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 Rectangle and Square

b) Both have diagonals that are equal and bisect each other.
[Solution Description]
The properties common to both rectangles and squares are:
- All angles are $90^\circ$.
- Opposite sides are equal and parallel.
- Diagonals are equal in length and bisect each other.
Only option b) satisfies all these conditions.

Key Concept: Rectangles, Diagonals

c) $10$ cm
[Solution Description]
To find the length of the diagonal of a rectangle, we can use the Pythagorean theorem. The diagonal divides the rectangle into two right-angled triangles. For a rectangle with sides $a$ and $b$, the diagonal $d$ is given by $d = \sqrt{a^2 + b^2}$. Here, $a = 8$ cm and $b = 6$ cm. So,
$d = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}.$
Thus, the diagonal is $10$ 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: Diagonals and Angles

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

Key Concept: Properties of a Square

c) 32 cm
[Solution Description]
Given the diagonal $d = 8\sqrt{2}$ cm. For a square, the relationship between the side $s$ and the diagonal $d$ is given by $d = s\sqrt{2}$. Solving for $s$, we get $s = \frac{d}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8$ cm. The perimeter of the square is $4s = 4 \times 8 = 32$ 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: Properties of Squares

d) Diagonals bisect each other at $90^\circ$
[Solution Description]
A square has all sides equal and diagonals bisecting at $90^\circ$, while a rectangle only requires opposite sides to be equal and diagonals to be equal and bisect each other. Thus, the property that is unique to squares is diagonals bisecting at $90^\circ$.

Key Concept: Diagonal Properties of Square and Rhombus, Conceptual Complexity

c) 32
[Solution Description]
Step 1: In a square, diagonals are equal and bisect each other at right angles. So, $PR = QS$ and $PT = TR = QT = TS = 4\sqrt{2}$. Thus, full diagonal $PR = 8\sqrt{2}$.
Step 2: Let side of square be s. By Pythagoras' theorem in triangle PQR: $PR^2 = PQ^2 + QR^2$. But since all sides are equal, $PQ = QR = s$. So, $(8\sqrt{2})^2 = s^2 + s^2 = 2s^2$.
Step 3: Simplify: $128 = 2s^2$ → $s^2 = 64$ → $s = 8$.
Step 4: Perimeter = $4 \times 8 = 32$.

Key Concept: Diagonals of Rectangle and Square

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

Key Concept: 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: Angles Between Diagonals

c) $90^\circ$
[Solution Description]
In a square, the diagonals bisect each other at right angles ($90^\circ$). This is a unique property of squares compared to other rectangles.

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 of a rectangle

a) $50^\circ, 50^\circ, 130^\circ$
[Solution Description]
In a rectangle, the diagonals bisect each other and form two pairs of equal angles. Given one angle as $x = 50^\circ$, the other angles will be $50^\circ$, $180^\circ - 50^\circ = 130^\circ$, and $130^\circ$ as they must add up to $360^\circ$ in total.

Key Concept: Angles between diagonals of a rectangle

d) Assertion is false, but Reason is true.
[Solution Description]
In a rectangle, diagonals are equal in length and bisect each other, forming two pairs of isosceles triangles. For any angle $x$ between the diagonals, the base angles of these isosceles triangles can be calculated as $a = 90^\circ - \frac{x}{2}$ and $b = \frac{x}{2}$. The sum of the angles around the point of intersection must satisfy $a + b = 90^\circ$, leading to all four angles being $90^\circ$. However, while the diagonals of a rectangle are equal and bisect each other (Reason), this property alone does not necessarily imply that they intersect at $90^\circ$ (Assertion is false). Only in special cases like squares or rhombuses do diagonals intersect at $90^\circ$.

Key Concept: Angles between diagonals in a rectangle

c) $90^\circ$
[Solution Description]
The diagonals of a rectangle bisect each other and are equal in length. When they intersect, they form four angles. Since the rectangle has all angles equal to $90^\circ$, the diagonals intersect at right angles. Thus, each angle at the intersection is $90^\circ$.

Key Concept: Construction 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: Square Construction, Diagonals

c) $90°$
[Solution Description]
In a square, the diagonals are equal in length, bisect each other, and intersect at right angles (90°). Thus, to construct a square with a diagonal of 6 cm, the angle between the diagonals must be 90°.

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: 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: Square Construction, Diagonal Properties

c) 8 cm
[Solution Description]
In a square, both diagonals are equal in length. Therefore, if one diagonal is 8 cm, the other diagonal must also be 8 cm. This is because squares have all properties of rectangles plus the additional property that all sides are equal, leading to equal diagonals.

Key Concept: Properties of Rectangle

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

Key Concept: Diagonals of 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: 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: 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: Square Properties

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

Key Concept: Properties of 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: 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 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: Properties of Rectangles

c) The diagonals are equal in length.
[Solution Description]
Since all angles are $90^\circ$, the quadrilateral is a rectangle. By Property 2, opposite sides must be equal and parallel (Property 3). Therefore, the correct statement is that its diagonals are equal in length.

Key Concept: Special Rectangle and Square Properties

c) Opposite sides are equal
[Solution Description]
We need to identify a property that is common to both squares and rectangles.
Property 1: All angles are $90^\circ$. This is true for both squares and rectangles.
Property 2: Opposite sides are equal. This is also true for both.
Property 3: Diagonals bisect each other. This is true for both.
However, only in a square are all sides equal and diagonals perpendicular. So, the correct statement must be one that applies to both but not exclusively to either.
The most general correct statement common to both is that "Opposite sides are equal."

Key Concept: Sum of angles in a quadrilateral

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: 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: Parallelogram angles

d) $120^\circ$
[Solution Description]
In a parallelogram, adjacent angles are supplementary, meaning they add up to $180^\circ$.
Given $\angle P = 60^\circ$, then $\angle Q = 180^\circ - 60^\circ = 120^\circ$.

Key Concept: 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: Parallelogram angles

c) $110^\circ$
[Solution Description]
In a parallelogram, adjacent angles are supplementary (add up to $180^\circ$). Given $∠A = 70^\circ$, then $∠B = 180^\circ - 70^\circ = 110^\circ$.

Key Concept: 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, diagonals creating triangles

b) 40°
[Solution Description]
Step 1: In triangle EFG, angles are 50°, 60°, and 70° (sum = 180°).
Step 2: Diagonal EG divides the quadrilateral into two triangles, EFG and EHG.
Step 3: Given one angle of triangle EHG is 40°. Let the other two angles be x and y ⇒ $40° + x + y = 180°$ ⇒ $x + y = 140°$.
Step 4: The quadrilateral's angles are the combination of both triangles' angles: 50°, 60°, 70°, 40°, x, y.
But since the quadrilateral has only four angles, they must be 50°, 60°, 70°, and 40° (the unique angles from both triangles). However, this leads to a sum of $50° + 60° + 70° + 40° = 220° ≠ 360°$, which is incorrect.
Upon reevaluating, the quadrilateral's angles are formed by combining adjacent angles from the two triangles. For example, $∠FEG + ∠HEG$ forms angle E, and similarly for others. Therefore, the quadrilateral's angles are:
$∠E = ∠FEG + ∠HEG$, $∠F = ∠EFG$, $∠G = ∠EGF + ∠EGH$, $∠H = ∠EHG$.
Given $∠F = 60°$ (from triangle EFG), $∠H = 40°$ (given in EHG), $∠FEG = 50°$, $∠EGF = 70°$, and let $∠HEG = z$, $∠EGH = w$.
Since $∠FEG + ∠HEG = ∠E ⇒ 50° + z = ∠E$.
And $∠EGF + ∠EGH = ∠G ⇒ 70° + w = ∠G$.
Also, in triangle EHG: $40° + z + w = 180° ⇒ z + w = 140°$.
Now, sum of quadrilateral angles: $∠E + ∠F + ∠G + ∠H = 360° ⇒ (50° + z) + 60° + (70° + w) + 40° = 360° ⇒ 220° + z + w = 360° ⇒ z + w = 140°$.
This confirms consistency but doesn't provide new information. To find the smallest angle, note that the quadrilateral's angles include $∠F = 60°$, $∠H = 40°$, and $∠E$ and $∠G$ are combinations of angles greater than or equal to their components. Thus, the smallest angle is $∠H = 40°$.

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: Diagonal properties, angle calculation

c) $95^\circ$
[Solution Description]
First, calculate the sum of given angles: $95^\circ + 85^\circ + 85^\circ = 265^\circ$.
Using angle sum property:
$265^\circ + \angle (\text{fourth angle}) = 360^\circ$
Thus, the fourth angle = $95^\circ$.
The information about diagonals intersecting at right angles is irrelevant for calculating the fourth angle in this case, as we already know three angles and can use the angle sum property directly.

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: 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: Definition of Parallelogram

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

Key Concept: Properties of Parallelograms

c) $115^\circ$
[Solution Description]
In a parallelogram, opposite angles are equal and adjacent angles are supplementary. Given $\angle A = 65^\circ$, $\angle B$ will be $180^\circ - 65^\circ = 115^\circ$. Since $\angle D$ is opposite to $\angle B$, $\angle D = 115^\circ$.

Key Concept: Properties of Parallelogram

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

Key Concept: 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: 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: 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: Opposite sides and angles of a parallelogram

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

Key Concept: Opposite sides 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: 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: 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: Properties, Angles

b) Rhombus
[Solution Description]
Step 1: Quadrilateral with diagonals bisecting each other is a parallelogram. So, $ABCD$ is initially a parallelogram.
Step 2: Since $\angle AOB = 90°$, the diagonals intersect at right angles. A parallelogram with perpendicular diagonals is either a rhombus or a square.
Step 3: However, without knowing whether all sides are equal or angles are $90°$, we cannot confirm if it is a square. But the given condition guarantees at least rhombus properties.
Step 4: Therefore, $ABCD$ must be a rhombus among the options provided.

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: 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 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: 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 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: 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

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: 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: Diagonals and side length relationship

b) $2\sqrt{5}$ cm
[Solution Description]
Area of rhombus = $\frac{d_1 \times d_2}{2}$. Let $d_1 = x$ cm, then $d_2 = 2x$ cm.
So, $\frac{x \times 2x}{2} = 16 \implies x^2 = 16 \implies x = 4$ cm.
Thus, $d_1 = 4$ cm, $d_2 = 8$ cm.
Now, half-diagonals are $2$ cm and $4$ cm.
Side length = $\sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$ cm.

Key Concept: 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: Adjacent Angles in a Rhombus

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

Key Concept: 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: Opposite angles, Diagonal properties

b) 60
[Solution Description]
In a rhombus, diagonals bisect each other at $90^\circ$ and also bisect the angles. So $\angle QPR = 2 \times \angle QPO = 60^\circ$. Since all sides are equal, triangle $PQR$ is isosceles with $PQ = QR$. Therefore, $\angle PQR = 180^\circ - 2(30^\circ) = 120^\circ$. The diagonals intersect at right angles, so in triangle $ROQ$, $\angle ROQ = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.

Key Concept: Adjacent Angles of Rhombus

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

Key Concept: 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

c) They are equal.
[Solution Description]
According to Property 3 of a rhombus, the opposite angles are equal. For example, if one angle is $50^\circ$, its opposite angle will also be $50^\circ$.

Key Concept: Properties of Quadrilaterals

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

Key Concept: Quadrilateral Identification

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

Key Concept: 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: Folding Paper to Form Quadrilaterals

d) Kite
[Solution Description]
Folding the sheet into half creates a straight line (crease). Folding it again into a quarter creates another perpendicular straight line. Making a triangular crease at the middle corner introduces diagonal folds. When these creases are opened, they form intersecting lines within the paper. Specifically:
- Two main creases from folding into quarters (vertical and horizontal).
- Diagonal creases from the triangular fold.
The resulting pattern forms a kite shape due to the intersection of the diagonal folds with the main creases, creating two pairs of adjacent sides of unequal length meeting at the central point.

Key Concept: 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: Geoboard Activity, Quadrilateral Properties

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

Key Concept: 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: 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: 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 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

c) Rhombus
[Solution Description]
When two equilateral triangles are joined along a common side, all sides of the resulting quadrilateral will be equal (8 cm). Also, since each angle of an equilateral triangle is $60^\circ$, the angles at the vertices where the triangles are joined will add up to $120^\circ$. The other two angles remain $60^\circ$ each. The quadrilateral formed is a rhombus but not a square because its angles are not $90^\circ$.

Key Concept: Kite Properties

b) Kite
[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 has adjacent sides equal (8 cm each). This satisfies the definition of a kite, where two pairs of adjacent sides are equal. Thus, the quadrilateral formed is a kite.

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: 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: Trapezium angles, Isosceles trapezium properties

c) 80$^\circ$
[Solution Description]
In an isosceles trapezium, the angles adjacent to each base are equal. Thus, $\angle P = \angle Q = 100^\circ$. Since the sum of angles on the same side of a transversal in a trapezium is $180^\circ$, we have $\angle P + \angle S = 180^\circ$. Substituting $\angle P = 100^\circ$, we get $\angle S = 80^\circ$.

Key Concept: Kite Properties

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

Key Concept: Property of Diagonals in Kite

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

Key Concept: Kite construction, side lengths

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

Key Concept: 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: Congruency in Kite

c) Both sides and angles are equal.
[Solution Description]
In kite ABCD, diagonal BD bisects $\angle ABC$ and $\angle ADC$. Hence, $\angle ABO = \angle CBO$. Also, $AB = BC$ by definition. Since BD is common to both $\Delta AOB$ and $\Delta COB$, they are congruent by SAS (Side-Angle-Side) rule. Therefore, corresponding parts like $AO = OC$ and $\angle AOB = \angle COB$ (both right angles).

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 Calculation

b) $\angle B = 75^\circ$, $\angle C = 105^\circ$
[Solution Description]
Since $AB \parallel CD$ and $AD$ is the transversal, the sum of the angles on the same side of the transversal is $180^\circ$. Therefore, $\angle A + \angle D = 105^\circ + 75^\circ = 180^\circ$ which is correct. Now, the sum of all angles in a quadrilateral is $360^\circ$. So, $\angle B + \angle C = 360^\circ - (\angle A + \angle D) = 360^\circ - 180^\circ = 180^\circ$. Also, since $AB \parallel CD$ and $BC$ is the transversal, $\angle B + \angle C = 180^\circ$. Hence, both $\angle B$ and $\angle C$ must be supplementary but not necessarily equal unless it's an isosceles trapezium.

Key Concept: Isosceles Trapezium Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An isosceles trapezium has two parallel sides (UV$\parallel$XW) and two non-parallel sides (UX = VW). According to Property 2 of the syllabus, in an isosceles trapezium, angles opposite to equal sides are equal ($\angle$U = $\angle$V). This equality arises because the trapezium can be divided into congruent triangles (e.g., $\triangle UXY \cong \triangle VWZ$), ensuring the base angles ($\angle$U and $\angle$V) are equal. The Reason correctly states the defining property of an isosceles trapezium (equal non-parallel sides), which directly leads to the Assertion.

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: 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: 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: Properties of Isosceles Trapezium - Angles

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