- 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 = 10^{2} + 24^{2}

=> 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 = a^{2} + 122

=> 225 = a^{2} + 144

=> a^{2} = 225 – 144

=> a^{2} = 81

=> a = √81

=> a = 9 cm

Thus, the distance of the foot of the ladder from the wall is 9 m.

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## Batch List

###### Medha Sharma

Course : CBSE Class 7

Start Date : 06.12.2022

End Date : 18.07.2023

Types of Batch : Classroom