Perimeter of a Triangle
    The sum of the lengths of the sides of a triangle is called its Perimeter.

In AC = b, BC = a, AB = c then perimeter is BC + CA + AB = a + b + c 

TRIANGULAR INEQUALITY PROPERTY

The sum of the lengths of any two sides of a triangle is always greater than the third side.
        AB + BC > AC
        AB + AC > BC
        AC + BC > AB
The difference between the length of any two sides of a triangle is smaller than the length of the third side.
        AB – BC < CA
        BC – CA < AB
        CA – AB < BC

Example

 

XY + YZ > XZ
        3 + 4 > 5
similary YZ + XZ > XY
        4 + 5 > 3
 similary XY + XZ > YZ
        3 + 5 > 4

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.