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:

  1. 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

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

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

  1. 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

  1. If we denote the length of a square by l, then the area of the square = l2
  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

  1. 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:

  1. 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.

  1. 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

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

        Class_7_Maths_Algebraic_Expressions_Example_Of_Algebraic_Expressions_3             

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 Class_7_Maths_Algebraic_Expressions_Example_Of_Algebraic_Expressions_4  .

 Solution:

 Class_7_Maths_Algebraic_Expressions_Example_Of_Algebraic_Expressions_5

 

(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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