Algebraic Expressions

Understand Factors and Coefficients of terms; Like and Unlike Terms, etc

Algebraic formulae have applications in many areas. They are also used extensively in geometry, where they are used to shorten the geometric formulae.

For example, we know that the perimeter of an equilateral triangle is three times the length of its side. If we denote the perimeter of the triangle by variable P and the length of the side of the triangle by l, then we can restate the formulae as:

P = 3l

Here, the value of P will change with a change in the value of l. In this formula, P and l are variables and 3 is a constant.

Similarly, the perimeter of a square is written as:

P = 4l

Here, the variables P and l represent the perimeter and length of the side of the square respectively, whereas 4 is a constant.

The perimeter of a regular pentagon is written as:

P = 5l

Again, the variables P and l represent the perimeter and length of the side of the pentagon respectively and 5 is a constant.

Let us now write the formula for the perimeter of a rectangle using variables. We know that the perimeter of a rectangle is twice the sum of its length and breadth.

If we denote variables P, l, and b for the respective perimeter, length, and breadth of the rectangle, then the formula of the perimeter of the rectangle in terms of algebraic expression will be written as:

P = 2 (l + b)

Here, P, l,and b are variables and 2 is a constant.

Now, let us try to denote the formulae of areas of some geometrical figures like triangle, rectangle, square, etc in terms of algebraic expression and variables. Let us start with the area of a rectangle.

We know that the area of a rectangle is the product of its length and breadth.

If we denote the variables A, l, and b for the area, length, and breadth of the rectangle, then the formula of area of rectangle in terms of variables is

A = l × b = lb

Here, A, l,and b are variables. This expression does not contain any constant.

We know that, area of a square = side × side

If we denote the variables A and l for the area and length of the side of the square, then the formula of area of the square in terms of variables is

A = l × l = l2

Here, A and l are variables. This expression also does not contain any constant.

We know that the area of a triangle equals one-half the product of the length of its base and its corresponding height.

If we denote the variables A, b, and h for the area, base, and height of the triangle, then the formula of area of triangle is

Here, A, b, and h are variables and is a constant.

Now, we can easily find the perimeter or area of a polygon by putting in the values of the variables in the given formula. For example, for a rectangle of length 7 cm and breadth 4 cm, the perimeter is obtained by putting the value l = 7 cm and b = 4 cm in the formula, P = 2 (l + b).

Now, the perimeter of the given rectangle, P = 2 (7 + 4) cm = 22 cm

We can also find the area of this rectangle by putting the value l = 7 cm and b = 4 cm in the formula, A = lb.

Now, the area of the given rectangle, A = lb = (7 × 4) cm2 = 28 cm2

Similarly, we can express the formulae used in different areas in terms of variables.

For example, the formula of loss in any transaction is as follows:

Loss = Cost price – Selling price

If we denote loss by l, cost price by c and selling price by s then the above formula can be expressed as follows:

l = c – s

Now, let us discuss some examples to have better understanding of use of variables in geometric formulae.

Example 1:

The perimeter of a regular octagon (an eight-sided polygon) is given by the formula

L = 8s, where L and s are the perimeter and length of the side of the octagon.

Identify the variables and constant in this formula.

Solution:

The formula for the perimeter of a regular octagon is L = 8s.

Here, the value of L changes with a change in the value of s. However, the value 8 does not change. Therefore, L and s are variables and 8 is a constant.

Example 2:

Derive the rule to find the perimeter of a regular pentagon by representing the length of one side by a variable l.

Solution:

The length of one side of a regular pentagon is l, where l is a variable.

Perimeter of a regular pentagon = 5 × length of one side

∴ Perimeter of a regular pentagon = 5 × l = 5l

Let p be the perimeter of the pentagon. Then, the following rule is obtained.

P = 5l

Example 3:

Derive the rule to find the diameter of a ball by taking the radius of the ball as the variable r.

Solution:

We know that the diameter of a ball is twice the radius of the ball.

∴ Diameter of the ball = 2 × radius of the ball = 2 × r = 2r

Let us take some matchsticks and join them to form some patterns as shown in the following figure. The number of matchsticks used to make each pattern is written alongside it.

Now, let us try to represent the number of matchsticks used to form each pattern with the help of an algebraic pattern. The given pattern is formed by repeating the shape, which is made of 5 matchsticks.

Thus, we require 5 matchsticks to form the first shape, 8 matchsticks to …

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