Class 7 Maths Algebraic Expressions | Using Algebraic Expressions – Formulas and Rules |

__Using Algebraic Expressions – Formulas and Rules__

Rules and formulas in mathematics are written in a concise and general form using algebraic

expressions.

**Perimeter Formula:**

- Let l be the length of the side of an equilateral triangle.

So, Perimeter of equilateral triangle = 3 * Length of side of equilateral triangle = 3l

- Let l be the length of the side of a square.

So, Perimeter of equilateral triangle = 4 * Length of side of square = 4l

- Let l be the length of the side of a regular pentagon.

So, Perimeter of equilateral triangle = 5 * Length of side of regular pentagon = 5l

**Area formulas**

- If we denote the length of a square by l, then the area of the square = l
^{2} - If we denote the length of a rectangle by l and its breadth by b, then the area of the rectangle

= l * b = lb

- If b stands for the base and h for the height of a triangle, then the area of the triangle

= (b * h)/2 = bh/2

**Rules for number patterns:**

- If a natural number is denoted by n, its successor is (n + 1).

Ex: If n = 20, its successor is n + 1 = 21, which is known.

- If a natural number is denoted by n, 2n is an even number and (2n + 1) an odd number.

Ex: Let n = 7; 2n = 2 * n = 2 * 7 = 14 is indeed an even number and 2n + 1 = 2 * 7 + 1 = 14 + 1 = 15 is indeed an odd number.

**Pattern in geometry**

- The number of diagonals we can draw from one vertex to another of a quadrilateral is 1.
- The number of diagonals we can draw from one vertex to another of a pentagon is 2.
- The number of diagonals we can draw from one vertex to another of a hexagon is 3.

Thus, the number of diagonals we can draw from one vertex of a polygon of n sides is (n – 3).

__Problem: __Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind .

Solution:

(i) 5n + 1

Putting n = 5, 5 * 5 + 1 = 25 + 1 = 26

Putting n = 10, 5 * 10 + 1 = 50 + 1 = 51

Putting n = 100, 5 * 100 + 1 = 500 + 1 = 501

(ii) 3n + 1

Putting n = 5, 3 * 5 + 1 = 15 + 1 = 16

Putting n = 10, 3 * 10 + 1 = 30 + 1 = 31

Putting n = 100, 3 * 100 + 1 = 300 + 1 = 301

(iii) 5n + 2

Putting n = 5, 5 * 5 + 2 = 25 + 2 = 27

Putting n = 10, 5 * 10 + 2 = 50 + 2 = 52

Putting n = 100, 5 * 100 + 2 = 500 + 2 = 502

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