Key Concept: Rectangle Diagonals, Square Diagonals

c) $147.48^\circ$
[Solution Description]
The diagonals of a rectangle bisect each other but are not perpendicular. To find the angle between them:
1. First, find the length of one diagonal using the Pythagorean theorem: $d = \sqrt{l^2 + w^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm.
2. The diagonals in a rectangle are equal and bisect each other, forming four congruent triangles.
3. In any of these triangles, two sides are half the length of the diagonal ($5$ cm each), and the third side is the width or length of the rectangle.
4. Using the cosine rule for one triangle with sides $\frac{d}{2} = 5$ cm, $\frac{d}{2} = 5$ cm, and the included side $w = 6$ cm:
$\cos(\theta) = \frac{5^2 + 5^2 - 6^2}{2 \times 5 \times 5} = \frac{25 + 25 - 36}{50} = \frac{14}{50} = 0.28$
5. Therefore, $\theta = \cos^{-1}(0.28) \approx 73.74^\circ$. Since the angle between the diagonals is twice this angle (as the diagonals form two such triangles back-to-back), the angle between the diagonals is $\approx 147.48^\circ$.

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: Squares, Diagonals

c) $40$ cm
[Solution Description]
In a square, the length of the diagonal $d$ is related to the side length $s$ by the formula $d = s \sqrt{2}$. Given $d = 10\sqrt{2}$ cm, we can solve for $s$:
$10\sqrt{2} = s \sqrt{2} \implies s = 10 \text{ cm}.$
The perimeter $P$ of a square is four times the side length:
$P = 4s = 4 \times 10 = 40 \text{ cm}.$
Thus, the perimeter is $40$ cm.

Key Concept: Definition of a Rectangle

c) A quadrilateral with all angles equal to $90^\circ$ and opposite sides equal
[Solution Description]
A rectangle is a quadrilateral with all angles equal to $90^\circ$ and opposite sides equal. This matches the definition provided in the syllabus.

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: 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: Diagonals of a Rectangle

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

Key Concept: 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 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: 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 Rhombus

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

Key Concept: Quadrilaterals, Diagonals of a Rectangle and Square

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

Key Concept: Diagonals, 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: Assertion-Reason

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

Key Concept: Properties of diagonals in a square

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

Key Concept: Properties of Rectangle

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

Key Concept: 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: 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 diagonals

c) Square
[Solution Description]
A quadrilateral with equal diagonals that bisect each other at $90^\circ$ is a square because all angles will be $90^\circ$ and all sides will be equal.

Key Concept: Angles Between Diagonals

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

Key Concept: Diagonals of Square

a) Diagonals must be perpendicular
[Solution Description]
For a quadrilateral to be a square, its diagonals must not only be equal in length and bisect each other but also intersect at 90\degree. This ensures all sides are equal and all angles are 90\degree.

Key Concept: 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: Definition of a Rectangle

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

Key Concept: Angle Sum Property of Quadrilateral

c) $115^\circ$
[Solution Description]
Sum of interior angles of a quadrilateral = $360^\circ$.
Let the fourth angle be $x$.
$$ 80^\circ + 95^\circ + 70^\circ + x = 360^\circ$$
$$ 245^\circ + x = 360^\circ$$
$$ x = 115^\circ$$

Key Concept: Diagonals of a Rectangle

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

Key Concept: Properties of Squares and Rectangles, Geometric Reasoning

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A square has diagonals that bisect each other at $90^\circ$ and are equal in length (Property 4 of squares). Since all sides of a square are equal (Property 1), the four triangles formed by the intersection of diagonals have two sides equal (half the diagonal) and the included angle as $90^\circ$. By SAS congruency, these triangles are congruent. Hence, both Assertion and Reason are true, and the Reason explains why the Assertion is true.

Key Concept: 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: Square and Rectangle Properties

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

Key Concept: Properties of Rectangles and Squares

b) 10 cm
[Solution Description]
To find the diagonal of a rectangle, we can use the Pythagorean theorem. The diagonal divides the rectangle into two right-angled triangles.
Step 1: Let the sides of the rectangle be $a = 6$ cm and $b = 8$ cm.
Step 2: The diagonal $d$ can be calculated using the formula:
$d = \sqrt{a^2 + b^2}$
Step 3: Substituting the values:
$d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}$
Thus, the diagonal is $10$ cm.

Key Concept: Square and Rectangle Relationship

b) Every square is a rectangle, but not every rectangle is a square
[Solution Description]
A square is a special type of rectangle where all four sides are of equal length. Therefore, every square is a rectangle, but not every rectangle is a square because rectangles do not necessarily have all sides equal.

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: 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: Quadrilateral Angle Sum, Triangle Angle Property

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

Key Concept: Quadrilateral Angle Sum, Algebraic Equation

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

Key Concept: Diagonals of a quadrilateral

b) Opposite sides are equal
[Solution Description]
If the diagonals of a quadrilateral bisect each other, then the opposite sides of the quadrilateral are equal in length. This property is characteristic of parallelograms, rectangles, rhombuses, and squares.

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: 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: Impossible quadrilateral configuration

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

Key Concept: 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: Properties of quadrilaterals with three right angles

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

Key Concept: Constructing quadrilaterals

c) All four angles must be right angles if three are already right angles.
[Solution Description]
A quadrilateral cannot have three right angles and one non-right angle because the sum of its angles must be $360^\circ$. Hence, if three angles are $90^\circ$, the fourth must also be $90^\circ$.

Key Concept: Quadrilateral Angle Sum Property

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

Key Concept: 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: Opposite Angles in Parallelogram

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

Key Concept: Angle Sum in Quadrilaterals

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

Key Concept: Diagonals of 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: 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: 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: Congruent triangles in a parallelogram

c) $\Delta ABD \cong \Delta BDC$
[Solution Description]
In parallelogram ABCD, opposite sides are equal ($AB = CD$ and $AD = CB$). By the SSS (Side-Side-Side) congruence condition, $\Delta ABD \cong \Delta CDB$ because $AB = CD$, $AD = CB$, and the common side BD is equal to itself. Hence, these two triangles are congruent.

Key Concept: 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: 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: 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: Diagonal Properties

b) 10 cm
[Solution Description]
In a rectangle, the diagonals are always equal in length. Therefore, if diagonal AC is 10 cm, then diagonal BD must also be 10 cm.

Key Concept: Diagonal Properties

c) Diagonals are equal and bisect each other.
[Solution Description]
The diagonals of a rectangle are equal in length and bisect each other. This is a key property that distinguishes rectangles from general parallelograms.

Key Concept: Rectangle as a Parallelogram

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
A rectangle satisfies the definition of a parallelogram because it has two pairs of opposite sides that are parallel and equal in length (Property 3 of Rectangle). Additionally, the diagonals of a rectangle bisect each other (Property 4 of Rectangle), which is also true for any parallelogram. However, the Reason does not explain why a rectangle is a "special" type of parallelogram (i.e., due to its all angles being $90^\circ$). Thus, while both Assertion and Reason are true, the Reason does not fully explain 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: Properties of Quadrilaterals

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

Key Concept: Properties of Quadrilaterals, Venn Diagram

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

Key Concept: Special Cases of Parallelograms

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

Key Concept: Properties of Rhombus, Diagonal properties

b) 20 cm
[Solution Description]
A rhombus has diagonals that bisect each other at $90^\circ$. Given $AC = 8$ cm, so $AO = CO = 4$ cm. Given $BD = 6$ cm, so $BO = DO = 3$ cm. Using the Pythagorean theorem in $\triangle AOB$:
$AB = \sqrt{AO^2 + BO^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ cm.
Since all sides of a rhombus are equal, perimeter $= 4 \times AB = 4 \times 5 = 20$ cm.

Key Concept: Adjacent angles, Opposite angles

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

Key Concept: 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: Diagonals of Rhombus

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

Key Concept: 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: Properties of Rhombus, Angles in Rhombus

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

Key Concept: Basic Properties of a Rhombus

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

Key Concept: Properties of a Rhombus

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

Key Concept: 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: Diagonal Properties

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

Key Concept: Properties of Rhombus

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

Key Concept: 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: 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: Folding Paper Activity

c) Kite

[Solution Description]

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

Key Concept: 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: 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: 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 Properties

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

Key Concept: Diagonals of Quadrilaterals, Rectangle Formation

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

Key Concept: Geoboard Activity: Properties of Quadrilaterals

c) Square
[Solution Description]
When two diagonals are equal in length and perpendicular to each other, joining their endpoints forms a square. This is because all sides become equal due to the properties of symmetry and the right angles between the sides.

Key Concept: Trapezium Angles

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

Key Concept: Construction of quadrilaterals using triangles

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

Key Concept: Constructing quadrilaterals by joining triangles of equal sides

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion (A) is true because two equilateral triangles of side length 8 cm can indeed be joined along one of their sides to form a quadrilateral (specifically, a rhombus). However, the Reason (R) is false because joining two congruent triangles along their equal sides does not necessarily result in a rectangle; it could also form other types of quadrilaterals such as a rhombus or parallelogram.

Key Concept: Constructing quadrilaterals using isosceles triangles

b) Kite
[Solution Description]
When two identical isosceles triangles (with sides 8 cm, 8 cm, 6 cm) are joined along their unequal sides (6 cm), the resulting quadrilateral will have two pairs of adjacent sides equal. This forms a kite because $AB = BC$ (8 cm each) and $CD = DA$ (8 cm each).

Key Concept: Isosceles Trapezium Properties

b) $\angle U = \angle V$
[Solution Description]
In an isosceles trapezium, the non-parallel sides are equal, and the angles opposite these sides are equal. Thus, $\angle U = \angle V$.

Key Concept: 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: Kite Properties

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

Key Concept: 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 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 construction, side lengths

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

Key Concept: Property of Diagonals in Kite

b) Bisects $\angle ABC$ and $\angle ADC$
[Solution Description]
According to the properties of a kite, diagonal BD bisects $\angle ABC$ and $\angle ADC$. Additionally, it bisects the other diagonal AC, making $AO = OC$, and is perpendicular to AC.

Key Concept: 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: Isosceles Trapezium: Property Verification

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

Key Concept: Trapezium 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: Isosceles trapezium properties

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

Key Concept: Trapezium 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: Properties of Trapezium - Angles

c) $110^\circ$
[Solution Description]
In a trapezium with one pair of sides parallel (PQ$\parallel$SR), the sum of consecutive angles on the same side of the legs is $180^\circ$. Therefore:
$\angle P + \angle S = 180^\circ$
Substituting the given value:
$70^\circ + \angle S = 180^\circ \implies \angle S = 110^\circ$
Hence, the measure of $\angle S$ is $110^\circ$.

Key Concept: Properties of Kite: Diagonals

c) Assertion is true, but Reason is false.
[Solution Description]
According to the property of a kite, one of the diagonals (the one connecting the vertices with equal adjacent sides) bisects the other diagonal at $90^\circ$ and also bisects the angles at those vertices. However, the diagonals of a kite do not necessarily bisect each other completely; only one diagonal is bisected by the other. Therefore, the Assertion is true, but the Reason is false.

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: Kite and Trapezium properties, Diagonals

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