Triangles Class 10 Notes Chapter 6

CBSE Class 10 Maths Triangles Notes:-Download PDF Here

Class 10 Maths Chapter 6 Triangle Notes

The triangles Class 10 notes Chapter 6 provided here, is one of the most crucial study resources for the students studying in class 10. These CBSE chapter 6 notes are concise and cover all the concepts from this chapter, from which questions might be included in the board exam. You will also come across theorems based on similar concepts. In your previous year’s classes, you must have learned about the basics of triangles, such as the area of a triangle and its perimeters, etc.

Triangles for class 10 -1

The main concepts from this chapter that are covered here are-

  • What is a triangle?
  • Similarity criteria of two polygons having the same number of sides
  • Similarity criteria of triangles
  • Proof of Pythagoras Theorem
  • Example Questions
  • Problems based on Triangles
  • Articles Related to Triangles

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

  • Triangles Short Notes
  • Triangles MCQ Practice Questions
  • Triangles MCQ Practice Solutions
  • What Is a Triangle?

    A triangle can be defined as a polygon which has three angles and three sides. The interior angles of a triangle sum up to 180 degrees, and the exterior angles sum up to 360 degrees. Depending upon the angle and its length, a triangle can be categorized into the following types-

    Different types of triangles

    1. Scalene Triangle – All three sides of the triangle are of different measure
    2. Isosceles Triangle – Any two sides of the triangle are of equal length
    3. Equilateral Triangle – All three sides of a triangle are equal and each angle measures 60 degrees
    4. Acute angled Triangle – All the angles are smaller than 90 degrees
    5. Right angle Triangle – Anyone of the three angles is equal to 90 degrees
    6. Obtuse-angled Triangle – One of the angles is greater than 90 degrees

    For more information on Triangles, watch the below video

    To know more about Triangles, visit here.

    Similarity Criteria of Two Polygons Having the Same Number of Sides

    Any two polygons which have the same number of sides are similar if the following two criteria are met-

    1. Their corresponding angles are equal, and
    2. Their corresponding sides are in the same ratio (or proportion)

    Similarity Criteria of Triangles

    Mathematical Expression for Similarity
    In Δ ABC and Δ DEF, if
    (i) ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and
    (ii)AB/DE = BC/ EF = CA/ FD, then the two triangles are similar.
    To find whether the given two triangles are similar or not, it has four criteria. They are:

    • Side-Side-Side (SSS) Similarity Criterion – When the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal, and the triangle will be considered similar triangles.
    • SSS

    • Angle Angle Angle (AAA) Similarity Criterion – When the corresponding angles of any two triangles are equal, then their corresponding side will be in the same ratio, and the triangles are considered to be similar.
    • Angle-Angle (AA) Similarity Criterion – When two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are considered similar.
    • Side-Angle-Side (SAS) Similarity Criterion – When one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are said to be similar.
    • SAS

    To know more about Similar Triangles, visit here.

    Basic Proportionality Theorem

    In Mathematics, the basic proportionality theorem states that “If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.”

    Basic proportionality theorem

    Now, let us understand the Basic proportionality theorem with the help of a diagram.
    Consider the triangle ABC, as depicted in the diagram. We draw a line PQ parallel to the side BC of ABC and intersect the sides AB and AC in P and Q, respectively.
    Thus, according to the Basic proportionality theorem,
    AP/PB = AQ/QC

    To know more about the Basic Proportionality Theorem, click here.

    Areas of Similar Triangles

    According to the area of similar triangles theorem, “If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.”

    Now, consider two triangles ΔABC and ΔPQR;

    Area of similar triangles

    Thus, according to the area of similar triangles, we can write:

    Area of triangle ABC/Area of Triangle PQR = (AB/PQ)2 = (BC/QR) = (CA/RP)2

    To know more about area of similar triangles, click here.

    Proof of Pythagoras Theorem

    Statement: As per Pythagoras theorem“In a right-angled triangle, the sum of squares of two sides of a right triangle is equal to the square of the hypotenuse of the triangle.”

    Pythagoras Theorem Formula

    According to the Pythagoras theorem, the formula is stated as follows:

    Hypotneuse2 = Base2 + Perpendicular2

    Now, consider the right triangle given below.

    Right triangle

    Here, the side AC is the hypotenuse, which is the opposite of the right angle. It is the longest side of a right triangle.

    The side AB is the base, which is the side adjacent to angle A.

    The BC is perpendicular, which is the side opposite to angle A.

    Proof –

    Consider the right triangle, right-angled at B.


    Draw BD ⊥ AC

    Triangles for class 10 -2
    Now,  △ADB ~ △ABC

    So, AD/AB = AB/AC

    or AD. AC = AB2 ……………(i)

    Also, △BDC ~ △ ABC

    So, CD/BC = BC/AC

    or, CD. AC = BC2 ……………(ii)

    Adding (i) and (ii),

    AD. AC + CD. AC = AB2 + BC2

    AC(AD + DC) = AB2 + BC2

    AC(AC) = AB2 + BC2

    ⇒ AC2 = AB2 + BC2

    Hence, proved.

    For more information on Pythagoras Theorem, watch the below video

    To know more about Pythagoras Theorem, visit here.

    Solved Example

    Example 1:

    In a right-angled triangle ABC, which is right-angled at A, where CM and BL are the medians of a triangle. Show that, 4(BL2 +CM2) = 5 BC2


    Given that,

    Triangles for class 10 -3

    Medians BL and CM,  ∠A = 90°

    From the triangle ABC, we can write it as:

    BC2 = AB2 + AC2 (Using Pythagoras Theorem) …(1)

    From the triangle, ABL,

    BL2 = AL2 +AB2

    or we can write the above equation as:

    BL2 =(AC/2)2 +AB2 (Where L is the midpoint of AC)

    BL2 = (AC2/4) + AB2

    4BL2 = AC2 + 4 AB2 ….(2)

    From triangle CMA,

    CM2 = AC2 + AM2

    CM2 = AC2 + (AB/2)2 (Where M is the midpoint of AB)

    CM2 = AC2 + AB2/4

    4CM = 4 AC +AB ….(3)

    Now, by adding (2) and (3), we  get,

    4(BL2+ CM2) = 5(AC2+ AB2)

    Using equation (1), we can write it as:

    4(BL2+ CM2) = 5 BC2

    Hence, proved.

    Example 2:

    In the given figure, DE is parallel to AC, and DF is parallel to AE. Prove that BF/FE = BE/EC.

    Triangles for class 10 - Example 2


    Given that DE || AC and DF||AE.

    To prove: BF/FE = BE/EC

    In a triangle ABC, DE || AC.

    We know that the line drawn parallel to one side of a triangle intersects the other two sides in distinct points, then it divides the other two sides in the same ratio.

    Therefore, BE/EC = BD/DA …(1)

    Now, consider the triangle AEB,

    DF || AE.

    Thus, we can say that

    BF/FE = BD/DA …(2)

    By comparing the equation (1) and (2), we can say

    BE/EC = BF/FE

    Hence, proved.

    Example 3:

    The sides AB and BC and median AD of a triangle ABC are respectively proportional to the sides PQ and QR and the median PM of a triangle PQR (as shown in the figure). Prove that △ABC ~ △PQR.

    Triangles for class 10 - Example 3


    Given that,

    In △ABC, AD is median

    In △PQR, PM is median

    Also, it is provided that AB/PQ = BC/QR = AD/PM

    Since AD is the median of △ABC,

    BD = DC = (½)BC

    Similarly, PM is the median of △PQR,

    QM = MR = (½)QR 

    Since, AB/PQ = BC/QR = AD/PM

    AB/PQ = 2BD/2QM = AD/PM

    AB/PQ = BD/QM = AD/PM ….(1)

    Since all three sides are proportional, by using the SSS similarity rule, we can write

    △ABD ~ △PQM

    Since the corresponding angles of similar triangles are equal

    ∠B = ∠Q …(2)

    Therefore, in △ABC and △PQR,

    ∠B = ∠Q [From (2)]

    AB/PQ = BC/QR [Given]

    Therefore, by using SAS similarity, we can conclude that

    △ABC ~ △PQR.

    Hence, proved.

    Problems Related to Triangles

    1. A girl having a height of 90 cm is walking away from a lamp-post’s base at a speed of 1.2 m/s. Calculate the length of that girl’s shadow after 4 seconds if the lamp is 3.6 m above the ground.
    2. S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS. Now, prove that triangle RPQ and triangle RTS are similar.
    3. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD
      at F. Show that triangles ABE and CFB are similar.

    Also Access:

    NCERT Solutions for Class 10 Maths Chapter 6 Triangles

    CBSE Class 10 Maths Chapter 5 Arithmetic Progression Notes

    CBSE Class 10 Maths Chapter 7 Coordinate Geometry Notes

    NCERT Exemplar Class 10 Maths Solutions for Chapter 6 Triangles

    RD Sharma Solutions for Class 10 Maths Triangles

    CBSE Sample Papers for Class 10 Maths

    Get more such class 10 maths and science notes at BYJU’S. Also, access class 10 question papers, sample papers, and other study materials to prepare for the board exam in a more effective way.

    Quiz on Triangles Class 10


    Leave a Comment

    Your Mobile number and Email id will not be published.



    1. Can u please provide me more questions of triangles chapters class 10th