Key Concept: Square Area, Rectangle Perimeter

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

Key Concept: Carpenter's Problem - Diagonal Length

c) 8 cm
[Solution Description]
In a rectangle, the diagonals are always equal in length. Therefore, if one diagonal is 8 cm, the other diagonal must also be 8 cm.

Key Concept: Properties of Rectangle

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

Key Concept: 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: Square vs Rectangle

a) A square has all sides equal, while a rectangle does not.
[Solution Description]
A square is a special type of rectangle where all sides are equal. While all squares are rectangles, not all rectangles are squares. Rectangles only require opposite sides to be equal and all angles to be $90^\circ$.

Key Concept: Diagonals of a Square

c) $4\sqrt{2}$ cm
[Solution Description]
In a square, the diagonal divides the square into two right-angled triangles. Using the Pythagorean theorem:
$\text{Diagonal}^2 = \text{Side}^2 + \text{Side}^2$
$8^2 = 2\text{Side}^2 \\ \\ 64 = 2\text{Side}^2 \\ \\ \text{Side}^2 = 32 \\ \\ \text{Side} = \sqrt{32} = 4\sqrt{2} \text{ cm}$

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: Square Properties, Diagonal

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

Key Concept: Diagonals of Rectangle and Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a rectangle always intersect at their midpoints. This is true because, in any rectangle, the diagonals are equal in length and they bisect each other at their midpoints. The reason states that a rectangle is a type of parallelogram where the diagonals bisect each other. This is also true as all rectangles are parallelograms with congruent diagonals that bisect each other. Moreover, the reason correctly explains why the assertion is true since it provides the underlying property (parallelogram diagonals bisecting each other) that leads to the assertion being valid.
Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: 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 Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because the diagonals of a rectangle are always equal in length, as proven by the properties of rectangles. The reason is also true because the diagonals of a rectangle bisect each other at their midpoints. Furthermore, the reason correctly explains why the assertion is true, as both properties (equality and bisecting each other) are fundamental to the definition and properties of a rectangle.

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: 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: Diagonal Intersection, Angle Calculation

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

Key Concept: Angles between diagonals of a 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: Construction and reasoning about squares

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

Key Concept: Properties of square diagonals

c) $90^\circ$
[Solution Description]
In a square, the diagonals bisect each other at $90^\circ$. Therefore, the angle between the diagonals is $90^\circ$.

Key Concept: Properties of 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: 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: Properties of Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
From the syllabus, we know that if the diagonals of a quadrilateral are equal and bisect each other, then it must be a rectangle. So, the assertion (A) is true. However, the reason (R) states that the diagonals bisect at $90^\circ$, which is only true for squares, not all rectangles. Therefore, while both statements are individually correct, the reason does not correctly explain the assertion.

Key Concept: Rectangle and Square Properties, Diagonal Congruence

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a quadrilateral to be a square, its diagonals must satisfy three conditions: 1) They must be of equal length, 2) They must bisect each other, and 3) They must intersect at right angles ($90^\circ$). The Assertion states that joining equal-length strips at their midpoints with a $90^\circ$ angle forms a square, which is correct because this setup ensures all three conditions are met. The Reason correctly identifies these properties as sufficient for a quadrilateral to be a square. Therefore, both Assertion and Reason are true, and Reason explains Assertion.

Key Concept: Rectangle Definition

c) 8 cm
[Solution Description]
In a rectangle, both diagonals are equal in length. Since one diagonal is given as 8 cm, the other diagonal must also be 8 cm.

Key Concept: Rectangle Construction

d) Connect sticks at midpoints and join ends with thread
[Solution Description]
To construct a right angle using two equal-length sticks and thread:
1) Join the sticks at their midpoints (5 cm from either end) forming an X-shape.
2) Use the thread to connect all four ends of the sticks sequentially - this forms a quadrilateral.
3) Since diagonals are equal and bisect each other, the quadrilateral is a rectangle (Property 4).
4) All angles in a rectangle are $90°$, so the angle formed by adjacent sides is a right angle.
Key points:
- Equal-length sticks act as diagonals.
- Bisecting them ensures rectangle formation.
- Rectangle guarantees right angles at vertices.

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: 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: 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: 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: 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: 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: 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

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

Key Concept: Opposite angles in a parallelogram

c) $20^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal. Hence, $\angle A = \angle C$.
So, $3x + 10^\circ = 5x - 30^\circ$.
Subtract $3x$ from both sides: $10^\circ = 2x - 30^\circ$.
Add $30^\circ$ to both sides: $40^\circ = 2x$.
Divide by 2: $x = 20^\circ$.

Key Concept: 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: Sum of angles in a quadrilateral, relationship between angles

d) 90°
[Solution Description]
Step 1: Given that angles P and R are supplementary, so $P + R = 180°$.
Step 2: Given $P = 110°$, so $R = 180° - 110° = 70°$.
Step 3: Angles Q and S are supplementary, so $Q + S = 180°$.
Step 4: Sum of all angles in quadrilateral PQRS is 360°.
Step 5: So, $P + Q + R + S = 360°$ ⇒ $110° + Q + 70° + S = 360°$.
Step 6: From step 3, substitute Q as $(180° - S)$ ⇒ $110° + (180° - S) + 70° + S = 360°$.
Step 7: Solve the equation: $110° + 180° - S + 70° + S = 360°$ ⇒ $360° = 360°$.
This shows that angle S can be any value as long as $Q + S = 180°$. However, since the options provided are specific, we need to consider possible constraints. Typically, for a valid quadrilateral, $S$ must be such that none of the angles exceed practical limits.

Key Concept: Rectangle properties

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

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

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

Key Concept: Sum of angles in a quadrilateral

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

Key Concept: Quadrilateral construction, angle sum

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

Key Concept: Quadrilateral 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: Sum of angles in a quadrilateral

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

Key Concept: Quadrilateral angles, diagonal properties

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

Key Concept: 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: Properties of Parallelograms and Rhombus

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a rhombus bisect each other at $90^\circ$, which is a known property of a rhombus. The reason states that a rhombus is a special type of parallelogram with all sides equal, which is also true. However, the reason does not directly explain why the diagonals bisect at $90^\circ$; it merely defines a rhombus. Therefore, both statements are true, but the reason is not the correct explanation of the assertion.

Key Concept: 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 Rhombus

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

Key Concept: Parallelogram Properties

d) Assertion is false, but Reason is true.
[Solution Description]
Assertion (A) claims that the diagonals of a parallelogram are equal in length. However, according to the syllabus, while the diagonals of a parallelogram bisect each other (Property 4), they need not be equal in length (as stated under "Note"). Therefore, the assertion is false.
Reason (R) states that the diagonals of a parallelogram bisect each other, which is true as per Property 4 in the syllabus.
Thus, Assertion is false while Reason is true.

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 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: 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: 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: 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: 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: Special Cases of Parallelograms

d) Square
[Solution Description]
A quadrilateral with all sides equal and all angles equal to $90^\circ$ is called a square. This is because a square is a special type of rectangle where all sides are equal.

Key Concept: Venn Diagram of Quadrilaterals

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

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

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

Key Concept: Angle Properties in Rhombus

d) $120^\circ$
[Solution Description]
Adjacent angles in a rhombus are supplementary (add up to $180^\circ$). Given $\angle A = 60^\circ$, we can find $\angle B$ using the property:
$\angle A + \angle B = 180^\circ$
$60^\circ + \angle B = 180^\circ$
$\angle B = 180^\circ - 60^\circ = 120^\circ$.
The opposite angles in a rhombus are equal, so $\angle D = \angle B = 120^\circ$.

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: Properties of Rhombus and Square

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, which is true because a square has all sides equal, satisfying the definition of a rhombus. The Reason states that a rhombus with all angles equal to $90^\circ$ is a square, which is also true as this defines a square. Moreover, the Reason correctly explains why the Assertion is true, since it describes the condition under which a rhombus becomes a square.

Key Concept: Types of Quadrilaterals

c) Rhombus
[Solution Description]
A rhombus has all sides equal, but its angles are not necessarily equal (unless it is a square). Therefore, a rhombus fits this description.

Key Concept: 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: 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: 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: Diagonals of a Rhombus

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

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: Diagonals of Rhombus

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

Key Concept: Angle 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: 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 Quadrilaterals

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

Key Concept: Quadrilateral Transformation

a) Kite

[Solution Description]

Extending one diagonal changes the property of equal diagonals. Now, the figure has diagonals that are perpendicular but not equal, which defines a kite.

Key Concept: Parallelograms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the definition of a parallelogram, which is correct. The reason states that rectangles and squares are special cases of parallelograms, which is also true. Furthermore, the reason correctly explains why the assertion holds because it provides additional classification information about parallelograms. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Construction of Parallelograms

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

Key Concept: Joining Triangles: Types of Quadrilaterals

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

Key Concept: 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: Diagonal Properties

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

Key Concept: Properties of Quadrilaterals

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

Key Concept: 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: Constructing quadrilaterals using scalene triangles

d) Parallelogram
[Solution Description]
When two identical scalene triangles (with sides 6 cm, 9 cm, and 12 cm) are joined along their 6 cm sides, the resulting quadrilateral will have sides measuring 9 cm, 12 cm, 12 cm, and 9 cm. Since both pairs of opposite sides are equal, it forms a parallelogram.

Key Concept: 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, Trapezium Properties

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

Key Concept: 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 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: Trapezium Angles

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

Key Concept: 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: Relationship Between Kite and Rhombus

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

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: Definition of Kite

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

Key Concept: Kite Properties

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

Key Concept: Sum of angles in a quadrilateral

c) $20^\circ$
[Solution Description]
The sum of the interior angles of any quadrilateral is $360^\circ$. Let the fourth angle be $x$. Then:
$105^\circ + 135^\circ + 100^\circ + x = 360^\circ$
Simplifying:
$105 + 135 = 240^\circ$
$240 + 100 = 340^\circ$
$x = 360^\circ - 340^\circ = 20^\circ$
So, the fourth angle is $20^\circ$.

Key Concept: 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: Trapezium angles, Isosceles trapezium properties

b) $105^\circ$
[Solution Description]
Property of isosceles trapezium states that angles adjacent to each base are equal. Here, $\angle A = \angle B = 75^\circ$. Sum of angles on same side of leg in trapezium is $180^\circ$, so $\angle D = 180^\circ - 75^\circ = 105^\circ$. Since it's isosceles, $\angle C = \angle D = 105^\circ$.

Key Concept: Trapezium angle sum, Supplementary angles

b) $110^\circ$ and $70^\circ$
[Solution Description]
Let trapezium be ABCD with AB $\parallel$ CD. Given $\angle A = 110^\circ$ and $\angle B = 70^\circ$. Angles on same side of leg are supplementary: $\angle A + \angle D = 180^\circ \Rightarrow \angle D = 70^\circ$; $\angle B + \angle C = 180^\circ \Rightarrow \angle C = 110^\circ$. Remaining angles are $110^\circ$ and $70^\circ$.

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.

Key Concept: Angle Relationships in Trapezium

b) $110^\circ$
[Solution Description]
A property of a trapezium states that the sum of adjacent angles on the same side of the non-parallel sides is $180^\circ$. Thus, $\angle P + \angle S = 180^\circ$. Given $\angle P = 70^\circ$, we can calculate $\angle S$ as follows:
$\angle S = 180^\circ - 70^\circ = 110^\circ$

Key Concept: Diagonal Properties of a Kite

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

Key Concept: 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, $UW = VX$.