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.
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
Statistics is the study of collection, organization, analysis and interpretation of data.
Data
A distinct piece of information in the form of fact or figures collected or represented for any specific purpose is called Data. In Latin, it is known as the Datum.
Collection of Data
Data are generally of two types
Primary Data
Secondary Data
Primary Data
Data collected from any firsthand experience for an explicit use or purpose is known as Primary Data
Secondary data
Data collected by any third party for a different purpose other than the user is known as Secondary Data.
Presentation of Data
After collecting data it is important to present it in a meaningful manner. There are many ways to present data.
1. Ungrouped Data
a. Raw Data– If there is no change in the data and it is in the same form as it is collected then it is said to be raw data.
Example
The marks obtained by 10 students in a Sanskrit test are
55 36 95 73 60 42 25 78 75 62
Range- The difference between the highest and the lowest number of data is called Range.
b. Frequency Distribution– When the number of items is large then we can convert it into the tabular form which is called a Frequency Distribution Table.
Frequency is the number of times the item comes in the table.
2. Grouped Data
To present the very large number of items in a data we use grouped distribution table.
a. Class Interval – The group used to classify the data is called the class interval i.e. 20 – 30, 30 – 40.
b. Upper Limit – In each class interval, the greatest number is the upper-class limit.
c. Lower Limit – In each class interval, the smallest number is the lower class limit.
d. Class Size – It is the difference between the upper limit and the lower limit i.e. 10.
e. Class Mark – The midpoint of each class interval is the class mark.
Grouped data could be of two types as below:–
Inclusive or discontinuous Frequency Distribution – If the upper limit of one class is different from the lower limit next class then it is said to be an Inclusive or discontinuous Frequency Distribution.
Exclusive or continuous Frequency Distribution – If the upper limit of one class is the same as the lower limit of the next class then it is said to be exclusive or continuous Frequency Distribution
Graphical Representation of Data
As you know a picture is better than thousand words so represent data in an easier way is to represent it graphically. Some of the methods of representing the data graphically are
1. Bar Graph
It is the easiest way to represent the data in the form of rectangular bars so it is called Bar graph.
The thickness of each bar should be the same.
The space between in bar should also be same.
The height of the bar should be according to the numerical data to be represented.
Example
Represent the average monthly rainfall of Nepal for the first six months in the year 2014.
Month
Jan
Feb
Mar
Apr
May
Jun
Average rainfall
45
65
40
60
75
30
Solution
On the x-axis mark the name of the months.
On the y-axis mark the class interval which we have chosen.
Then mark the average rainfall respective to the name of the month by the vertical bars.
The bars could be of any width but should be same.
This is the required bar graph.
2. Histogram
It is like the Bar graph only but it is used in case of a continuous class interval.
The class intervals are to be taken along an x-axis.
The height represents the frequencies of the respective class intervals.
Example
Draw the histogram of the following frequency distribution.
Daily earnings (in Rs)
700 – 750
750 – 800
800 – 850
850 – 900
900 – 950
950 – 1000
No. of stores
6
9
2
7
11
5
Solution
Mark the daily earnings on the x-axis.
Mark the no. of stores on the y-axis.
As the scale is starting from 700 so we will mark the zigzag to show the break.
Mark the daily earnings through the vertical bars.
3. Frequency Polygon
To draw the frequency polygon
First, we need to draw a histogram
Then join the midpoint of the top of the bars a line segment and the figure so obtained is required frequency polygon.
The midpoint of the first bar is to be joined with the midpoint of the imaginary interval of the x-axis
The midpoint of the last bar is to be joined with the midpoint of the next interval of the x-axis.
If we need to draw the frequency polygon without drawing the histogram then first we need to calculate the class mark of each interval and these points will make the frequency polygon.
Example
Draw the frequency polygon of a city in which the following weekly observations were made in a study on the cost of living index without histogram.
Step 1: First of all we need to calculate the class mark of each class interval.
Step 2: Take the suitable scale and represent the class marks on the x-axis.
Step 3: Take the suitable scale and represent the frequency distribution on the y-axis.
Step 4: To complete the frequency polygon we will join it with the x-axis before the first class and after the last interval.
Step 5: Now plot the respective points and join to make the frequency polygon.
Measures of Central Tendency
To make all the study of data useful, we need to use measures of central tendencies. Some of the tendencies are
1. Mean
The mean is the average of the number of observations. It is calculated by dividing the sum of the values of the observations by the total number of observations.
It is represented by x bar or.
The meanof n values x1, x2, x3, …… xn is given by
Mean of Grouped Data (Without Class Interval)
If the data is organized in such a way that the frequency is given but there is no class interval then we can calculate the mean by
where, x1, x2, x3,…… xn are the observations
f1, f2, f3, …… fn are the respective frequencies of the given observations.
Example
Here, x1, x2, x3, x4, and x5 are 20, 40, 60, 80,100 respectively.
and f1 , f2 , f3 , f4, f5 are 40, 60, 30, 50, 20 respectively.
2. Median
The median is the middle value of the given number of the observation which divides into exactly two parts.
For median of ungrouped data, we arrange it in ascending order and then calculated as follows
If the number of the observations is odd then the median will beAs in the above figure the no. of observations is 7 i.e. odd, so the median will beterm.
= 4th term.
The fourth term is 44.
If the number of observations is even then the median is the average of n/2 and (n/2) +1 term.
Example
Find the median of the following data.
1. First, we need to arrange it in ascending order.
4, 6, 7,8,10,12,12,13
2. The no. of observation is 8. As the no. of observation is even the median is the average of n/2 and (n/2)+1.
3.
4. 4th term is 8 and the 5th term is 10.
5. So the median
3. Mode
The mode is the value of the observation which shows the number that occurs frequently in data i.e. the number of observations which has the maximum frequency is known as the Mode.
Example
Find the Mode of the following data:
15, 20, 22, 25, 30, 20,15, 20,12, 20
Solution
Here the number 20 appears the maximum number of times so
Mode = 20.
Remark: The empirical relation between the three measures of central tendency is
There are lot many objects in our life which are round in shape. Few examples are the clock, dart board, cartwheel, ring, Vehicle wheel, Coins, etc.
Circles
Any closed shape with all points connected at equidistance from centre forms a Circle.
Any point which is at equidistance from anywhere from its boundary is known as the Centre of the Circle.
Radius is a Latin word which means ‘ray’ but in the circle it is the line segment from the centre of the Circle to its edge. So any line starting or ending at the centre of the circle and joining anywhere on the border on the circle is known as the Radius of Circle.
Interior and Exterior of a Circle
In a flat surface, the interior of a circle is the line whose distance from the centre is less than the radius.
The exterior of a circle is the line in the plane whose distance from the centre is larger than the radius.
Terms related to circle
Chord: Any straight line segment that’s both endpoints falls on the boundary of the circle is known as Chord. In Latin, it means ‘bowstring’.
Diameter: Any straight line segment or Chord which passes through the centre of the Circle and its endpoints connects on the boundary of the Circle is known as the Diameter of Circle. So in a circle Diameter is the longest chord possible in a circle.
Arc: Any smooth curve joining two points is known as Arc. So in Circle, we can have two possible Arcs, the bigger one is known as Major Arc and the smaller one is known as Minor Arc.
Circumference: It is the length of the circle if we open and straightened out to make a line segment.
Segment and Sector of the Circle
A segment of the circle is the region between either of its arcs and a chord. It could be a major or minor segment.
Sector of the circle is the area covered by an arc and two radii joining the centre of the circle. It could be the major or minor sector.
Angle Subtended by a Chord at a Point
If in a circle AB is the chord and is making ∠ACB at any point of the circle then this is the angle subtended by the chord AB at a point C.
Likewise, ∠AOB is the angle subtended by chord AB at point O i.e. at the centre and ∠ADB is also the angle subtended by AB at point D on the circle.
Theorem 1: Any two equal chords of a circle subtend equal angles at the centre.
Here in the circle, the two chords are given and PQ = RS with centre O.
So OP = OS = OQ = OR (all are radii of the circle)
∆POQ ≅ ∆SOR
∠POQ = ∠SOR
This shows that the angles subtended by equal chords to the centre are also equal.
Theorem 2: If the angles made by the chords of a circle at the centre are equal, then the chords must be equal.
This theorem is the reverse of the above Theorem 1.
Perpendicular from the Centre to a Chord
Theorem 3: If we draw a perpendicular from the centre of a circle to any chord then it bisects the chord.
If we draw a perpendicular from the centre to the chord of the circle then it will bisect the chord. And the bisector will make 90° angle to the chord.
Theorem 4: The line which is drawn from the centre of a circle to bisect a chord must be perpendicular to the chord.
If we draw a line OB from the centre of the circle O to the midpoint of the chord AC i.e. B, then OB is the perpendicular to the chord AB.
If we join OA and OC, then
In ∆OBA and ∆OBC,
AB = BC (B is the midpoint of AC)
OA = OC (Both are the radii of the same circle)
OB = OB (same side)
Hence, ΔOBA ≅ ΔOBC (both are congruent by SSS congruence rule)
⇒ ∠OBA = ∠OBC (respective angles of congruent triangles)
∠OBA + ∠OBC = ∠ABC = 180° [Linear pair]
∠OBC + ∠OBC = 180° [Since ∠OBA = ∠OBC]
2 x ∠OBC = 180°
∠OBC = 90o
∠OBC = ∠OBA = 90°
∴ OB ⊥ AC
Circle through Three Points
Theorem 5: There is one and only one circle passing through three given non-collinear points.
In this figure, we have three non-collinear points A, B and C. Let us join AB and BC and then make the perpendicular bisector of both so that RS and PQ the perpendicular bisector of AB and BC respectively meet each other at Point O.
Now take the O as centre and OA as the radius draw the circle which passes through the three points A, B and C.
This circle is known as Circumcircle. Its centre and radius are known as the Circumcenter and Circumradius.
Equal Chords and Their Distances from the Centre
Theorem 6: Two equal chords of a circle are at equal distance from the centre.
AB and CD are the two equal chords in the circle. If we draw the perpendicular bisector of these chords then the line segment from the centre to the chord is the distance of the chord from the centre.
If the chords are of equal size then their distance from the centre will also be equal.
Theorem 7: Chords at equal distance from the centre of a circle are also equal in length. This is the reverse of the above theorem which says that if the distance between the centre and the chords are equal then they must be of equal length.
Angle Subtended by an Arc of a Circle
The angle made by two different equal arcs to the centre of the circle will also be equal.
There are two arcs in the circle AB and CD which are equal in length.
So ∠AOB = ∠COD.
Theorem 8: The angle subtended by an arc at the centre is twice the angle subtended by the same arc at some other point on the remaining part of the circle.
In the above figure ∠POQ = 2∠PRQ.
Theorem 9: Angles from a common chord which are on the same segment of a circle are always equal.
If there are two angles subtended from a chord to any point on the circle which are on the same segment of the circle then they will be equal.
∠a = (1/2) ∠c (By theorem 8)
∠b = (1/2) ∠c
∠a = ∠b
Cyclic Quadrilaterals
If all the vertex of the quadrilateral comes on a circle then it is said to be a cyclic quadrilateral.
Theorem 10: Any pair of opposite angles of a cyclic quadrilateral has the sum of 180º.
∠A + ∠B + ∠C + ∠D = 360º (angle sum property of a quadrilateral)
∠A + ∠C = 180°
∠B + ∠D = 180º
Theorem 11: If the pair of opposite angles of a quadrilateral has sum of 180º, then the quadrilateral will be cyclic.
The part of any plane which is enclosed with the closed figure is known as the Planar Region. And the measure of this region is called the Area of that figure. This is expressed in the form of numbers using any unit.
Properties of the Area of a figure
If the shape and size of the two figures are the same then these are said to be congruent. And if the two figures are congruent then their area will also be the same. If ∆ABC and ∆DEF are two congruent figures then ar (∆ABC) = ar (∆DEF).
But if the two figures have the same area they need not be congruent.
If the two non-overlapping plane regions form a new planner region. Let Area of the square be ar(S) and Area of Triangle be ar (T). So Area of the new figure is Ar (P) = ar(S) + ar (T)
Figures on the Same Base and Between the Same Parallels
If the two figures have the same base and the vertices opposite to the base is also on the line parallel to the base then the two figures are said to be on the same base and between the same parallels.
∆ABC and ∆BDC have the same base and the opposite vertex is on the parallel line.
Parallelograms on the same Base and between the same Parallels
If the two parallelograms have the same base and are between the same parallel lines then these two parallelograms must have equal area.
Here, ABCD and ABGH are the two parallelograms having common base i.e. AB and between the two parallel lines i.e. AB and HC.
ar (ABCD) = ar (ABGH)
Remark: The parallelograms having the same base and equal area than these two parallelograms must lie between the same parallels.
Area of Parallelogram
Area of parallelogram = base × height
Height is the perpendicular on the base.
If the Area is given and one of the height or base is missing then we can find it as
Remark: The formula of area of the parallelogram is base × height that’s why the two parallelograms having the same base and between the same parallel lines have equal area.
Example:
Calculate the Area of the parallelogram if the base is 15 ft and the height is 3 ft.
Solution:
Given b = 15 ft
h = 3 ft
Area of parallelogram = b × h
= 15 × 3
= 45 ft2
Triangles on the same Base and between the same Parallels
If the two triangles are on the same base and their opposite vertex is on the parallel line then their area must be equal.
Here, ABC and DBC are the two triangles having common base i.e. BC and between the two parallel lines i.e. XY and BC.
ar (ABC) = ar (DBC)
Remark: If the triangles have the same base and equal area then these two triangles must lie between the same parallels.
Area of Triangle
Median of a Triangle
The line segment from any vertex of the triangle to the midpoint of the opposite side is the Median.
There are three medians of a triangle and the intersection of all the three medians is known as the Centroid.
The median divides the triangle into two equal parts.
In ∆ABC AE, CD and BF are the three medians and the centroid is the point O.
AE divides the triangle into two equal parts i.e. ∆ACE and ∆AEB,
CD divides the triangle into two equal parts i.e. ∆CBD and ∆CDA
BF divides the triangle into two equal parts i.e. ∆BFA and ∆BFC.
A Parallelogram and a Triangle on the same base and also between same parallel
If a triangle is on the base which is same with a parallelogram and between the same parallel line then the area of the triangle is half of the area of the parallelogram.
Here ∆ ABC and parallelogram ABCE are on the same base and between same parallel lines i.e. XY and BC so
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’. 
.
As in the above figure the no. of observations is 7 i.e. odd, so the median will be
If we join OA and OC, then
If we draw a diagonal in the quadrilateral, it divides it into two triangles. 
Theorem 1: In a given triangle if two sides are unequal then the angle opposite to the longer side will be larger.
