1. Median of a triangle

Chapter 6  

Triangle and its properties

Median of a triangle

Median of a triangle is a line segment joining a vertex to the midpoint of the opposing side, bisecting it.

A median connects a vertex of a triangle to the mid-point of the opposite side.
In the ∆ ABC, the line segment AD joining the mid-point of BC to its opposite vertex A is called a median of the triangle.

Properties of Median of a Triangle

Every triangle has exactly three medians one from each vertex and they all intersect each other at the triangle's centroid.

  • The 3 medians always meet at a single point, no matter what the shape of the triangle is.
  • The point where the 3 medians meet is called the centroid of the triangle. Point O is the centroid of the triangle ABC.
  • Each median of a triangle divides the triangle into two smaller triangles which have equal area.
  • In fact, the 3 medians divide the triangle into 6 smaller triangles of equal area. 

In ∆ ABC, three medians are AD, CE and BF and they are intersecting the point O which is centroid of the triangle.

2. Altitude of a triangle

Altitude of a triangle

An altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side opposite the vertex).

In ∆ ABC, AD is the altitude of triangle ABC.
Through each vertex, an altitude can be drawn. So, there are at most three altitudes in a triangle.
Properties of Altitudes of a Triangle

  • Every triangle has 3 altitudes, one from each vertex. AE, BF and CD are the 3 altitudes of the triangle ABC.
  • The altitude is the shortest distance from the vertex to its opposite side.
  • The 3 altitudes always meet at a single point, no matter what the shape of the triangle is.
  • The point where the 3 altitudes meet is called the ortho-centre of the triangle. Point O is the ortho-centre of the triangle ABC.
  • The altitude of a triangle may lie inside or outside the triangle. 

3. Exterior angles of a triangle

Exterior angles of a triangle

An exterior angle of a triangle is equal to the sum of the opposite interior angles.   

In the above figure, ACD is the exterior angle of the Δ ABC.
So, ACD = CAB + CBA
At each vertex of a triangle, an exterior angle of the triangle may be formed by extending one side of the triangle.

4. Apply angle sum property of a triangle

Apply angle sum property of a triangle

The sum of the measures of the three angles of a triangle is 180°

 

In Δ ABC,
A + B + C = 180°

5. Special triangles

Special triangles

Two Special Triangles: Equilateral and Isosceles

A triangle in which all the three sides are of equal lengths and each angle has measure 60o is called an equilateral triangle.

In an equilateral triangle,
AB = BC = CA

And ÐA = ÐB = ÐC = 600

A triangle in which two sides are of equal lengths is called an isosceles triangle

In triangle XYZ,
XY = XZ

Sum of the Lengths of Two Sides of a Triangle
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

 

 In the above figure,
AB + BC > AC
Also, the difference between the lengths of any two sides of a triangle is smaller than the length of the third side.

6. Sides of a triangle

Sides of a triangle

Right-Angled Triangles and Pythagoras Property

In a right angle triangle, the side opposite to the right angle is called the hypotenuse and the other two sides are known as the base and perpendicular of the right-angled triangle.   

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of base and perpendicular.
(Hypotenuse)2 = (Base)2 + (Perpendicular)2]
If a triangle holds Pythagoras property, then the triangle must be right-angled.
Problem 1: PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:

Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2                       [By Pythagoras theorem]

=> (QR)2 = (PQ)2 + (PR)   
=> x = 102 + 242
=> x = 100 + 576
=> x= 100 + 576 = 676
=> x = 676
=> x = √676
=> x = 26 cm
Thus, the length of QR is 26 cm.

Problem 2: A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:

Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2                      [By Pythagoras theorem]
=> (AC)2 = (CB)2 + (AB)2
=> 152 = a2 + 122
=> 225 = a2 + 144
=> a2 = 225 – 144
=> a2 = 81
=> a = √81
=> a = 9 cm
Thus, the distance of the foot of the ladder from the wall is 9 m.                                                                  

Related Chapter Name