Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7  >  Chapter Notes: The Triangles & Its Properties

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Triangles

A triangle is a basic shape you see in many places. It is a simple closed figure made up of three straight line segments. 

Triangle we see in everyday lifeTriangle we see in everyday life

So, in small-small things around us we see triangles. Triangle is a very common geometrical shape we use in day to day life.

A triangle has:

  • Three Vertices: The points where the sides meet (like points A, B, and C in triangle ∆ABC).
  • Three Sides: The line segments that form the triangle (AB, BC, and CA).
  • Three Angles: The space between the sides (∠BAC, ∠ABC, and ∠BCA).

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Triangles can be classified in two main ways:

  1. Based on Sides:

    • Scalene Triangle: All sides are different lengths.
    • Isosceles Triangle: Two sides are of equal length.
    • Equilateral Triangle: All three sides are equal in length.
  2. Based on Angles:

    • Acute-Angled Triangle: All angles are less than 90º
    • Obtuse-Angled Triangle: One angle is greater than 90º
    • Right-Angled Triangle: One angle is exactly 90º

So, in small-small things around us we see triangles. Triangle is a very common geometrical shape we use in day to day life.

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:
What is the classification of triangles based on their sides?
View Solution

Medians of a Triangle

Let's understand with an example; Suppose there are 6 members in Kartik’s family and he has only one triangular piece of pizza. 

 Now, Kartik wants to divide this piece of pizza equally among all the family members. So, how will Kartik divide this piece of pizza?

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

So, Kartik will divide the piece of pizza into six equal pieces by cutting along the lines as shown in the above figure.

What are these lines called?

These lines are nothing but the medians of a triangle. A line segment drawn from any vertex of the triangle to the mid-point of the opposite sides of a triangle is called a median.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

In triangle ABC, segment AD is the median of the triangle because one of its endpoints A is the vertex of the triangle and his other endpoint D is the midpoint of its opposite side BC.

Triangle has three vertices. So, we can draw three medians to each triangle from each vertex.

Perpendicular bisector: A perpendicular bisector is a segment, line, or ray that is perpendicular to a segment at its midpoint.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

In triangle ABC, DX is the perpendicular bisector of segment BC because it is perpendicular to segment BC at its midpoint X.

Altitudes of a Triangle

Altitude of a triangle is a perpendicular line segment that joins a vertex of the triangle to the opposite side.

  • Triangle ABC has segment AD as an altitude.
  • Endpoint A is the vertex of the triangle.
  • Endpoint D of segment AD is on the opposite side of the triangle.
  • Segment AD is perpendicular to the opposite side BC.
  • Each triangle has a total of three vertexes.
  • We can draw a maximum of three altitudes in each triangle.
  • The altitudes are shown in the adjacent figure.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

An altitude of a triangle can be inside or may lie outside the triangle.
Note:  These all perpendicular are nothing but the altitudes of the triangle.

Example 1: Draw the altitude and median in the given triangle from vertex A.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Sol: In the given triangle ABC to draw a median first we find the mid-point of segment BC then join it with vertex A.

Also, we know the altitude of a triangle is a perpendicular line segment that joins a vertex of the triangle to the opposite side.

Hence, to draw altitude first we extend the side BC then we draw the perpendicular from vertex A.


The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:
What is a median in a triangle?
View Solution

Exterior Angle of Triangle and Its Properties

In the earlier chapter, we see angles that are present inside the triangle. Now, we talk about angles that are present outside of the triangle, and they are called exterior angle.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

In the above ΔABC we see, ∠1, ∠2 and ∠3 is present inside the triangle, so we called it as interior angles of a triangle. 

So, if we extend the BC to D observe ∠4 is formed at point C. This angle lies in the exterior of ΔABC. We call it the exterior angle of the ΔABC.

Angle is formed between one side of a triangle and extended adjacent side, so we say that an exterior angle is an angle between one side of a triangle and an extended adjacent side.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

We know there are three sides of the triangle and when we extend these sides six exterior angles are formed.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

To prove: Sum of interior opposite angles are equal to an exterior angle.

i.e ∠1 + ∠2 = ∠4
In the figure we see, ∠3 and ∠4 are adjacent angles. ∠4 is exterior angle and ∠1 and ∠2 are Interior opposite angles of ∠4

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Draw a line EF parallel to AB which is passing through C. Line EF divide ∠4 into two angles ∠ECA marked as ∠5 and ∠ECD marked as ∠6

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∠5 + ∠6 = ∠4 ... (i)
∠1 = ∠5 AB||EF and AC is a transversal line
(Alternate interior angles are equal)... (ii)
∠2 = ∠6 AB||EF and BD is a transversal line
(Corresponding angles are equal)... (iii)
By adding (ii) and (iii) we get
∠1 + ∠2 = ∠5 + ∠6
From equation (i)
∠1 + ∠2 = ∠4

Example: In the given figure, find ∠ACD

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Sol: For ΔABC, ∠ACD, is an exterior angle. ∠ABC, ∠BAC are interior opposite angles.
We know, an exterior angle of a triangle is equal to the sum of its interior opposite angles.
∴ ∠ACD = ∠ABC + ∠BAC
∴ ∠ACD = 50° + 35°
∴ ∠ACD = 85°

Example: Exterior angle ∠ACD = 100° and ∠ABC and ∠ACB are in the ratio of 3: 2. Find ∠ABC and ∠BAC of the triangle.
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Sol: Let, ∠ABC = 3x and ∠ACB = 2x
We know, an exterior angle of a triangle is equal to the sum of its interior opposite angles.
∴ ∠ACD = ∠ABC + ∠ACB
∴ 100° = 3x + 2x
∴ 100° = 5x

∴ x = 100°/5
∴ x = 20°

→ ∠ABC = 3x = 3 × 20° = 60°
→ ∠ACB = 2x = 2 × 20° = 40°

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:In a triangle, if an exterior angle is 120 and one of the interior opposite angles is 45, what is the measure of the other interior opposite angle?
View Solution

Angle Sum Property of Triangle


The sum of the interior angles of a triangle is 180°. This property of a triangle is true for all triangles.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

In ΔABC, ∠1, ∠2, and ∠3 are the interior angles of a triangle.


To prove:   Sum of the interior angles of a triangle is 180°. 

 ∠1 + ∠2 + ∠3 = 180°
Construction: Extend side BC to D we get exterior ∠4

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∠1, ∠2 and ∠3are angles of ΔABC. ∠4is the exterior angle when BC is extended to D.
∠1 + ∠2 = ∠4 ... (By exterior angle property)
Adding ∠3 to both the side
∠1 + ∠2 + ∠3 = ∠4 + ∠3
But, ∠4 and ∠3 form a linear pair
So,
∠4 + ∠3 = 180°... (Linear pair)
Therefore,
∠1 + ∠2 + ∠3 = 180°
Hence, proved

Example: In ∆ABC, find the measure of ∠A if ∠C = 25° and ∠B = 100° 

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Sol:

We know that the sum of angles of a triangle is 180°
∴ ∠A + ∠B + ∠C = 180°
→  ∠A + 100° + 25° = 180°
→ 125° + ∠A = 180°
→ ∠A = 180° − 125°
→ ∠A = 55°
Hence, the measure of ∠A is 55°

Example: Find the measure of a triangle which are in the ratio of 2:1:1

Sol: Let the common ratio be x°
The measure of given angles of the triangle be 2x°, x°, and x° respectively.
We know that the sum of angles of a triangle is 180°
2x° + x° + x° = 180°
4x° = 180°
x° = 180°/4
x° = 45°
So the measure of the angles ( (2 × 45°), 45°, and 45°) i.e.,90°, 45°and  45°

Example: If the angles of a triangle have measures (3A + 40°),(A + 20°), 2A. Find the value of A.

Sol: Let∆ABC,
∠BAC = (3A + 40°), ∠ACB = (A + 20°), and∠ABC = 2A
We know that the sum of angles of a triangle is 180°
∴ ∠BAC + ∠ACB + ∠ABC = 180°
→ (3A + 40°) + (A + 20°) + 2A = 180°
→ 3A + 40° + A + 20° + 2A = 180°
→ 3A + A + 2A + 40° + 20° = 180°
→ 6A + 60° = 180°
→ 6A + 60° − 60° = 180° − 60°
→ 6A = 120°
→ A = 120/6°
→ A = 20°
Hence, the value of A is 20°

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:In triangle XYZ, if angle X = 50 and angle Y = 70, what is the measure of angle Z?
View Solution

Two Special Triangles: Equilateral And Isosceles

Equilateral Triangle

A triangle in which all three sides are of equal length and three angles are equal. Each angle is 60°.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∆ABC is an equilateral triangle in which we see all the three sides of the triangle are of equal length and the measure of all the angles of the triangle is also equal.

Isosceles Triangle:

A triangle in which two sides are of equal length and angles opposite to equal sides are equal. 

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∆ABC is an isosceles triangle in which we see two sides of the triangle are of equal length and the angles opposite to equal sides are equal.

1. Find angle x in the following figure:
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∆ABC is an isosceles triangle.

So, AB = AC
∠ABC = ∠ACB

x = ∠ACB
∠ACB + ∠ACD = 180°... (Linear pair)
x + 130° = 180°
x + 130° − 130° = 180° − 130°
x = 50°

2. Find angle x in the following figure:

Sol: ∆ABC is an isosceles triangle.
So, AB = BC
∠ABC = ∠CAB
∠CAB = x
Hence,
∠ABC = ∠CAB = x
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

∠ABC + ∠BCA + ∠CAB = 180°...(angle sum property)
x + 90° + x = 180°
2x + 90° = 180°
2x + 90° − 90° = 180° − 90°
2x = 90°
x = 90°/2
x = 45°

Sum of the Lengths of Two Sides of a Triangle

In a school, Math’s teacher tells students to draw a triangle. After some time, teacher asked the students, have you drawn the triangle? Students: yes. So, firstly teacher asked Ben, how have you drawn the triangle? And what is the length of sides of the triangle you took to draw the triangle?
Ben said I have drawn a triangle in which I took the length of the sides of the triangle is 5 cm, 7 cm, and 12 cm.
So, the teacher said to students to check if we can draw a triangle from this length of the sides of the triangle. 

To check this, “The sum of any two sides of the triangle is greater than the length of the third side.”

So, let’s find out. We have 5 cm, 7 cm, and 12 cm three sides of the triangle. So,
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Here, we see the sum of the two sides is equal to third sides.

It means the triangle with sides 5 cm, 7 cm, and 12 cm cannot be formed.


Now, the teacher asked to Nobita, how has she drawn the triangle?
She told that to draw a triangle she took the length of the sides of the triangle is 3 cm, 4 cm, and 5 cm. The teacher said let’s check these sides are correct or not?
According to Nobita, The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Here, we see the sum of the two sides is greater than the third side. Can Nobita draw a triangle from it? No, because this is only one case.

Now, we see the second combination,
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Now, we see the third combinations
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Here, all three cases can satisfy the condition. It means the triangle with sides 3 cm, 4 cm, and 5 cm can be formed.

Another way we can also find the given sides can form a triangle or not.
If the difference of any two sides of the triangle is less than the third side then we are able to draw a triangle.

Example: Let us consider the sides of the triangle be 5 cm, 10 cm, and 12 cm. So,
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Here, we see the difference of the two sides is less than the third side. It means according to the theorem, the triangle can be formed with sides 5 cm, 10 cm, and 12 cm.

Example: Is it possible to have a triangle with the following sides?
(i) 3 cm, 5 cm, and 7 cm

Sol: We know, the sum of two sides of a triangle is greater than the third side.

We check this property for the given sides of the triangle.
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Here, we see the sum of two sides of the triangle is greater than the third side. Hence, it follows triangle inequality. It means the triangle with sides 3 cm, 5 cm, and 7 cm can be formed.

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:
Which of the following statements is true about an equilateral triangle?
View Solution

Right-Angled Triangles and Pythagoras Theorem

Pythagoras theorem is related to the right-angled triangle. So, let us consider a right-angled triangle


The Triangle and Its Properties Class 7 Notes Maths Chapter 6

In a right-angled triangle, the side opposite to the right angle is called the hypotenuse, and the other two sides are known as legs of the right-angled triangle. So, perpendicular, base, and hypotenuse are the three sides of the right-angled triangle. Now, according to this theorem, if you have a right-angled triangle in which

a→ Leg of a triangle (Perpendicular side)

b→ Leg of a triangle (Base side)

c→ Hypotenuse

Then let’s try this if we draw a square on the perpendicular side that means each side of the square would be equal to a and the area of that square is equal to a2.
If we draw a square on the base side that means each side of the square would be equal to b and the area of that square is equal to b2.
In the same way, if we draw a square on hypotenuse side that means each side of the square would be equal to c and the area of that square is equal to c2.
The Triangle and Its Properties Class 7 Notes Maths Chapter 6So, according to Pythagoras theorem,

In a right-angled triangle, the square of the hypotenuse equals the sum of the square of its two sides.
∴ (Perpendicular)2 + (Base)2 = (Hypotenuse )2
∴ (a)2 + (b)2 = (c )2
This is called the Pythagoras theorem.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

There is another way we prove Pythagoras theorem

Suppose we have eight right angle triangles and two identical squares, and the side of both squares is (a + b).
The Triangle and Its Properties Class 7 Notes Maths Chapter 6

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Now, we place four triangles in the first square and remaining four triangles in the second square in a different way as shown below

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Now, we know the squares are identical and the eight triangles inserted in the squares are also identical.
Uncovered area of 1st square = Uncovered area of 2nd square
Area of inner square of first square = The total area of two uncovered squares in 2nd square
(c )2 = (a)2 + (b)2
Area of a square with side (a + b) = Area of inside square of side c + Area of four inside triangles
Area of a square with side (a + b) = (a + b)2
Area of inside square of side (c) = (c)2
Area of four inside triangles = 4 × Area of one triangle
Area of four inside triangles = 4 × 1/2 × ab = 2ab
 (a + b)2 = c2 + 2ab
a2 + b2 + 2ab = c2 + 2ab
a2 + b2 + 2ab − 2ab = c2 + 2ab − 2ab
a2 + b2 = c2
Hence, Pythagoras theorem proved in another way.

Example 1: Verify Pythagoras theorem in triangle ABC

Sol: In ∆ABC,
AC = 13 (Hypotenuse)
AB = 12 (Perpendicular)
BC = 5 (Base)
By Pythagoras theorem
(AB)2 + (BC)2 = (AC)2

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Take LHS: = (AB)2 + (BC)2
= (12)2 + (5)2 = 144 + 25 = 169
Take RHS: = (AC)2S

= (13)2 = 169
Here, LHS = RHS
Hence, the Pythagoras theorem is verified for this triangle.

Example 2 : A 15 m long ladder is placed against a wall in such a way that the foot of the ladder is 9 m away from the wall. So, up to what height does the ladder reach the wall?

Sol: 

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

Consider BC be the wall and AC be the ladder.
Then, AB = 9 m and AC = 15 m

Now, ∆ABC is a right-angled triangle.
Hence,
(AC)2 = (AB)2 + (BC)2
⇒ (BC)2 = (AC)2 − (AB)2
= (15)2 − (9)2
= 225 − 81 = 144
⇒ BC = √144 = 12 m
Hence, the ladder reaches the wall up to 13 m.

Example 3: Find the perimeter of the rectangle whose length is 5 cm and its diagonal is 13 cm.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6
Sol: In ∆ABC
(AC)2 = (AB)2 + (BC)2
(13)2 = (AB)2 + (5)2
169 = (AB)2 + 25
169 − 25 = (AB)2 + 25 − 25
144 = (AB)2
(12)2 = (AB)2
12 = AB
Now, we know the perimeter of the rectangle = 2 (l + b)
= 2 (5 + 12)
= 2 (17)
= 34 cm
Hence, the perimeter of the rectangle is 34 cm.

Example 4: A 7 m high pole is 9 m away from a building. If the distance between the top of the pole and top of the building is 25 m then find the height of the building.

Sol: Let AC be the height of the building. ED be the height of the pole. AE be the distance between their tops and CD be the distance between the foot of the pole and

the foot of the building. Now, from E draw EB ⊥ AC
AE = 25 m
BE = CD = 9 m
In the right-angled triangle ∆ABE,

(AE)2 = (AB)2 + (BE)2
(25)2 = (AB)2 + (9)2
625 = (AB)2 + 81
Subtracting 81 from both side
625 − 81 = (AB)2 + 81 − 81
576 = (AB)2
∴ AB = 24 m
Since AC = AB + BC and BC = ED = 9 m
∴ AC = 24 + 9 = 33 m
Hence, the height of the building is 33 m.

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

The converse of Pythagoras theorem
If any triangle obeys the Pythagoras theorem, then the triangle is right-angled.
Let’s explain this theorem by using the example

Example 5: Δ ABC is right-angled at C. If AC = 2.5 cm and BC = 6 cm find the length of AB

Sol: 

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

By Pythagoras theorem,
In ∆ABC, (AB)2 = (AC)2 + (BC)2
(AB)2 = (2. 5)2 + (6)2
(AB)2 = 6. 25 + 36
(AB)2 = 42. 25
∴ AB = √42. 25 = 6. 5 cm
Hence, the length of AB is 6. 5 cm

Question for Chapter Notes: The Triangles & Its Properties
Try yourself:Which of the following triangles does not satisfies the Pythagoras theorem?
View Solution

The document The Triangle and Its Properties Class 7 Notes Maths Chapter 6 is a part of the Class 7 Course Mathematics (Maths) Class 7.
All you need of Class 7 at this link: Class 7
76 videos|345 docs|39 tests

Top Courses for Class 7

FAQs on The Triangle and Its Properties Class 7 Notes Maths Chapter 6

1. What is the relationship between the medians of a triangle and its centroid?
Ans. The medians of a triangle intersect at a point called the centroid, which divides each median into two segments in a 2:1 ratio.
2. How can the altitudes of a triangle help determine the area of the triangle?
Ans. The altitudes of a triangle are perpendicular line segments from each vertex to the opposite side. The area of a triangle can be calculated by multiplying the base by the corresponding altitude and dividing by 2.
3. What is the exterior angle of a triangle and what are its properties?
Ans. The exterior angle of a triangle is formed when one side of the triangle is extended. The exterior angle is equal to the sum of the two interior angles not adjacent to it.
4. What is the angle sum property of a triangle and how is it useful in geometry?
Ans. The angle sum property of a triangle states that the sum of the interior angles of a triangle is always 180 degrees. This property is essential in various geometric calculations and proofs.
5. How does the Pythagoras theorem apply to right-angled triangles and what is its significance?
Ans. The Pythagoras theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is fundamental in geometry and trigonometry for solving problems involving right-angled triangles.
76 videos|345 docs|39 tests
Download as PDF
Explore Courses for Class 7 exam

Top Courses for Class 7

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

ppt

,

Summary

,

video lectures

,

Exam

,

Free

,

MCQs

,

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

,

pdf

,

mock tests for examination

,

Objective type Questions

,

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

,

The Triangle and Its Properties Class 7 Notes Maths Chapter 6

,

Previous Year Questions with Solutions

,

Extra Questions

,

Viva Questions

,

shortcuts and tricks

,

Important questions

,

practice quizzes

,

study material

,

Semester Notes

,

Sample Paper

,

past year papers

;