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Class 6 Mathematics Chapter 15 of 18

Chapter 15 — Triangles and their Properties

Open the lesson Play with this chapter — pictures, sound and practice.

Overview

This unit introduces triangles, one of the most important shapes in geometry. Students learn how to name and draw triangles, identify their parts (vertices, sides and angles), and classify triangles by sides and by angles. The unit explains key properties such as the angle-sum property (angles of a triangle add to 180°), the exterior angle relation, and special facts about isosceles and equilateral triangles. Basic facts about right triangles and how to find the perimeter are included. The unit develops visual reasoning by using simple constructions, measuring with a ruler and protractor, and verifying properties with examples. These ideas matter because triangles form the basis of many geometric constructions and real-life structures (roofs, bridges, maps). Learning about triangles also builds skills in careful drawing, reasoning with angles and lengths, and solving simple geometry problems — skills that will be used throughout higher classes in mathematics and in everyday problem solving.

Learning Objectives

  • Identify and name the parts of a triangle: vertices, sides and angles.
  • Classify triangles by their sides and by their angles with proper reasoning.
  • Draw triangles using a ruler and a divider or compass and measure their sides and angles.
  • Explain and use the angle-sum property of a triangle to find a missing angle.
  • State and apply the exterior angle property of a triangle to solve angle problems.
  • Describe properties of isosceles and equilateral triangles and use them to find unknown angles.
  • Understand the concept of right triangles and recognise right, acute and obtuse triangles.
  • Compute the perimeter of a triangle and apply the basic triangle inequality concept.

Topics in this chapter

13 topics · tap a topic title to jump straight to it.

📐1

Introduction to triangles

A triangle is a closed figure formed by joining three non-collinear points by straight lines. The three straight lines are called the sides of the triangle and the three points are called the vertices. Triangles are the simplest polygons with three edges and three corners.

Every triangle has three interior angles, each located at a vertex and measured in degrees. The lines where two sides meet are called vertices, and we name a triangle by its vertices, for example △ABC. Triangles are rigid shapes — if the lengths of the three sides are fixed, the shape of the triangle is fixed too. This is why triangles are used in structures where stability is needed.

In this chapter we will learn to draw triangles carefully using a ruler and compass or a divider, measure sides and angles with a ruler and protractor, and use simple rules to find unknown angles or decide types of triangles. Observing small hands-on examples — cutting a triangle from paper, folding, and measuring — helps build an intuitive understanding. Working with triangles involves combining measurement skills (ruler and protractor) and logical steps to apply properties like angle sum and relationships between angles.

Triangles appear in many everyday objects: roof shapes, triangular road signs, and supports in bridges. Learning triangles gives a foundation for understanding more complex shapes and for solving geometry problems using proofs and calculations in later classes.

📌 Examples
  • Draw a triangle from three non-collinear points A, B and C and label sides AB, BC and CA.
  • Cut a triangle from paper, fold it so that two vertices coincide and observe the shape changes.
  • Identify triangles in the classroom: a triangular roof corner, a triangular signboard and a wedge-shaped bookend.
  • Show with a model that fixing three side lengths fixes the shape of a triangle.
🧮 Formulas
  1. Triangle name: △ABC where A, B, C are vertices
  2. Sides are written as AB, BC, CA
  3. Angles at vertices are denoted ∠A, ∠B, ∠C
📊 Visual ideas
A clear labelled triangle showing vertices A, B, C; sides AB, BC, CA; and interior angles ∠A, ∠B, ∠C that a student should draw
🧫2

Elements of a triangle: vertices, sides and angles

In every triangle there are three main elements: vertices, sides and angles. The three points where sides meet are the vertices; typically we label them A, B and C. The straight lines joining these points are the sides and are named by the two endpoints, for example AB means the side joining vertex A and vertex B.

The angles inside the triangle are made by two sides meeting at a vertex. The angle at vertex A is written as ∠A, and similarly ∠B and ∠C. Angles are measured in degrees using a protractor. Each triangle also has three exterior angles: these are formed by extending one side and measuring the angle outside the triangle. Understanding which lines form which angles is important when solving problems.

To draw a triangle from given data we often use tools: a ruler to measure side lengths; a compass or divider to copy lengths; and a protractor to measure or construct angles. For example, if we are given two sides and the included angle (SAS), we can draw the triangle by drawing one side, constructing the given angle at one end, and then marking the length of the second side with the compass to find the third vertex.

Careful labelling prevents mistakes: always mark equal sides or equal angles with small dashes or arc marks when needed. Practice distinguishing interior and exterior angles, and mark vertices clearly. This attention to detail helps in later proofs and constructions.

📌 Examples
  • Label a drawn triangle as △PQR. Mark sides PQ, QR, RP and angles ∠P, ∠Q, ∠R.
  • Construct a triangle when two sides and the included angle are given: draw side AB, construct ∠A, and cut an arc for the second side to locate C.
🧮 Formulas
  1. Vertex notation: A, B, C
  2. Side notation: AB, BC, CA
  3. Angle notation: ∠A, ∠B, ∠C
📊 Visual ideas
Triangle △PQR with vertices P, Q, R labelled, sides PQ, QR, RP and interior angles ∠P, ∠Q, ∠R shown
🔢3

Classification by sides: equilateral, isosceles and scalene

Triangles can be classified by comparing their side lengths. There are three types by sides: equilateral, isosceles and scalene. An equilateral triangle has all three sides equal. Because the sides are equal, all interior angles are also equal (each 60°). An isosceles triangle has two equal sides and a third side of different length. The angles opposite the equal sides are equal. A scalene triangle has all three sides of different lengths and therefore all three interior angles are different.

Recognising types helps in solving problems: equal sides suggest equal opposite angles, and equal angles suggest equal opposite sides. We usually mark equal sides with short parallel dashes and equal angles with matching arc marks so a diagram becomes easy to read. For example, in an isosceles triangle with sides AB = AC, we mark AB and AC with the same dash; then ∠B and ∠C are equal and we may mark them with the same arc symbol.

Practical tasks include measuring sides with a ruler to decide the type, or using a compass to copy lengths and check equality. These classifications also guide constructions: drawing an equilateral triangle from a given side is simple using a compass: draw two circles of that radius with centres at the endpoints of the side; their intersection gives the third vertex.

Understanding these side-based types prepares students for angle relationships and for using symmetry in later geometry problems.

📌 Examples
  • Measure sides of triangle ABC: if AB = BC = CA, state it is equilateral and each angle is 60°.
  • Given triangle DEF with DE = DF, mark DE and DF and conclude that ∠E = ∠F.
🧮 Formulas
  1. Equilateral: AB = BC = CA; ∠A = ∠B = ∠C = 60°
  2. Isosceles: two sides equal, e.g. AB = AC ⇒ ∠B = ∠C
  3. Scalene: AB ≠ BC ≠ CA; ∠A ≠ ∠B ≠ ∠C
📊 Visual ideas
An equilateral △ABC with AB = BC = CA and all angles 60°
An isosceles △ABC with AB = AC and equal base angles at B and C
A scalene △PQR with all sides different
📐4

Classification by angles: acute, right and obtuse triangles

Triangles are also classified by their interior angles. If all three interior angles are less than 90°, the triangle is acute. If one interior angle is exactly 90°, we call the triangle a right triangle. If one interior angle is greater than 90°, the triangle is obtuse. These classes are useful because different angle relationships and methods apply: right triangles have special properties related to perpendicular sides, while obtuse triangles behave differently with respect to altitude placement.

To identify the type, measure each angle with a protractor. If you find one angle equal to 90° exactly, mark the corner with a small square to show a right angle. If one angle is greater than 90°, the triangle is obtuse — the other two angles must then be acute. For an acute triangle all vertices will show angles less than a right angle when measured.

Right triangles are especially important since the two sides forming the right angle are called perpendicular and are often referred to as legs; the side opposite the right angle is the longest side and is called the hypotenuse (this name will be useful later in class 8 and above). Practical exercises include drawing triangles to match a given angle description and checking by measurement and classification. Remember that classification by sides and by angles are independent — for example a triangle can be isosceles and acute at the same time.

📌 Examples
  • Measure angles of triangle ABC: if ∠A = 90°, name it a right triangle and mark a square at A.
  • If ∠P = 110° in triangle PQR, label the triangle as obtuse and note that other angles are acute.
🧮 Formulas
  1. Right triangle: one angle = 90°
  2. Acute triangle: all angles < 90°
  3. Obtuse triangle: one angle > 90°
📊 Visual ideas
A right triangle with a small square at the right angle; legs and hypotenuse labelled
An obtuse triangle showing one angle greater than 90°
An acute triangle with all angles less than 90°
📐5

How to draw and construct triangles

Drawing triangles accurately uses three tools: a ruler for straight lines and measuring lengths, a compass or divider to copy or transfer distances, and a protractor for measuring angles. There are standard ways to construct a triangle when some data is given: for example three sides (SSS), two sides and included angle (SAS), two angles and included side (ASA), or right triangle with two sides or one side and one angle. Learning these constructions helps students make correct diagrams for solving problems.

For SSS construction: draw one side AB with the ruler, then with compass set to the length of the second side draw an arc with centre A, and with the compass set to the length of the third side draw an arc with centre B. The intersection of arcs gives point C; join AC and BC to complete the triangle. For SAS: draw side AB, then at A construct the given angle using the protractor. From the ray that makes that angle, mark off the second side length with compass; this locates the third vertex. For ASA: use the given side, construct both angles at the ends and draw their rays; the intersection gives the third vertex.

When constructing triangles, be careful to draw light construction lines and then darken the final triangle. Label vertices clearly. Practise several constructions so you become comfortable with matching a given combination of data to the correct method. Accurate drawing is essential before measuring or reasoning about angles and lengths in exercises.

📌 Examples
  • Construct △ABC given AB = 6 cm, BC = 5 cm and CA = 4 cm using SSS method.
  • Construct △PQR given PQ = 5 cm and ∠P = 50°, and PR = 4 cm using SAS method.
🧮 Formulas
  1. Construction methods: SSS, SAS, ASA (practical rules for drawing triangles)
📊 Visual ideas
Stepwise SSS construction: draw AB, arcs from A and B meeting at C, join AC and BC
SAS construction: draw AB, construct angle at A, mark length on ray to locate C
📐6

Angle-sum property of a triangle

One of the most important facts about triangles is the angle-sum property: the sum of the three interior angles of any triangle is 180 degrees. That is ∠A + ∠B + ∠C = 180°. This property helps find a missing angle when the other two are known and is used in many geometry problems.

You can see the idea by a simple paper activity: cut out any triangle, tear off a small corner at each vertex so each corner becomes a small triangle piece, and place the three corner pieces together at a point. The three corners fit to form a straight line, which measures 180°. This gives an intuitive demonstration why the sum of interior angles is 180°.

In practical exercises you will often be given two angles and asked to find the third. For example, if ∠A = 50° and ∠B = 60°, then ∠C = 180° − (50° + 60°) = 70°. Always write the step showing subtraction from 180° so the reasoning is clear. Also remember that angle measures are numbers that add; they are not lengths. Use a protractor only to check or to measure angles when needed.

The angle-sum property is a foundation for other rules such as the exterior angle relationship and for classifying triangles by angle sizes. Practice several problems to become fluent in using this property and in setting up the correct arithmetic to find missing angles.

📌 Examples
  • If ∠A = 40° and ∠B = 70°, find ∠C. Solution: ∠C = 180° − (40° + 70°) = 70°.
  • Given an isosceles triangle with two equal angles 55° each, find the third angle: 180° − (55° + 55°) = 70°.
🧮 Formulas
  1. Angle-sum property: ∠A + ∠B + ∠C = 180°
📊 Visual ideas
A triangle with angles ∠A, ∠B, ∠C and a straight line showing the three corner pieces forming 180°
📐7

Exterior angle property

An exterior angle of a triangle is formed when one side is extended. For example, if side BC of triangle ABC is extended beyond C to point D, then ∠ACD is an exterior angle. The key property is: an exterior angle equals the sum of the two opposite interior angles. So ∠ACD = ∠A + ∠B. This gives an easy method to find an interior angle when one exterior and one interior are known.

To understand this, note that the interior angle at C and the exterior angle at C are supplementary (they add to 180°) because they form a straight line. Using the angle-sum property, ∠A + ∠B + ∠C = 180°. Replace 180° by ∠ACD + ∠C (since they are supplementary) and cancel ∠C to get ∠ACD = ∠A + ∠B. This short logical step is important to remember and apply in problems.

In practice, students use the exterior angle property to solve problems where an outside angle is easier to measure or where it is given. For instance, if an exterior angle is 120° and one opposite interior angle is 50°, the other opposite interior angle is 70° (since 120° = 50° + unknown). Mark diagrams clearly to identify which angles are opposite the exterior angle.

Try several problems where an exterior angle is given and use subtraction to find the missing interior angles. This strengthens understanding of linear pairs and supplementary angles as well.

📌 Examples
  • In triangle ABC, exterior angle at C is 130°, and ∠A = 50°. Find ∠B. Solution: ∠B = 130° − 50° = 80°.
  • If ∠ACB (interior) = 30° and exterior ∠BCD = 150°, verify exterior property: ∠BCD = ∠A + ∠B = 150°.
🧮 Formulas
  1. Exterior angle property: exterior angle = sum of two opposite interior angles
📊 Visual ideas
Triangle ABC with side BC extended to D showing exterior angle ∠ACD and opposite interior angles ∠A and ∠B
📐8

Properties of isosceles triangles

An isosceles triangle has two equal sides. Suppose in △ABC, AB = AC. Then AB and AC are called the equal sides and BC is called the base. The angles opposite the equal sides—∠B and ∠C—are equal. This is the fundamental property used most often in solving problems. Mark the equal sides with identical short dashes and the equal angles with matching arc marks to make the symmetry visible in diagrams.

More properties follow from this symmetry. The line drawn from the apex A (the vertex between the equal sides) to the midpoint of the base BC has three special roles: it is an altitude (perpendicular to BC), a median (dividing BC into two equal parts) and an angle bisector (splitting ∠A into two equal angles). These three roles coincide in an isosceles triangle because the triangle is symmetric about the altitude from A. When you draw the altitude AD to BC at its midpoint D, you can show BD = DC, ∠BAD = ∠DAC and AD ⟂ BC. These results can be checked by construction and measurement in class.

Using these facts we can solve many kinds of questions. For example, if the base angles are given, the vertex angle is 180° minus their sum. If the vertex angle is given, each base angle is (180° − vertex)/2. Also if you know the length of the base and one equal side, you can often find the altitude using Pythagorean ideas later; here, at class 6 level, you can estimate lengths by construction and measurement. Exercises often ask to locate the midpoint, draw the altitude, and show equal parts on the base. Practise constructing isosceles triangles from a given base and apex angle and verify that the altitude from the apex bisects the base and the apex angle.

Remember these consequences: equal sides ⇒ equal opposite angles; the altitude from the apex in an isosceles triangle is also a median and an angle bisector. These make isosceles triangles easier to study and are used many times in solving geometric problems.

📌 Examples
  • In △ABC, AB = AC and ∠B = 40°. Then ∠C = 40° and ∠A = 180° − (40° + 40°) = 100°.
  • If AB = AC and the altitude from A meets BC at D, then BD = DC and ∠BAD = ∠DAC.
🧮 Formulas
  1. Isosceles triangle property: AB = AC ⇒ ∠B = ∠C
  2. Altitude from apex in an isosceles triangle is also median and angle bisector
📊 Visual ideas
An isosceles triangle showing equal sides AB and AC with dashes, equal base angles marked, and an altitude from A to midpoint of BC
📐9

Properties of equilateral triangles

An equilateral triangle has all three sides equal. Let △ABC be equilateral with AB = BC = CA. This equality of sides leads to strong and useful properties. First, all interior angles are equal. Since the sum of angles in a triangle is 180°, each angle measures 60°. Thus ∠A = ∠B = ∠C = 60°. Second, because of the complete symmetry, medians, altitudes and angle bisectors from any vertex coincide: the line from a vertex to the opposite side both divides that side into equal parts and splits the vertex angle into two equal halves while being perpendicular to the opposite side.

Construction of an equilateral triangle is simple and important to practise. Given a segment AB as one side, place the compass at A and draw a circle of radius AB; then place the compass at B and draw a circle of the same radius. The two circles intersect at two points; choose one intersection as C and join AC and BC to form △ABC. All sides will be equal by construction. You can verify the equal sides with the compass and measure the angles with a protractor to see that each is 60°.

Because medians and altitudes coincide, each median also has a special length and the three medians meet at one point (the centroid), which in an equilateral triangle is the same as the orthocentre and incenter. These deeper facts are useful later, but at this level focus on symmetry: any line from vertex to midpoint of opposite side is perpendicular and bisects the angle. Equilateral triangles also tessellate neatly in patterns and are used in many design problems. Practise drawing, labelling, and measuring to gain confidence: construct from one side, check its three equal sides with compass, and measure angles to confirm 60° each. Understanding equilateral triangles gives an easy example of perfect symmetry in geometry.

📌 Examples
  • Given side AB = 5 cm, construct equilateral △ABC by drawing circles centre A and B with radius 5 cm; their intersection is C.
  • In an equilateral triangle each angle = 60°. If one angle is marked, the other two are automatically 60° each.
🧮 Formulas
  1. Equilateral triangle: AB = BC = CA and ∠A = ∠B = ∠C = 60°
📊 Visual ideas
Equilateral triangle △ABC with all sides equal and medians/altitudes shown meeting at a common point
📐10

Right triangles: basics and recognition

A right triangle has one angle equal to 90°. The two sides that meet at the right angle are called legs; the side opposite the right angle is the hypotenuse and is the longest side of the triangle. Right triangles appear often in practical situations where perpendiculars are involved, such as ramps, ladders and building corners.

To recognise a right triangle, measure angles with a protractor or use a set-square for drawing or checking right angles. In diagrams a right angle is usually marked by a small square at the vertex. If you know two sides in a right triangle, you can compare their lengths: the hypotenuse is longer than either leg. Although detailed relations like Pythagoras are taught in later classes, at this level you should be comfortable identifying right triangles and using the fact that the sum of the other two angles equals 90° (because all three sum to 180° and one is 90°).

Problems may ask to find an acute angle when the other is given, for example if one acute angle is 35°, the other acute angle is 55° since 35° + 55° = 90°. Practice recognitions and simple computations involving right triangles. Also try drawing a right triangle by constructing a right angle at a point and marking lengths along the legs and finishing with the hypotenuse.

📌 Examples
  • Draw a right triangle with one leg 4 cm, other leg 3 cm. Mark the right angle with a square.
  • In △ABC, if ∠B = 90° and ∠A = 30°, find ∠C = 60° because ∠A + ∠C = 90°.
📊 Visual ideas
A right triangle showing the right angle marked with a small square, legs labelled and hypotenuse named
📐11

Perimeter of a triangle

The perimeter of a triangle is the total length around it, found by adding the lengths of its three sides. If a triangle has sides a, b and c then its perimeter P = a + b + c. This simple rule lets us find the boundary length when side lengths are known and is useful in practical tasks such as measuring fencing required around a triangular garden or the length of a ribbon to go round a triangular gift.

When side lengths are given in centimetres or metres, add them keeping units same and write the result with the same unit. If some side is unknown, use given information (for example two equal sides in an isosceles triangle) to express the unknown and calculate the perimeter. For instance, in an isosceles triangle with equal sides 7 cm and base 10 cm, perimeter = 7 + 7 + 10 = 24 cm.

Sometimes perimeter problems combine integer and fractional lengths; add carefully and convert mixed units when needed (for example cm and mm) before adding. Also check diagrams: lengths marked on figures are often the ones to use. Practice word problems: if three children each walk along one side of a triangular park, how far does each walk and what is the total? These exercises build facility with addition and unit consistency in geometry questions.

📌 Examples
  • Find perimeter of triangle with sides 5 cm, 7 cm and 8 cm. P = 5 + 7 + 8 = 20 cm.
  • An isosceles triangle has equal sides 6 cm and base 9 cm. Perimeter = 6 + 6 + 9 = 21 cm.
🧮 Formulas
  1. Perimeter P = a + b + c (where a, b, c are side lengths)
📊 Visual ideas
A triangle with sides labelled a, b, c showing that perimeter is the sum of side lengths
📐12

Triangle inequality (simple form)

The triangle inequality gives a basic but important rule about the lengths of sides of any triangle: the sum of lengths of any two sides must be greater than the third side. That is, for sides a, b and c we must have a + b > c, b + c > a and c + a > b. If these inequalities do not hold, you cannot make a triangle with those three lengths.

This rule is easy to check with small examples: try to place two segments end to end and see whether their combined length reaches beyond the third segment. If the sum equals exactly the third side, the three segments would lie in a straight line and not form a triangle. For practical construction, always check the triangle inequality before attempting to draw from three given lengths.

Use the triangle inequality to test possibilities in problems. For example, sides 2 cm, 3 cm and 6 cm cannot form a triangle because 2 + 3 = 5 which is not greater than 6. On the other hand 4 cm, 5 cm and 8 cm do form a triangle since each pair sum exceeds the remaining side. This simple check prevents waste of time trying to construct impossible triangles and helps in problem solving when side lengths are unknown or variable.

📌 Examples
  • Check: can 2 cm, 4 cm and 6 cm form a triangle? 2 + 4 = 6, not greater, so NO.
  • Check: 3 cm, 4 cm and 5 cm → 3 + 4 > 5, 4 + 5 > 3, 5 + 3 > 4 so YES; a triangle is possible.
🧮 Formulas
  1. Triangle inequality: a + b > c, b + c > a, c + a > b
📊 Visual ideas
A diagram showing two segments placed end-to-end compared with the third segment to visualise the inequality
📐13

Using equal angles to find sides and vice versa (intro)

There is a direct relation between equal sides and equal angles in a triangle. If two sides of a triangle are equal, then the angles opposite those sides are equal; conversely, if two angles are equal, the sides opposite those angles are equal. This two-way relation helps to determine unknown sides or angles when some equalities are given.

For example, in △ABC if AB = AC, then ∠B = ∠C. This is easy to see from symmetry: the triangle is reflected across the altitude from A. The converse is also true: if ∠B = ∠C, then AB = AC. At class 6 level you should learn both directions as tools for solving problems. When given equal angles, mark them with identical arc marks and then conclude the opposite sides are equal; when given equal sides, mark the sides with dashes to conclude the opposite angles are equal.

Combine these relations with the angle-sum property to compute unknown angles. For instance, if ∠B = ∠C and ∠A is known, then ∠B + ∠C = 180° − ∠A, and since ∠B = ∠C each equals (180° − ∠A)/2. Similarly, if AB = AC and a base angle is known, the other base angle is the same and the third angle is found by subtraction from 180°. These are practical methods for solving many class-level problems.

Use measurements and constructions to check these relations: draw a triangle with two equal angles and measure sides with a ruler to confirm equality; or draw two sides equal with the compass and measure the opposite angles with a protractor. Practise both directions until you can move quickly from equal sides to equal angles and back. This skill links measurement and reasoning and is widely used in geometry problems and constructions.

📌 Examples
  • Given in △ABC, ∠B = ∠C = 45°. Then ∠A = 180° − (45° + 45°) = 90° and AB = AC.
  • If in △PQR, PQ = PR, then angles opposite them are ∠R = ∠Q.
🧮 Formulas
  1. If two sides are equal ⇒ angles opposite them are equal
  2. If two angles are equal ⇒ sides opposite them are equal
📊 Visual ideas
An isosceles triangle showing equal sides and equal base angles, and an example with two equal angles showing corresponding equal sides

Key Concepts

Triangle
A polygon with three sides and three vertices formed by joining three non-collinear points.
Vertex
A point where two sides of a triangle meet.
Side
A straight segment joining two vertices of a triangle.
Interior angle
The angle formed inside the triangle at a vertex by two sides.
Exterior angle
An angle formed by extending one side of the triangle beyond a vertex.
Equilateral triangle
A triangle with all three sides equal and all angles 60°.
Isosceles triangle
A triangle with two equal sides and two equal base angles.
Scalene triangle
A triangle with all three sides of different lengths and all angles different.
Right triangle
A triangle with one interior angle equal to 90°.
Hypotenuse
The side opposite the right angle in a right triangle; it is the longest side.
Angle-sum property
The sum of the three interior angles of a triangle is 180°.
Exterior angle theorem
An exterior angle of a triangle equals the sum of the two opposite interior angles.
Perimeter
The total length around the triangle equal to the sum of its three sides.
Triangle inequality
For any triangle, the sum of any two side lengths is greater than the third side.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Name the three parts of a triangle and label them on a sketch. / त्रिभुज के तीन भागों का नाम बताइए और एक रेखाचित्र पर उन्हें चिन्हित कीजिए।
    Show answer

    Answer: The three parts are vertices (points A, B, C), sides (segments AB, BC, CA) and interior angles (∠A, ∠B, ∠C). On a sketch label three points A, B, C; draw lines AB, BC, CA; and mark the angles at A, B, C. / उत्तर: तीनों भाग हैं शिखर बिंदु (A, B, C), भुजाएँ (AB, BC, CA) और अंतर्निहित कोण (∠A, ∠B, ∠C)। रेखाचित्र पर तीन बिंदु A, B, C अंकित करें; AB, BC, CA खींचें; और A, B, C पर कोण अंकित करें।

  2. Given ∠A = 50° and ∠B = 60° in triangle ABC, find ∠C. / त्रिभुज ABC में यदि ∠A = 50° और ∠B = 60° हैं, तो ∠C क्या होगा?
    Show answer

    Answer: Using angle-sum property, ∠C = 180° − (50° + 60°) = 70°. So ∠C = 70°. / उत्तर: कोण-योग गुण के अनुसार ∠C = 180° − (50° + 60°) = 70°. अतः ∠C = 70°।

  3. Classify a triangle with sides 5 cm, 5 cm and 8 cm. / 5 cm, 5 cm और 8 cm वाली त्रिभुज को किस प्रकार से वर्गीकृत करेंगे?
    Show answer

    Answer: Two sides are equal (5 cm and 5 cm), so the triangle is isosceles. The angles opposite the equal sides are equal. / उत्तर: दो भुजाएँ समान हैं (5 cm और 5 cm), अतः त्रिभुज समद्विबाहु (isosceles) है। समान भुजाओं के विपरीत कोण समान होंगे।

  4. If an exterior angle of a triangle is 110° and one opposite interior angle is 40°, find the other opposite interior angle. / यदि किसी त्रिभुज का बाह्य कोण 110° है और उसके विपरीत का एक अंतर्निहित कोण 40° है, तो दूसरा विपरीत कोण क्या होगा?
    Show answer

    Answer: Exterior angle equals sum of the two opposite interior angles. So other opposite interior angle = 110° − 40° = 70°. / उत्तर: बाह्य कोण = दोनों विपरीत अंतर्निहित कोणों का योग, अतः दूसरा कोण = 110° − 40° = 70°।

  5. Can lengths 2 cm, 3 cm and 6 cm form a triangle? Explain. / क्या 2 cm, 3 cm और 6 cm की लंबाइयाँ त्रिभुज बना सकती हैं? स्पष्ट कीजिए।
    Show answer

    Answer: No. Triangle inequality requires sum of any two sides to be greater than the third. Here 2 + 3 = 5 which is not greater than 6, so these lengths cannot form a triangle. / उत्तर: नहीं। त्रिभुज-असमानता के अनुसार किसी भी दो भुजाओं का योग तीसरी भुजा से बड़ा होना चाहिए। यहां 2 + 3 = 5 जो 6 से छोटा है, इसलिए त्रिभुज नहीं बन सकता।

  6. Draw an equilateral triangle with side 6 cm. What is each angle? / 6 cm भुजा वाला समभुज त्रिभुज बनाइए। प्रत्येक कोण कितना होगा?
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    Answer: Construction: Draw a straight line segment AB = 6 cm using a ruler. Place the compass point at A, open it to 6 cm, draw an arc. Without changing compass width, place the compass at B and draw another arc intersecting the first arc; call the intersection C. Join AC and BC to form △ABC; AB = BC = CA by construction. Each angle equals 60°. So every interior angle ∠A, ∠B and ∠C is 60°. / उत्तर: निर्माण: रूलर से AB = 6 cm की रेखा खींचिए। कम्पास की नोक A पर रखकर 6 cm की दूरी पर चाप बनाइए। कम्पास की चौड़ाई नहीं बदलकर B पर चाप बनाइए; दोनों चापों का छेदन बिंदु C लीजिए। AC तथा BC को जोड़िए, इससे △ABC बनेगा और AB = BC = CA निर्माण द्वारा समान होंगी। प्रत्येक कोण 60° होगा। अतः ∠A, ∠B और ∠C प्रत्येक 60° हैं।

  7. Find the perimeter of a triangle with sides 9 cm, 12 cm and 15 cm. / 9 cm, 12 cm और 15 cm भुजाओं वाले त्रिभुज का परिमाप ज्ञात कीजिए।
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    Answer: Perimeter = sum of sides = 9 + 12 + 15 = 36 cm. Therefore the perimeter is 36 cm. / उत्तर: परिमाप = भुजाओं का योग = 9 + 12 + 15 = 36 cm। अतः परिमाप 36 cm है।

  8. In isosceles triangle ABC, AB = AC and ∠B = 55°. Find ∠A. / समद्विबाहु त्रिभुज ABC में AB = AC और ∠B = 55° है। ∠A ज्ञात कीजिए।
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    Answer: In an isosceles triangle with AB = AC, base angles ∠B and ∠C are equal. So ∠C = 55°. Then ∠A = 180° − (55° + 55°) = 70°. Thus ∠A = 70°. / उत्तर: AB = AC होने पर ∠B = ∠C = 55°. अतः ∠A = 180° − (55° + 55°) = 70°. इसलिए ∠A = 70°।

  9. A triangle has angles 30° and 60°. What is the third angle and what type by angles is the triangle? / एक त्रिभुज के कोण 30° और 60° हैं। तीसरा कोण क्या होगा और कोणों के अनुसार त्रिभुज किस प्रकार का होगा?
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    Answer: Third angle = 180° − (30° + 60°) = 90°. Since one angle is 90°, the triangle is a right triangle. / उत्तर: तीसरा कोण = 180° − (30° + 60°) = 90°. एक कोण 90° होने पर त्रिभुज समकोण (right triangle) है।

  10. If in △PQR, ∠P = ∠Q, and ∠R = 40°, find ∠P and ∠Q. / यदि △PQR में ∠P = ∠Q और ∠R = 40° है, तो ∠P और ∠Q ज्ञात कीजिए।
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    Answer: Let ∠P = ∠Q = x. Using angle-sum property, x + x + 40° = 180° ⇒ 2x = 140° ⇒ x = 70°. So ∠P = ∠Q = 70°. / उत्तर: मान लीजिए ∠P = ∠Q = x। कोण-योग के अनुसार 2x + 40° = 180° ⇒ 2x = 140° ⇒ x = 70°. अतः ∠P = ∠Q = 70°।

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