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Triangles – CBSE NCERT Study Resources

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9th

9th - Mathematics

Triangles

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Overview of the Chapter: Triangles

This chapter introduces students to the fundamental concepts related to triangles, their properties, types, and congruence criteria as per the CBSE Grade 9 Mathematics curriculum. The chapter covers the basic definitions, angle sum property, inequalities, and congruence rules, along with practical applications.

Key Concepts

Triangle: A closed figure formed by three intersecting lines, consisting of three sides, three angles, and three vertices.

Types of Triangles

  • Based on Sides:
    • Equilateral Triangle: All sides and angles are equal (each angle = 60°).
    • Isosceles Triangle: Two sides and two angles are equal.
    • Scalene Triangle: All sides and angles are unequal.
  • Based on Angles:
    • Acute-angled Triangle: All angles less than 90°.
    • Right-angled Triangle: One angle is exactly 90°.
    • Obtuse-angled Triangle: One angle is greater than 90°.

Properties of Triangles

Angle Sum Property: The sum of the interior angles of a triangle is always 180°.

Exterior Angle Property: An exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Congruence of Triangles

Two triangles are congruent if their corresponding sides and angles are equal. The congruence criteria are:

  • SSS (Side-Side-Side): All three sides are equal.
  • SAS (Side-Angle-Side): Two sides and the included angle are equal.
  • ASA (Angle-Side-Angle): Two angles and the included side are equal.
  • AAS (Angle-Angle-Side): Two angles and a non-included side are equal.
  • RHS (Right-Hypotenuse-Side): Right-angled triangles with hypotenuse and one side equal.

Inequalities in Triangles

Triangle Inequality Theorem: The sum of any two sides of a triangle is always greater than the third side.

Other inequalities include:

  • The side opposite the larger angle is longer.
  • The angle opposite the longer side is larger.

Summary

This chapter provides a comprehensive understanding of triangles, their classifications, properties, and congruence rules. Students learn to apply these concepts to solve problems and prove geometric theorems.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
In a right-angled triangle, if one angle is 45°, what is the measure of the third angle?
Answer:
45°
Question 2:
What is the Pythagorean theorem formula for a right-angled triangle?
Answer:

Hypotenuse² = Base² + Height²

Question 3:
If two sides of a triangle are 3 cm and 4 cm, what is the third side for it to be right-angled?
Answer:
5 cm
Question 4:
What is the congruency rule if two triangles have all sides equal?
Answer:

SSS Congruency

Question 5:
In ΔABC, if ∠A = 60° and ∠B = 70°, what is ∠C?
Answer:
50°
Question 6:
What is the sum of interior angles of a triangle?
Answer:
180°
Question 7:
If two triangles are congruent by ASA rule, what must be equal?
Answer:

Two angles and included side

Question 8:
What is the area of a triangle with base 6 cm and height 4 cm?
Answer:
12 cm²
Question 9:
In an equilateral triangle, what is each angle measure?
Answer:
60°
Question 10:
What is the perimeter of a triangle with sides 5 cm, 6 cm, and 7 cm?
Answer:
18 cm
Question 11:
If two triangles have two angles and one side equal, which congruency rule applies?
Answer:

AAS Congruency

Question 12:
What is the hypotenuse of a right triangle with legs 8 cm and 6 cm?
Answer:
10 cm
Question 13:
State the Angle Sum Property of a triangle.
Answer:

The Angle Sum Property states that the sum of the three interior angles of a triangle is always 180°.

Question 14:
If two angles of a triangle are 45° and 55°, find the third angle.
Answer:

Using the Angle Sum Property:
Third angle = 180° - (45° + 55°)
Third angle = 180° - 100°
Third angle = 80°.

Question 15:
What is the name of a triangle with all sides equal?
Answer:

A triangle with all sides equal is called an equilateral triangle.

Question 16:
In a right-angled triangle, one acute angle is 30°. What is the measure of the other acute angle?
Answer:

Using the Angle Sum Property:
Other acute angle = 180° - (90° + 30°)
Other acute angle = 180° - 120°
Other acute angle = 60°.

Question 17:
Define the Pythagoras Theorem.
Answer:

The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Question 18:
If the sides of a triangle are 6 cm, 8 cm, and 10 cm, is it a right-angled triangle?
Answer:

Check using the Pythagoras Theorem:
6² + 8² = 36 + 64 = 100
10² = 100
Since 6² + 8² = 10², the triangle is right-angled.

Question 19:
State the Exterior Angle Property of a triangle.
Answer:

The Exterior Angle Property states that the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Question 20:
If one angle of an isosceles triangle is 80°, what are the other two angles?
Answer:

In an isosceles triangle, two angles are equal.
If the given angle is the vertex angle:
Other two angles = (180° - 80°) ÷ 2 = 50° each.

Question 21:
What is the difference between a scalene and an isosceles triangle?
Answer:

Scalene triangle has all sides and angles unequal.
Isosceles triangle has two sides and two angles equal.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
If two sides of a triangle are equal, what type of triangle is it?
Answer:

It is an isosceles triangle because an isosceles triangle has two sides equal and the angles opposite to these sides are also equal.

Question 2:
What is the measure of each angle in an equilateral triangle?
Answer:

Each angle in an equilateral triangle measures 60° because all sides and angles are equal, and the sum of angles is 180°.

Question 3:
In a right-angled triangle, what is the side opposite to the right angle called?
Answer:

The side opposite to the right angle is called the hypotenuse, which is the longest side of the right-angled triangle.

Question 4:
If two triangles have their corresponding sides equal, what can you say about their angles?
Answer:

If two triangles have their corresponding sides equal (SSS congruence), then their corresponding angles are also equal.

Question 5:
What is the Pythagorean theorem?
Answer:

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Formula: Hypotenuse² = Base² + Perpendicular².

Question 6:
Can a triangle have two right angles? Justify your answer.
Answer:

No, a triangle cannot have two right angles because the sum of all three angles would exceed 180°, violating the Angle Sum Property.

Question 7:
What is the difference between the altitude and median of a triangle?
Answer:

Altitude is a perpendicular line segment from a vertex to the opposite side.
Median is a line segment joining a vertex to the midpoint of the opposite side.

Question 8:
If two angles of a triangle are 45° and 55°, what is the measure of the third angle?
Answer:

Using the Angle Sum Property:
Third angle = 180° - (45° + 55°)
Third angle = 80°.

Question 9:
What is the condition for two triangles to be congruent by the SAS rule?
Answer:

Two triangles are congruent by the SAS rule if two sides and the included angle of one triangle are equal to the corresponding sides and angle of the other triangle.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
In a triangle ABC, if ∠A = 60° and ∠B = 70°, find the measure of ∠C. Justify your answer using the angle sum property of a triangle.
Answer:

According to the angle sum property of a triangle, the sum of all three interior angles is 180°.
Given: ∠A = 60° and ∠B = 70°.
Let ∠C be the third angle.
Using the property: ∠A + ∠B + ∠C = 180°.
Substituting the values: 60° + 70° + ∠C = 180°.
Simplifying: 130° + ∠C = 180°.
Thus, ∠C = 180° - 130° = 50°.
Therefore, the measure of ∠C is 50°.

Question 2:
Prove that the angles opposite to equal sides of an isosceles triangle are equal. Draw a diagram to support your answer.
Answer:

Consider an isosceles triangle ABC with AB = AC.
We need to prove that ∠B = ∠C.

Steps:
1. Draw the angle bisector of ∠A meeting BC at D.
2. In triangles ABD and ACD:
- AB = AC (given).
- AD is common.
- ∠BAD = ∠CAD (by construction).
3. By the SAS congruence rule, △ABD ≅ △ACD.
4. Therefore, ∠B = ∠C (corresponding parts of congruent triangles).

Thus, the angles opposite to equal sides are equal.

Question 3:
In a right-angled triangle, one of the acute angles is 40°. Find the measure of the other acute angle.
Answer:

In a right-angled triangle, one angle is 90°, and the sum of the other two angles is 90° (by the angle sum property).
Given: One acute angle is 40°.
Let the other acute angle be x.
Thus, 40° + x = 90°.
Solving: x = 90° - 40° = 50°.
Therefore, the other acute angle measures 50°.

Question 4:
If two sides of a triangle are 5 cm and 7 cm, what can be the possible length of the third side? State the triangle inequality theorem used to justify your answer.
Answer:

According to the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side.
Let the third side be x cm.
The possible conditions are:
1. 5 + 7 > x ⇒ x < 12 cm.
2. 5 + x > 7 ⇒ x > 2 cm.
3. 7 + x > 5 ⇒ x > -2 cm (always true since lengths are positive).

Combining the first two conditions: 2 cm < x < 12 cm.
Thus, the third side must be greater than 2 cm and less than 12 cm.

Question 5:
In △ABC and △DEF, AB = DE, BC = EF, and ∠B = ∠E. Are the two triangles congruent? If yes, state the congruence rule applied.
Answer:

Yes, △ABC and △DEF are congruent.
Given:
1. AB = DE (one pair of equal sides).
2. BC = EF (another pair of equal sides).
3. ∠B = ∠E (included angle between the two sides).

By the SAS (Side-Angle-Side) congruence rule, if two sides and the included angle of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent.
Thus, △ABC ≅ △DEF by SAS.

Question 6:
In a triangle ABC, if angle A = 50° and angle B = 70°, find angle C. Justify your answer using the Angle Sum Property of a triangle.
Answer:

According to the Angle Sum Property of a triangle, the sum of all three interior angles is always 180°.


Given: ∠A = 50° and ∠B = 70°.
Let ∠C be the third angle.
Using the property: ∠A + ∠B + ∠C = 180°.
Substituting the values: 50° + 70° + ∠C = 180°.
Simplifying: 120° + ∠C = 180°.
Thus, ∠C = 180° - 120° = 60°.
Therefore, angle C measures 60°.
Question 7:
State the Pythagoras Theorem and verify it for a right-angled triangle with sides 6 cm, 8 cm, and 10 cm.
Answer:

The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.


Given sides: 6 cm, 8 cm, and 10 cm (hypotenuse).
According to the theorem: (Hypotenuse)² = (Side 1)² + (Side 2)².
Substituting the values: (10)² = (6)² + (8)².
Calculating: 100 = 36 + 64.
Simplifying: 100 = 100.
Since both sides are equal, the theorem is verified for this triangle.
Question 8:
Explain why a triangle cannot have two right angles.
Answer:

A triangle cannot have two right angles because:

  • The sum of all three interior angles in a triangle must be 180° (Angle Sum Property).
  • If two angles are right angles (90° each), their sum alone would be 180° (90° + 90° = 180°).
  • This leaves no room for a third angle, violating the definition of a triangle.

Thus, a triangle with two right angles is impossible.

Question 9:
In two triangles, ABC and PQR, if AB = PQ, BC = QR, and ∠B = ∠Q, which congruence rule applies here? Write the congruent parts.
Answer:

The given conditions satisfy the SAS (Side-Angle-Side) congruence rule because:

  • Two sides (AB = PQ and BC = QR) and the included angle (∠B = ∠Q) are equal.

The congruent parts are:

  • AB ≅ PQ
  • BC ≅ QR
  • ∠B ≅ ∠Q
  • By SAS rule, ∆ABC ≅ ∆PQR.
Question 10:
If the sides of a triangle are in the ratio 3:4:5 and its perimeter is 24 cm, find the lengths of its sides.
Answer:

Let the sides of the triangle be 3x, 4x, and 5x (given ratio).


Perimeter = Sum of all sides = 24 cm.
Thus, 3x + 4x + 5x = 24.
Simplifying: 12x = 24.
Solving for x: x = 24 / 12 = 2.
Now, the sides are:
First side = 3x = 3 × 2 = 6 cm.
Second side = 4x = 4 × 2 = 8 cm.
Third side = 5x = 5 × 2 = 10 cm.
Question 11:
In a triangle ABC, if angle A = 60° and angle B = 70°, find angle C. Justify your answer using the Angle Sum Property of a triangle.
Answer:

According to the Angle Sum Property of a triangle, the sum of all three interior angles is 180°.


Given: ∠A = 60° and ∠B = 70°.
Let ∠C be the third angle.
Using the property: ∠A + ∠B + ∠C = 180°.
Substituting the values: 60° + 70° + ∠C = 180°.
Simplifying: 130° + ∠C = 180°.
Thus, ∠C = 180° - 130° = 50°.

Therefore, angle C measures 50°.

Question 12:
Prove that the angles opposite to equal sides of an isosceles triangle are equal. Use a diagram if necessary.
Answer:

Consider an isosceles triangle ABC with AB = AC.


To prove: ∠B = ∠C.

Construction: Draw the angle bisector AD of ∠A, meeting BC at D.


Proof:
In triangles ABD and ACD:
1. AB = AC (Given).
2. AD is common.
3. ∠BAD = ∠CAD (AD is the angle bisector).
By the SAS congruence rule, ΔABD ≅ ΔACD.
Thus, ∠B = ∠C (Corresponding parts of congruent triangles).

Hence, the angles opposite to equal sides are equal.

Question 13:
If two sides of a triangle are 5 cm and 7 cm, what can be the possible length of the third side? Justify using the Triangle Inequality Theorem.
Answer:

According to the Triangle Inequality Theorem, the sum of any two sides must be greater than the third side.


Let the third side be x cm.
Case 1: 5 + 7 > x ⇒ 12 > x ⇒ x < 12.
Case 2: 5 + x > 7 ⇒ x > 2.
Case 3: 7 + x > 5 ⇒ x > -2 (always true for lengths).

Combining the inequalities: 2 < x < 12.


Thus, the third side must be greater than 2 cm and less than 12 cm.
Question 14:
In a right-angled triangle PQR, right-angled at Q, if PQ = 6 cm and PR = 10 cm, find the length of QR using the Pythagoras Theorem.
Answer:

Given: Right-angled triangle PQR with ∠Q = 90°, PQ = 6 cm, and PR = 10 cm.


Using the Pythagoras Theorem: PR² = PQ² + QR².
Substituting the values: 10² = 6² + QR².
100 = 36 + QR².
QR² = 100 - 36 = 64.
QR = √64 = 8 cm.

Thus, the length of QR is 8 cm.

Question 15:
State and prove the ASA congruence rule for triangles with a suitable example.
Answer:

The ASA Congruence Rule states that if two angles and the included side of one triangle are equal to the corresponding angles and included side of another triangle, the triangles are congruent.


Example: Consider triangles ABC and DEF where:
∠B = ∠E, BC = EF, and ∠C = ∠F.

Proof:


1. ∠B = ∠E (Given).
2. BC = EF (Given).
3. ∠C = ∠F (Given).
By the ASA rule, ΔABC ≅ ΔDEF.

Thus, the triangles are congruent.

Question 16:
State and prove the Basic Proportionality Theorem (Thales Theorem) for triangles.
Answer:

The Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.


Given: A triangle ABC with DE || BC, intersecting AB at D and AC at E.
To Prove: AD/DB = AE/EC.
Proof:
1. Since DE || BC, by AA similarity, ∆ADE ~ ∆ABC.
2. Therefore, AD/AB = AE/AC (corresponding sides of similar triangles).
3. Rewriting, AD/(AD + DB) = AE/(AE + EC).
4. Cross-multiplying gives AD·EC = AE·DB.
5. Hence, AD/DB = AE/EC (Proved).
Question 17:
Explain why the sum of any two sides of a triangle is greater than the third side.
Answer:

This property is derived from the triangle inequality theorem, which states that for any triangle with sides a, b, and c, the following must hold true:


1. a + b > c
2. a + c > b
3. b + c > a

The reason is geometric: the shortest distance between two points is a straight line. If the sum of two sides were ≤ the third side, they wouldn't meet to form a closed shape (triangle). For example, if a + b ≤ c, sides a and b would be too short to connect when laid end-to-end.

Question 18:
In a right-angled triangle, the hypotenuse is 13 cm and one side is 5 cm. Find the third side using the Pythagoras Theorem.
Answer:

Given a right-angled triangle with:


Hypotenuse (c) = 13 cm
One side (a) = 5 cm
Let the third side be b.

Applying Pythagoras Theorem:
a² + b² = c²
5² + b² = 13²
25 + b² = 169
b² = 169 - 25
b² = 144
b = √144
b = 12 cm

Thus, the third side is 12 cm.

Question 19:
Differentiate between congruent and similar triangles with examples.
Answer:

Congruent triangles are identical in shape and size (all corresponding sides and angles equal), whereas similar triangles have the same shape but not necessarily the same size (angles equal, sides proportional).


  • Example of Congruence: Two triangles with sides 3 cm, 4 cm, 5 cm each are congruent by SSS rule.
  • Example of Similarity: A 3-4-5 triangle and a 6-8-10 triangle are similar (sides in ratio 1:2) but not congruent.
Question 20:
Prove that the angles opposite to equal sides of an isosceles triangle are equal.
Answer:

Given: An isosceles ∆ABC with AB = AC.
To Prove: ∠B = ∠C.
Proof:
1. Draw the angle bisector AD of ∠A, meeting BC at D.
2. In ∆ABD and ∆ACD:
- AB = AC (given)
- ∠BAD = ∠CAD (by construction)
- AD is common.
3. Thus, ∆ABD ≅ ∆ACD by SAS congruence.
4. Therefore, ∠B = ∠C (corresponding angles of congruent triangles).

This confirms that equal sides subtend equal angles in an isosceles triangle.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Prove that the angles opposite to equal sides of an isosceles triangle are equal. Use the SAS congruence rule to justify your answer.
Answer:
Introduction

We studied that an isosceles triangle has two equal sides. Our textbook shows that angles opposite these sides are also equal.


Argument 1
  • Let △ABC be isosceles with AB = AC.
  • Draw AD as the angle bisector of ∠A, meeting BC at D.

Argument 2
  • In △ABD and △ACD, AB = AC (given), AD is common, and ∠BAD = ∠CAD (by construction).
  • By SAS rule, △ABD ≅ △ACD, so ∠B = ∠C.

Conclusion

Thus, angles opposite equal sides are equal, as proved using SAS congruence.

Question 2:
In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides. Derive this using a Pythagorean theorem example from NCERT.
Answer:
Introduction

Our textbook shows the Pythagorean theorem for right-angled triangles. Let’s verify it with an example.


Argument 1
  • Consider △ABC with ∠B = 90°, AB = 3 cm, BC = 4 cm.
  • By Pythagoras, AC² = AB² + BC².

Argument 2
  • Substituting values: AC² = 3² + 4² = 9 + 16 = 25.
  • Thus, AC = 5 cm, which matches our measurement.

Conclusion

The theorem holds true, proving AC² = AB² + BC² in right-angled triangles.

Question 3:
Explain how the ASA congruence rule can be used to prove two triangles congruent. Give an example from real-life applications.
Answer:
Introduction

We learned that ASA rule requires two angles and the included side to be equal for congruence.


Argument 1
  • Example: Two roof trusses with ∠A = ∠P, ∠B = ∠Q, and included side AB = PQ.
  • By ASA, △ABC ≅ △PQR, ensuring structural symmetry.

Argument 2
  • NCERT shows △DEF and △LMN where ∠D = ∠L, ∠E = ∠M, and DE = LM.
  • Thus, △DEF ≅ △LMN by ASA.

Conclusion

ASA rule is vital in construction and design for ensuring identical triangular shapes.

Question 4:
Prove that the angles opposite to equal sides of an isosceles triangle are equal. Use the ASA congruence rule to justify your answer.
Answer:
Introduction

We studied that an isosceles triangle has two equal sides. Our textbook shows that angles opposite these sides are also equal.


Argument 1
  • Draw △ABC with AB = AC.
  • Construct AD as the angle bisector of ∠A, meeting BC at D.

Argument 2
  • In △ABD and △ACD, AB = AC (given), AD is common, and ∠BAD = ∠CAD (by construction).
  • By ASA congruence rule, △ABD ≅ △ACD, so ∠B = ∠C.

Conclusion

Hence, angles opposite equal sides are equal, verified using ASA rule.

Question 5:
In △PQR, PQ = PR and ∠Q = 50°. Find ∠R and ∠P. Explain using angle sum property and isosceles triangle properties.
Answer:
Introduction

Given △PQR with PQ = PR, it is an isosceles triangle. We use the angle sum property to find missing angles.


Argument 1
  • Since PQ = PR, ∠Q = ∠R = 50° (angles opposite equal sides).

Argument 2
  • By angle sum property, ∠P + ∠Q + ∠R = 180°.
  • Substituting, ∠P + 50° + 50° = 180° ⇒ ∠P = 80°.

Conclusion

Thus, ∠R = 50° and ∠P = 80°, derived using isosceles properties and angle sum.

Question 6:
A ladder 10 m long reaches a window 8 m above the ground. How far is the ladder's foot from the wall? Use the Pythagoras theorem and show steps.
Answer:
Introduction

This is a real-life application of the Pythagoras theorem. The ladder, wall, and ground form a right triangle.


Argument 1
  • Let the distance from the wall be x meters.
  • By Pythagoras theorem: x² + 8² = 10².

Argument 2
  • Simplifying, x² + 64 = 100 ⇒ x² = 36 ⇒ x = 6 m.

Conclusion

The ladder's foot is 6 m from the wall, as calculated using the theorem.

Question 7:
In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides. State and prove this theorem using Pythagoras' theorem.
Answer:
Introduction

Pythagoras' theorem is a fundamental concept in triangles. We will prove it step-by-step.


Argument 1

Let ∆ABC be right-angled at B. Construct squares on all sides. The area of the square on hypotenuse AC = sum of areas on AB and BC.


Argument 2

Using algebra, AC² = AB² + BC². Our textbook verifies this with a 3-4-5 triangle example, showing 5² = 3² + 4².


Conclusion

Thus, Pythagoras' theorem is validated, useful in real-life measurements like ladder problems.

Question 8:
Show that the sum of any two sides of a triangle is greater than the third side. Explain with an inequality and a practical example.
Answer:
Introduction

We know triangles have three sides. Let’s prove their inequality property.


Argument 1

For ∆ABC, AB + BC > AC, BC + AC > AB, and AC + AB > BC. Our textbook derives this using the shortest path between two points.


Argument 2

Example: If sides are 5cm, 7cm, and 10cm, 5 + 7 > 10, 7 + 10 > 5, and 5 + 10 > 7. This holds true.


Conclusion

This inequality is vital in construction to ensure triangular stability.

Question 9:
In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides. Verify this using a Pythagorean triplet example.
Answer:
Introduction

We learned the Pythagoras theorem, which states that in a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.


Argument 1

Let’s take the Pythagorean triplet (3, 4, 5). Here, hypotenuse (c) = 5, and the other sides (a, b) = 3 and 4.


Argument 2

Calculating, a² + b² = 3² + 4² = 9 + 16 = 25. Also, c² = 5² = 25. Thus, a² + b² = c². Our textbook uses this example in Chapter 6.


Conclusion

This verifies the Pythagoras theorem for the given triplet.

Question 10:
Show that the sum of any two sides of a triangle is greater than the third side. Use a construction method to explain.
Answer:
Introduction

We studied the triangle inequality theorem, which states that the sum of any two sides of a triangle is greater than the third side.


Argument 1

Let’s take ∆ABC. To prove AB + AC > BC, we extend BA to D such that AD = AC. Now, ∠ACD = ∠ADC (isosceles triangle property).


Argument 2

In ∆BCD, ∠BCD > ∠BDC. Thus, BD > BC (side opposite to larger angle). But BD = AB + AD = AB + AC. Hence, AB + AC > BC. Our textbook explains this in Chapter 7.


Conclusion

This proves the triangle inequality theorem using construction.

Question 11:
Prove that the angles opposite to equal sides of a triangle are equal. Use the isosceles triangle property to explain.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, let us consider an isosceles triangle ABC where AB = AC.


Steps:


1. Draw the angle bisector of ∠A that meets BC at point D.


2. In triangles ABD and ACD:
- AB = AC (Given)
- ∠BAD = ∠CAD (AD is the angle bisector)
- AD = AD (Common side)


3. By the SAS (Side-Angle-Side) congruence rule, △ABD ≅ △ACD.


4. Therefore, ∠B = ∠C (Corresponding parts of congruent triangles).


This proves that the angles opposite to equal sides are equal. This property is fundamental in solving problems related to isosceles triangles.

Question 12:
In a triangle ABC, if ∠A = 60° and ∠B = 80°, find ∠C. Also, state the type of triangle based on its angles.
Answer:

To find ∠C in triangle ABC, we use the angle sum property of a triangle, which states that the sum of all interior angles is 180°.


Steps:


1. Given: ∠A = 60° and ∠B = 80°.


2. Using the angle sum property: ∠A + ∠B + ∠C = 180°.


3. Substitute the given values: 60° + 80° + ∠C = 180°.


4. Simplify: 140° + ∠C = 180°.


5. Therefore, ∠C = 180° - 140° = 40°.


Since all angles (60°, 80°, and 40°) are less than 90°, the triangle is an acute-angled triangle.


This classification is important as it helps in understanding the properties and behavior of triangles in geometry.

Question 13:
Prove that the angles opposite to equal sides of a triangle are equal. Use the isosceles triangle property to explain your answer.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, let us consider an isosceles triangle ABC where AB = AC.


Steps:


1. Draw the angle bisector of ∠A that meets BC at D.


2. In triangles ABD and ACD:


  • AB = AC (Given)
  • ∠BAD = ∠CAD (AD is the angle bisector)
  • AD = AD (Common side)

3. By the SAS congruence rule, ΔABD ≅ ΔACD.


4. Therefore, ∠B = ∠C (Corresponding parts of congruent triangles).


This proves that the angles opposite to equal sides are equal. This property is fundamental in solving problems related to isosceles triangles.

Question 14:
Prove that the angles opposite to equal sides of a triangle are equal. Use the Isosceles Triangle Theorem and provide a step-by-step proof with a diagram.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, we will use the Isosceles Triangle Theorem. Let us consider an isosceles triangle ABC where AB = AC.


Step 1: Draw triangle ABC with AB = AC.
Step 2: Draw the angle bisector AD of ∠BAC, meeting BC at D.
Step 3: In triangles ABD and ACD:
- AB = AC (Given)
- ∠BAD = ∠CAD (AD is the angle bisector)
- AD = AD (Common side)
Step 4: By the SAS (Side-Angle-Side) congruence rule, ΔABD ≅ ΔACD.
Step 5: Therefore, ∠B = ∠C (Corresponding parts of congruent triangles).

This proves that the angles opposite to equal sides are equal. A diagram would show triangle ABC with AB = AC and AD as the angle bisector, clearly marking the congruent parts.

Question 15:
In a triangle, if one angle is 90°, prove that the sum of the other two angles is also 90°. Use the Angle Sum Property of a Triangle and justify your answer.
Answer:

To prove that the sum of the other two angles in a right-angled triangle is 90°, we will use the Angle Sum Property of a Triangle.


Step 1: Let the triangle be ABC with ∠A = 90°.
Step 2: According to the Angle Sum Property, the sum of all three angles in a triangle is 180°.
So, ∠A + ∠B + ∠C = 180°.
Step 3: Substitute ∠A = 90° into the equation:
90° + ∠B + ∠C = 180°.
Step 4: Subtract 90° from both sides:
∠B + ∠C = 180° - 90° = 90°.

Thus, the sum of the other two angles (∠B and ∠C) in a right-angled triangle is 90°. This property is fundamental in solving problems related to right-angled triangles.

Question 16:
In a triangle ABC, the sides AB and AC are equal. If the measure of ∠B is 50°, find the measures of the other two angles. Justify your answer with proper reasoning.
Answer:

Given: In triangle ABC, AB = AC and ∠B = 50°.

Since AB = AC, triangle ABC is an isosceles triangle with ∠B = ∠C (angles opposite to equal sides are equal).


Thus, ∠C = 50°.


Now, using the angle sum property of a triangle:


∠A + ∠B + ∠C = 180°


Substituting the known values:


∠A + 50° + 50° = 180°


∠A = 180° - 100° = 80°


Therefore, the measures of the other two angles are:

  • ∠A = 80°
  • ∠C = 50°

This solution adheres to the properties of an isosceles triangle and the angle sum property, ensuring accuracy.

Question 17:
Prove that the angles opposite to equal sides of a triangle are equal. Use a diagram and provide a step-by-step proof.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, let us consider a triangle ABC where AB = AC.


Given: In ∆ABC, AB = AC.


To prove: ∠B = ∠C.


Construction: Draw the angle bisector of ∠A meeting BC at D.


Proof:


1. In ∆ABD and ∆ACD,
- AB = AC (Given)
- AD = AD (Common side)
- ∠BAD = ∠CAD (By construction, AD is the angle bisector)


2. Therefore, by the SAS (Side-Angle-Side) congruence rule, ∆ABD ≅ ∆ACD.


3. Since corresponding parts of congruent triangles are equal, ∠B = ∠C.


Conclusion: Hence, the angles opposite to equal sides of a triangle are equal.


Diagram: (Draw triangle ABC with AB = AC and angle bisector AD meeting BC at D.)


Value-added information: This property is known as the Isosceles Triangle Theorem and is widely used in solving geometric problems involving triangles with two equal sides.

Question 18:
Prove that the angles opposite to equal sides of a triangle are equal. Use the isosceles triangle property and provide a step-by-step proof with a diagram.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, let us consider an isosceles triangle ABC where AB = AC.


Given: In ∆ABC, AB = AC.
To prove: ∠B = ∠C.


Construction: Draw the angle bisector AD of ∠A, meeting BC at D.


Proof:
1. In ∆ABD and ∆ACD:
- AB = AC (Given)
- ∠BAD = ∠CAD (By construction, AD is the angle bisector)
- AD = AD (Common side)
2. By the SAS (Side-Angle-Side) congruence rule, ∆ABD ≅ ∆ACD.
3. Since corresponding parts of congruent triangles are equal, ∠B = ∠C.


Conclusion: Hence, the angles opposite to the equal sides of a triangle are equal.


Diagram: (Draw ∆ABC with AB = AC and angle bisector AD, labeling all parts clearly.)


Value-added note: This property is foundational in geometry and is used to solve problems involving isosceles and equilateral triangles. It also helps in proving other theorems, such as the Base Angles Theorem.

Question 19:
In a triangle ABC, the sides AB and AC are equal. If the measure of angle B is 50°, find the measures of the other two angles. Justify your answer with proper reasoning.
Answer:

Given: In triangle ABC, AB = AC and ∠B = 50°.

Since AB = AC, triangle ABC is an isosceles triangle with AB and AC as equal sides. In an isosceles triangle, the angles opposite to the equal sides are equal.


Therefore, ∠B = ∠C = 50°.

Now, using the angle sum property of a triangle:


∠A + ∠B + ∠C = 180°
∠A + 50° + 50° = 180°
∠A = 180° - 100° = 80°

Thus, the measures of the other two angles are:

  • ∠A = 80°
  • ∠C = 50°
Question 20:
Prove that the angles opposite to equal sides of a triangle are equal. Use this result to show that in an isosceles triangle, the altitude from the vertex bisects the base.
Answer:

To prove that the angles opposite to equal sides of a triangle are equal, consider a triangle ABC where AB = AC.


1. Draw the angle bisector of ∠A meeting BC at D.
2. In triangles ABD and ACD:
- AB = AC (given)
- ∠BAD = ∠CAD (by construction)
- AD = AD (common side)
3. By the SAS congruence rule, ΔABD ≅ ΔACD.
4. Therefore, ∠B = ∠C (by CPCT).


Now, to show that the altitude from the vertex bisects the base in an isosceles triangle:
1. Let AD be the altitude from A to BC.
2. In triangles ABD and ACD:
- AB = AC (given)
- ∠ADB = ∠ADC (both 90°)
- AD = AD (common side)
3. By the RHS congruence rule, ΔABD ≅ ΔACD.
4. Hence, BD = DC (by CPCT), proving that the altitude bisects the base.

Question 21:
In a triangle ABC, if ∠A = 60° and ∠B = 80°, find ∠C. Also, verify the Angle Sum Property of a triangle using these angles.
Answer:

Given: In triangle ABC, ∠A = 60° and ∠B = 80°.


1. To find ∠C, use the Angle Sum Property:
∠A + ∠B + ∠C = 180°
60° + 80° + ∠C = 180°
140° + ∠C = 180°
∠C = 180° - 140° = 40°.


2. Verification of the Angle Sum Property:
Sum of all angles = 60° + 80° + 40° = 180°.
This matches the property, confirming its validity.


Additional Insight: The Angle Sum Property is fundamental in geometry and helps in solving various problems related to triangles, such as finding missing angles or proving congruence.

Question 22:
Prove that the angles opposite to the equal sides of an isosceles triangle are equal. Use a diagram to support your proof.
Answer:

To prove: In an isosceles triangle, the angles opposite to the equal sides are equal.

Let us consider an isosceles triangle ABC with AB = AC.

We need to prove that ∠B = ∠C.


Construction: Draw the angle bisector of ∠A, meeting BC at D.


Proof:


In triangles ABD and ACD:
1. AB = AC (Given)
2. AD = AD (Common side)
3. ∠BAD = ∠CAD (AD is the angle bisector)

By the SAS (Side-Angle-Side) congruence rule, ΔABD ≅ ΔACD.

Since corresponding parts of congruent triangles are equal, ∠B = ∠C.

Hence, the angles opposite to the equal sides of an isosceles triangle are equal.

Diagram: (Visualize triangle ABC with AB = AC and AD as the angle bisector, dividing ∠A into two equal parts and meeting BC at D.)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
In ΔABC, sides AB = 5 cm, BC = 7 cm, and AC = 8 cm. Verify if it satisfies the triangle inequality property. Also, classify the triangle based on angles.
Answer:
Problem Interpretation

We need to check if the sum of any two sides exceeds the third side and determine the angle type.


Mathematical Modeling
  • AB + BC = 12 > AC
  • BC + AC = 15 > AB
  • AB + AC = 13 > BC

Solution

Since all sums satisfy the inequality, ΔABC exists. Using the Pythagoras theorem, 5² + 7² ≠ 8², so it is an acute-angled triangle.

Question 2:
A ladder 10 m long leans against a wall, touching it at 8 m above the ground. Prove that the ladder, wall, and ground form a right-angled triangle. Calculate the distance from the ladder's base to the wall.
Answer:
Problem Interpretation

We must verify the right-angle condition and find the base distance.


Mathematical Modeling
  • Hypotenuse (ladder) = 10 m
  • Height (wall) = 8 m
  • Let base distance = x

Solution

Using Pythagoras theorem, x² + 8² = 10² ⇒ x = 6 m. Since 6² + 8² = 10², it is a right-angled triangle.

Question 3:
In a right-angled triangle ABC, angle B = 90°, AB = 6 cm, and BC = 8 cm. Using the Pythagoras theorem, find the length of AC. Also, state whether a triangle with sides 6 cm, 8 cm, and 10 cm is right-angled or not.
Answer:
Problem Interpretation

We need to find the hypotenuse AC in right-angled ΔABC and verify if given sides form a right triangle.


Mathematical Modeling
  • Using Pythagoras theorem: AC² = AB² + BC²

Solution
  • AC² = 6² + 8² = 100 ⇒ AC = 10 cm
  • For sides 6 cm, 8 cm, 10 cm: 6² + 8² = 10² ⇒ It is right-angled.
Question 4:
A ladder 5 m long leans against a wall. Its foot is 3 m away from the wall. Using triangle properties, find how high the ladder reaches. Also, determine if the ladder, wall, and ground form a Pythagorean triplet.
Answer:
Problem Interpretation

We must find the ladder's height on the wall and check for Pythagorean triplet.


Mathematical Modeling
  • Let height be h. Using Pythagoras theorem: h² + 3² = 5²

Solution
  • h² = 25 - 9 = 16 ⇒ h = 4 m
  • The sides 3 m, 4 m, 5 m form a triplet as 3² + 4² = 5².
Question 5:
In a right-angled triangle ABC, angle B = 90°, AB = 6 cm, and BC = 8 cm. Using the Pythagoras theorem, find the length of AC. Also, state whether a triangle with sides 6 cm, 8 cm, and 10 cm is right-angled or not.
Answer:
Problem Interpretation

We are given a right-angled triangle ABC with AB = 6 cm, BC = 8 cm, and angle B = 90°. We need to find AC.


Mathematical Modeling

Using Pythagoras theorem: AC² = AB² + BC².


Solution
  • AC² = 6² + 8² = 36 + 64 = 100 ⇒ AC = 10 cm.
  • For sides 6 cm, 8 cm, 10 cm: 6² + 8² = 10² ⇒ It is right-angled.
Question 6:
A ladder 5 m long leans against a wall. The foot of the ladder is 3 m away from the wall. Find the height on the wall where the ladder touches. Also, explain if the ladder slips to 4 m away from the wall, how the height changes.
Answer:
Problem Interpretation

A ladder forms a right-angled triangle with the wall and ground. Initial distance = 3 m, length = 5 m.


Mathematical Modeling

Using Pythagoras theorem: Height² + Base² = Hypotenuse².


Solution
  • Height² = 5² - 3² = 25 - 9 = 16 ⇒ Height = 4 m.
  • If base = 4 m: Height² = 5² - 4² = 9 ⇒ Height = 3 m. The height decreases.
Question 7:
In a park, two ladders leaning against a wall form right-angled triangles with the ground. The first ladder is 5 m long and reaches 3 m up the wall. The second ladder is 10 m long. Using the Pythagoras theorem, find how high the second ladder reaches.
Answer:
Problem Interpretation

We need to find the height reached by the second ladder using the Pythagoras theorem.

Mathematical Modeling
  • First ladder: Base = √(5² - 3²) = 4 m (ground distance).
  • Second ladder: Same ground distance (4 m).
Solution

Height = √(10² - 4²) = √84 ≈ 9.17 m. Our textbook shows similar right-angled triangle problems.

Question 8:
A triangular garden has sides 6 m, 8 m, and 10 m. Prove it is a right-angled triangle using the converse of Pythagoras theorem. Also, calculate its area.
Answer:
Problem Interpretation

We must verify if the triangle is right-angled and find its area.

Mathematical Modeling
  • Check 6² + 8² = 10² → 36 + 64 = 100 (holds true).
Solution

It is right-angled. Area = ½ × 6 × 8 = 24 m². We studied such examples in NCERT.

Question 9:
In a park, two ladders lean against opposite walls. The first ladder is 5 m long and reaches 4 m up the wall. The second ladder is 6 m long and reaches 4.8 m up the other wall. Using the Pythagorean theorem, prove whether the distance between the walls is equal to the sum of the bases of the two triangles formed.
Answer:
Problem Interpretation

We need to verify if the distance between the walls matches the sum of the bases of the right triangles formed by the ladders.

Mathematical Modeling
  • First ladder: Hypotenuse = 5 m, height = 4 m. Base = √(5² - 4²) = 3 m.
  • Second ladder: Hypotenuse = 6 m, height = 4.8 m. Base = √(6² - 4.8²) = 3.6 m.
Solution

Total base = 3 + 3.6 = 6.6 m. If walls are 6.6 m apart, the condition holds. Our textbook shows similar applications of the Pythagorean theorem.

Question 10:
A triangular garden has sides 6 m, 8 m, and 10 m. Using the converse of the Pythagorean theorem, determine if it is a right-angled triangle. Also, find its area.
Answer:
Problem Interpretation

We must check if the garden's sides satisfy the Pythagorean theorem and calculate its area.

Mathematical Modeling
  • Check: 6² + 8² = 36 + 64 = 100 = 10². Hence, it is right-angled.
Solution

Area = ½ × base × height = ½ × 6 × 8 = 24 m². We studied such problems in NCERT where sides verify right angles.

Question 11:

In a park, two triangular flower beds ABC and PQR are designed such that AB = PQ, BC = QR, and ∠B = ∠Q. A gardener claims that both triangles are congruent. Using the SAS congruence rule, verify his claim and justify your answer.

Answer:

The gardener's claim is correct because the two triangles satisfy the SAS (Side-Angle-Side) congruence rule.


Given:
AB = PQ (Side),
∠B = ∠Q (Angle),
BC = QR (Side).


Explanation:
According to the SAS congruence rule, if two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, then the triangles are congruent.


Here, AB = PQ, ∠B = ∠Q, and BC = QR.
Since the angle is included between the two sides, △ABC ≅ △PQR by SAS rule.


Conclusion:
Thus, the gardener is right, and the two flower beds are congruent.

Question 12:

A student draws two triangles, △XYZ and △LMN, where XY = LM, YZ = MN, and XZ = LN. She states that both triangles are congruent by the SSS congruence rule. Explain whether her statement is correct with proper reasoning.

Answer:

The student's statement is correct because the given conditions satisfy the SSS (Side-Side-Side) congruence rule.


Given:
XY = LM,
YZ = MN,
XZ = LN.


Explanation:
According to the SSS congruence rule, if all three sides of one triangle are equal to the corresponding sides of another triangle, then the triangles are congruent.


Here, all corresponding sides are equal:
XY = LM,
YZ = MN,
XZ = LN.
Thus, △XYZ ≅ △LMN by SSS rule.


Conclusion:
The student's claim is valid, and the triangles are congruent.

Question 13:

In a park, three children are standing such that the distance between Child A and Child B is 5 meters, Child B and Child C is 12 meters, and Child A and Child C is 13 meters. They form a triangle ABC.

(i) Using the Pythagoras theorem, determine if triangle ABC is a right-angled triangle.

(ii) If Child B moves 3 meters closer to Child A, how will it affect the angles of the triangle? Justify your answer.

Answer:

Solution:

(i) To check if triangle ABC is right-angled, we verify the Pythagoras theorem:


AB = 5 m, BC = 12 m, AC = 13 m
AB² + BC² = 5² + 12² = 25 + 144 = 169
AC² = 13² = 169
Since AB² + BC² = AC², triangle ABC is right-angled at ∠B.

(ii) If Child B moves 3 meters closer to Child A, the new distance AB becomes 2 m (5 m - 3 m).


New sides: AB = 2 m, BC = 12 m, AC = ?
Using Pythagoras theorem: AC² = AB² + BC² = 4 + 144 = 148 → AC ≈ 12.17 m
The angle at ∠B will no longer be 90° because the sides no longer satisfy the Pythagorean condition. The triangle becomes acute-angled as all angles are now less than 90°.
Question 14:

Two triangles, PQR and XYZ, are drawn such that PQ = XY, QR = YZ, and ∠Q = ∠Y.

(i) By which congruency rule are these triangles congruent? State the rule and explain.

(ii) If ∠P = 50° and ∠R = 70°, find the measure of ∠X and ∠Z in triangle XYZ.

Answer:

Solution:

(i) Triangles PQR and XYZ are congruent by the SAS (Side-Angle-Side) rule because:

  • PQ = XY (given)
  • ∠Q = ∠Y (given)
  • QR = YZ (given)

Two sides and the included angle are equal, so the triangles are congruent.


(ii) Since ΔPQR ≅ ΔXYZ, corresponding angles are equal:


∠P = ∠X = 50°
∠R = ∠Z = 70°

Thus, ∠X = 50° and ∠Z = 70°.

Question 15:
In a park, two triangular flower beds ABC and PQR are designed such that AB = PQ, BC = QR, and ∠B = ∠Q. A gardener claims that both flower beds are congruent. Is he correct? Justify your answer using the appropriate congruence rule.
Answer:

Yes, the gardener is correct. The two triangles ABC and PQR are congruent by the SAS (Side-Angle-Side) congruence rule.


Justification:


Given:

  • AB = PQ (Side)
  • BC = QR (Side)
  • ∠B = ∠Q (Included Angle)

Since two sides and the included angle of one triangle are equal to the corresponding parts of the other triangle, the triangles are congruent by SAS rule.

Question 16:
A ladder LM of length 5 m is placed against a wall such that its foot is 3 m away from the wall. Another ladder XY of the same length is placed such that its foot is 4 m away from the wall. Are the triangles formed by the ladders, wall, and ground congruent? Explain using the Pythagoras theorem and a suitable congruence criterion.
Answer:

No, the triangles formed by the ladders are not congruent.


Explanation:


For ladder LM:


Using the Pythagoras theorem, the height (h1) reached on the wall is:


h1 = √(5² - 3²) = √(25 - 9) = √16 = 4 m


For ladder XY:


The height (h2) reached on the wall is:


h2 = √(5² - 4²) = √(25 - 16) = √9 = 3 m


Since the corresponding sides of the two triangles (formed by the ladders) are not equal, they do not satisfy any congruence criterion (such as SSS, SAS, or RHS). Hence, the triangles are not congruent.

Question 17:
In a park, three children are standing at points A, B, and C such that AB = AC and ∠B = 50°. A fourth child joins them and stands at point D, making AD the angle bisector of ∠BAC. Prove that △ABD ≅ △ACD using the appropriate congruence criterion.
Answer:

Given: AB = AC and ∠B = 50° in △ABC. AD is the angle bisector of ∠BAC.

To prove: △ABD ≅ △ACD.


Step 1: Since AB = AC, △ABC is an isosceles triangle with ∠B = ∠C = 50°.
Step 2: Using the angle sum property of a triangle, ∠BAC = 180° - (50° + 50°) = 80°.
Step 3: AD bisects ∠BAC, so ∠BAD = ∠CAD = 40°.
Step 4: In △ABD and △ACD:
  • AB = AC (given)
  • AD is common
  • ∠BAD = ∠CAD (proved above)

By the SAS congruence criterion, △ABD ≅ △ACD.
Question 18:
A ladder 10 m long leans against a wall, touching it at a height of 8 m. The foot of the ladder is moved 2 m away from the wall. Using the properties of triangles, find the new height at which the ladder touches the wall and justify your answer.
Answer:

Given: A ladder forms a right-angled triangle with the wall and ground. Initial height (h) = 8 m, ladder length (L) = 10 m.


Step 1: Using Pythagoras' theorem, initial distance (d) from the wall:
d = √(L² - h²) = √(100 - 64) = √36 = 6 m.
Step 2: The foot is moved 2 m away, so new distance (d') = 6 + 2 = 8 m.
Step 3: New height (h') = √(L² - d'²) = √(100 - 64) = √36 = 6 m.

Thus, the ladder now touches the wall at a height of 6 m. This shows that as the base increases, the height decreases to maintain the ladder's length.

Question 19:
In a park, two triangular flower beds ABC and PQR are such that AB = PQ, BC = QR, and ∠B = ∠Q. A gardener wants to confirm if both flower beds are congruent. Using the SAS congruence rule, explain whether the triangles are congruent and justify your answer.
Answer:

The triangles ABC and PQR are congruent by the SAS (Side-Angle-Side) congruence rule.


Given:
1. AB = PQ (Side)
2. BC = QR (Side)
3. ∠B = ∠Q (Included Angle)


Explanation:
The SAS rule states that if two sides and the included angle of one triangle are equal to the corresponding sides and included angle of another triangle, then the triangles are congruent.


Here, the sides AB and BC in ABC correspond to PQ and QR in PQR, and the included angle ∠B equals ∠Q. Thus, the conditions for SAS congruence are satisfied.


Conclusion:
Both flower beds are congruent, meaning they have the same shape and size.

Question 20:
A ladder leaning against a wall forms a right-angled triangle with the ground and the wall. The foot of the ladder is 3 meters away from the wall, and the top touches the wall at 4 meters height. Later, the ladder is shifted such that its foot is now 6 meters away from the wall. Using the Pythagorean theorem, determine whether the new triangle formed is similar to the original one and justify your answer.
Answer:

The new triangle formed is not similar to the original one.


Original Triangle:
Using the Pythagorean theorem, the length of the ladder (hypotenuse) is calculated as:
√(3² + 4²) = √(9 + 16) = √25 = 5 meters.


New Triangle:
When the ladder is shifted, the new base is 6 meters, and the height (h) can be found using the same ladder length (5 meters):
6² + h² = 5²
36 + h² = 25
h² = -11 (Not possible, as height cannot be imaginary).


Explanation:
The calculation shows that with the ladder's length fixed at 5 meters, it cannot form a right-angled triangle with a base of 6 meters. Thus, the new configuration is invalid, and the triangles cannot be similar.


Conclusion:
The original and new triangles are not similar because the new measurements do not satisfy the Pythagorean theorem.

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