- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
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
- Books Name
- class 7 Mathematics Book
- Publication
- ReginaTagebücher
- Course
- CBSE Class 7
- Subject
- Mathmatics
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)2
=> 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.