A Tale of Three Intersecting Lines

Given: Side lengths 4 cm, 4 cm, 6 cm.

Check Triangle Inequality:
- 4 + 4 = 8 > 6 ✓
- 4 + 6 = 10 > 4 ✓
- 4 + 6 = 10 > 4 ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 6$ cm.

Step 2: With A as centre, draw an arc of radius 4 cm.

Step 3: With B as centre, draw an arc of radius 4 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $AC = 4$ cm, $BC = 4$ cm.

This is an isosceles triangle (two equal sides of 4 cm).

Given: Side lengths 3 cm, 4 cm, 5 cm.

Check Triangle Inequality:
- 3 + 4 = 7 > 5 ✓
- 3 + 5 = 8 > 4 ✓
- 4 + 5 = 9 > 3 ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 5$ cm.

Step 2: With A as centre, draw an arc of radius 3 cm.

Step 3: With B as centre, draw an arc of radius 4 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 5$ cm, $AC = 3$ cm, $BC = 4$ cm.

This is a scalene triangle (all sides different). It is also a right-angled triangle since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.

Given: Side lengths 1 cm, 5 cm, 5 cm.

Check Triangle Inequality:
- 1 + 5 = 6 > 5 ✓
- 1 + 5 = 6 > 5 ✓
- 5 + 5 = 10 > 1 ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 1$ cm.

Step 2: With A as centre, draw an arc of radius 5 cm.

Step 3: With B as centre, draw an arc of radius 5 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 1$ cm, $AC = 5$ cm, $BC = 5$ cm.

This is an isosceles triangle (two equal sides of 5 cm).

Given: Side lengths 4 cm, 6 cm, 8 cm.

Check Triangle Inequality:
- 4 + 6 = 10 > 8 ✓
- 4 + 8 = 12 > 6 ✓
- 6 + 8 = 14 > 4 ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 8$ cm.

Step 2: With A as centre, draw an arc of radius 4 cm.

Step 3: With B as centre, draw an arc of radius 6 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 8$ cm, $AC = 4$ cm, $BC = 6$ cm.

This is a scalene triangle (all sides different).

Given: Side lengths 3.5 cm, 3.5 cm, 3.5 cm.

Check Triangle Inequality:
- 3.5 + 3.5 = 7 > 3.5 ✓
- 3.5 + 3.5 = 7 > 3.5 ✓
- 3.5 + 3.5 = 7 > 3.5 ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 3.5$ cm.

Step 2: With A as centre, draw an arc of radius 3.5 cm.

Step 3: With B as centre, draw an arc of radius 3.5 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required equilateral triangle with all sides equal to 3.5 cm.

Concept: In a circle, all radii are equal. Any two radii and the chord joining their endpoints form an isosceles triangle.

Method:
- Let O be the centre of the circle and let A, B be any two points on the circle.
- Then $OA = OB =$ radius.
- Triangle OAB has two equal sides (both equal to the radius), so $\triangle OAB$ is an isosceles triangle.

Conclusion: By choosing any two points on the circle and joining them to the centre O, we always get an isosceles triangle. We can form as many isosceles triangles as we wish by choosing different pairs of points on the circle.

Case 1: Two circles of the same size with centres A and B.

Let the two equal circles intersect at points P and Q.

- Since both circles have the same radius $r$, we have $AP = AQ = BP = BQ = r$.
- Triangle APB: $AP = BP = r$ (both radii of their respective circles), so $\triangle APB$ is isosceles.
- Triangle AQB: Similarly, $AQ = BQ = r$, so $\triangle AQB$ is isosceles.
- Triangle APQ: $AP = AQ = r$, so $\triangle APQ$ is isosceles.
- Triangle BPQ: $BP = BQ = r$, so $\triangle BPQ$ is isosceles.
- Triangle APBQ (quadrilateral), but focusing on $\triangle APB$: if the circles are drawn so that each passes through the other's centre (i.e., $AB = r$), then $AP = BP = AB = r$, making $\triangle APB$ equilateral.

Case 2: Three circles of the same size with centres A, B, C.

If each circle passes through the centres of the other two (i.e., $AB = BC = CA = r$), then:
- $\triangle ABC$ is equilateral (all sides equal to $r$).
- The intersection points of the circles along with the centres can also form isosceles triangles.
- For example, if P is the intersection of circles centred at A and B, then $AP = BP = r = AB$, so $\triangle APB$ is equilateral.

Conclusion: By choosing the two intersection points of any two same-sized circles as two vertices and one of the centres as the third vertex, we get isosceles triangles. When each circle passes through the other's centre, we get equilateral triangles.

Given: Side lengths 3 cm, 4 cm, 8 cm.

Check Triangle Inequality:
- $3 + 4 = 7$, but 7 < 8. ✗ (The sum of the two smaller sides is less than the largest side.)

Construction attempt:
Draw base $AB = 8$ cm. Draw an arc of radius 3 cm from A and an arc of radius 4 cm from B. The two arcs do not intersect because the maximum distance they can reach toward each other is $3 + 4 = 7$ cm, which is less than 8 cm (the distance between A and B).

Conclusion: The triangle cannot be constructed. The two arcs do not meet, so no third vertex C exists. This is because the lengths 3, 4, 8 do not satisfy the triangle inequality (3 + 4 < 8).

Given: Side lengths 2 cm, 3 cm, 6 cm.

Check Triangle Inequality:
- $2 + 3 = 5$, but 5 < 6. ✗

The sum of the two smaller sides (2 + 3 = 5) is less than the largest side (6).

Conclusion: The triangle is not possible. The lengths 2, 3, 6 do not satisfy the triangle inequality, so no triangle can be constructed with these side lengths.

Examples of sets for which a triangle is impossible:

1. 1, 2, 5: 1 + 2 = 3 < 5 ✗
2. 2, 3, 7: 2 + 3 = 5 < 7 ✗
3. 1, 4, 6: 1 + 4 = 5 < 6 ✗
4. 5, 6, 15: 5 + 6 = 11 < 15 ✗
5. 10, 15, 30: 10 + 15 = 25 < 30 ✗

Pattern observed: In each case, the sum of the two smaller (or any two) lengths is less than or equal to the third (largest) length. That is, the triangle inequality is violated: the longest side is greater than or equal to the sum of the other two sides.

Rule: A triangle is impossible when the largest side $\geq$ sum of the other two sides.

Yes, we can determine this without construction by checking the triangle inequality.

For 3, 4, 8:
3 + 4 = 7 < 8
The sum of the two smaller sides is less than the largest side. Triangle inequality is violated. So no triangle is possible.

For 2, 3, 6:
2 + 3 = 5 < 6
Again, the sum of the two smaller sides is less than the largest side. Triangle inequality is violated. So no triangle is possible.

Conclusion: We only need to check whether the sum of the two smaller lengths is greater than the largest length. If not, the triangle does not exist — no construction needed.

Concept: Check the triangle inequality — the sum of any two sides must be greater than the third side. It is sufficient to check: sum of the two smaller sides > largest side.

(a) 10 km, 10 km, 25 km:
10 + 10 = 20 < 25
Triangle inequality is violated. Triangle does not exist.

(b) 5 mm, 10 mm, 20 mm:
5 + 10 = 15 < 20
Triangle inequality is violated. Triangle does not exist.

(c) 12 cm, 20 cm, 40 cm:
12 + 20 = 32 < 40
Triangle inequality is violated. Triangle does not exist.

Conclusion: None of the three sets can form a triangle.

Exploration:

Let the three lengths be $a \leq b \leq c$ (arranged in increasing order).

Comparison 1: a < b + c? Since b \geq a > 0 and c \geq a > 0, we have b + c \geq a + a > a. So a < b + c is always true.

Comparison 2: b < a + c? Since $c \geq b$ and a > 0, we have a + c > c \geq b. So b < a + c is always true.

Comparison 3: c < a + b? This is not always true — this is the critical comparison.

Conclusion:
- The two smaller lengths are always less than the sum of the other two. No calculation needed for them.
- The only comparison that matters is whether the largest side is less than the sum of the other two sides.
- So to check for the existence of a triangle, we only need to verify: \text{largest side} < \text{sum of the other two sides}

For example, for 10, 15, 30: 30 > 10 + 15 = 25, so no triangle exists.

Rule: A triangle exists if and only if the sum of the two smaller sides is greater than the largest side.

(a) 2, 2, 5:
2 + 2 = 4 < 5
Triangle inequality violated. Triangle NOT possible.

(b) 3, 4, 6:
3 + 4 = 7 > 6 ✓
Triangle inequality satisfied. Triangle IS possible.

(c) 2, 4, 8:
2 + 4 = 6 < 8
Triangle inequality violated. Triangle NOT possible.

(d) 5, 5, 8:
5 + 5 = 10 > 8 ✓
Triangle inequality satisfied. Triangle IS possible.

(e) 10, 20, 25:
10 + 20 = 30 > 25 ✓
Triangle inequality satisfied. Triangle IS possible.

(f) 10, 20, 35:
10 + 20 = 30 < 35
Triangle inequality violated. Triangle NOT possible.

(g) 24, 26, 28:
24 + 26 = 50 > 28 ✓
Triangle inequality satisfied. Triangle IS possible.

Background: When constructing a triangle with sides $a$, $b$, $c$ (where $c$ is the base), we draw circles of radii $a$ and $b$ from the two ends of the base $c$.

(a) Circles touch each other at exactly one point (degenerate triangle):
This happens when $a + b = c$ exactly — the two arcs just touch.

Examples:
1. $2, 3, 5$: $2 + 3 = 5$ — circles touch at one point.
2. $4, 6, 10$: $4 + 6 = 10$ — circles touch at one point.
3. $1, 4, 5$: $1 + 4 = 5$ — circles touch at one point.

In each case, the three points are collinear — a degenerate (flat) triangle.

(b) Circles do not intersect (no triangle possible):
This happens when a + b < c — the two arcs are too far apart.

Examples:
1. $2, 3, 8$: 2 + 3 = 5 < 8 — circles do not intersect.
2. $1, 2, 6$: 1 + 2 = 3 < 6 — circles do not intersect.
3. $5, 6, 15$: 5 + 6 = 11 < 15 — circles do not intersect.

Procedure to check existence of a triangle for given lengths $a$, $b$, $c$:

Step 1: Identify the three given lengths.

Step 2: Find the largest of the three lengths. Let it be $c$.

Step 3: Find the sum of the other two lengths: $a + b$.

Step 4: Compare:
- If a + b > c: The triangle inequality is satisfied and the triangle exists.
- If $a + b = c$: The three points are collinear (degenerate case) — a proper triangle does not exist.
- If a + b < c: The triangle inequality is violated — the triangle does not exist.

Note: It is sufficient to check only the largest side against the sum of the other two, since the other two comparisons are automatically satisfied.

Rule: Triangle exists if and only if the largest side < sum of the other two sides.

(a) 1, 100, 100:
Largest side = 100.
1 + 100 = 101 &gt; 100 ✓
Triangle inequality satisfied. Triangle EXISTS.

(b) 3, 6, 9:
Largest side = 9.
$$3 + 6 = 9$$
$9 = 9$, not strictly greater. Triangle inequality is not satisfied (degenerate case).
Triangle does NOT exist (the three points would be collinear).

(c) 1, 1, 5:
Largest side = 5.
1 + 1 = 2 &lt; 5
Triangle inequality violated. Triangle does NOT exist.

(d) 5, 10, 12:
Largest side = 12.
5 + 10 = 15 &gt; 12 ✓
Triangle inequality satisfied. Triangle EXISTS.

For sides 50, 50, 50:
Check triangle inequality:
50 + 50 = 100 &gt; 50 ✓

The triangle inequality is satisfied. Yes, an equilateral triangle with sides 50, 50, 50 exists.

In general, for any equilateral triangle with side length $a$ (where a &gt; 0):
a + a = 2a &gt; a ✓

The triangle inequality is always satisfied for any positive value of $a$.

Conclusion: An equilateral triangle exists for any positive side length. There is no restriction on the side length of an equilateral triangle.

Rule: If two sides are $a$ and $b$ ($a \leq b$), the third side $x$ must satisfy:
b - a &lt; x &lt; a + b

(a) Sides: 1 and 100:
100 - 1 &lt; x &lt; 100 + 1
99 &lt; x &lt; 101

All numbers strictly between 99 and 101 are valid.

5 possible values: $99.5,\ 99.8,\ 100,\ 100.2,\ 100.5$

(b) Sides: 5 and 5:
5 - 5 &lt; x &lt; 5 + 5
0 &lt; x &lt; 10

All numbers strictly between 0 and 10 are valid.

5 possible values: $1,\ 3,\ 5,\ 7,\ 9$

(c) Sides: 3 and 7:
7 - 3 &lt; x &lt; 7 + 3
4 &lt; x &lt; 10

All numbers strictly between 4 and 10 are valid.

5 possible values: $4.5,\ 5,\ 6,\ 8,\ 9.5$

Given: Two sides 3 cm and 7 cm with included angle 75°.

Steps of Construction:

Step 1: Draw a line segment $AB = 7$ cm (taking the longer side as base).

Step 2: At point A, construct an angle of $75°$ using a protractor, drawing the other arm of the angle.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 7$ cm, $AC = 3$ cm, and $\angle A = 75°$.

Note: A triangle always exists for any two positive side lengths and any included angle strictly between $0°$ and $180°$, so this triangle is always constructible.

Given: Two sides 6 cm and 3 cm with included angle 25°.

Steps of Construction:

Step 1: Draw a line segment $AB = 6$ cm.

Step 2: At point A, construct an angle of $25°$ using a protractor.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $AC = 3$ cm, and $\angle A = 25°$.

Given: Two sides 3 cm and 8 cm with included angle 120°.

Steps of Construction:

Step 1: Draw a line segment $AB = 8$ cm.

Step 2: At point A, construct an angle of $120°$ using a protractor.

Step 3: On this arm, mark point C such that $AC = 3$ cm.

Step 4: Join BC.

$\triangle ABC$ is the required triangle with $AB = 8$ cm, $AC = 3$ cm, and $\angle A = 120°$.

Answer: No, a triangle is always possible when two sides and their included angle are given, as long as the angle is strictly between $0°$ and $180°$ (exclusive).

Justification:
- Draw one side as the base, say $AB$.
- At A, draw the included angle. The second side $AC$ is drawn along this angle arm.
- Join B to C. The triangle $ABC$ is always formed.
- The two sides and the included angle uniquely determine the third side BC by the construction.
- There is no condition like the triangle inequality that can fail here, because once two sides and the included angle are fixed, the third side is automatically determined and is always positive (as long as the angle is between $0°$ and $180°$).

Conclusion: For any two positive lengths and any angle strictly between $0°$ and $180°$, a triangle always exists.

Given: $\angle A = 75°$, $AB = 5$ cm (included side), $\angle B = 75°$.

Check existence: \angle A + \angle B = 75° + 75° = 150° &lt; 180° ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 5$ cm.

Step 2: At A, construct an angle of $75°$ (above AB).

Step 3: At B, construct an angle of $75°$ (above AB, on the same side).

Step 4: The point C where the two arms intersect is the third vertex.

$\triangle ABC$ is the required triangle with $AB = 5$ cm, $\angle A = 75°$, $\angle B = 75°$.

The third angle: $\angle C = 180° - 75° - 75° = 30°$.

Given: $\angle A = 25°$, $AB = 3$ cm (included side), $\angle B = 60°$.

Check existence: \angle A + \angle B = 25° + 60° = 85° &lt; 180° ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 3$ cm.

Step 2: At A, construct an angle of $25°$ above AB.

Step 3: At B, construct an angle of $60°$ above AB.

Step 4: The intersection point of the two arms is vertex C.

$\triangle ABC$ is the required triangle with $AB = 3$ cm, $\angle A = 25°$, $\angle B = 60°$.

The third angle: $\angle C = 180° - 25° - 60° = 95°$.

Given: $\angle A = 120°$, $AB = 6$ cm (included side), $\angle B = 30°$.

Check existence: \angle A + \angle B = 120° + 30° = 150° &lt; 180° ✓. Triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 6$ cm.

Step 2: At A, construct an angle of $120°$ above AB.

Step 3: At B, construct an angle of $30°$ above AB.

Step 4: The intersection point of the two arms is vertex C.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $\angle A = 120°$, $\angle B = 30°$.

The third angle: $\angle C = 180° - 120° - 30° = 30°$.

Answer: No, a triangle does not always exist for every combination of two angles and an included side.

Condition for existence: A triangle exists if and only if the sum of the two given angles is strictly less than $180°$.

Examples where triangle is NOT possible:

1. $\angle A = 90°$, $\angle B = 90°$: Sum $= 180°$. The two arms from A and B are parallel and never meet. No triangle.

2. $\angle A = 100°$, $\angle B = 100°$: Sum = 200° &gt; 180°. No triangle.

3. $\angle A = 40°$, $\angle B = 140°$: Sum $= 180°$. The arms are parallel. No triangle.

4. $\angle A = 60°$, $\angle B = 130°$: Sum = 190° &gt; 180°. No triangle.

Rule: A triangle exists when \angle A + \angle B &lt; 180°.

Given: $\angle A = 40°$ (acute angle at A).

For the line from B to not meet the line $l$ (the arm from A), the line from B must be parallel to $l$ or inclined further to the right (away from $l$).

When the line from B is parallel to $l$:
- AB is the transversal cutting two parallel lines.
- Co-interior angles (same-side interior angles) add up to $180°$.
- So $\angle A + \angle B = 180°$, giving $\angle B = 180° - 40° = 140°$.

For the lines to not meet, $\angle B \geq 140°$.

A possible value: $\angle B = 150°$ (the lines will not meet).

Given: $\angle A = 40°$.

The smallest value of $\angle B$ for which the lines do not meet is when the line from B is parallel to the arm from A.

Using the property of parallel lines (co-interior angles sum to $180°$):
$$\angle A + \angle B = 180°$$
$$40° + \angle B = 180°$$
$$\angle B = 140°$$

Conclusion: The smallest value of $\angle B$ for the lines to not meet is $\mathbf{140°}$. For $\angle B \geq 140°$, the lines are parallel or diverge, and no triangle is formed.

Rule: Given one angle $\alpha$, a triangle is possible with another angle $\beta$ if \alpha + \beta &lt; 180°, i.e., \beta &lt; 180° - \alpha. A triangle is not possible if $\alpha + \beta \geq 180°$.

(a) Given angle: 30°
For triangle to be possible: \beta &lt; 180° - 30° = 150°
- Possible: $\beta = 60°$ (since 30° + 60° = 90° &lt; 180°); $\beta = 100°$ (since 30° + 100° = 130° &lt; 180°)
- Not possible: $\beta = 150°$ (since $30° + 150° = 180°$); $\beta = 160°$ (since 30° + 160° = 190° &gt; 180°)

(b) Given angle: 70°
For triangle to be possible: \beta &lt; 180° - 70° = 110°
- Possible: $\beta = 50°$ (since 70° + 50° = 120° &lt; 180°); $\beta = 80°$ (since 70° + 80° = 150° &lt; 180°)
- Not possible: $\beta = 110°$ (since $70° + 110° = 180°$); $\beta = 120°$ (since 70° + 120° = 190° &gt; 180°)

(c) Given angle: 54°
For triangle to be possible: \beta &lt; 180° - 54° = 126°
- Possible: $\beta = 70°$ (since 54° + 70° = 124° &lt; 180°); $\beta = 90°$ (since 54° + 90° = 144° &lt; 180°)
- Not possible: $\beta = 126°$ (since $54° + 126° = 180°$); $\beta = 140°$ (since 54° + 140° = 194° &gt; 180°)

(d) Given angle: 144°
For triangle to be possible: \beta &lt; 180° - 144° = 36°
- Possible: $\beta = 20°$ (since 144° + 20° = 164° &lt; 180°); $\beta = 30°$ (since 144° + 30° = 174° &lt; 180°)
- Not possible: $\beta = 36°$ (since $144° + 36° = 180°$); $\beta = 50°$ (since 144° + 50° = 194° &gt; 180°)

Rule: Two angles $\alpha$ and $\beta$ can be angles of a triangle if and only if \alpha + \beta &lt; 180° (so that the third angle = 180° - \alpha - \beta &gt; 0°).

(a) 35°, 150°:
35° + 150° = 185° &gt; 180°
Third angle would be $180° - 185° = -5°$ (negative). Cannot be angles of a triangle.

(b) 70°, 30°:
70° + 30° = 100° &lt; 180°
Third angle = 180° - 100° = 80° &gt; 0°. Can be angles of a triangle.

(c) 90°, 85°:
90° + 85° = 175° &lt; 180°
Third angle = 180° - 175° = 5° &gt; 0°. Can be angles of a triangle.

(d) 50°, 150°:
50° + 150° = 200° &gt; 180°
Third angle would be $180° - 200° = -20°$ (negative). Cannot be angles of a triangle.

Concept (Angle Sum Property): The sum of all three angles of a triangle is $180°$.
$$\angle A + \angle B + \angle C = 180°$$

(a) 36°, 72°:
$$\text{Third angle} = 180° - 36° - 72° = 180° - 108° = 72°$$

(b) 150°, 15°:
$$\text{Third angle} = 180° - 150° - 15° = 180° - 165° = 15°$$

(c) 90°, 30°:
$$\text{Third angle} = 180° - 90° - 30° = 180° - 120° = 60°$$

(d) 75°, 45°:
$$\text{Third angle} = 180° - 75° - 45° = 180° - 120° = 60°$$

If all three angles are 70°:
$$70° + 70° + 70° = 210° \neq 180°$$
This violates the angle sum property. A triangle with all angles equal to 70° cannot be constructed.

If two angles are 70°:
$$\text{Third angle} = 180° - 70° - 70° = 40°$$
So the third angle would be $40°$.

If all three angles are equal:
Let each angle $= x$.
$$x + x + x = 180°$$
$$3x = 180°$$
$$x = 60°$$

Conclusion: For all three angles to be equal, each must be $60°$. Such a triangle is an equilateral triangle. All equilateral triangles have each angle equal to $60°$.

Given: $\angle A = 50°$, $\angle B = \angle C$.

Using Angle Sum Property:
$$\angle A + \angle B + \angle C = 180°$$
$$50° + \angle B + \angle C = 180°$$

Since $\angle B = \angle C$:
$$50° + 2\angle B = 180°$$
$$2\angle B = 180° - 50° = 130°$$
$$\angle B = \frac{130°}{2} = 65°$$

Therefore, $\angle C = 65°$.

Answer: $\angle B = \angle C = 65°$.

Given: $BC = 5$ cm, $AB = 6$ cm, $CA = 5$ cm.

Check Triangle Inequality:
5 + 5 = 10 &gt; 6 ✓; 5 + 6 = 11 &gt; 5 ✓. Triangle exists. (It is isosceles with $BC = CA = 5$ cm.)

Steps of Construction:

Step 1: Draw base $BC = 5$ cm.

Step 2: With B as centre, draw an arc of radius 6 cm.

Step 3: With C as centre, draw an arc of radius 5 cm intersecting the first arc at A.

Step 4: Join AB and AC. $\triangle ABC$ is formed.

Constructing Altitude from A to BC:

Step 5: With A as centre, draw an arc that cuts BC (or its extension) at two points, say P and Q.

Step 6: With P and Q as centres, draw arcs of equal radius (more than half PQ) on the opposite side of A. Let them intersect at point D.

Step 7: Join AD. The foot of the perpendicular from A to BC is the point where AD meets BC. This segment is the altitude from A.

Note: Since the triangle is isosceles ($BC = CA$), the altitude from A bisects BC.

Given: $RY = 4$ cm, $TR = 7$ cm, $\angle R = 140°$ (included angle between TR and RY).

Steps of Construction:

Step 1: Draw base $RY = 4$ cm.

Step 2: At R, construct an angle of $140°$ using a protractor.

Step 3: On this arm, mark point T such that $RT = 7$ cm.

Step 4: Join TY. $\triangle TRY$ is formed.

Constructing Altitude from T to RY:

Step 5: Since $\angle R = 140°$ is obtuse, the altitude from T will fall outside the triangle (on the extension of RY beyond R).

Step 6: Extend RY beyond R. From T, draw a perpendicular to the line RY (extended). Use a set square or compass construction to drop a perpendicular from T to line RY.

Step 7: The foot of the perpendicular (let's call it H) lies on the extension of RY. TH is the required altitude from T.

Given: $\angle B = 90°$, $AC = 5$ cm (hypotenuse).

Analysis: We know one angle ($\angle B = 90°$) and the hypotenuse ($AC = 5$ cm). The other two angles $\angle A$ and $\angle C$ must satisfy:
$$\angle A + \angle C = 90°$$

For any value of $\angle A$ strictly between $0°$ and $90°$, we get a valid $\angle C = 90° - \angle A$, also strictly between $0°$ and $90°$. Each such choice gives a different triangle.

Steps of Construction (for one example, say ∠A = 30°, ∠C = 60°):

Step 1: Draw $AC = 5$ cm as the base.

Step 2: At A, construct $\angle A = 30°$.

Step 3: At C, construct $\angle C = 60°$.

Step 4: The intersection of the two arms is B, where $\angle B = 90°$.

Answer: There are infinitely many different triangles with $\angle B = 90°$ and $AC = 5$ cm, because $\angle A$ can take any value strictly between $0°$ and $90°$, giving a different triangle each time.

Equilateral Triangle:
In an equilateral triangle, all three angles are equal to $60°$.

(i) Equilateral and right-angled:
A right-angled triangle has one angle of $90°$. But in an equilateral triangle, all angles are $60°$. Since $60° \neq 90°$, an equilateral triangle cannot be right-angled.

(ii) Equilateral and obtuse-angled:
An obtuse-angled triangle has one angle greater than $90°$. But all angles in an equilateral triangle are 60° &lt; 90°. So an equilateral triangle cannot be obtuse-angled.

Conclusion: An equilateral triangle is always acute-angled (all angles = 60°). It can be neither right-angled nor obtuse-angled.

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Isosceles Triangle:

(i) Isosceles and right-angled:
This is possible. Example: An isosceles right-angled triangle has angles $90°, 45°, 45°$.

Construction: Draw $AB = BC = 5$ cm with $\angle B = 90°$. Join AC. $\triangle ABC$ is isosceles ($AB = BC$) and right-angled ($\angle B = 90°$).

(ii) Isosceles and obtuse-angled:
This is possible. Example: An isosceles triangle with angles $120°, 30°, 30°$.

Construction: Draw base $BC = 4$ cm. At B and C, construct angles of $30°$ each. The apex angle at A will be $180° - 30° - 30° = 120°$ (obtuse). Since $\angle B = \angle C = 30°$, we have $AB = AC$ (sides opposite equal angles are equal), making it isosceles.

Conclusion: An isosceles triangle can be both right-angled and obtuse-angled.