It refers to the length of the outline of the enclosed figure.
Area
It refers to the surface of the enclosed figure.
Area and Perimeter of Square
Square is a quadrilateral, with four equal sides.
Area = Side × Side
Perimeter = 4 × Side
Example
Find the area and perimeter of a square-shaped cardboard whose length is 5 cm.
Solution
Area of square = (side)2
= (5)2
= 25 cm2
Perimeter of square = 4 × side
= 4 × 5
= 20 cm
Area and Perimeter of Rectangle
The rectangle is a quadrilateral, with equal opposite sides.
Area = Length × Breadth
Perimeter = 2(Length + Breadth)
Example
What is the length of a rectangular field if its width is 20 ft and Area is 500 ft2?
Solution
Area of rectangular field = length × width
500 = l × 20
l = 500/20
l = 25 ft
Note: Perimeter of a regular polygon = Number of sides × length of one side
Triangles as Parts of Rectangles
If we draw a diagonal of a rectangle then we get two equal sizes of triangles. So the area of these triangles will be half of the area of a rectangle.
The area of each triangle = 1/2 (Area of the rectangle)
Likewise, if we draw two diagonals of a square then we get four equal sizes of triangles .so the area of each triangle will be one-fourth of the area of the square.
The area of each triangle = 1/4 (Area of the square)
Example
What will be the area of each triangle if we draw two diagonals of a square with side 7 cm?
Solution
Area of square = 7 × 7
= 49 cm2
The area of each triangle = 1/4 (Area of the square)
= 1/4 × 49
= 12.25 cm2
Congruent Parts of Rectangles
Two parts of a rectangle are congruent to each other if the area of the first part is equal to the area of the second part.
Example
The area of each congruent part = 1/2 (Area of the rectangle)
= 1/2 (l × b) cm2
=1/2 (4 × 3) cm2
= 1/2 (12) cm2
= 6 cm2
Parallelogram
It is a simple quadrilateral with two pairs of parallel sides.
Also denoted as ∥ gm
Area of parallelogram = base × height
Or b × h (bh)
We can take any of the sides as the base of the parallelogram. And the perpendicular drawn on that side from the opposite vertex is the height of the parallelogram.
Example
Find the area of the figure given below:
Solution
Base of ∥ gm = 8 cm
Height of ∥ gm = 6 cm
Area of ∥ gm = b × h
= 8 × 6
= 48 cm
Area of Triangle
Triangle is a three-sided closed polygon.
If we join two congruent triangles together then we get a parallelogram. So the area of the triangle will be half of the area of the parallelogram.
Area of Triangle = 1/2(Area of ∥ gm)
= 1/2 (base × height)
Example
Find the area of the figure given below:
Solution
Area of triangle = 1/2 (base × height)
= 1/2 (12 × 5)
= 1/2 × 60
= 30 cm2
Note: All the congruent triangles are equal in area but the triangles equal in the area need not be congruent.
Circles
It is a round, closed shape.
The circumference of a Circle
The circumference of a circle refers to the distance around the circle.
Radius: A straight line from the Circumference till the centre of the circle.
Diameter: It refers to the line from one point of the Circumference to the other point of the Circumference.
π (pi): It refers to the ratio of a circle’s circumference to its diameter.
Circumference(c) = π × diameter
C = πd
= π × 2r
Note: diameter (d) = twice the radius (r)
d = 2r
Example
What is the circumference of a circle of diameter 12 cm (Take π = 3.14)?
Solution
C = πd
C = 3.14 × 12
= 37.68 cm
Area of Circle
Area of the circle = (Half of the circumference) × radius
= πr2
Example
Find the area of a circle of radius 23 cm (use π = 3.14).
Solution
R = 23 cm
π = 3.14
Area of circle = 3.14 × 232
= 1,661 cm2
Conversion of Units
Sometimes we need to convert the unit of the given measurements to make it similar to the other given units.
Unit
Conversion
1 cm
10 mm
1 m
100 cm
1 km
1000 m
1 hectare(ha)
100 × 100 m
Unit
Conversion
1 cm2
100 mm2
1 m2
10000 cm2
1 km2
1000000 m2 (1e + 6)
1 ha
10000 m2
Example: 1
Convert 70 cm2 in mm2
Solution:
1 cm = 10 mm
1 cm2 = 10 × 10
1 cm2 = 100 mm2
70 cm2 = 700 mm2
Example: 2
Convert 3.5 ha in m2
Solution:
1 ha = 10000 m2
3.5 ha = 10000 × 3.5
ha = 35000 m2
Applications
We can use these concepts of area and perimeter of plane figures in our day to day life.
If we have a rectangular field and want to calculate that how long will be the length of the fence required to cover that field, then we will use the perimeter.
If a child has to decorate a circular card with the lace then he can calculate the length of the lace required by calculating the circumference of the card, etc.
Example:
A rectangular park is 35 m long and 20 m wide. A path 1.5 m wide is constructed outside the park. Find the area of the path.
Solution
Area of rectangle ABCD – Area of rectangle STUV
AB = 35 + 2.5 + 2.5
= 40 m
AD = 20 + 2.5 + 2.5
= 25 m
Area of ABCD = 40 × 25
= 1000 m2
Area of STUV = 35 × 20
= 700 m2
Area of path = Area of rectangle ABCD – Area of rectangle STUV
A line segment is a part of a line with two endpoints.
A line perpendicular to a line segment
Any line which is perpendicular to a line segment makes an angle of 90°.
Construction of a line parallel to a given line, through a point not on the line
We need to construct it using ruler and compass only.
Step 1: Draw a line PQ and take a point R outside it.
Step 2: Take a point J on the line PQ and join it with R.
Step 3: Take J as a centre and draw an arc with any radius which cuts PQ at C and JR at B.
Step 4: Now with the same radius, draw an arc taking R as a centre.
Step 5: Take the measurement of BC with compass and mark an arc of the same measurement from R to cut the arc at S.
Step 6: Now join RS to make a line parallel to PR.
∠ARS = ∠BJC, hence RS ∥ PQ because of equal corresponding angles.
This concept is based on the fact that a transversal between two parallel lines creates a pair of equal corresponding angles.
Remark: This can be done by taking alternate interior angles instead of corresponding angles.
Construction of triangles
The construction of triangles is based on the rules of congruent triangles. A triangle can be drawn if-
Three sides are given (SSS criterion).
Two sides and an included angle are given (SAS criterion).
Two angles and an included side are given. (ASA criterion).
A hypotenuse and a side are given for right angle triangle (RHS criterion).
Construction of a triangle with three given sides (SSS criterion)
Example
Draw a triangle ABC with the sides AB = 6 cm, BC = 5 cm and AC = 9 cm.
Solution
Step 1: First of all draw a rough sketch of a triangle, so that we can understand how to go ahead.
Step 2: Draw a line segment AB = 6 cm.
Step 3: From point A, C is 9 cm away so take A as a centre and draw an arc of 9 cm.
Step 4: From point B, C is 5 cm away so take B as centre and draw an arc of 5 cm in such a way that both the arcs intersect with each other.
Step 5: This point of intersection of arcs is the required point C. Now join AC and BC.
ABC is the required triangle.
Construction of a triangle if two sides and one included angle is given (SAS criterion)
Example
Construct a triangle LMN with LM = 8 am, LN = 5 cm and ∠NLM = 60°.
Solution
Step 1: Draw a rough sketch of the triangle according to the given information.
Step 2: Draw a line segment LM = 8 cm.
Step 3: draw an angle of 60° at L and make a line LO.
Step 4: Take L as a centre and draw an arc of 5 cm on LO.
Step 5: Now join NM to make a required triangle LMN.
Construction of a triangle if two angles and one included side is given (ASA criterion)
Example
Draw a triangle ABC if BC = 8 cm, ∠B = 60°and ∠C = 70°.
Solution
Step 1: Draw a rough sketch of the triangle.
Step 2: Draw a line segment BC = 8 cm.
Step 3: Take B as a centre and make an angle of 60° with BC and join BP.
Step 4: Now take C as a centre and draw an angle of 70° using a protractor and join CQ. The point where BP and QC intersects is the required vertex A of the triangle ABC.
ABC is the required triangle ABC.
Construction of a right angle triangle if the length of the hypotenuse and one side is given (RHScriterion)
Example
Draw a triangle PQR which is right angled at P, with QR =7 cm and PQ = 4.5 cm.
Solution
Step 1: Draw a rough sketch of the triangle.
Step 2: Draw a line segment PQ = 4.5 cm.
Step 3: At P, draw PS ⊥ PQ. This shows that R must be somewhere on this perpendicular.
Step 4: Take Q as a centre and draw an arc of 7 cm which intersects PS at R.
Revision Notes on Practical Geometry
Line segment
A line segment is a part of a line with two endpoints.
A line perpendicular to a line segment
Any line which is perpendicular to a line segment makes an angle of 90°.
Construction of a line parallel to a given line, through a point not on the line
We need to construct it using ruler and compass only.
Step 1: Draw a line PQ and take a point R outside it.
Step 2: Take a point J on the line PQ and join it with R.
Step 3: Take J as a centre and draw an arc with any radius which cuts PQ at C and JR at B.
Step 4: Now with the same radius, draw an arc taking R as a centre.
Step 5: Take the measurement of BC with compass and mark an arc of the same measurement from R to cut the arc at S.
Step 6: Now join RS to make a line parallel to PR.
∠ARS = ∠BJC, hence RS ∥ PQ because of equal corresponding angles.
This concept is based on the fact that a transversal between two parallel lines creates a pair of equal corresponding angles.
Remark: This can be done by taking alternate interior angles instead of corresponding angles.
Construction of triangles
The construction of triangles is based on the rules of congruent triangles. A triangle can be drawn if-
Three sides are given (SSS criterion).
Two sides and an included angle are given (SAS criterion).
Two angles and an included side are given. (ASA criterion).
A hypotenuse and a side are given for right angle triangle (RHS criterion).
Construction of a triangle with three given sides (SSS criterion)
Example
Draw a triangle ABC with the sides AB = 6 cm, BC = 5 cm and AC = 9 cm.
Solution
Step 1: First of all draw a rough sketch of a triangle, so that we can understand how to go ahead.
Step 2: Draw a line segment AB = 6 cm.
Step 3: From point A, C is 9 cm away so take A as a centre and draw an arc of 9 cm.
Step 4: From point B, C is 5 cm away so take B as centre and draw an arc of 5 cm in such a way that both the arcs intersect with each other.
Step 5: This point of intersection of arcs is the required point C. Now join AC and BC.
ABC is the required triangle.
Construction of a triangle if two sides and one included angle is given (SAS criterion)
Example
Construct a triangle LMN with LM = 8 am, LN = 5 cm and ∠NLM = 60°.
Solution
Step 1: Draw a rough sketch of the triangle according to the given information.
Step 2: Draw a line segment LM = 8 cm.
Step 3: draw an angle of 60° at L and make a line LO.
Step 4: Take L as a centre and draw an arc of 5 cm on LO.
Step 5: Now join NM to make a required triangle LMN.
Construction of a triangle if two angles and one included side is given (ASA criterion)
Example
Draw a triangle ABC if BC = 8 cm, ∠B = 60°and ∠C = 70°.
Solution
Step 1: Draw a rough sketch of the triangle.
Step 2: Draw a line segment BC = 8 cm.
Step 3: Take B as a centre and make an angle of 60° with BC and join BP.
Step 4: Now take C as a centre and draw an angle of 70° using a protractor and join CQ. The point where BP and QC intersects is the required vertex A of the triangle ABC.
ABC is the required triangle ABC.
Construction of a right angle triangle if the length of the hypotenuse and one side is given (RHScriterion)
Example
Draw a triangle PQR which is right angled at P, with QR =7 cm and PQ = 4.5 cm.
Solution
Step 1: Draw a rough sketch of the triangle.
Step 2: Draw a line segment PQ = 4.5 cm.
Step 3: At P, draw PS ⊥ PQ. This shows that R must be somewhere on this perpendicular.
Step 4: Take Q as a centre and draw an arc of 7 cm which intersects PS at R.
Rational Numbers are the numbers that can be expressed in the form p/q where p and q are integers (q ≠ 0). It includes all natural, whole numbers, fractions and integers.
Equivalent Rational Numbers
By multiplying or dividing the numerator and denominator of a rational number by the same integer, we can obtain another rational number equivalent to the given rational number.
Numbers are said to be equivalent if they are proportionate to each other.
Example
Therefore 1/2, 2/4, 4/8 are equivalent to each other as they are equal to each other.
Positive and Negative Rational Numbers
1. Positive Rational Numbers are the numbers whose both the numerator and denominator are positive.
Example: 3/4, 12/24 etc.
2. Negative Rational Numbers are the numbers whose one of the numerator or denominator is negative.
Example: (-2)/6, 36/(-3) etc.
Remark: The number 0 is neither a positive nor a negative rational number.
Rational Numbers on the Number Line
Representation of whole numbers, natural numbers and integers on a number line is done as follows
Rational Numbers can also be represented on a number line like integers i.e. positive rational numbers are on the right to 0 and negative rational numbers are on the left of 0.
Representation of rational numbers can be done on a number line as follows
Rational Numbers in Standard Form
A rational number is in the standard form if its denominator is a positive integer and there is no common factor between the numerator and denominator other than 1.
If any given rational number is not in the standard form then we can reduce it to its standard form or the lowest form by dividing its numerator and denominator by their HCF ignoring its negative sign.
Example
Find the standard form of 12/18
Solution
2/3 is the standard or simplest form of 12/18
Comparison of Rational Numbers
1. To compare the two positive rational numbers we need to make their denominator same, then we can easily compare them.
Example
Compare 4/5 and 3/8 and tell which one is greater.
Solution
To make their denominator same, we need to take the LCM of the denominator of both the numbers.
LCM of 5 and 8 is 40.
2. To compare two negative rational numbers, we compare them ignoring their negative signs and then reverse the order.
Example
Compare – (2/5) and – (3/7) and tell which one is greater.
Solution
To compare, we need to compare them as normal numbers.
LCM of 5 and 7 is 35.
by reversing the order of the numbers.
3. If we have to compare one negative and one positive rational number then it is clear that the positive rational number will always be greater as the positive rational number is on the right to 0 and the negative rational numbers are on the left of 0.
Example
Compare 2/5 and – (2/5) and tell which one is greater.
Solution
It is simply that 2/5 > – (2/5)
Rational Numbers between Rational Numbers
To find the rational numbers between two rational numbers, we have to make their denominator same then we can find the rational numbers.
Example
Find the rational numbers between 3/5 and 3/7.
Solution
To find the rational numbers between 3/5 and 3/7, we have to make their denominator same.
LCM of 5 and 7 is 35.
Hence the rational numbers between 3/5 and 3/7 are
These are not the only rational numbers between 3/5 and 3/7.
If we find the equivalent rational numbers of both 3/5 and 3/7 then we can find more rational numbers between them.
Hence we can find more rational numbers between 3/5 and 3/7.
Remark: There are “n” numbers of rational numbers between any two rational numbers.
Operations on Rational Numbers
1. Addition
a. Addition of two rational numbers with the same denominator
i. We can add it using a number line.
Example:
Add 1/5 and 2/5
Solution:
On the number line we have to move right from 0 to 1/5 units and then move 2/5 units more to the right.
ii. If we have to add two rational numbers whose denominators are same then we simply add their numerators and the denominator remains the same.
Example
Add 3/11 and 7/11.
Solution
As the denominator is the same, we can simply add their numerator.
b. Addition of two Rational Numbers with different denominator
If we have to add two rational numbers with different denominators then we have to take the LCM of denominators and find their equivalent rational numbers with the LCM as the denominator, and then add them.
Example
Add 2/5 and 3/7.
Solution
To add the two rational numbers, first, we need to take the LCM of denominators the find the equivalent rational numbers.
LCM of 5 and 7 is 35.
c. Additive Inverse
Like integers, the additive inverse of rational numbers is also the same.
This shows that the additive inverse of 3/7 is – (3/7) This shows that
2. Subtraction
If we have to subtract two rational numbers then we have to add the additive inverse of the rational number that is being subtracted to the other rational number.
a – b = a + (-b)
Example
Subtract 4/21 from 8/21.
Solution
i. In the first method, we will simply subtract the numerator and the denominator remains the same.
ii. In the second method, we will add the additive inverse of the second number to the first number.
3. Multiplication
a. Multiplication of a Rational Number with a Positive Integer.
To multiply a rational number with a positive integer we simply multiply the integer with the numerator and the denominator remains the same.
Example
b. Multiply of a Rational Number with a Negative Integer
To multiply a rational number with a negative integer we simply multiply the integer with the numerator and the denominator remains the same and the resultant rational number will be a negative rational number.
Example
c. Multiply of a Rational Number with another Rational Number
To multiply a rational number with another rational number we have to multiply the numerator of two rational numbers and multiply the denominator of the two rational numbers.
Example
Multiply 3/7 and 5/11.
Solution
4. Division
a. Reciprocal
Reciprocal is the multiplier of the given rational number which gives the product of 1.
Reciprocal of a/b is b/a
Product of Reciprocal
If we multiply the reciprocal of the rational number with that rational number then the product will always be 1.
Example
b. Division of a Rational Number with another Rational Number
To divide a rational number with another rational number we have to multiply the reciprocal of the rational number with the other rational number.
Example
Divide
Solution
Rational Numbers
Rational Numbers are the numbers that can be expressed in the form p/q where p and q are integers (q ≠ 0). It includes all natural, whole numbers, fractions and integers.
Equivalent Rational Numbers
By multiplying or dividing the numerator and denominator of a rational number by the same integer, we can obtain another rational number equivalent to the given rational number.
Numbers are said to be equivalent if they are proportionate to each other.
Example
Therefore 1/2, 2/4, 4/8 are equivalent to each other as they are equal to each other.
Positive and Negative Rational Numbers
1. Positive Rational Numbers are the numbers whose both the numerator and denominator are positive.
Example: 3/4, 12/24 etc.
2. Negative Rational Numbers are the numbers whose one of the numerator or denominator is negative.
Example: (-2)/6, 36/(-3) etc.
Remark: The number 0 is neither a positive nor a negative rational number.
Rational Numbers on the Number Line
Representation of whole numbers, natural numbers and integers on a number line is done as follows
Rational Numbers can also be represented on a number line like integers i.e. positive rational numbers are on the right to 0 and negative rational numbers are on the left of 0.
Representation of rational numbers can be done on a number line as follows
Rational Numbers in Standard Form
A rational number is in the standard form if its denominator is a positive integer and there is no common factor between the numerator and denominator other than 1.
If any given rational number is not in the standard form then we can reduce it to its standard form or the lowest form by dividing its numerator and denominator by their HCF ignoring its negative sign.
Example
Find the standard form of 12/18
Solution
2/3 is the standard or simplest form of 12/18
Comparison of Rational Numbers
1. To compare the two positive rational numbers we need to make their denominator same, then we can easily compare them.
Example
Compare 4/5 and 3/8 and tell which one is greater.
Solution
To make their denominator same, we need to take the LCM of the denominator of both the numbers.
LCM of 5 and 8 is 40.
2. To compare two negative rational numbers, we compare them ignoring their negative signs and then reverse the order.
Example
Compare – (2/5) and – (3/7) and tell which one is greater.
Solution
To compare, we need to compare them as normal numbers.
LCM of 5 and 7 is 35.
by reversing the order of the numbers.
3. If we have to compare one negative and one positive rational number then it is clear that the positive rational number will always be greater as the positive rational number is on the right to 0 and the negative rational numbers are on the left of 0.
Example
Compare 2/5 and – (2/5) and tell which one is greater.
Solution
It is simply that 2/5 > – (2/5)
Rational Numbers between Rational Numbers
To find the rational numbers between two rational numbers, we have to make their denominator same then we can find the rational numbers.
Example
Find the rational numbers between 3/5 and 3/7.
Solution
To find the rational numbers between 3/5 and 3/7, we have to make their denominator same.
LCM of 5 and 7 is 35.
Hence the rational numbers between 3/5 and 3/7 are
These are not the only rational numbers between 3/5 and 3/7.
If we find the equivalent rational numbers of both 3/5 and 3/7 then we can find more rational numbers between them.
Hence we can find more rational numbers between 3/5 and 3/7.
Remark: There are “n” numbers of rational numbers between any two rational numbers.
Operations on Rational Numbers
1. Addition
a. Addition of two rational numbers with the same denominator
i. We can add it using a number line.
Example:
Add 1/5 and 2/5
Solution:
On the number line we have to move right from 0 to 1/5 units and then move 2/5 units more to the right.
ii. If we have to add two rational numbers whose denominators are same then we simply add their numerators and the denominator remains the same.
Example
Add 3/11 and 7/11.
Solution
As the denominator is the same, we can simply add their numerator.
b. Addition of two Rational Numbers with different denominator
If we have to add two rational numbers with different denominators then we have to take the LCM of denominators and find their equivalent rational numbers with the LCM as the denominator, and then add them.
Example
Add 2/5 and 3/7.
Solution
To add the two rational numbers, first, we need to take the LCM of denominators the find the equivalent rational numbers.
LCM of 5 and 7 is 35.
c. Additive Inverse
Like integers, the additive inverse of rational numbers is also the same.
This shows that the additive inverse of 3/7 is – (3/7) This shows that
2. Subtraction
If we have to subtract two rational numbers then we have to add the additive inverse of the rational number that is being subtracted to the other rational number.
a – b = a + (-b)
Example
Subtract 4/21 from 8/21.
Solution
i. In the first method, we will simply subtract the numerator and the denominator remains the same.
ii. In the second method, we will add the additive inverse of the second number to the first number.
3. Multiplication
a. Multiplication of a Rational Number with a Positive Integer.
To multiply a rational number with a positive integer we simply multiply the integer with the numerator and the denominator remains the same.
Example
b. Multiply of a Rational Number with a Negative Integer
To multiply a rational number with a negative integer we simply multiply the integer with the numerator and the denominator remains the same and the resultant rational number will be a negative rational number.
Example
c. Multiply of a Rational Number with another Rational Number
To multiply a rational number with another rational number we have to multiply the numerator of two rational numbers and multiply the denominator of the two rational numbers.
Example
Multiply 3/7 and 5/11.
Solution
4. Division
a. Reciprocal
Reciprocal is the multiplier of the given rational number which gives the product of 1.
Reciprocal of a/b is b/a
Product of Reciprocal
If we multiply the reciprocal of the rational number with that rational number then the product will always be 1.
Example
b. Division of a Rational Number with another Rational Number
To divide a rational number with another rational number we have to multiply the reciprocal of the rational number with the other rational number.
The ratio is used to compare two quantities. These quantities must have the same units.
The ratio is represented by “:”, which is read as “to”. We can write it in the form of “fraction”.
Example
Write the ratio of the height of Sam to John, where Sam’s height is 175 cm and john’s height is 125 cm.
Solution
The ratio of Sam’s height to John’s height is 175: 125 = 7: 5.
We can write it in fraction as 7/5.
Equivalent Ratios
The equivalent ratio is like the equivalent fractions so to find the equivalent ratio we need to write it in the form of a fraction. To find the equivalent ratio we need to multiply or divide the numerator and denominator with the same number.
Example
Find the two equivalent ratios of 5: 20.
Solution
First multiply it by 2.
So the two equivalent ratios are 10:40 and 1: 4.
To compare that the two ratios are equivalent or not we need to convert them in the form of like a fraction. Like fractions are the fractions with the same denominator.
Example
Check whether the ratios 2: 3 and 3: 4 are equivalent are not?
Solution
To check, first, we need to make their denominator same.
Hence the ratio 2:3 is not equivalent to 3:4.
Proportion
Proportion shows the equality between two ratios. If two ratios are in proportion then these must be equal.
How to solve proportion problems?
Example
If the cost of 8 strawberries is Rs. 64 then what will be the cost of 25 strawberries.
Solution
Using Unitary Method
Solution using proportion
Let the cost of 25 strawberries = Rs. x
Hence the cost of 25 strawberries is Rs. 200
Percentage
The percentage is another way of comparisons. In ratios we have to make the denominator same then only we can compare them but in percentage, we can compare by calculating the percentage of the given quantity.
The percentage is the numerator of the fraction with the 100 denominators.
Symbol of Percentage
Example
What is the percentage of boys and girls in the class of 100 students if the number of boys is 55 and the number of girls is 45?
Solution
Percentage if the total is not a hundred
If the total number of quantity is not hundred i.e. the denominator is not hundred then to find the percentage we need to make the denominator 100.
Example
Out of 4 bees, 2 are going right and 2 are going left. So what percentage of bees is going right?
Solution
Unitary Method
Out of 4 bees, the number of bees going right are 2. Hence, out of 100, the number of bees going right is
By making denominator 100
Out of 4 bees, the number of bees going right is 2.
Converting fractional numbers to percentage
Fractional numbers have different denominator and to convert them into percentage we have to multiply the fraction with the 100%.
Example
Out of 15 fishes, 5 are red. What is the percentage of the red fishes?
Solution
Converting decimals to percentage
To convert the decimal into a percentage, first, we need to convert the decimal into fraction then multiply it by 100%.
Example
Convert 0.65 into a percentage.
Solution
Multiply the decimal with the 100%.
Converting Percentage to fractions or Decimals
We can reverse the above process to convert the percentage into fraction or decimal.
Parts always add to give a whole
If we know the one part of a whole then we can find the other part because all the parts together form a whole or 100%.
Example
If there are 25 men in the office of 100 employees then the remaining 75 would be women.
This means that if 25% are men the (100% – 25%) = 75% are women.
Fun with estimation
With the help of percentage, we can estimate the parts of an area.
Example
What percent of the given figure is shaded?
Solution
First, we have to find the fraction of the shaded portion.
Use of Percentages
Interpreting percentages
To use the percentages in real life we must be able to interpret the percentage.
Example
If we say that Seema is spending 20% of her income then it means that Seema is spending Rs. 25 out of every Rs. 100 she earns.
Converting percentages to “How many”.
Example
If 20% of students get a distinction out of 45 students in a class, then how many students got the distinction?
Solution
The number of students got distinction = [20/100] × 45 = 9.
Hence, 9 students out of 45 got the distinction.
Ratios to percent
Example
If the profit of Rs. 2500 is divided among three partners in such a way that A, B and C got the two parts, three parts and five parts of profit respectively. How much money will each get? What percent of the profit do they get?
Solution
The three partners are getting profit in the ratio of 2: 3: 5, so the total of the parts is
2 + 3 + 5 = 10
Increase or decrease as Percent
Sometimes we have to find the increase or decrease in certain quantities as a percentage. Like the increase in population, decrease in sale etc.
Example
The total marks of Charlie increased from 365 to 380 from last year’s result. Find the increase in percentage.
Solution
Original amount = Marks of Charlie last year = 365
Amount of change = increase in the number of marks = 380 – 365 = 15.
Therefore,
Buying and Selling
Cost Price
Cost price is the price at which you buy some product. It is written as CP.
Selling Price
Selling price is the price at which you sell something. It is written as SP.
These are the factors which tell us that the sale of some product is profitable or not.
CP < SP
Profit
Profit = SP – CP
CP = SP
No profit no loss
–
CP > SP
Loss
Loss = CP – SP
Example
If the buying price (or CP) of a table is Rs 700 and the selling price (or SP) is Rs 820, then find the profit or loss.
Solution
As the SP is more than CP, so the seller earns the profit in the table.
Profit made = SP – CP
= Rs 820 – Rs 700
= Rs 80
Profit or loss percentage
The profit and loss can be converted into a percentage. It is always calculated on the cost price.
Example
If the cost price of a laptop is Rs.45000 and the selling price is Rs. 50000, then what is the profit or loss percentage?
Solution
How to find SP if CP and profit or loss % is given?
Example
If the cost of a TV is Rs.25000 and shopkeeper sells it at a loss % of 5% then what is the selling price of the TV?
Solution
Hence, the shopkeeper sells it at the price of Rs. 23750
How to find CP if SP and profit or loss % is given?
Example
If the Selling price of a bookshelf is Rs 750 and the profit made by the seller is 10% then what is the cost price of the bookshelf?
Solution
Hence the seller bought the bookshelf at the cost of Rs. 682.
Simple Interest
When we borrow some money from the bank then we have to pay some interest to the bank.
The money which we borrow is called the Principal.
The amount which we have to pay to the bank to use that money is called interest.
At the end of the year we return the money to the bank with interest, that money is called Amount.
Amount = Principal + interest
Where,
SI = Simple interest
P = Principal
R = Rate of Interest
T = time period
Example
Sunita borrows a loan of Rs 5,0000 at 15% per year as the rate of interest. Find the interest she has to pay at end of one year.
Solution
Total amount to be paid by Sunita at the end of one year = Rs.50000 + Rs. 7500 = Rs.57500.
Interest for multiple years
If we have to calculate the interest for more than one year then we have to change the time period only.
Example
In the above example if Sunita takes the loan for 3 years then what will be the total amount after 3 years?
Solution
Total amount to be paid by Sunita at the end of 3 years = Rs.50000 + Rs. 22500 = Rs.72500.
If we superpose one figure over other and they fit into each other then they must be congruent shapes. They must have the same shape and size.
Symbol of Congruence
Congruence of 2-dimensional Shapes
In the case of 2D shapes, the two shapes will be congruent if they have the same shape and size. You cannot bend, stretch or twist the image.
Congruence among Line Segments
To check whether the line segments are congruent or not, we can superpose one line segment over another and if they completely cover each other then they must be congruent.
Two line segments are congruent if they have equal length and if two line segments have equal length then they must be congruent.
Congruence of Angles
Two angles of the same measurement are congruent and if two angles are congruent then their measurement must be the same.
Here, ∠R ≅ ∠Q
Congruence of Triangle
If we superpose one triangle over other triangle and they cover each other properly, then they must be congruent triangles.
In case of congruent triangles-
All the sides of one triangle must be equal to the corresponding sides of another triangle.
All the angles of one triangle must be equal to the corresponding angles of another triangle.
All the vertices of one triangle must be corresponding to the vertices of another triangle.
In the above triangles,
If, ∆ABC ≅ ∆FDE then
Corresponding vertices are – ∠A ↔ ∠F, ∠B ↔ ∠D and ∠C ↔ ∠E
Corresponding angles are – ∠A ↔ ∠F, ∠B ↔ ∠D and ∠C ↔ ∠E
Corresponding sides are – AB ↔ FD, BC ↔ DE and AC ↔ FE
Remark: It is the order of the letters in the names of congruent triangles which tells the corresponding relationships between two triangles. If we change it from ∆ABC ≅ ∆FDE to ∆BCA ≅ ∆FDE, then it is not necessary that the two triangles are congruent as it is important that all the corresponding sides, angles and vertices are same.
The Criterion for Congruence of Triangles
1. SSS Criterion(Side-Side-Side)
This criterion says that the two triangles will be congruent if their corresponding sides are equal.
If Side AB = DE
Side BC = EF
Side AC = DF
Then, ∆ABC ≅ ∆DEF
Example
In the two given triangles, ∆ABC and ∆DEF AB = 7 cm, BC = 5 cm, AC = 9 cm, DE = 7 cm, DF = 9 cm and EF = 5 cm. Check whether the two triangles are congruent or not.
Solution
In ∆ABC and ∆DEF,
AB = DE = 7 cm,
BC = EF = 5 cm,
AC = DF = 9 cm
This show that all the three sides of ∆ABC are equal to the sides of ∆DEF.
Hence with the SSS criterion of congruence, the two triangles are congruent.
∆ABC ≅ ∆DEF
2. SAS Criterion(Side-Angle-Side)
This criterion says that the two triangles will be congruent if their corresponding two sides and one included angle are equal.
If Side AB = DE
Angle ∠B = ∠E
Side BC = EF
Then, ∆ABC ≅ ∆DEF
Example
In ∆JKN, JK = KN and AK is the bisector of ∠JKN, then
1. Find the three pairs of equal parts in triangles JKA and AKN.
2. Is ΔJKA ≅ ΔNKA? Give reasons.
Is ∠J = ∠N? Give reasons.
Solution
1. The three pairs of equal parts are:
JK = KN (Given)
∠JKA = ∠NKA (KA bisects ∠JKN)
AK = AK (common)
2. Yes, ΔJKA ≅ ΔNKA (By SAS congruence rule)
3. ∠J = ∠N(Corresponding parts of congruent triangles)
3. ASA criterion(Angle-Side-Angle)
This criterion says that the two triangles are congruent if the two adjacent angles and one included side of one triangle are equal to the corresponding angles and one included side of another triangle.
If Angle ∠B = ∠B’
Side BC = EF
Angle ∠C = ∠C’
Then, ∆ABC ≅ ∆A’B’C’
Example
In ∆LMN and ∆OPN, if LMN = ∆NPO = 60°, LNM = 35° and LM = PO = 4 cm.Then check whether the triangle LMN is congruent to triangle PON or not.
Solution
In the two triangles ∆LMN and ∆OPN,
Given,
LMN = NPO = 60°
LNM = ∠PNO = 35° (vertically opposite angles)
So, ∠L of ΔLMN = 180° – (60° + 35°) = 85° (by angle sum property of a triangle) similarly,
∠O of ΔOPN =180° – (60° + 35°) = 85°
Thus, we have ∠L = ∠O, LM = PO and ∠M = ∠P
Now, side LM is between ∠L and ∠M and side PO is between ∠P and ∠O.
Hence, by ASA congruence rule,
∆LMN ≅ ∆OPN.
4. RHS Criterion(Right angle-Hypotenuse –Side)
This criterion says that the two right-angled triangles will be congruent if the hypotenuse and one side of one triangle are equal to the corresponding hypotenuse and one side of another triangle.
If Right angle ∠B = ∠E
Hypotenuse AC = DF
Side BC = EF
Then, ∆ABC ≅ ∆DEF
Example
Prove that ∆RSV ≅ ∆RKV, if RS = RK = 7 cm and RV = 5 cm and is perpendicular to SK.
Solution
In ∆RSV and ∆RKV,
Given
RS = RK = 7 cm
RV = RV = 5 cm (common side)
If RV is perpendicular to SK then
∠RVS = ∠RVK = 90°.
Hence, ∆RSV ≅ ∆RKV
As in the two right-angled triangles, the length of the hypotenuse and one side of both the sides are equal.
Remark: AAA is not the criterion for the congruent triangles because if all the angles of two triangles are equal then it is not compulsory that their sides are also equal. One of the triangles could be greater in size than the other triangle.
In the above figure, the two triangles have equal angles but their length of sides is not equal so they are not congruent triangles.
Triangle is a closed curve made up of three line segments. It has three vertices, sides and angles.
Here, in ∆ABC,
AB, BC and CA are the three sides.
A, B and C are three vertices.
∠A, ∠B and ∠C are the three angles.
Types of Triangle on the basis of sides
Types of Triangle on the basis of angles
Medians of a Triangle
Median is the line segment which made by joining any vertex of the triangle with the midpoint of its opposite side. Median divides the side into two equal parts.
Every triangle has three medians like AE, CD and BF in the above triangle.
The point where all the three medians intersect each other is called Centroid.
Altitudes of a Triangle
Altitude is the line segment made by joining the vertex and the perpendicular to the opposite side. Altitude is the height if we take the opposite side as the base.
The altitude form angle of 90°.
There are three altitudes possible in a triangle.
The point of intersection of all the three altitudes is called Orthocenter.
The Exterior Angle of a Triangle
If we extend any side of the triangle then we get an exterior angle.
An exterior angle must form a linear pair with one of the interior angles of the triangle.
There are only two exterior angles possible at each of the vertices.
Here ∠4 and ∠5 are the exterior angles of the vertex but ∠6 is not the exterior angle as it is not adjacent to any of the interior angles of the triangle.
Exterior Angle Property of the Triangle
An Exterior angle of a triangle will always be equal to the sum of the two opposite interior angles of the triangle.
Here, ∠d = ∠a + ∠b
This is called the Exterior angle property of a triangle.
Example
Find the value of “x”.
Solution
x is the exterior angle of the triangle and the two given angles are the opposite interior angles.
Hence,
x = 64°+ 45°
x = 109°
Angle Sum Property of a Triangle
This property says that the sum of all the interior angles of a triangle is 180°.
Example
Find the value of x and y in the given triangle.
Solution
x + 58° = 180° (linear pair)
x = 180° – 58°
x = 122°
We can find the value of y by two properties-
1. Angle sum property
60° + 58° + y = 180°
y = 180°- (60° + 58)
y = 62°
2. Exterior angle property
x = 60°+ y
122° = 60° + y
y = 122° – 60°
y = 62°
Two Special Triangles
1. Equilateral Triangle
It is a triangle in which all the three sides and angles are equal.
2. Isosceles Triangle
It is a triangle in which two sides are equal and the base angles opposite to the equal sides are also equal.
Sum of the length of the two sides of a triangle
Sum of the length of the two sides of a triangle will always be greater than the third side, whether it is an equilateral, isosceles or scalene triangle.
Example
Check whether it is possible to make a triangle using these measurements or not?
1. 3 cm, 4 cm, 7 cm
We have to check whether the sum of two sides is greater than the third side or not.
4 + 7 = 11
3 + 7 = 10
3 + 4 =7
Here the sum of the two sides is equal to the third side so the triangle is not possible with these measurements.
2. 2 cm, 5 cm, 6 cm
2 + 5 = 7
6 + 5 =11
6 + 2 = 8
Here the sum of the two sides is greater than the third side so the triangle could be made with these measurements.
Right Angled Triangle
A right-angled triangle is a triangle which has one of its angles as 90° and the side opposite to that angle is the largest leg of the triangle which is known as Hypotenuse .the other two sides are called Legs.
Pythagoras theorem
In a right angle triangle,
(Hypotenuse)2 = (base)2 + (height)2
The reverse of Pythagoras theorem is also applicable, i.e. if the Pythagoras property holds in a triangle then it must be a right-angled triangle.
Example
Find the value of x in the given triangle if the hypotenuse is 5 cm and height is 4 cm.
A point is a geometrical element which has no dimensions.
Line
A line is a straight path which has no endpoints.
Line Segment
A line segment is a straight path which has two endpoints.
Ray
A ray is a line which has one endpoint and endless from another side.
Angles
The corners made by the intersection of two lines or line segments are called Angles.
We write angle as ∠ABC in first figure and ∠XOY, ∠ZOW, ∠YOW and ∠XOZ are angles in the second figure.
Related Angles
1. Complementary Angles
If the sum of two angles is 90° then they are said to be complementary angles.
Or you can say that two angles which make up a right angle are called Complementary Angle.
2. Supplementary Angles
If the sum of two angles is 180° then they are said to be supplementary angles. If two angles are supplementary then they are the supplement to each other.
3. Adjacent Angles
It is the pair of two angles which are placed next to each other.
Adjacent angles have-
A common vertex.
A common arm.
A non-common arm could be on either side of the common arm.
4. Linear Pair
A pair of adjacent angles whose non-common arm makes a single line i.e. they are the opposite rays.
A linear pair is also a pair of supplementary angles as their sum is 180°.
The above pair of angles is –
Adjacent, as they have one common arm.
Supplementary, as the sum of two angles, is 180°.
The linear pair, as the sum is 180° and the non – common arms are opposite rays.
5. Vertically Opposite Angles
When two lines intersect each other then they form four angles. So that
∠a and ∠b is pair of vertically opposite angles.
∠n and ∠m is pair of vertically opposite angles.
Vertically opposite angles are equal.
Pairs of Lines
1. Intersecting Lines
If two lines touch each other in such a way that there is a point in common then these lines are called intersecting lines.
That common point is called a Point of Intersection.
Here, line l and m intersect each other at point C.
2. Transversal
If a line intersects two or more lines at different points then that line is called Transversal Line.
3. Angles made by a transversal
When a transversal intersects two lines then they make 8 angles.
Some of the angles made by transversal-
Types of Angles
Angles shown in figure
Interior Angles
∠6, ∠5, ∠4, ∠3
Exterior Angles
∠7,∠8,∠1,∠2
Pairs of Corresponding Angles
∠1 and ∠5,∠2 and ∠6, ∠3 and ∠7,∠4 and ∠8
Pairs of Alternate Interior Angles
∠3 and ∠6,∠4 and ∠5
Pairs of Alternate Exterior Angles
∠1 and ∠8,∠2 and ∠7
Pairs of Interior Angles on the same side of the transversal
∠3 and ∠5,∠4 and ∠6
Transversal of Parallel Lines
The two lines which never meet with each other are called Parallel Lines. If we have a transversal on two parallel lines then-
a. All the pairs of corresponding angles are equal.
∠3 = ∠7
∠4 = ∠8
∠1 = ∠5
∠2 = ∠6
b. All the pairs of alternate interior angles are equal.
∠3 = ∠6
∠4 = ∠5
c. The two Interior angles which are on the same side of the transversal will always be supplementary.
∠3 + ∠5 = 180°
∠4 + ∠6 = 180°
Checking for Parallel Lines
This is the inverse of the above properties of the transversal of parallel lines.
If a transversal passes through two lines so that the pairs of corresponding angles are equal, then these two lines must be parallel.
If a transversal passes through two lines in so that the pairs of alternate interior angles are equal, then these two lines must be parallel.
If a transversal passes through two lines so that the pairs of interior angles on the same side of the transversal are supplementary, then these two lines must be parallel.
Example: 1
If AB ∥ PQ, Find ∠W.
Solution:
We have to draw a line CD parallel to AB and PQ passing through ∠W.
∠QPW = ∠PWC = 50° (Alternate Interior Angles)
∠BAW =∠CWA = 46°(Alternate Interior Angles)
∠PWA = ∠PWC +∠CWA
= 50°+ 46°= 96°
Example: 2
If XY ∥ QR with ∠4 = 50° and ∠5 = 45°, then find all the three angles of the ∆PQR.
Solution:
Given: XY ∥ QR
∠4 = 50° and ∠5 = 45°
To find: ∠1, ∠2 and ∠3
Calculation: ∠1 + ∠4 + ∠5 = 180° (sum of angles making a straight angle)
∠1 = 180°- 50°- 45°
∠1 = 85°
PQ is the transversal of XY and QR, so
∠4 = ∠2 (Alternate interior angles between parallel lines)
∠2 = 50°
Pr is also the transversal of XY and QR, so
∠5 = ∠3 (Alternate interior angles between parallel lines)
It is an expression involving constant, variable and some operations like addition, multiplication etc.
Variable
Variable is an unknown number which could have a different numerical value. It is called Variable as it can vary.
It is represented by different letters like x, y, a, b etc.
Equation
An equation is a condition on a variable. It says that two expressions are equal.
Important Points Related to the Equation
One of the expressions must have a variable.
LHS of the equation is equal to the RHS of the equation.
An expression does not have equality sign but an equation always has an equality sign.
If we interchange the position of the expression from LHS to RHS or vice versa, the equation remains the same.
5x + 7 = 2
2 = 5x + 7
Both the above equations are same.
How to form equations using statements?
1. The sum of four times of x and 12 is equal to 35.
4x + 12 = 35
2. Half of a number is 3 more than 8.
Balanced Equation
When the LHS = RHS of an equation, then it is said to be a balanced equation.
1. If we add the same number to both the sides.
We can add the same number on both the sides of a balanced equation, the equation will remain the same.
12 – 8 = 3 + 1
If we add 3 to both the sides,
12 – 8 + 3 = 7
3 + 1 + 3 = 7
LHS = RHS
2. If we subtract the same number from both sides.
We can subtract the same number from both the sides of a balanced equation, the equation will remain the same.
12 – 8 = 3 + 1
If we subtract 3 from both the sides,
12 – 8 – 3 = 1
3 + 1 -3 = 1
LHS = RHS
3. If we multiply the same number to both the sides.
We can multiply the same number on both the sides of a balanced equation, the equation will remain the same.
12 – 8 = 3 + 1
If we multiply 3 to both the sides,
(12 – 8) × 3 = 36 – 24 = 12
(3 + 1) × 3 = 9 + 3 = 12
LHS = RHS
4. If we divide the same number from both sides.
We can divide the same number from both the sides of a balanced equation, the equation will remain the same.
12 – 8 = 3 + 1
If we divide both the sides by 2,
(12 – 8) ÷ 2 = 4 ÷ 2 = 2
(3 + 1) ÷ 2 = 4 ÷ 2 = 2
LHS = RHS
The Solution of an Equation
Any value of the variable which satisfies the equation is the solution of the equation.
There are two methods to solve an equation
1. By adding or subtracting the same number to both the sides of the equation as wehave above seen that the equation will remain the same.
Example: 1
x + 11 = 35
Solution:
Subtract 11 from both the sides.
x + 11 – 11 = 35 – 11
x = 24
Here, x = 24 is the solution of the given equation.
Example: 2
25y = 125
Solution:
Divide both the sides by 25.
y = 5
1. Transposing Method
In this method, we transpose the numbers from one side of the equation to the other side so that all the terms with variable come on one side and all the constants come on another side.
While transposing the numbers the sign of the terms will get changed. i.e. Negative will become positive and positive will become negative.
Example
x + 11 = 35
Solution
Now we will transfer 11 from LHS to RHS and its sign will get reversed.
x = 35 – 11
x = 24
From a Solution to the Equation
As we solve the equation to get the solution, we can get the equation also if we have the solution.
Any equation has only one solution but if we make an equation from a solution then there could be many equations.
Example
Given x = 7
3x = 21 (multiply both sides by 3)
3x + 8 = 29 (add 8 to both the sides)
This is not the only possible equation. There could be other equations also.
Applications of Simple Equations to Practical Situations
If we have statements related to a practical situation then first we have to convert it in the form of the equation then solve it to find the solution.
Example: 1
Radha’s Mother’s age is 5 years more than three times Shikha’s age. Find Shikha’s age, if her mother is 44 years old.
Solution:
Let Shikha’s age = y years
Her mother’s age is 3y + 5 which is 44.
Hence, the equation for Shikha’s age is 3y + 5 =44
3y + 5 = 44
3y = 44 – 5 (by transposing 5)
3y = 39
y = 13 (by dividing both sides by 3)
Hence, Shikha’s age = 13 years
Example: 2
A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 18 be subtracted from the number, the digits are reversed. Find the number.
Solution:
Let the digit at units place = y
So, the digit in the tens place = 2y
So, the number is (2y) y.
As it is given that if 18 is subtracted from the number, the digits are reversed.
Any raw information which we collect to know about it and to compare certain things is called Data. This information is in the form of facts and figures and is collected for some specific purpose.
Example
Number of students in the class
The temperature in a state on daily basis.
Data Handling
We need to collect, organize and represent that data to draw inferences from it. This is called Data Handling.
Collecting Data
Collection of data depends upon the further requirement of the data. Before collecting any data we must know that what will be the use of data.
If we have to compare the marks of the toppers in different classes then we need the data of all the classes not only one class having any topper.
Organization of Data
Before using any data, first, we need to organize it in a systematic manner so that it could be understood easily. Generally, data is organized in tabular form as it is easy to read and understand.
In this tabular form, we can easily understand that how many students get how much mark.
Representative Values
There must be a particular value which represents the complete data. This is the average of the data. The average lies between the smallest and the largest number of data so it is called Central tendency of the group of data.
There are three types of central tendency of data-
1. Arithmetic Mean
The mean is the average of the number of observations. To calculate mean we have to divide the sum of the values of the observations by the total number of observations.
Example
The score of 8 students in science is given. Find the average score of the students.
25, 28, 23, 24, 29, 35, 42, 48
Solution
Remark: This is not necessary that the value of mean will be from one of the observations.
Range
Arithmetic mean lies between the smallest and the largest observation. A range is a difference between the largest and the smallest observation.
Range = Largest Observation – Smallest Observation
Example
If the age of the students in a class is given then what will be the range of the given students?
2. Mode
For a different type of requirements different central tendencies are used.
Mode tells us the number of observation which occurs more frequently. The observation which occurs most of the time is called the Mode of that group.
Example
If we have the observation of average temperature in New Delhi for 12 months then find the month in which it has the maximum average temperature? What is the mode of the given observation?
Month
Average Temperature
January
18
February
22
March
24
April
25
May
25
June
29
July
27
August
27
September
27
October
25
November
21
December
19
Solution
As you can see that the maximum average temperature is in the month of June.
Its mode will be 27 as it occurs more frequently i.e. three times.
Mode of a Large Number of Observations
If the number of observations is very large then we can convert the data in the tabular form using frequencies and tally marks. Then it will be easy to find the mode of the given data.
So you can check the number of observation which has a large number of frequency is the mode of that group.
Example
Find the mode of the data of scores obtained by students of class 7 in Sanskrit given below.
Solution
As you can see that the maximum average temperature is in the month of June.
Its mode will be 27 as it occurs more frequently i.e. three times.
Mode of a Large Number of observations
If the number of observations is very large then we can convert the data in the tabular form using frequencies and tally marks. Then it will be easy to find the mode of the given data.
So you can check the number of observation which has a large number of frequency is the mode of that group.
Example
Find the mode of the data of scores obtained by students of class 7 in Sanskrit given below.
Score
Number of Students
2
4
4
2
8
3
9
2
11
5
13
4
15
6
18
8
Solution
The mode is the 18 as the maximum number of students i.e. 8 students score 18.
3. Median
The middle value of the given number of the observations which divides it into exactly two parts is called Median.
To find the median, we have to arrange the data in ascending or descending order then find the middle value of the given number of observations that is the median of that group.
a. If the number of observation is odd
b. If the number of observation is even
Example
Use of Bar Graphs with a Different Purpose
Bar graphs can be used for finding the measures of central tendency also, as we know the observation with the more frequency is the mode hence the bar with the tallest height must be the mode of the data.
Choosing a Scale
It is important to choose the scale according to the given data as the length of the bar depends upon the scale we choose.
Bar Graph
It is the representation of data with the use of bars of the same width and the length of bars depends upon the number of frequency.
Here we can see that the highest number is 14 and the lowest number is 7 so we can take the scale of one.
By the graph, we can observe that jazz is the most preferred form of music by the students.
Double Bar Graph
This is the same as the bar graph just the two bars are joined off to represent two data on the same graph. This is used to compare certain information.
Example
Represent the number of wild animals found in two states given below in double bar graph.
Wild Animals
Karnataka
Tamil Nadu
Lion
20
18
Tiger
16
20
Elephant
30
25
Rhino
15
22
Zebra
25
28
Solution
Here we have chosen a scale of 5.
The x-axis represents the name of wild animals.
Y-axis represents the number of wild animals.
The blue bar represents the number of animals in Karnataka.
Pink bars represent the number of animals in Tamil Nadu.
This double bar graph is used to compare the number of animals in different states.
Chance and Probability
Chance
In our day to day life, there are so many situations when we say that this is impossible, or this is possible, or this may or may not possible. So the situations which may or may not happen have the chance to happen.
This shows that it is not possible to throw 14 in the combination of two dices.
It is certain that the sun will rise.
It may or may not happen that a head has come when we flip a coin. As both, the head and tail have an equal chance.
Probability
Probability is the study of uncertainty The uncertainty of any doubtful situation is measured by means of Probability.
This tells us the chance of happening some outcomes from the total possible outcomes.
Example
If we throw a dice then what is the probability that we will get a 5?
Solution
Favourable outcome = 1 (there is only one possibility of getting 5)
Total no. of possible outcomes = 6 (total six numbers are there on a dice)
Probability is the study of the uncertainty. The uncertainty of any doubtful situation is measured by means of Probability.
Uses of Probability
Probability is used in many fields like Mathematics, Physical Sciences, Commerce, Biological Sciences, Medical Sciences, Weather Forecasting, etc.
Basic terms related to Probability
1. Randomness
If we are doing an experiment and we don’t know the next outcome of the experiment to occur then it is called a Random Experiment.
2. Trial
A trial is that action whose result is one or more outcomes. Like –
Throw of a dice
Toss of a coi
3. Independent trial
A trial will be independent if it does not affect the outcome of any other random trial. Like throwing a dice and tossing a coin are independent trials as they do not impact each other.
4. Event
While doing an experiment, an event will be the collection of some outcomes of that experiment.
Example
If we are throwing a dice then the possible outcome for even number will be three i.e. 2, 4, 6. So the event would consist of three outcomes.
Probability – An Experimental Approach
Experimental probability is the result of probability based on the actual experiments. It is also called the Empirical Probability.
In this probability, the results could be different, every time you do the same experiment. As the probability depends upon the number of trials and the number of times the required event happens.
If the total number of trials is ‘n’ then the probability of event D happening is
Examples
1. If a coin is tossed 100 times out of which 49 times we get head and 51 times we get tail.
a. Find the probability of getting head.
b. Find the probability of getting tail.
c. Check whether the sum of the two probabilities is equal to 1 or not.
Solution
a. Let the probability of getting head is P(H)
b. Let the probability of getting tail is P(T)
c. The sum of two probability is
= P(H) + P(T)
Impossible Events
While doing a test if an event is not possible to occur then its probability will be zero. This is known as an Impossible Event.
Example
You cannot throw a dice with number seven on it.
Sure or Certain Event
While doing a test if there is surety of an event to happen then it is said to be the sure probability. Here the probability is one.
Example: 1
It is certain to draw a blue ball from a bag contain a blue ball only.
This shows that the probability of an event could be
0 ≤ P (E) ≤ 1
Example: 2
There are 5 bags of seeds. If we select fifty seeds at random from each of 5 bags of seeds and sow them for germination. After 20 days, some of the seeds were germinated from each collection and were recorded as follows:
Bag
1
2
3
4
5
No. of seeds germinated
40
48
42
39
41
What is the probability of germination of
(i) more than 40 seeds in a bag?
(ii) 49 seeds in a bag?
(iii) more than 35 seeds in a bag?
Solution:
(i) The number of bags in which more than 40 seeds germinated out of 50 seeds is 3.
P (germination of more than 40 seeds in a bag) =3/5 = 0.6
(ii) The number of bags in which 49 seeds germinated = 0.
P (germination of 49 seeds in a bag) = 0/5 = 0.
(iii) The number of bags in which more than 35 seeds germinated = 5.
So, the required probability = 5/5 = 1.
Elementary Event
If there is only one possible outcome of an event to happen then it is called an Elementary Event.
Remark
If we add all the elementary events of an experiment then their sum will be 1.
The general form
P (H) + P (T) = 1
P (H) + P= 1 (whereis ‘not H’.
P (H) – 1 = P
P (H) and Pare the complementary events.
Example
What is the probability of not hitting a six in a cricket match, if a batsman hits a boundary six times out of 30 balls he played?
Solution
Let D be the event of hitting a boundary.
So the probability of not hitting the boundary will be
a. All the pairs of corresponding angles are equal.
We have to draw a line CD parallel to AB and PQ passing through ∠W.
A trial is that action whose result is one or more outcomes. Like –
a. Find the probability of getting head.
= 1 (where
is ‘not H’. 
