Key Concept: Triangle Naming, Vertices, Sides

d) 6
[Solution Description]
To name a triangle using its three vertices, the order of the vertices does not matter. The number of permutations of 3 distinct vertices is $3! = 6$. Therefore, there are 6 different ways to name the triangle.

Key Concept: Sum of Angles in a Triangle, Parallel Lines and Alternate Angles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given triangle ABC with line XY parallel to BC passing through vertex A. According to the property of parallel lines, the alternate angles are equal when a transversal cuts them. Hence:
$\angle$XAB = $\angle$ABC (Alternate angles since XY $\parallel$ BC and AB is the transversal)
Similarly, $\angle$YAC = $\angle$ACB (Alternate angles since XY $\parallel$ BC and AC is the transversal)
Now, on straight line XY at point A, the sum of angles around a point is $180^{\circ}$:
$\angle$XAB + $\angle$BAC + $\angle$YAC = $180^{\circ}$
Substituting the values from the alternate angle property:
$\angle$ABC + $\angle$BAC + $\angle$ACB = $180^{\circ}$
This proves the assertion as it matches the angle sum property of a triangle. The reason is true and correctly explains the assertion.

Key Concept: Triangle Definition \& Naming

c) $∠ZYP$
[Solution Description]
The angles of a triangle are formed by its vertices. For $∆XYZ$, the angles are $∠YXZ$, $∠XYZ$, and $∠XZY$. These can also be written as $∠X$, $∠Y$, and $∠Z$ respectively. The option that does not represent any of these angles is incorrect.

Key Concept: Triangle inequality, classifications

c) 14
[Solution Description]
For a triangle to exist, the sum of any two sides must be greater than the third side. So:
$$5 + 12 > x \Rightarrow x 12 \Rightarrow x > 7$$
$$12 + x > 5 \Rightarrow x > -7$$
Combining these, $7 < x < 17$.
For an obtuse-angled triangle, one of the following must hold (using cosine rule):
$$5^2 + 12^2 < x^2 \Rightarrow 25 + 144 169 \Rightarrow x > 13$$
OR
$$5^2 + x^2 < 12^2 \Rightarrow 25 + x^2 < 144 \Rightarrow x^2 < 119 \Rightarrow x < 10.9$$
OR
$$12^2 + x^2 < 5^2 \Rightarrow 144 + x^2 < 25 \Rightarrow x^2 7$ and $x$ must make the triangle scalene (all sides unequal), we consider only $x > 13$ (since for $x < 10.9$, $x$ could be 8 or 9 but would not satisfy all conditions).
The only valid integer values are $x = 14, 15, 16$.

Key Concept: Naming triangles, vertices ordering

c) PQR
[Solution Description]
A triangle can be named using its vertices in any order. So all permutations of P, Q, R are valid names. The symbol $\Delta$ must precede the vertex names to denote a triangle. Here, "PQR" without $\Delta$ is not a valid triangle name.

Key Concept: Identifying Triangle Angles

b) $\angle Y$
[Solution Description]
In a triangle, the angle at a vertex is denoted by the symbol $\angle$ followed by the vertex name. For $\Delta XYZ$, the angle at vertex Y is $\angle XYZ$ or simply $\angle Y$.

Key Concept: Vertices on a Straight Line

b) No, because all three points are collinear.
[Solution Description]
For three points to form a triangle, they must not lie on the same straight line. If all three vertices are collinear (lie on a straight line), the figure formed will not enclose any area and thus cannot be a triangle.

Key Concept: Basic Properties

c) 3
[Solution Description]
A triangle is defined as a three-sided polygon. Therefore, it has exactly three sides.

Key Concept: Existence of Triangle

d) 2 cm, 3 cm, 6 cm
[Solution Description]
We use the Triangle Inequality Theorem to check each option:
Option A: 7 cm, 7 cm, 14 cm
1. $7 + 7 > 14$: $14 > 14$ (False; equal, not greater)
This violates the theorem, so it cannot form a triangle.
Option B: 3 cm, 4 cm, 5 cm
1. $3 + 4 > 5$: $7 > 5$ (True)
2. $3 + 5 > 4$: $8 > 4$ (True)
3. $4 + 5 > 3$: $9 > 3$ (True)
Valid triangle.
Option C: 5 cm, 6 cm, 7 cm
1. $5 + 6 > 7$: $11 > 7$ (True)
2. $5 + 7 > 6$: $12 > 6$ (True)
3. $6 + 7 > 5$: $13 > 5$ (True)
Valid triangle.
Option D: 2 cm, 3 cm, 6 cm
1. $2 + 3 > 6$: $5 > 6$ (False)
Invalid triangle.
The correct answer is D.

Key Concept: Special Case: Collinear Vertices

d) No shape
[Solution Description]
When three vertices lie on a straight line, they are collinear. This means they cannot form a triangle since a triangle requires three non-collinear vertices. Hence, no shape is formed in this case.

Key Concept: Types of Triangles

c) Obtuse-angled
[Solution Description]
An equilateral triangle has all sides equal and all angles equal to $60^\circ$.
- It cannot be right-angled because a right-angled triangle requires one angle to be $90^\circ$.
- It cannot be obtuse-angled because an obtuse-angled triangle requires one angle to be greater than $90^\circ$.
- It is always acute-angled since all angles are $60^\circ$ (less than $90^\circ$).

Key Concept: Construction, Properties

b) $(-2, 2\sqrt{3})$
[Solution Description]
In an equilateral triangle, the centroid divides the median in a 2:1 ratio. Given that the centroid is at the origin and P is at $(4, 0)$:
1. The distance from the centroid to P is 4 units.
2. The centroid divides the median in a 2:1 ratio, so the entire median length is 6 units (since 4/2 * 3 = 6).
For an equilateral triangle, the other two vertices will be symmetric about the x-axis. Using rotational symmetry of 120 degrees around the origin:
$Q = (4\cos(120°), 4\sin(120°)) = (-2, 2\sqrt{3})$
Alternatively, another vertex could be rotated by $-120°$:
$Q = (4\cos(-120°), 4\sin(-120°)) = (-2, -2\sqrt{3})$
Thus, the possible coordinates of Q are $(-2, 2\sqrt{3})$ or $(-2, -2\sqrt{3})$.

Key Concept: Verification of Construction

c) By confirming all sides are equal and angles are $60°$
[Solution Description]
Verification steps for the construction of an equilateral triangle:
1. Measure all three sides (AB, BC, AC) using a ruler. All should be equal to the radius used (4 cm).
2. Check all three angles using a protractor. Each angle in an equilateral triangle should be $60°$.
3. Ensure the point of intersection (vertex C) was correctly marked by the intersection of the two arcs.
Thus, verifying both side lengths and angles confirms the correctness of the construction.

Key Concept: Construction of Equilateral Triangles, Properties of Equilateral Triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, recall the definition of an equilateral triangle: all three sides are equal, and all three internal angles are equal. Since the sum of angles in any triangle is $180^\circ$, each angle in an equilateral triangle must be $\frac{180^\circ}{3} = 60^\circ$.
The Assertion states that it is not possible to construct an equilateral triangle with a $90^\circ$ angle. This is correct because if one angle were $90^\circ$, the other two angles would have to sum up to $90^\circ$. However, since all angles in an equilateral triangle must be equal ($60^\circ$ each), having one angle as $90^\circ$ violates this property. Thus, such a construction is impossible.
The Reason states that all angles in an equilateral triangle are equal to $60^\circ$, which is true and directly explains why the Assertion is correct.
Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Constructing Equilateral Triangle using Ruler

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states that constructing an equilateral triangle with a ruler alone involves multiple trials because ensuring all sides are exactly equal is challenging without additional tools like a compass. The reason explains that a ruler lacks the ability to fix a radius or angle, which makes it difficult to measure and adjust the positions of points accurately. Thus, both statements are true, and the reason correctly explains why the method described in the assertion is inefficient.

Key Concept: Triangle Inequality

b) 5 mm, 10 mm, 20 mm
[Solution Description]
For lengths to form a triangle, each length must be less than the sum of the other two. Checking option b: 5 + 10 = 15 which is less than 20, so these lengths cannot form a triangle.

Key Concept: Equilateral Triangle Properties, Construction Accuracy

c) Scalene triangle with sides 5 cm, 4 cm, and an unknown side
[Solution Description]
For an accurate equilateral triangle construction:
1. Both arcs must have radius equal to the side length (5 cm).
2. Here, the second arc has radius 4 cm, which does not match the required side length.
3. The intersection point will give sides $AC = 5$ cm and $BC = 4$ cm, making it a scalene triangle.

Key Concept: Compass Construction

c) Draw the base $AB = 4$ cm
[Solution Description]
The first step involves drawing the base of the triangle. According to the syllabus, we start by drawing the base $AB = 4$ cm using a straightedge or ruler.

Key Concept: Construction, Properties of Equilateral Triangle

d) From both A and B with radius 4 cm each
[Solution Description]
To construct point C, first draw the base AB = 4 cm. Using a compass, set its width to 4 cm. Place the compass pointer at A and draw an arc above AB. Then place the compass pointer at B and draw another arc intersecting the first arc. The intersection point is C. Ensure both arcs have radius 4 cm so AC = BC = 4 cm.

Key Concept: Construction of an Equilateral Triangle, Verification of Angles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that constructing a triangle with all sides equal to 4 cm ensures each angle measures $60^\circ$. This is correct because an equilateral triangle has all sides and angles equal. The reason states that the sum of the angles in any triangle is $180^\circ$ and that an equilateral triangle has three equal angles. Both statements are true, and the reason logically explains why the assertion holds. Since each angle must be $60^\circ$ ($180^\circ/3$), the reasoning is correct.

Key Concept: Tool Usage

c) Compass
[Solution Description]
When constructing a triangle with given side lengths, arcs are drawn to locate the third vertex. A compass is the primary tool used for drawing these arcs because it allows you to accurately measure and draw a circle or arc of a specific radius from a given point. This ensures that the distances from the endpoints of the base to the third vertex are exactly equal to the given side lengths.

Key Concept: Constructing Triangles Given Sides

c) Drawing both arcs from the endpoints of the base
[Solution Description]
When constructing a triangle using three given sides, the key steps involve:
1. Drawing one side as the base (e.g., 4 cm).
2. Constructing arcs from both ends of the base with radii corresponding to the other two sides (5 cm and 6 cm).
3. The intersection point of these two arcs determines the third vertex.
The critical step is ensuring that the two arcs (with radii 5 cm and 6 cm) intersect correctly, which guarantees the accurate positioning of the third vertex. Without this step, the triangle cannot be formed.

Key Concept: Triangle Inequality

b) 3 cm, 4 cm, 8 cm
[Solution Description]
To determine if lengths can form a triangle, we apply the triangle inequality theorem which states that the sum of any two sides must be greater than the third side.
For option a) 2, 4, 5:
2+4 >5 (6>5), 2+5>4 (7>4), 4+5>2 (9>2) - valid
For option b) 3, 4, 8:
3+4 >8 (7>8 is false) - invalid
For option c) 1, 5, 5:
1+5>5 (6>5), 1+5>5 (6>5), 5+5>1 (10>1) - valid
For option d) 3.5, 3.5, 3.5:
3.5+3.5>3.5 (7>3.5) - valid
Therefore, the set that cannot form a triangle is 3, 4, 8.

Key Concept: Triangle Inequality

d) Assertion is false, but Reason is true.
[Solution Description]
To check if a triangle can be constructed with sides of length 3 cm, 4 cm, and 8 cm, we need to verify the triangle inequality condition: the sum of any two sides must be greater than the third side.
Step 1: Check $3 + 4 > 8$. Here, $7 > 8$ is false.
Step 2: Check $3 + 8 > 4$. Here, $11 > 4$ is true.
Step 3: Check $4 + 8 > 3$. Here, $12 > 3$ is true.
Since one of the conditions (Step 1) is false, the triangle cannot be constructed. Therefore, the Assertion is false. However, the Reason correctly states the triangle inequality rule, so it is true.

Key Concept: Triangle Inequality, Construction of Triangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
1. First, we check if the given sides satisfy the triangle inequality. The sides are 7 cm, 24 cm, and 25 cm.
2. Check: $7 + 24 = 31 > 25$, $7 + 25 = 32 > 24$, and $24 + 25 = 49 > 7$. All conditions hold true.
3. Since all sums are greater than the third side, the triangle can indeed be constructed as per the steps for constructing a triangle when sides are given.
4. The Reason correctly explains why the Assertion is true because the triangle inequality must be satisfied for construction to be possible.

Key Concept: Construction Steps, Triangle Inequality

d) The sum of two sides is less than the third side
[Solution Description]
Before constructing a triangle, we must verify if the given side lengths satisfy the triangle inequality. For sides $a = 3$ cm, $b = 4$ cm, and $c = 8$ cm, we check:
$a + b > c \rightarrow 3 + 4 > 8 \rightarrow 7 > 8 \text{ (False)}$
$a + c > b \rightarrow 3 + 8 > 4 \rightarrow 11 > 4 \text{ (True)}$
$b + c > a \rightarrow 4 + 8 > 3 \rightarrow 12 > 3 \text{ (True)}$
Since the first condition ($a + b > c$) is false, the triangle inequality is not satisfied. Hence, no triangle can be formed with these side lengths.

Key Concept: Existence of Triangle

b) 4, 6, 10
[Solution Description]
We need to verify each set using the triangle inequality condition ($a + b > c$):
1. For 7, 9, 11: $7 + 9 > 11$ → $16 > 11$ (True).
2. For 4, 6, 10: $4 + 6 > 10$ → $10 > 10$ (False, degenerate case).
3. For 15, 20, 30: $15 + 20 > 30$ → $35 > 30$ (True).
4. For 2, 3, 4: $2 + 3 > 4$ → $5 > 4$ (True).
The set where a triangle does not exist is 4, 6, 10.

Key Concept: Triangle Inequality, Existence of Triangles

d) Assertion is false, but Reason is true.
[Solution Description]
To determine if the triangle exists, we first identify the longest side among 5, 12, and 7. Here, $c = 12$. Now, check if the sum of the other two sides is greater than 12: $5 + 7 = 12$. Since $12 \ngtr 12$, the condition $a + b > c$ fails. Therefore, a triangle with these sides does not exist.
The Reason states that for a triangle to exist, $a + b > c$ must hold true, which is correct based on the Triangle Inequality Theorem. However, the Assertion claims that a triangle with sides 5, 12, 7 exists, which is false.
Thus, Assertion is false, but Reason is true.

Key Concept: Combining Concepts: Triangle Inequality, Variable Constraints

b) $x$ must equal 9 units exactly
[Solution Description]
1) Perimeter gives $8 + 13 + x = 30$ $\Rightarrow$ $x = 9$ units
2) Verify triangle inequalities:
- $8 + 13 > 9$ $\Rightarrow$ $21 > 9$ ✓
- $8 + 9 > 13$ $\Rightarrow$ $17 > 13$ ✓
- $13 + 9 > 8$ $\Rightarrow$ $22 > 8$ ✓
Additionally, check boundary conditions from the other way:
- Minimum possible $x > |13 - 8| = 5$
- Maximum possible $x < 13 + 8 = 21$
9 units satisfies both.

Key Concept: Triangle Inequality

b) 3, 4, 5
[Solution Description]
To determine if a triangle can be formed, we apply the Triangle Inequality Theorem which states that the sum of any two sides must be greater than the third side.
1. For option (a): 5, 10, 15
$5 + 10 = 15$ is not greater than 15. Hence, no triangle.
2. For option (b): 3, 4, 5
$3 + 4 = 7 > 5$, $3 + 5 = 8 > 4$, and $4 + 5 = 9 > 3$. All conditions satisfy.
3. For option (c): 2, 3, 6
$2 + 3 = 5$ is not greater than 6. Hence, no triangle.
4. For option (d): 1, 1, 3
$1 + 1 = 2$ is not greater than 3. Hence, no triangle.
Therefore, only option (b) satisfies all conditions.

Key Concept: Triangle Inequality Theorem, Existence of Triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we analyze the possible range for $x$ using the Triangle Inequality Theorem:
1. First inequality: $5 + 12 > x \Rightarrow 17 > x \Rightarrow x 12 \Rightarrow x > 7$.
3. Third inequality: $12 + x > 5 \Rightarrow x > -7$, which is always true for positive $x$.
Combining these results, $x$ must satisfy $7 < x < 17$. Therefore, the minimum possible value of $x$ is strictly greater than 7 cm, proving the assertion true.
The reason correctly states the Triangle Inequality Theorem, which forms the basis for verifying the assertion. Thus, both are true, and the reason explains the assertion.

Key Concept: Triangle Inequality Theorem

d) Assertion is false, but Reason is true.
[Solution Description]
To determine if a triangle with sides 5 cm, 12 cm, and 7 cm can be constructed, we check the Triangle Inequality Theorem conditions:
1. Check if $5 + 12 > 7$: $17 > 7$ is true.
2. Check if $5 + 7 > 12$: $12 > 12$ is false (since it must be strictly greater).
3. Check if $7 + 12 > 5$: $19 > 5$ is true.
Since one of the conditions fails ($5 + 7 = 12$, not greater), the Assertion (A) is false. The Reason (R) correctly states the Triangle Inequality Theorem, which is always true.

Key Concept: Triangle Inequality, Multiple Comparisons

c) x must be between 7 cm and 17 cm
[Solution Description]
For three lengths to form a triangle, the sum of any two sides must be greater than the third side. Let's analyze each case:
1) $5 + 12 > x$ → $17 > x$ or $x 12$ → $x > 7$
3) $12 + x > 5$ → $x > -7$ (always true since lengths are positive)
Combining the meaningful conditions, we get $7 < x < 17$. The value of x must be strictly between 7 cm and 17 cm.

Key Concept: Triangle Inequality

c) 1, 100, 100
[Solution Description]
To check if a set of lengths can form a triangle, we need to verify the triangle inequality: sum of any two sides must be greater than the third side.
Option 1: 5 mm, 10 mm, 20 mm
5 + 10 = 15 is not greater than 20 → No triangle possible
Option 2: 12 cm, 20 cm, 40 cm
12 + 20 = 32 is not greater than 40 → No triangle possible
Option 3: 1, 100, 100
1 + 100 > 100, 100 + 100 > 1 → Triangle possible
Option 4: 3, 6, 9
3 + 6 = 9 is not greater than 9 → No triangle possible
Therefore, only option with lengths 1, 100, 100 satisfies the triangle inequality.

Key Concept: Triangle Inequality

a) No, because $5 + 7 = 12$
[Solution Description]
To determine if the triangle exists, we check the triangle inequality conditions: $a + b > c$, $b + c > a$, and $c + a > b$. Here, $a = 5$ cm, $b = 7$ cm, $c = 12$ cm.
First condition: $5 + 7 > 12 \Rightarrow 12 > 12$ (False, since 12 is not greater than 12).
Second condition: $7 + 12 > 5 \Rightarrow 19 > 5$ (True).
Third condition: $12 + 5 > 7 \Rightarrow 17 > 7$ (True).
Since one condition fails, the triangle cannot exist.

Key Concept: Side Length Verification

d) 4, 5, 6
[Solution Description]
For each option, we apply the triangle inequality: sum of any two sides > third side.
Option a) 5, 5, 10:
5 + 5 = 10 (not greater), invalid.
Option b) 7, 8, 15:
7 + 8 = 15 (not greater), invalid.
Option c) 2, 3, 6:
2 + 3 = 5 6 (9>6),
4 + 6 > 5 (10>5),
5 + 6 > 4 (11>4). All conditions satisfied.

Key Concept: Triangle Inequality, Side Length Comparison

b) The sum of the two shorter sides must exceed the longest side.
[Solution Description]
To check if a triangle can be formed, we must verify that the sum of any two sides is greater than the third side.
Step 1: Check $7 + 10 > 18$
$17 > 18$? False.
Since one inequality fails, no triangle can be formed with these lengths.

Key Concept: Triangle Inequality

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To check if a triangle exists for given side lengths, we use the triangle inequality theorem. The theorem states that the sum of any two sides of a triangle must be greater than the third side. Here, the side lengths are 10 cm, 15 cm, and 30 cm.
Step 1: Check if $10 < 15 + 30$. Yes, because $10 < 45$.
Step 2: Check if $15 < 10 + 30$. Yes, because $15 < 40$.
Step 3: Check if $30 25$.
Since one of the conditions fails, the triangle cannot exist. Therefore, the Assertion is true. The Reason correctly explains why the triangle does not exist, as it states the necessary condition for the existence of a triangle.

Key Concept: Triangle Inequality and Circle Intersection

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that a triangle with side lengths 3 cm, 4 cm, and 8 cm can be constructed because the circles centered at points A and B with radii 3 cm and 4 cm respectively will intersect internally when AB = 8 cm. However, this is not true. To verify, let's check the triangle inequality conditions:
1. $3 + 4 > 8$ → $7 > 8$ (False)
2. $3 + 8 > 4$ → $11 > 4$ (True)
3. $4 + 8 > 3$ → $12 > 3$ (True)
Since one of the conditions fails, a triangle cannot be constructed with these side lengths. The reason provided is correct and states a fundamental property of triangles, but it does not explain why the circles would intersect internally in this case because they do not satisfy the triangle inequality. Thus, the Assertion is false, but the Reason is true.

Key Concept: Triangle Inequality

b) 5, 6, 10
[Solution Description]
To check if a set of lengths can form a triangle, we apply the Triangle Inequality Theorem. The theorem states that for any three lengths to form a triangle, the sum of any two lengths must be greater than the third length. Let's evaluate each option:
1. For 3 cm, 4 cm, 8 cm: $3 + 4 = 7$ which is not greater than 8. Hence, no triangle.
2. For 5 cm, 6 cm, 10 cm: $5 + 6 = 11 > 10$, $5 + 10 = 15 > 6$, and $6 + 10 = 16 > 5$. Triangle possible.
3. For 2 cm, 7 cm, 9 cm: $2 + 7 = 9$ which is equal to the third side. So, no triangle.
4. For 12 cm, 18 cm, 30 cm: $12 + 18 = 30$ which is equal to the third side. Hence, no triangle.
Therefore, only the second set satisfies the Triangle Inequality Theorem.

Key Concept: Triangle Inequality, Constructing Circles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states the condition for three lengths to form a triangle, which is the triangle inequality. This is correct because only when the sum of the two smaller lengths is greater than the longest length can the three lengths form a triangle. The reason discusses one specific case where the sum of the two radii equals the distance between the centers, leading to the circles touching externally. In this scenario, no triangle can be formed as all three points lie on a straight line. However, the reason does not fully explain the assertion since it focuses only on one specific condition (circles touching externally) while the assertion covers all conditions under which a triangle can or cannot be formed. Therefore, both are true but the reason is not the complete explanation of the assertion.

Key Concept: Circle Intersection and Triangle Formation

c) A triangle can be formed
[Solution Description]
From the syllabus, when two circles intersect internally (Case 3), the sum of the two radii (smaller lengths) is greater than the base AB (longest length). This directly implies that the triangle inequality is satisfied, and thus, a triangle can be formed with these lengths.

Key Concept: Triangle Inequality and Circle Intersection

c) A triangle exists because the sum of the two smaller sides is greater than the longest side.
[Solution Description]
To check if a triangle exists, we verify the triangle inequality: the sum of any two sides must be greater than the third side. Here, the longest side is 10 cm.
Step 1: Check $6 + 8 > 10$ → $14 > 10$ (true)
Step 2: Check $6 + 10 > 8$ → $16 > 8$ (true)
Step 3: Check $8 + 10 > 6$ → $18 > 6$ (true)
Since all conditions are satisfied, a triangle exists. Moreover, since the circles centered at A and B with radii 6 cm and 8 cm respectively will intersect internally (as $6 + 8 > 10$), confirming the triangle's formation.

Key Concept: Triangle Inequality and Circle Intersection

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if a triangle can exist with side lengths 7 cm, 10 cm, and 18 cm, we apply the triangle inequality theorem. The theorem states that for any three lengths to form a triangle, the sum of any two sides must be greater than the third side.
Let's check:
- 7 + 10 = 17 > 18? No, 17 10? Yes, 25 > 10.
- 10 + 18 = 28 > 7? Yes, 28 > 7.
Since one of the conditions (7 + 10 > 18) fails, no such triangle can exist. Therefore, the Assertion is true. The Reason correctly explains why the triangle does not exist, as it directly quotes the triangle inequality theorem.

Key Concept: Degenerate Equilateral Triangle

b) No, because the condition $2a = a$ implies $a = 0$
[Solution Description]
An equilateral triangle has all sides equal, i.e., $a = b = c$. For a degenerate triangle, the sum of any two sides must equal the third side. So, $a + a = a \Rightarrow 2a = a \Rightarrow a = 0$. However, lengths cannot be zero, so a degenerate equilateral triangle is not possible.

Key Concept: Triangle Inequality and Circles

b) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let us analyze the given assertion and reason step by step.
1. **Assertion Analysis**:
The lengths are 3, 6, and 9.
Check the triangle inequality:
$3 + 6 = 9$ (sum of the two smaller lengths equals the longest length).
Therefore, the circles will touch externally, and no proper triangle can be formed. This is a degenerate case.
2. **Reason Analysis**:
The reason states that when the sum of two smaller lengths equals the longest length, the circles touch externally, leading to a degenerate triangle.
This is correct because in such cases, the circles will have a single point of contact, and the three points (the centers and the point of contact) lie on a straight line, forming no area.
3. **Conclusion**:
Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Possible Triangle Lengths

c) 7
[Solution Description]
Let the third side be $x$. According to the triangle inequality:
1. $5 + 8 > x$ → $13 > x$ → $x 8$ → $x > 3$
3. $8 + x > 5$ → $x > -3$ (always true for positive lengths)
Combining these, $3 < x < 13$.
Among the options, only 7 falls within this range.

Key Concept: Triangle Inequality and Circle Intersection

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side. Here, the side lengths are 3, 6, and 9. We check if the sum of the two smaller sides (3 + 6 = 9) is greater than the largest side (9). However, 9 is not greater than 9; it is equal. Therefore, a triangle with these side lengths cannot be formed. The reason correctly explains why the assertion is true.

Key Concept: Two Angles and the Included Side

c) Sum must be less than $180^\circ$
[Solution Description]
For a triangle to exist when two angles and the included side are given, the sum of the two given angles must be less than $180^\circ$ because the sum of all three angles in a triangle is always $180^\circ$. The third angle will be $180^\circ$ minus the sum of the given two angles.

Key Concept: Two Sides and Included Angle, Sum of Angles in Triangle

b) $2\sqrt{13}$ cm
[Solution Description]
To find the third side of the triangle when two sides and the included angle are given, we can use the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab\cos(C)$
Here, $a = 6$ cm, $b = 8$ cm, and $C = 60^\circ$. Plugging in the values:
$c^2 = 6^2 + 8^2 - 2 \times 6 \times 8 \cos(60^\circ)$
Calculate each term:
$6^2 = 36$
$8^2 = 64$
$2 \times 6 \times 8 = 96$
$\cos(60^\circ) = 0.5$
So,
$c^2 = 36 + 64 - 96 \times 0.5 = 100 - 48 = 52$
Taking square root:
$c = \sqrt{52} = 2\sqrt{13} \text{ cm}$

Key Concept: Two sides and included angle

b) Draw AB = 5 cm, construct $\angle$A = 30\textdegree, mark C such that AC = 4 cm, join BC
[Solution Description]
First, draw one side AB = 5 cm. Next, at point A, construct an angle of 30\textdegree using a protractor. From point A, mark a point C on the new arm such that AC = 4 cm. Finally, join points B and C to complete the triangle ABC.

Key Concept: Construction steps

c) Draw base AB = 5 cm
[Solution Description]
The construction process begins by drawing the base side AB of given length before constructing the angle at point A and marking point C.

Key Concept: Two Sides and Included Angle

b) Construct $\angle$A = $60^\circ$
[Solution Description]
The correct order of construction is:
1. Draw the base side AB = 3 cm.
2. Using a protractor, draw an angle of $60^\circ$ at point A.
3. Measure AC = 5 cm on the new arm of the angle from point A and mark point C.
4. Join points B and C to complete the triangle ABC.
The step that comes after drawing the base side AB is constructing $\angle$A = $60^\circ$.

Key Concept: Construction of Triangle with Given Sides and Angle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if the assertion is true, we check the triangle inequality conditions for the given sides:
- 3 cm + 4 cm > 5 cm (\$7 \text{ cm} > 5 \text{ cm}\$) → True
- 3 cm + 5 cm > 4 cm (\$8 \text{ cm} > 4 \text{ cm}\$) → True
- 4 cm + 5 cm > 3 cm (\$9 \text{ cm} > 3 \text{ cm}\$) → True
Since all conditions are satisfied, the assertion is true. The reason correctly explains why the triangle can be constructed by stating the triangle inequality theorem.

Key Concept: Existence of Triangle

b) $50^\circ$, $130^\circ$ with included side 4 cm
[Solution Description]
A triangle does not exist if the sum of the given two angles is greater than or equal to $180^\circ$. Let's check each pair:
1. $50^\circ$, $130^\circ$: Sum = $180^\circ$ → No triangle exists.
2. $60^\circ$, $70^\circ$: Sum = $130^\circ < 180^\circ$ → Triangle exists.
3. $45^\circ$, $85^\circ$: Sum = $130^\circ < 180^\circ$ → Triangle exists.
4. $30^\circ$, $60^\circ$: Sum = $90^\circ < 180^\circ$ → Triangle exists.
Therefore, the pair $50^\circ$, $130^\circ$ does not form a valid triangle.

Key Concept: Sum of Angles in a Triangle

c) $80^\circ$
[Solution Description]
We know that the sum of all angles in a triangle is $180^\circ$. Given $\angle A = 40^\circ$ and $\angle B = 60^\circ$, we can find $\angle C$ as follows:
$\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (40^\circ + 60^\circ) = 80^\circ$

Key Concept: Sum of Two Angles in a Triangle

a) A triangle does not exist
[Solution Description]
According to the syllabus, when the sum of two given angles is less than $180^\circ$, a triangle exists with these angles. If the sum is greater than or equal to $180^\circ$, there is no triangle with these angles.

Key Concept: Sum of two angles determining existence of a triangle

c) The triangle exists with the third angle being $20°$
[Solution Description]
The sum of given angles $P$ and $Q$ is $70° + 90° = 160°$. Since this sum is less than $180°$, a triangle with these angles is possible. The third angle $R$ would be $180° - 160° = 20°$. Hence, such a triangle exists.

Key Concept: Angle Sum Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, let's verify the assertion. We calculate the sum of the given angles: $100^\circ + 40^\circ + 40^\circ = 180^\circ$. Since the sum is equal to $180^\circ$, the assertion is true. The reason states that the sum of the angles in a triangle must be $180^\circ$, which is also true. Moreover, the reason correctly explains why the given angles form a valid triangle. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Angle Sum Property

c) $60°$
[Solution Description]
Using the angle sum property of a triangle, the sum of all three angles is $180°$. Given two angles as $30°$ and $90°$, the third angle is calculated as follows:
Step 1: Add the two given angles.
$30° + 90° = 120°$
Step 2: Subtract the sum from $180°$ to find the third angle.
$180° - 120° = 60°$
Thus, the measure of the remaining angle is $60°$.

Key Concept: Angle Sum Property, Parallel Lines

b) 80°
[Solution Description]
Since XY is parallel to BC, alternate angles are equal. So, $∠XAB = ∠B = 45°$ and $∠YAC = ∠C = 55°$. The angles around point A add up to 180°, so:
$∠XAB + ∠BAC + ∠YAC = 180°$
$45° + ∠BAC + 55° = 180°$
$∠BAC = 180° - 100° = 80°$

Key Concept: Isosceles Triangle, Angle Sum Property

b) 60° each
[Solution Description]
Let $∠P = ∠Q = x$. The sum of angles in triangle PQR is $x + x + 60° = 180°$. So:
$2x + 60° = 180°$
$2x = 120°$
$x = 60°$
Thus, $∠P = ∠Q = 60°$.

Key Concept: Using Parallel Lines to Find Third Angle

b) 72°
[Solution Description]
To find $\angle A$, construct a line XY parallel to BC through vertex A. Since XY is parallel to BC, $\angle XAB = \angle B = 36°$ and $\angle YAC = \angle C = 72°$ (alternate angles). The sum of angles on a straight line is $180°$, so:
$36° + 72° + \angle BAC = 180°$
$108° + \angle BAC = 180°$
Therefore, $\angle BAC = 180° - 108° = 72°$.

Key Concept: Triangle Angle Sum Property

b) No, the third angle is $40°$
[Solution Description]
The sum of all angles in a triangle must be $180°$. If all three angles are $70°$, their sum would be $210°$, which is greater than $180°$. Hence, such a triangle cannot exist. If two angles are $70°$ each, the third angle is:
$180° - (70° + 70°) = 40°$.

Key Concept: Exterior Angle

c) $60°$
[Solution Description]
The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Let the other interior angle be $x$.
Therefore, $50° + x = 110°$.
Solving for $x$, we get $x = 110° - 50° = 60°$.

Key Concept: Angle Sum Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Using the angle sum property, we know that $∠A + ∠B + ∠C = 180°$. Given $∠B = 60°$ and $∠C = 70°$, substituting these values gives:
$∠A + 60° + 70° = 180°$
Simplifying, $∠A + 130° = 180°$,
Thus, $∠A = 180° - 130° = 50°$.
The assertion correctly calculates $∠A$ using the angle sum property, which is stated in the reason. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Sum of Angles in a Triangle

c) 75°
[Solution Description]
In any triangle, the sum of all three angles is always 180°. Given two angles as 45° and 60°, we find the third angle by subtracting their sum from 180°.
Step 1: Add the given angles → $45° + 60° = 105°$
Step 2: Subtract from total angle sum → $180° - 105° = 75°$
Hence, the third angle is 75°.

Key Concept: Sum of Angles in a Triangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the third angle of a triangle, we subtract the sum of the two given angles from $180°$. Here, $\text{Third Angle} = 180° - (60° + 70°) = 50°$. Thus, the Assertion is true. The Reason states that the sum of all interior angles of a triangle is $180°$, which is also true and correctly explains why the third angle must be $50°$.

Key Concept: Exterior Angles

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the angle sum property of a triangle, $\angle A + \angle B + \angle C = 180^\circ$. The exterior angle at vertex C, say $\angle ACD$, is formed by extending side BC. Since $\angle C$ and $\angle ACD$ lie on a straight line, $\angle C + \angle ACD = 180^\circ$. Comparing these two equations, we get $\angle A + \angle B = \angle ACD$. This shows that the exterior angle is indeed equal to the sum of the two opposite interior angles. The reason correctly states that the exterior angle is formed by extending one side and is supplementary to the adjacent interior angle, which is true but does not directly explain why the exterior angle equals the sum of the opposite interior angles.

Key Concept: Exterior Angle Property

b) $100^\circ$
[Solution Description]
The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Here, the opposite interior angles to the exterior angle at $C$ are $\angle A$ and $\angle B$. So, the exterior angle at $C$ can be calculated as follows:
Exterior angle at $C = \angle A + \angle B = 30^\circ + 70^\circ = 100^\circ$.

Key Concept: Exterior Angle Theorem, Unknown Angles

c) 75$^\circ$
[Solution Description]
Let $\angle PQR$ be the interior angle adjacent to the exterior angle at Q.
Since the exterior angle and $\angle PQR$ form a straight line:
Exterior angle + $\angle PQR = 180^\circ$
$125^\circ + \angle PQR = 180^\circ$
$\angle PQR = 55^\circ$
Now, use the angle sum property in $\triangle PQR$:
$\angle P + \angle PQR + \angle R = 180^\circ$
$50^\circ + 55^\circ + \angle R = 180^\circ$
$\angle R = 180^\circ - 105^\circ = 75^\circ$
Alternatively, using the exterior angle theorem:
Exterior angle = $\angle P + \angle R$
$125^\circ = 50^\circ + \angle R$
$\angle R = 75^\circ$

Key Concept: Exterior Angle Property

d) 70° and 65°
[Solution Description]
Step 1: Since the exterior angle at A is 135°, the interior angle at A is:
$180° - 135° = 45°$
Step 2: Let the angles at B and C be $x$ and $y$ respectively. From the angle sum property:
$x + y + 45° = 180°$
$x + y = 135°$
Any pair of angles that add up to 135° is valid.

Key Concept: Exterior Angles, Angle Sum Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Given $\angle A = 40^\circ$ and $\angle B = 70^\circ$. According to the angle sum property of a triangle, $\angle A + \angle B + \angle C = 180^\circ$.
Step 2: Substitute the given values:
$40^\circ + 70^\circ + \angle C = 180^\circ$.
Step 3: Calculate $\angle C$:
$110^\circ + \angle C = 180^\circ$
$\Rightarrow \angle C = 180^\circ - 110^\circ = 70^\circ$.
Step 4: The exterior angle at vertex C is supplementary to $\angle C$ because they form a linear pair. So, exterior angle at C $= 180^\circ - \angle C = 180^\circ - 70^\circ = 110^\circ$.
Step 5: Alternatively, according to the property mentioned in Reason (R), exterior angle at C $= \angle A + \angle B = 40^\circ + 70^\circ = 110^\circ$.
Conclusion: Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Exterior Angle Sum

b) 110$^\circ$
[Solution Description]
To find $\angle ZYW$, we use the property that the exterior angle is equal to the sum of the two opposite interior angles.
Step 1: Identify the given angles. $\angle X = 60^\circ$ and $\angle Z = 50^\circ$.
Step 2: Apply the exterior angle theorem. $\angle ZYW = \angle X + \angle Z$.
Step 3: Substitute the values. $\angle ZYW = 60^\circ + 50^\circ = 110^\circ$.
Therefore, the measure of $\angle ZYW$ is $110^\circ$.

Key Concept: Angle Sum Property, Exterior Angle Theorem

b) $80^\circ$
[Solution Description]
First, find $\angle C$ using the angle sum property of a triangle:
$\angle A + \angle B + \angle C = 180^\circ$.
$35^\circ + 45^\circ + \angle C = 180^\circ$.
Adding $35^\circ$ and $45^\circ$ gives $80^\circ$.
So, $80^\circ + \angle C = 180^\circ$.
Therefore, $\angle C = 180^\circ - 80^\circ = 100^\circ$.
Now, using the exterior angle theorem, $\angle ACD = \angle A + \angle B$.
Substituting the given values: $\angle ACD = 35^\circ + 45^\circ = 80^\circ$.
Alternatively, since $\angle ACD$ and $\angle C$ are supplementary (they form a straight angle), $\angle ACD = 180^\circ - 100^\circ = 80^\circ$.

Key Concept: Exterior Angle Calculation

b) 110°
[Solution Description]
First, find the third angle of the triangle using the angle sum property.
$40° + 70° + \angle C = 180°$
$110° + \angle C = 180°$
$\angle C = 70°$
The exterior angle at vertex C is supplementary to angle C since they form a straight line.
$\text{Exterior angle} = 180° - 70° = 110°$

Key Concept: Equilateral Triangle Altitudes

c) Three, all equal in length
[Solution Description]
An equilateral triangle has three sides equal and all angles equal to $60^{\circ}$. Since all sides are equal, the altitudes drawn from each vertex to the opposite side are also equal in length. Moreover, in an equilateral triangle, the altitudes coincide with the medians and angle bisectors, and they intersect at a common point known as the centroid, orthocenter, or circumcenter due to the symmetry of the triangle. Thus, there are three equal altitudes in an equilateral triangle.

Key Concept: Altitudes in Right-Angled Triangles

b) One of the legs
[Solution Description]
In a right-angled triangle, one of the legs (the sides forming the right angle) can also be an altitude because it is perpendicular to the other leg. For example, if the right angle is at C, the side AC is perpendicular to BC, so AC is both a side and an altitude.

Key Concept: Extending the Base

d) In obtuse-angled triangles where the foot of the perpendicular lies outside the base.
[Solution Description]
Extending the base is necessary when the foot of the perpendicular from the vertex does not lie on the base itself but lies outside it. This typically happens in obtuse-angled triangles where one of the angles is greater than $90^\circ$. For such triangles, the altitude from the vertex opposite the obtuse angle will fall outside the triangle, requiring the base to be extended to meet the perpendicular line drawn from the vertex.

Key Concept: Construction of Altitudes Using Ruler and Set Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that an accurate altitude cannot be constructed without a set square, which is true because freehand drawing may not ensure exact perpendicularity. The reason explains why a set square is necessary—it guarantees the perpendicular condition required for altitudes. Therefore, both are true, and the reason correctly justifies the assertion.

Key Concept: Properties of altitudes

d) Infinitely many
[Solution Description]
Given the hypotenuse $AC = 5$ cm, there are infinitely many right-angled triangles possible because the other two sides can vary in length as long as they satisfy Pythagoras' theorem ($AB^2 + BC^2 = AC^2$). The angles $\angle A$ and $\angle C$ can also vary, but their sum must always be $90^\circ$.

Key Concept: Right-angled triangles, Altitude properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Consider a right-angled triangle ABC with right angle at C. Draw the altitude CD from C to the hypotenuse AB.
1) Since the altitude is perpendicular to the hypotenuse, it forms two right angles at D.
2) This splits triangle ABC into two smaller right-angled triangles: ADC and BDC.
3) Both these smaller triangles share one angle with the original triangle ABC and have a right angle, so by AA similarity criterion, all three triangles are similar.
4) However, the assertion states the altitude is "always shorter than the legs". While this is generally true in most cases, there exists a special case of an isosceles right-angled triangle where the two legs are equal. Here, the altitude will be exactly half the hypotenuse length, but the legs will be $x$ and the hypotenuse will be $x\sqrt{2}$. So the altitude = $\frac{x\sqrt{2}}{2}$ which is less than $x$ (since $\sqrt{2}/2 ≈ 0.707 < 1$). Thus the assertion is actually correct in all cases of right-angled triangles.
The reason about similarity is completely true and explains why certain properties hold, but doesn't directly explain the length relationship stated in the assertion. Therefore, while both statements are true, the reason doesn't correctly explain the assertion.

Key Concept: Altitude Definition

c) A perpendicular from A to BC
[Solution Description]
The altitude from a vertex to the opposite side (base) is defined as the perpendicular line segment drawn from that vertex to the line containing the base. Therefore, it must be perpendicular to BC.

Key Concept: Altitude Construction

d) Set square and ruler
[Solution Description]
A ruler alone cannot guarantee precise right angles. A set square has fixed right angles (90°) which ensures accuracy when used with a ruler for alignment. Protractors measure but don't fix angles precisely, while compasses help draw arcs rather than straight altitudes.

Key Concept: Identification of Altitudes

c) Two sides are themselves altitudes, and the third lies inside the triangle.
[Solution Description]
In a right-angled triangle, the two sides forming the right angle are altitudes to each other. The third altitude is drawn from the right-angled vertex to the hypotenuse. Here, $PQ$ is the altitude from $P$ to $QR$, $QR$ is the altitude from $Q$ to $PR$, and the third altitude is from $Q$ perpendicular to $PR$.

Key Concept: Right-angled triangle properties, Altitude construction

b) 6 cm
[Solution Description]
Let the altitude from $B$ meet $AC$ at point $D$. According to the property of right-angled triangles, the altitude to the hypotenuse satisfies $BD^2 = AD \times DC$. Given $AD = 4$ cm and $DC = 9$ cm, we can calculate:
$BD^2 = AD \times DC = 4 \times 9 = 36$
Taking the square root of both sides gives:
$BD = \sqrt{36} = 6 \, \text{cm}$
Thus, the length of the altitude is 6 cm.

Key Concept: Angle Sum Property

a) 30° and 30°
[Solution Description]
An isosceles triangle has two equal angles. Let the equal angles be $x$. Using the angle sum property:
$120^\circ + x + x = 180^\circ \Rightarrow 2x = 60^\circ \Rightarrow x = 30^\circ.$
Thus, the other two angles are each $30^\circ$.

Key Concept: Angle sum property, Right-angled triangles

c) Scalene right-angled triangle
[Solution Description]
Let the smaller acute angle be $x$. Then the other acute angle is $2x$.
Since it's a right-angled triangle:
$x + 2x + 90° = 180°$ → $3x = 90°$ → $x = 30°$
The angles are 30°, 60° and 90°.
This makes it a scalene triangle as all angles are different which implies all sides are different (since sides opposite to unequal angles are unequal).
So it's a right-angled scalene triangle.

Key Concept: Classification by sides

d) Scalene triangle
[Solution Description]
A scalene triangle is defined as a triangle where all three sides have different lengths. Here, since 5 cm $\neq$ 7 cm $\neq$ 9 cm, the triangle is scalene.

Key Concept: Classification by sides

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a triangle with sides 5 cm, 5 cm, and 6 cm is an isosceles triangle. Let's verify this:
- By definition, an isosceles triangle has exactly two sides of equal length.
- Here, two sides are equal (5 cm each) while the third side (6 cm) is different. This meets the condition for being an isosceles triangle.
The reason correctly defines an isosceles triangle and also explains why the given triangle in the assertion is isosceles.
Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Classification by sides

c) Equilateral triangle
[Solution Description]
All three sides of the triangle are equal in length (12 cm). By definition, this is an equilateral triangle.

Key Concept: Triangle Existence, Angle Sum Property

c) Yes, it's a right-angled triangle
[Solution Description]
To check if these angles form a valid triangle, their sum should be $180^{\circ}$. Here, $45^{\circ} + 45^{\circ} + 90^{\circ} = 180^{\circ}$. So, such a triangle exists. It's a right-angled triangle because one angle is exactly $90^{\circ}$.

Key Concept: Construction, Triangle Classification

b) Right-angled triangle
[Solution Description]
Check the type of triangle using the Pythagorean theorem. The largest side is 10 cm. If $6^2 + 8^2 = 10^2$ (36 + 64 = 100), it satisfies the condition for a right-angled triangle.

Key Concept: Classification by angles

b) Right-angled triangle
[Solution Description]
To determine the type of triangle PQR, we first find the measure of angle R. The sum of angles in a triangle is $180^\circ$. So, angle R = $180^\circ - 30^\circ - 60^\circ = 90^\circ$. Since one angle is $90^\circ$, triangle PQR is a right-angled triangle.

Key Concept: Types of Triangles based on Angles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- **Assertion Evaluation**: An equilateral triangle has all three sides equal and all three angles equal to $60^\circ$. Since none of its angles measure $90^\circ$ (right angle) or exceed $90^\circ$ (obtuse angle), the assertion is correct.
- **Reason Evaluation**: By definition, an equilateral triangle has three identical angles summing to $180^\circ$, so each angle must be $60^\circ$. The reason is factually accurate.
- **Explanation Link**: The reason directly explains why the assertion is true because if all angles are $60^\circ$, the triangle cannot have a right or obtuse angle.

Key Concept: Types of Triangles

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In an equilateral triangle, all three sides are equal, and consequently, all three angles are equal. The sum of the angles in any triangle is $180^\circ$. Therefore, each angle is $\frac{180^\circ}{3} = 60^\circ$. Since $60^\circ < 90^\circ$, all angles in an equilateral triangle are acute. Thus, an equilateral triangle is always acute-angled, and the reason correctly explains this by stating that each angle is $60^\circ$ and hence acute.

Key Concept: Right-angled, Angle Sum Property

d) 55
[Solution Description]
Given the angles are $x^\circ$, $(x+10)^\circ$, and $(2x-20)^\circ$. One of them is $90^\circ$. We consider three cases:
Case 1: $x = 90^\circ$
Then other angles are $100^\circ$ and $160^\circ$. But sum exceeds $180^\circ$. Invalid.
Case 2: $x+10 = 90^\circ$
So, $x = 80^\circ$. Other angles are $80^\circ$ and $140^\circ$. Sum is $80^\circ + 90^\circ + 140^\circ = 310^\circ > 180^\circ$. Invalid.
Case 3: $2x - 20 = 90^\circ$
$2x = 110^\circ$
$x = 55^\circ$. Other angles are $55^\circ$ and $65^\circ$. Sum is $55^\circ + 65^\circ + 90^\circ = 210^\circ > 180^\circ$. Wait, incorrect calculation. Correct sum should be $55^\circ + 65^\circ + 90^\circ = 210^\circ > 180^\circ$. No valid case found. Rechecking:
Actually, initial assumption leads to no valid solution. Alternatively, set sum first: $x + (x+10) + (2x-20) = 180^\circ$
$4x - 10 = 180^\circ$
$4x = 190^\circ$
$x = 47.5^\circ$. Then angles are $47.5^\circ$, $57.5^\circ$, and $75^\circ$. Here, none is $90^\circ$. Hence, inconsistent data. However, assuming the problem implies one angle must be $90^\circ$, then no solution exists as shown. But per options:

Key Concept: Equilateral Triangle Angle Classification

c) Acute-angled
[Solution Description]
In an equilateral triangle, all sides are equal, and all angles are also equal.
The sum of interior angles of a triangle is 180°. So, each angle in an equilateral triangle is $\frac{180°}{3} = 60°$.
Since 60° is an acute angle, all three angles are acute.
Therefore, the triangle is acute-angled.

Key Concept: Angle Classification

c) Right-angled triangle
[Solution Description]
Based on angle classification, a triangle with exactly one right angle is known as a right-angled triangle.

Key Concept: Side Classification, Angle Classification

c) Obtuse-angled isosceles triangle
[Solution Description]
Since two sides are equal (PQ = PR = 5 cm) and the third side is different (QR = 8 cm), it is an isosceles triangle. The angle opposite the unequal side is 100 degrees, which is obtuse, so it's an obtuse-angled triangle. Therefore, the correct classification is obtuse-angled isosceles triangle.