Mensuration on Circle

Measuring Circumference of  a Circle

In our daily life, we come across circular shapes in various places.  The diameter of a circle is twice the length of its radius (2r). i.e., d = 2r.  For any circle with a given ‘r’ or ‘d’, we can calculate its area and circumference using the formula given.

Area of the circle is calculated using the formula πr², whereas the circumference is calculated using 2πr. The value of π is ²²/₇.

Using the formula for finding the circumference of the circle, we can derive the following:

Circumference = 2πr
            C = 2r x π
            C = πd (since 2r = d)
            d = ^C/π or
            π = ^C/d or
            π = ^C/2r

From this, we can conclude that the ratio of the circumference to that of diameter is a constant (π).

Circumference / Diameter = π [²²/₇] or ^C/d = π 

Example 1: The diameter of a circular well is 4.2 m. What is its circumference?

Example 2: What is the circumference and area of the circular disc of radius 14 cm?

Given:
radius  = 14 cm

Solution:
circumference = 2πr

Thus, The circumference of the circular disk is 88 cm and the area is 616 square cm.

Example 3:  The diameter of a circular room is 6.3 m.  Find its circumference and area.

Given:
diameter  = 6.3 cm

Solution:

Example 4: If the circumference of the circle is 132 m. Then calculate the radius and diameter.

Given:
Circumference = 132 m

Solution:

d = 2r
d = 2 x 21
d = 42 cm

Thus, the radius and diameter are 21 cm and 42 cm respectively.

Example 5.  What is the distance travelled by the tip of the seconds hand of a clock in 1 minute, if the length of the hand is 56 mm.

Given:
radius = 56 mm

Solution:
circumference = 2πr
c = 2 x π x 56
c = 2 x 22 x 8 [π = 22/7]
c = 352 mm

Thus the distance travelled by the tip of the seconds hand of a clock in 1 minute is 352mm.

Example 6.  The radius of a tractor wheel is 77 cm.  Calculate the distance covered by it in 35 rotations?

Given:
radius = 77 cm

Solution:
The distance covered in one rotation (circumference) = 2πr
c = 2 x π x 77
c = 2 x 22 x 11 [π = 22/7]
c = 484 cm

The distance covered in 35 rotations = 35 x 484 = 16940 cm

Example 7:  A farmer wants to fence his circular poultry farm with barbed wire whose radius is 420 m.  The cost of fencing is ₹12 per meter.  How much more amount will be needed to fence his farm? 

Given:
radius = 420 m

Solution:
The length of the fence (circumference) = 2πr
c = 2 x π x 420
c = 2 x 22 x 60 [π = 22/7]
c = 2640 m

Cost of fencing 1 meter = 12
Cost of fencing 2640 meter = 12 x 2640 = 31,680.

Thus the cost of fencing the farm is RS. 31,680.

Example 8:  A ground is in the form of a circle whose diameter is 350 m. An athlete makes 4 revolutions. Find the distance covered by the athlete.

Given:
diameter (d) = 350 m

Solution:

Now, calculate the distance covered by the athlete:
Distance covered of 1 revolution = 1100 m
Distance covered of 4 revolution = 4 x 1100 = 4400

Thus, the distance covered by the athlete is 4400 m.

Example 9:  The diameter of the bullock cart wheel is 1.4 m.  Find the distance covered by it in 150 rotations?

Practice Questions - Set 1 on Algebra

MCQ on Algebraic Expressions

Choose the Right Answer:

Forming Expressions using Terms in Algebra

Forming Expression Using Factors & Terms

        An algebraic expression can have one term, two terms or more than two terms. An expression with one term is called a monomial, two terms is called a binomial and three terms is called a trinomial.  An expression with one or more terms is called a polynomial. 

        For example, the expression 2x is a monomial, 2x + 3y is a binomial,  and 2x + 3y + 4z is a  trinomial. All the expressions given above are polynomials.

A term may be any one of the following: 

        i)      a constant such as 8, −11, 7, −1,… 

        ii)     a variable such as x,a,p,y,… 

        iii)    a product of two or more variables such as xy, pq, abc, ... 

        iv)    a product of constant and a variable/variables such as 5x, −7pq, 3abc,…

Questions for Practice

1.  Identify the variables, terms and number of terms in each of the following expressions: 

            (i) 12−x   (ii) 7 + 2y   (iii) 29+3x+5y    (iv) 3x–5+7z 

2.  Find the numerical co-efficient of the following terms.  Also, find the co efficient of x and y in each of the term: 

            3x,  - 5xy,  - yz, 7xyz, y, 16yx. 

3.  If x  = 3, y = 2 find the value of  (i) 4x + 7y  (ii)  3x + 2y − 5  (iii)  x − y

4.  Find the value of  

        (i) 3m + 2n          (ii) 2m − n          (iii) mn − 1,     given that m = 2, n =  − 1. 

5.  Fill in the blanks: 

        (i) The variable in the expression 16x − 7 is ________. 

        (ii) The constant term of the expression 2y − 6 is _________. 

        (iii) In the expression 25m + 14n, the type of the terms are ________ terms. 

        (iv) The number of terms in the expression 3ab + 4c –9 is ________. 

        (v) The numerical co-efficient of the term  −xy is ________.

6.  Say True or False. 

        (i) x + ( −x) = 0 

        (ii) The co-efficient of ab in the term 15abc is 15. 

        (iii) 2pq and  − 7qp are like terms 

        (iv) When y = −1, the value of the expression 2y − 1 is 3 

7.  Find the numerical coefficient of each of the following terms: 

            −3yx, 12k, y, 121bc, − x, 9pq, 2ab 

8.  Write the variables, constants and terms of the following expressions. 

        (i)18 + x − y          (ii) 7p − 4q + 5           (iii) 29x + 13y         (iv) b + 2 5. 

9.  Identify the like terms among the following : 

            7x, 5y, −8x, 12y, 6z, z, −12x, −9y, 11z. 

10.  An algebraic statement which is equivalent to the verbal statement “Three times the sum of x and y” is 

        (i) 3(x + y)         (ii)  3 + x + y         (iii)  3x + y         (iv) 3 + xy 

Identifying Terms and Co-efficient of an Expression

Terms and Co-efficients 

(VII CBSE Maths)

When terms have the same algebraic factors, they are like terms. When terms have different algebraic factors, they are unlike terms. 

For example, in the expression 2xy – 3x + 5xy – 4, look at the terms 2xy and 5xy. The factors of 2xy are 2, x and y. The factors of 5xy are 5, x and y.  Thus their algebraic (i.e., those which contain variables) factors are the same and hence they are like terms.

 On the other hand the terms 2xy and –3x, have different algebraic factors. They are unlike terms. Similarly, the terms, 2xy and 4, are unlike terms. Also, the terms –3x and 4 are unlike terms.

Terms and Co-efficient of an Algebraic Expression

An expression with only one term is called a monomial; for example, 7xy, – 5m, 3z2, 4 etc. 

An expression which contains two unlike terms is called a binomial; for example, x + y, m – 5, mn + 4m, a2 – b2 are binomials. 

An expression which contains three terms is called a trinomial; for example, the expressions x + y + 7, ab + a +b, 3x2 – 5x + 2, m + n + 10 are trinomials.

The expression ab + a + b + 5 is, however not a trinomial; it contains four terms and not three. The expression x +  y + 5x is not a trinomial as the terms x and 5x are like terms. 

In general, an expression with one or more terms is called a polynomial. Thus a monomial, a binomial and a trinomial are all polynomials. 

Introduction to Chapter 12 - Algebraic Expressions

 Algebraic Expression

(VII CBSE Maths)

 A variable can take various values. Its value is not fixed. We use letters x, y, l, m, ... etc. to denote variables.  On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, –17, etc.  We combine variables and constants to make algebraic expressions. For this, we use the operations of addition, subtraction, multiplication and division.

Factors and Terms in an Algebraic Expression

For instance,  the expression 4x + 5 is obtained from the variable x, first by multiplying x by the constant 4 and then adding the constant 5 to the product. Similarly, 10y – 20 is obtained by first multiplying y by 10 and then subtracting 20 from the product. 

We can also obtain expressions by combining variables with themselves or with other variables.  Look at how the following expressions are obtained: x2, 2y2, 3x2 – 5, xy, 4xy + 7 

 TERMS OF AN EXPRESSION

Consider the expression (4x + 5). In forming this expression, we first formed 4x separately as a product of 4 and x and then added 5 to it. 

Similarly consider the expression (3x2 + 7y). Here we first formed 3x2 separately as a product of 3, x and x. We then formed 7y separately as a product of 7 and y. Having formed 3x2 and 7y separately, we added them to get the expression.  Such parts of an expression which are formed separately first and then added are known as terms. 

Look at the expression (4x2 – 3xy). We say that it has two terms, 4x2 and –3xy. The term 4x2 is a product of 4, x and x, and the term (–3xy) is a product of (–3), x and y. 

Terms are added to form expressions. Just as the terms 4x and 5 are added to form the expression (4x + 5), the terms 4x2 and (–3xy) are added to give the expression (4x2 – 3xy). This is because 4x2 + (–3xy) = 4x2 – 3xy.

Note, the minus sign (–)  is included in the term. In the expression 4x2 –3xy, we took the term as (–3xy) and not as (3xy). That is why we do not need to say that terms are ‘added or subtracted’ to form an expression; just ‘added’ is enough.

Factors of a term :  The term 4x2 is a product of 4, x and x; we say that 4, x and x are the factors of the term 4x2. A term is a product of its factors. The term –3xy is a product of the factors –3, x and y.