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: Diagonal Properties

c) The quadrilateral must be a rectangle.
[Solution Description]
By definition, a quadrilateral with equal and bisecting diagonals must be a rectangle. If the angle between the diagonals is also $90^\circ$, it becomes a square, but otherwise, it remains a rectangle. Hence, option c) is correct.

Key Concept: Square as a Rectangle

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

Key Concept: Rectangle properties, Diagonal congruence

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that $\triangle AOB \cong \triangle COD$ by SAS congruence in rectangle $ABCD$.
Step 1: Properties of rectangle diagonals – Diagonals are equal in length ($AC = BD$) and bisect each other at point $O$. So, $AO = OC$ and $BO = OD$.
Step 2: Since opposite sides of rectangle are equal, $AB = CD$.
Step 3: Included angle between sides $AO$ and $BO$ is $\angle AOB$, which is equal to $\angle COD$ (vertically opposite angles).
Thus, by SAS criteria ($AO = OC$, $\angle AOB = \angle COD$, $BO = OD$), $\triangle AOB \cong \triangle COD$.
The reason correctly states that diagonals are equal and bisect each other, but this property alone does not directly explain why SAS congruence holds because it doesn’t mention the vertically opposite angles or the equality of opposite sides. Hence, both assertion and reason are true, but reason is not the correct explanation.

Key Concept: Properties of Squares

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a square bisect each other at $90^\circ$, which is true because the diagonals of a square are perpendicular bisectors. The reason states that the diagonals are equal in length and bisect the angles, which is also true as it’s a property of squares. However, the reason does not correctly explain why the diagonals bisect each other at $90^\circ$ (it only mentions their equality and angle bisection). Thus, both statements are true, but the reason is not the correct explanation for the assertion.

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: Properties of Rectangles, Diagonals

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

Key Concept: Rectangle Angles, Parallel Sides

a) 30$^\circ$
[Solution Description]
Since $EFGH$ is a rectangle, all angles are $90^\circ$ and opposite sides are equal and parallel.
Draw $EM$ to midpoint $M$ of $FG$.
Points $E(0,0)$, $F(6,0)$, $G(6,8)$, $H(0,8)$ can represent the rectangle.
Midpoint $M$ of $FG$ is at $(6,4)$.
Slope of $EF$ is $m_{EF} = 0$.
Slope of $EM$ is $m_{EM} = \frac{4-0}{6-0} = \frac{2}{3}$.
The angle between two lines with slopes $m_1$ and $m_2$ is given by $\tan\theta = \left|\frac{m_2 – m_1}{1 + m_1m_2}\right|$.
Here, $m_1 = 0$, $m_2 = \frac{2}{3}$.
$\theta = \tan^{-1}\left(\frac{\frac{2}{3}}{1 + 0}\right) = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ$
However, the exact answer can be derived more simply since the tangent ratio defines the angle. Thus, $\angle FEM$ has a tangent of $\frac{2}{3}$.
But we need an exact value. Alternatively, consider that the angle formed by the median from the right angle to the hypotenuse is not straightforward.
Recalculating, let’s position $EFGH$ as $E(0,0)$, $F(a,0)$, $G(a,b)$, $H(0,b)$. Then, $M\left(a,\frac{b}{2}\right)$. The slope of $EM$ is $\frac{b/2}{a}$, so $\tan\theta = \frac{b}{2a}$.
The options do not require exact trigonometric values; hence we choose the closest logical option.
(This question is complex; the closest reasonable answer is $30^\circ$ if we assume a specific aspect ratio.)

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: Diagonals of Rectangle

c) Are equal in length and bisect each other
[Solution Description]
The diagonals of a rectangle are equal in length and bisect each other at their midpoints. They do not necessarily intersect at right angles unless the rectangle is a square.

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 of a square

c) $10\ \text{cm}$
[Solution Description]
For a square with diagonal $d$, the side length $s$ is related by $d = s\sqrt{2}$. Given $d = 10\sqrt{2}\ \text{cm}$, solving for $s$ gives:
$s = \frac{d}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10\ \text{cm}$

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: Diagonal Intersection, Angle Calculation

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

Key Concept: 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: Construction of Square

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

Key Concept: Diagonals of a rectangle

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

Key Concept: Properties of Rectangles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In a rectangle, the diagonals are always equal in length. This is a proven property of rectangles. The diagonals also bisect each other, but this property alone does not explain why they are equal in length. Both statements are true, but the Reason does not correctly explain the Assertion.

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: Properties of Rectangle Diagonals

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

Key Concept: 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: Rectangle and Square Properties, Diagonal Congruence

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

Key Concept: Quadrilateral Construction

c) Square
[Solution Description]
When the diagonals of a quadrilateral are equal in length and bisect each other at $90^\circ$, the quadrilateral must be a square. This is because such properties are specific to squares and not to other quadrilaterals like rectangles or rhombuses unless all sides are also equal.

Key Concept: Properties of a Square

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the diagonals of a square bisect each other at right angles. This is true as per Property 4 of a square mentioned in the syllabus. The reason given is that a square is a special type of rectangle where all sides are equal and all angles are $90^\circ$. This is also true as per the definition and properties of a square. However, the reason does not directly explain why the diagonals bisect each other at right angles; it only defines what a square is. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Properties of a Square

c) The diagonals are not equal in length.
[Solution Description]
Properties of a square:
1. All sides are equal.
2. All angles are $90^\circ$.
3. Diagonals are equal, bisect each other at $90^\circ$, and also bisect the angles.
The statement “The diagonals are not equal in length” is false for a square.

Key Concept: Properties of Rectangle

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

Key Concept: Properties of 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: Angles in a Rectangle

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

Key Concept: Construction of Parallelograms and Relation to Rectangles

a) 6.24 cm
[Solution Description]
To find the shorter diagonal of the parallelogram, we use the Law of Cosines.
Step 1: Let the sides be $a = 5$ cm and $b = 7$ cm, with the included angle $\theta = 60^\circ$.
Step 2: The shorter diagonal $d_1$ can be calculated using:
$d_1 = \sqrt{a^2 + b^2 – 2ab \cos(\theta)}$
Step 3: Substituting the values:
$d_1 = \sqrt{5^2 + 7^2 – 2 \times 5 \times 7 \cos(60^\circ)} = \sqrt{25 + 49 – 70 \times 0.5} = \sqrt{74 – 35} = \sqrt{39} \approx 6.24 \text{ cm}$
Therefore, the length of the shorter diagonal is approximately $6.24$ cm.

Key Concept: Special Case: Square

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

Key Concept: Sum of angles in a quadrilateral

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

Key Concept: Angles in a Quadrilateral

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The sum of the given angles is calculated as follows:
$70^\circ + 80^\circ + 100^\circ + 110^\circ = 360^\circ$.
Since the total equals $360^\circ$, the quadrilateral is valid, confirming the Assertion (A).
The Reason (R) states the fundamental property that the sum of interior angles in any quadrilateral is always $360^\circ$, which justifies why the given quadrilateral is valid. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Sum of angles in a quadrilateral

c) 360$^\circ$
[Solution Description]
The sum of all interior angles of any quadrilateral is always $360^\circ$, as per the angle sum property of quadrilaterals.

Key Concept: Quadrilateral Angle Sum, Algebraic Equation

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

Key Concept: Sum of angles in a quadrilateral

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the sum of angles in any quadrilateral is 360°, which is true based on the syllabus. The reason explains that when a quadrilateral is split into two triangles via a diagonal, the sum of angles in each triangle is 180°, leading to a total of 360° for the quadrilateral. Therefore, both the assertion and reason are correct, and the reason justifies the assertion.

Key Concept: Sum of angles in a quadrilateral, properties of diagonals

c) 110°
[Solution Description]
Step 1: The sum of all angles in a quadrilateral is 360°.
Step 2: Given angles A, B, and C are 80°, 100°, and 70° respectively.
Step 3: Calculate the sum of known angles: $80° + 100° + 70° = 250°$.
Step 4: Subtract the sum from 360° to find angle D: $360° – 250° = 110°$.

Key Concept: Parallelogram angles

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

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

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

Key Concept: Properties of quadrilaterals with three right angles

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

Key Concept: 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: 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: Angles in a Quadrilateral

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

Key Concept: 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: Angle Sum in Quadrilaterals

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

Key Concept: Properties of Parallelograms and Rhombus

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

Key Concept: Properties of Parallelograms, Supplementary Angles

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

Key Concept: Adjacent Angles in Parallelogram

c) $120°$
[Solution Description]
Property 3 of parallelograms states that adjacent angles add up to $180°$. So, if one angle is $60°$, the adjacent angle will be $180° – 60° = 120°$.

Key Concept: Definition of Parallelogram

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

Key Concept: 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: Assertion-Reason Question

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the opposite sides of a parallelogram are equal, which is one of the fundamental properties of a parallelogram as given in the syllabus. The Reason states that a parallelogram has its opposite sides parallel, which is the definition of a parallelogram. While both statements are true, the Reason does not explain why the opposite sides are equal; it merely defines what a parallelogram is. Therefore, the correct option is b).

Key Concept: Opposite Sides, Diagonals

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

Key Concept: 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: 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: Venn Diagram, Classification

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

Key Concept: Venn Diagram, Properties of Quadrilaterals

d) Parallelogram
[Solution Description]
The quadrilateral has both pairs of opposite sides equal and parallel, which means it is a parallelogram. Since the diagonals are not perpendicular and do not bisect the angles, it cannot be a rhombus or square. It also does not have all angles equal to $90^\circ$, so it is not a rectangle. Therefore, it is a general parallelogram.

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

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

Key Concept: Venn Diagram Relationships

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

Key Concept: Properties of Rhombus and Square

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

Key Concept: 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: 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: Properties of Rhombus, Angle Relationships

c) $180^\circ$
[Solution Description]
Opposite angles in a rhombus are equal, so $\angle PQR = \angle PSR = 110^\circ$. Adjacent angles in a rhombus are supplementary, so $\angle QPS = 180^\circ – \angle PQR = 70^\circ$. Therefore, the sum is:
$\angle QPS + \angle RSP = 70^\circ + 110^\circ = 180^\circ$

Key Concept: Opposite Angles of a Rhombus

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

Key Concept: Relationship with Other Quadrilaterals

d) A quadrilateral with all sides equal
[Solution Description]
We need to identify the quadrilateral that is always a rhombus but not necessarily a rectangle. Let’s evaluate the options:
1) “Square” is incorrect because a square is both a rhombus and a rectangle.
2) “Parallelogram” is incorrect because not all parallelograms have all sides equal (only rhombuses do).
3) “Rectangle” is incorrect because a rectangle does not necessarily have all sides equal unless it is a square.
4) “A quadrilateral with all sides equal” is correct because it defines a rhombus, which may or may not be a rectangle (unless it is a square).
The correct answer is option d).

Key Concept: 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: Side length, Diagonal relationship

c) 6 cm
[Solution Description]
In a rhombus, the diagonals bisect each other at right angles and divide the rhombus into four right-angled triangles. Let’s denote half of diagonal 1 as $d_1/2 = 4$ cm and half of diagonal 2 as $d_2/2 = x$ cm. Using Pythagoras theorem: $(d_1/2)^2 + (d_2/2)^2 = \text{side}^2$ → $4^2 + x^2 = 5^2$ → $16 + x^2 = 25$ → $x^2 = 9$ → $x = 3$ cm. Thus the full diagonal length is $6$ cm.

Key Concept: Angle Calculation in Rhombus

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

Key Concept: Properties of Rhombus – Diagonals and Angles

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

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

Key Concept: Properties of Rhombus

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

Key Concept: Quadrilateral Transformation

a) Kite

[Solution Description]

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

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

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

Key Concept: Folding Paper to Form Quadrilaterals

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

Key Concept: Geoboard Activity: 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: 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: 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: 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: 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: 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: 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 equilateral triangles

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

Key Concept: Trapezium Angles

d) $120^\circ$
[Solution Description]
One property of a trapezium states that the sum of angles on the same side of a leg is $180^\circ$. Here, $\angle P + \angle S = 180^\circ$. Given $\angle P = 60^\circ$, we calculate $\angle S = 180^\circ – 60^\circ = 120^\circ$.

Key Concept: Properties of a Kite

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A kite is defined as a quadrilateral with two distinct pairs of adjacent sides equal. One of its properties is that the diagonals intersect at right angles (perpendicular). Therefore, both the Assertion and Reason are true, and the Reason correctly explains why the Assertion is true.

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 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: 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: Kite Properties

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the diagonals of a kite bisect each other at right angles. According to the properties of a kite, one diagonal bisects the other and is perpendicular to it, but only one diagonal gets bisected (not both). Hence, the assertion is false.
The reason states that a kite has two pairs of adjacent sides equal, which is true as per the definition of a kite. Therefore, the assertion is false, but the reason is true.

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: 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: 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 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: Sum of angles in a quadrilateral

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

Key Concept: 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: Kite properties, Angle relationships

b) $100^\circ$
[Solution Description]
First, recall that in a kite, one diagonal bisects the angles at the vertices it connects. So, diagonal $BD$ bisects $\angle ABC$ and $\angle ADC$.
Given $\angle ABC = 120^\circ$, so $\angle ABO = \angle CBO = \frac{120^\circ}{2} = 60^\circ$.
Similarly, $\angle ADC = 80^\circ$, so $\angle ADO = \angle CDO = \frac{80^\circ}{2} = 40^\circ$.
Now, consider triangle $AOD$. The sum of angles in a triangle is $180^\circ$.
We know $\angle OAD$ can be found using the property that diagonals of a kite are perpendicular. Thus, $\angle AOD = 90^\circ$.
However, let’s verify this by calculating $\angle OAD$:
In quadrilateral $ABCD$, the sum of all angles is $360^\circ$.
So, $\angle BAD + \angle ABC + \angle BCD + \angle CDA = 360^\circ$.
Since $AB = BC$ and $CD = DA$, we can find $\angle BAD = \angle BCD$ as follows:
Let $\angle BAD = x$, then $\angle BCD = x$ (because $AB = BC$ and $CD = DA$).
So, $x + 120^\circ + x + 80^\circ = 360^\circ$.
Which gives $2x + 200^\circ = 360^\circ$.
So, $2x = 160^\circ$ and $x = 80^\circ$.
Now, $\angle OAD = \frac{\angle BAD}{2} = \frac{80^\circ}{2} = 40^\circ$.
In triangle $AOD$, $\angle AOD + \angle OAD + \angle ODA = 180^\circ$.
Substituting known values: $\angle AOD + 40^\circ + 40^\circ = 180^\circ$.
So, $\angle AOD = 100^\circ$.

Key Concept: Angle Relationships in Trapezium

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

Key Concept: 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 $115^\circ$.

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.