqrt{2}$. Given $d = 10\sqrt{2}$ cm, we can solve for $s$:
$10\sqrt{2} = s\sqrt{2} \implies s = 10 \, \text{cm}$
The perimeter of the square is $4s = 4 \times 10 = 40$ cm.

Key Concept: Square Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that all sides of a square are equal, which is true as per the syllabus. The reason states that a square is a special type of rectangle with all angles equal to $90^\circ$, which is also true. However, the reason does not explain why all sides of a square are equal; it only describes another property of squares.

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: Properties of Rectangles, Diagonals

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us analyze the assertion and reason step by step:
1. The assertion states that if diagonals of a quadrilateral are equal and bisect each other at $90^\circ$, it must be a square. From the syllabus, we know this is true because the square has diagonals meeting these conditions.
2. The reason states that in a rectangle, diagonals are equal and bisect each other but may not intersect at $90^\circ$. This is also true as only squares (special rectangles) have diagonals intersecting at $90^\circ$, while general rectangles do not satisfy this condition.
3. However, the reason correctly explains why the assertion is true: A quadrilateral with diagonals equal and bisecting at $90^\circ$ must be a square because no other rectangle satisfies this property.
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion.

Key Concept: Diagonals of Rectangle and Square, Multi-step Solution

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

Key Concept: 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 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: Angles between diagonals of a rhombus

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

Key Concept: 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: 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: 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: 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: Diagonals of Quadrilaterals

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

Key Concept: Construction and reasoning about squares

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

Key Concept: 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: 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: Angles Between Diagonals

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

Key Concept: Properties of Rectangle Diagonals

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

Key Concept: Rectangle Properties

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

Key Concept: 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: Square Properties

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

Key Concept: Properties of Rectangle

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

Key Concept: Properties of a 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: Venn Diagram Representation

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

Key Concept: Properties of Rectangles and Squares

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that every square is a rectangle. This is true because a square satisfies all the properties of a rectangle: all angles are $90^\circ$, opposite sides are equal and parallel, and diagonals bisect each other.
The reason given is that in a square, all angles are $90^\circ$ and opposite sides are equal and parallel. This is also true as it aligns with the properties of both squares and rectangles. However, while the reason correctly describes properties of a square, it does not explicitly link these properties to why a square is a rectangle. The key observation from the syllabus mentions this relationship directly ("Every square is a rectangle"), but the reason does not provide an explanation for the assertion; it merely lists properties. Hence, both the assertion and reason are true, but the reason is NOT the correct explanation of the assertion.

Key Concept: Angles in a Rectangle

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

Key Concept: Quadrilateral Angle Sum, Triangle Angle Property

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

Key Concept: 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: Quadrilateral Angles

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

Key Concept: Angles formed by diagonals in a quadrilateral

c) 360°
[Solution Description]
The sum of angles in the first triangle is $40° + 60° + 80° = 180°$.
The sum of angles in the second triangle is $50° + 70° + 60° = 180°$.
Adding these gives the total sum of angles in the quadrilateral: $180° + 180° = 360°$.

Key Concept: Sum of angles in a quadrilateral

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: Sum of angles in a quadrilateral

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, let's calculate the fourth angle:
Sum of given angles = $80° + 100° + 90° = 270°$.
Since the total sum of angles in a quadrilateral is $360°$, the fourth angle must be $360° - 270° = 90°$. This matches the assertion.
The reason states that the sum of interior angles in any quadrilateral is $360°$, which is true and correctly explains why the fourth angle must be $90°$ in this case.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: 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: Sum of angles in a quadrilateral

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

Key Concept: Construction of quadrilaterals

c) The fourth angle must be a right angle
[Solution Description]
The sum of angles in any quadrilateral is $360^\circ$. If a quadrilateral has three right angles ($90^\circ$ each), their sum would be $270^\circ$. Therefore, the fourth angle must be:
$360^\circ - 270^\circ = 90^\circ$
Hence, if a quadrilateral has three right angles, the fourth angle must also be a right angle.

Key Concept: 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: Angles in a Quadrilateral

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

Key Concept: Properties of Parallelograms, Supplementary Angles

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

Key Concept: Diagonals of Parallelogram, Bisection Property

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

Key Concept: Adjacent Angles in Parallelogram

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

Key Concept: Special Cases: Rhombus, Rectangle, Square

c) All sides are equal
[Solution Description] A rhombus has all sides equal, but its angles are not necessarily $90^\circ$. A rectangle has all angles equal to $90^\circ$, but its sides are not necessarily equal. Therefore, the property that applies to all rhombuses but not necessarily to all rectangles is "All sides are equal".

Key Concept: Opposite sides and diagonals of parallelogram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us verify the assertion using the properties of parallelograms. Consider a parallelogram with adjacent sides $a$ and $b$. Let $d_1$ and $d_2$ be its diagonals. By the property of parallelograms, the diagonals bisect each other. Using the law of cosines in one of the triangles formed by the diagonals, we get:
For triangle formed by half-diagonals and side $a$:
$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 - 2 \cdot \frac{d_1}{2} \cdot \frac{d_2}{2} \cos(\theta) = a^2$
Similarly for side $b$:
$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 - 2 \cdot \frac{d_1}{2} \cdot \frac{d_2}{2} \cos(180^\circ - \theta) = b^2$
Adding these two equations and simplifying:
$2\left(\frac{d_1^2}{4} + \frac{d_2^2}{4}\right) + \frac{d_1 d_2 \cos(\theta)}{2} - \frac{d_1 d_2 \cos(\theta)}{2} = a^2 + b^2$
Which simplifies to:
$\frac{d_1^2 + d_2^2}{2} = a^2 + b^2 \implies d_1^2 + d_2^2 = 2(a^2 + b^2)$
Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Definition of Parallelogram

c) A quadrilateral in which opposite sides are parallel
[Solution Description]
A parallelogram is defined as a quadrilateral in which opposite sides are parallel.

Key Concept: 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: 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: 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: 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: Classification of Quadrilaterals

c) Every rectangle is a parallelogram, but not vice versa.
[Solution Description]
Every rectangle is a parallelogram since it has opposite sides parallel and equal. However, not every parallelogram is a rectangle because parallelograms do not necessarily have all angles equal to $90^{\circ}$.

Key Concept: Rectangles as a type of parallelogram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A rectangle is a quadrilateral with all angles equal to $90^\circ$ and opposite sides parallel and equal, which satisfies the definition of a parallelogram. The reason correctly states that a rectangle has opposite sides parallel and equal. Since both statements are true, and the Reason correctly explains why the Assertion is true, the correct answer is option a).

Key Concept: Venn Diagram Relationships

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

Key Concept: Venn Diagram, Classification

c) Every square is a rhombus.
[Solution Description]
A square is a special type of rhombus where all angles are $90^\circ$. However, not all rhombuses are squares because a rhombus does not require all angles to be $90^\circ$. Therefore, every square is a rhombus, but not every rhombus is a square.

Key Concept: Venn Diagram: Relationship between 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: Venn Diagram of Quadrilaterals

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

Key Concept: Properties of Rhombus

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

Key Concept: 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 a Rhombus

a) All sides are equal
[Solution Description]
A rhombus has all sides equal, opposite sides parallel, adjacent angles summing to $180^\circ$, and diagonals that bisect each other. From Property 1, we know all sides are equal.

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: Opposite Angles of a Rhombus

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

Key Concept: 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: Diagonals of a Rhombus

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

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: 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 - Diagonals and Angles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]

The assertion is true because the diagonals of a rhombus bisect each other at right angles, that is, $90^\circ$. The reason is also true because the diagonals of a rhombus bisect the opposite angles. However, the angle-bisecting property does not directly explain why the diagonals intersect at $90^\circ$. Therefore, both statements are true, but R is not the correct explanation of A.

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: Angle Calculation in Rhombus

d) $120^\circ$
[Solution Description]
In a rhombus, adjacent angles add up to $180^\circ$ and opposite angles are equal. Given $\angle A = 60^\circ$, its adjacent angle $\angle B$ can be found as:
$\angle B = 180^\circ - \angle A = 180^\circ - 60^\circ = 120^\circ$

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: 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: Properties of Quadrilaterals

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

Key Concept: Properties of Quadrilaterals, Constructing Quadrilaterals

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

Key Concept: Diagonal Properties

c) It is a square
[Solution Description]
A quadrilateral with diagonals that are equal and perpendicular must be a square because it satisfies all properties of a square: equal sides, equal angles, and diagonals that bisect each other at 90\textdegree.

Key Concept: Quadrilateral Properties, Angle Sum Property

c) $95^\circ$
[Solution Description]
The sum of the interior angles of a quadrilateral is always $360^\circ$. Given three angles as $90^\circ$, $90^\circ$, and $85^\circ$, we can find the fourth angle by subtracting the sum of these angles from $360^\circ$.
Calculation: $360^\circ - (90^\circ + 90^\circ + 85^\circ) = 95^\circ$.

Key Concept: Quadrilateral Transformation

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

Key Concept: 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: Joining Triangles, Kite Properties

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

Key Concept: Constructing quadrilaterals 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: 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: 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: 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 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 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: Isosceles Trapezium

d) Both $\angle U = \angle V$ and $\angle W = \angle X$
[Solution Description]
An isosceles trapezium has equal non-parallel sides and equal base angles. Therefore, angles opposite to the equal sides are equal.

Key Concept: Kite diagonals, congruence

b) $10$ cm[Solution Description]The area of a kite is given by half the product of its diagonals. Here, $PR = PO + RO = 5 + 5 = 10$ cm and the area is 50 cm\textsuperscript{2}.
Using the formula: $\text{Area} = \frac{1}{2} \times PR \times QS$.
So, $50 = \frac{1}{2} \times 10 \times QS \implies 100 = 10 \times QS \implies QS = 10$ cm.

Key Concept: Classification of Quadrilaterals

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

Key Concept: Properties of Kite

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

Key Concept: 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: Kite: Diagonal Properties

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

Key Concept: 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: Kite diagonals, Congruency

b) 6 cm
[Solution Description]
Diagonals of a kite intersect at right angles, bisecting one diagonal. Here, PR is bisected by QS, so PO = OR = 4 cm. Using Pythagoras' theorem in $\triangle$PQO: $5^2 = 4^2 + QO^2 \Rightarrow QO = 3$ cm. Thus, full diagonal QS = 2 $\times$ QO = 6 cm.

Key Concept: Trapezium 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: Diagonals of an Isosceles Trapezium

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

Key Concept: Properties of Kite - Diagonal Bisection

b) $40^\circ$
[Solution Description]
Given that diagonal BD bisects $\angle ABC$, this means it splits $\angle ABC$ into two equal angles. Since $\angle ABC = 80^\circ$, $\angle ABD$ will be half of $\angle ABC$.
Calculation:
$\angle ABD = \frac{\angle ABC}{2} = \frac{80^\circ}{2} = 40^\circ$
Thus, the measure of $\angle ABD$ is $40^\circ$.

Key Concept: Kite 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: Kite Diagonal Properties, Trapezium Angle Relationships

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