Class 8th

A quadrilateral has some measurements like – 4 sides, 4 angles and 2 diagonals.

We can construct a unique quadrilateral if we know the five measurements.

1. If the four sides and a diagonal of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which AB = 5 cm, BC = 7 cm, CD = 6 cm, DA = 6.5 cm and AC = 8 cm.

Solution

Step 1: ∆ABC can be constructed using SSS criterion of the construction of triangle.

Step 2: Here we can see that AC is diagonal, so D will be somewhere opposite to B with reference to AC.

AD = 6.5 cm so draw an arc from A as the centre with radius 6.5 cm.

Step 3: Now draw an arc with C as the centre and by taking radius 6 cm so that it intersects the above arc.

Step 4: The point of intersection of the two arcs is point D. Now join AD and DC to complete the quadrilateral.

Hence, ABCD is the required quadrilateral.

2. If  two diagonals and three sides of the quadrilateral are given

Example

Construct a quadrilateral ABCD if the two diagonals are AC = 6.5 cm and BD = 8 cm. The other sides are BC = 5.5 cm, AD = 6.5 cm and CD = 6 cm.

Solution

First of all, draw a rough sketch of the quadrilateral by using the given measurements. Then start constructing the real one.

Step 1: We can see that AD, AC and DC are given so we can construct a triangle ΔACD by using SSS criterion.

Step 2: Now, we know that BD is given so we can draw the point B keeping D as the centre and draw an arc of radius 8 cm just opposite to the point D with reference to AC.

Step 3: BC is given so we can draw an arc keeping C as centre and radius 5.5 cm so that it intersects the other arc.

Step 4: That point of intersection of the arcs is point B. Join AB and BC to complete the quadrilateral.

ABCD is the required quadrilateral.

3. If three angles and two adjacent sides of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which the two adjacent sides are AB = 4.5 cm and BC = 7.5 cm. The given three angles are ∠A = 75ᵒ, ∠B = 105ᵒ and ∠C = 120ᵒ.

Solution

Draw a rough sketch so that we can construct easily.

Step 1: Draw AB = 4.5 cm. Then measure ∠B = 105° using protractor and draw BC = 7.5 cm.

Step 2: Draw ∠C = 120°.

Step 3: Measure ∠A = 75° and make a line until it touches the line coming from point C.

ABCD is the required quadrilateral.

4. If the three sides with two included angles of the quadrilateral are given.

Example

Construct a quadrilateral ABCD in which the three sides are AB = 5 cm, BC = 6 cm and CD = 7.5 cm. The two included angles are ∠B = 105° and ∠C = 80°.

Solution

Draw a rough sketch.

Step 1: Draw the line BC = 6 cm. Then draw ∠B = 105° and mark the length of AB = 5 cm.

Step 2: Draw ∠C = 80° using protractor towards point B.

Step 3: Mark the length of CD i.e.7.5 cm from C to make CD = 7.5 cm.

Step 4: Join AD which will complete the quadrilateral ABCD.

Hence ABCD is the required quadrilateral.

Some Special Cases

There are some special cases in which we can construct the quadrilateral with less number of measurements also.

Example

Construct a square READ with RE = 5.1 cm.

Solution

Given Re = 5.1 cm.

As it is a special quadrilateral called square, we can get more details out of it.

a. All sides of square are equal, so RE = EA = AD = RD = 5.1 cm.

b. All the angles of a square are 90°, so ∠R = ∠E = ∠A = ∠D = 90°

Step 1: Draw a rough sketch of the square.

Step 2: To construct a square, draw a line segment RE = 5.1 cm. Then draw the angle of 90° at both ends R and E of the line segment RE.

Step 3: As all the sides of the square READ are equal, draw the arc of 5.1 cm from the vertex R and E to cut the lines RD and EA respectively.

Step 4: Join A and D to make a line segment AD.

READ is the required square.

Algebraic Expressions

Any expression involving constant, variable and some operations like addition, multiplication etc is called Algebraic Expression.

• A variable is an unknown number and generally, it is represented by a letter like x, y, n etc.
• Any number without any variable is called Constant.
• A number followed by a variable is called Coefficient of that variable.
• A term is any number or variable separated by operators.

Equation

A statement which says that the two expressions are equal is called Equation.

Linear Expression

A linear expression is an expression whose highest power of the variable is one only.

Example

2x + 5, 3y etc.

The expressions like x² + 1, z² + 2z + 3 are not the linear expressions as their highest power of the variable is greater than 1.

Linear Equations

The equation of a straight line is the linear equation. It could be in one variable or two variables.

Linear Equation in One Variable

If there is only one variable in the equation then it is called a linear equation in one variable.

The general form is

ax + b = c, where a, b and c are real numbers and a ≠ 0.

Example

x + 5 = 10

y – 3 = 19

These are called linear equations in one variable because the highest degree of the variable is one and there is only one variable.

Some Important points related to Linear Equations

• There is an equality sign in the linear equation. The expression on the left of the equal sign is called the LHS (left-hand side) and the expression on the right of the equal sign is called the RHS (right-hand side).
• In the linear equation, the LHS is equal to RHS but this happens for some values only and these values are the solution of these linear equations.

Graph of the Linear Equation in One Variable

We can mark the point of the linear equation in one variable on the number line.

x = 2 can be marked on the number line as follows-

Solving Equations which have Linear Expressions on one Side and Numbers on the other Side

There are two methods to solve such type of problems-

1. Balancing Method

In this method, we have to add or subtract with the same number on both the sides without disturbing the balance to find the solution.

Example

Find the solution for 3x – 10 = 14

Solution

Step 1: We need to add 10 to both the sides so that the numbers and variables come on the different sides without disturbing the balance.

3x – 10 +10 =10+14

3x = 24

Step 2: Now to balance the equation, we need to divide by 3 into both the sides.

3x/3 = 24/3

x = 8

Hence x = 8 is the solution of the equation.

We can recheck our answer by substituting the value of x in the equation.

3x – 10 = 14

3(8) – 10 = 14

24-10 = 14

14 = 14

Here, LHS = RHS, so our solution is correct.

2. Transposing Method

In this method, we need to transpose or transfer the constants or variables from one side to another until we get the solution. When we transpose the terms the sign will get changed.

Example

Find the solution for 2z +10 = 4.

Solution

Step 1: We transpose 10 from LHS to RHS so that all the constants come in the same side.

2z = 4 -10 (sign will get changed)

2z = -6

Step 2: Now divide both the sides by 2.

2z/2 = – 6/2

z = – 3

Here z = -3 is the solution of the equation.

Some Applications of Linear Equation

We can use the concept of linear equations in our daily routine also. There are some situations where we need to use the variable to find the solution. Like,

• What number should be added to 23 to get 75?
• If the sum of two numbers is 100 and one of the no. is 63 then what will be the other number?

Example

What is the height of the rectangle whose perimeter is 96 cm² and the length is 12 cm?

Solution

Let the height of the rectangle be ‘s’.

Area of rectangle = Length × Breadth

96 = S × 12

Now, this is a linear equation with variable s.

We need to divide both sides by 12 to find the solution.

96/12 = 12s/12

s = 8

Hence the height of the rectangle is 8 cm.

Solving Equations having the Variable on both Sides

As the equation can have the variable on both the sides also so we should know how to solve such problems.

In this type of problems, we need to bring all the constants on one side and all the terms having variables on the other side. Then they can be solved easily.

Example

Find the solution of 2x−3 = 6 − x.

Solution

Step 1: Bring all the terms including variable x on LHS and the constants on the RHS.

2x + x = 6 + 3 (sign will change while changing the position of the terms)

Step 2: Solve the equation

3x = 9

Step 3: Divide both the sides by 3 to get the solution.

3x/3 = 9/3

x = 3

Hence the solution of the equation is x = 3.

Some More Applications

Example

Renu’s age is four times that of her younger brother. Five years back her age was 9 times her brother’s age. Find their present ages.

Solution

Let the Renu’s brother age = x

Renu’s age = 4x (as her age is 4 times that of her younger brother)

Five years back her age was = 9(x – 5) which is equal to 4x – 5

9(x – 5) = 4x – 5

9x – 45 = 4x – 5

9x – 4x = – 5 + 45 (by transferring the variable and constants on different sides)

5x = 40

x = 40/5 = 8

Renu’s brother age = x = 8 years

Renu’s age = 4x = 4(8) = 32 years.

Reducing Equations to Simpler Form

When linear equations are in fractions then we can reduce them to a simpler form by-

• Taking the LCM of the denominator
• Multiply the LCM on both the sides, so that the number will reduce without the denominator and we can solve them by the above methods.

Example
Solve the linear equation

Solution
As the equation is in complex form, we have to reduce it into a simpler form.

Step 1: Take the L.C.M. of the denominators, 2, 3, 4, and 5, which is 60.
Step 2: Multiply both the sides by 60,

30x −12 = 20x + 15 + 60

Step 3. Bring all the variables on the LHS and all the constants on the RHS

30x − 20x = 15 + 12 + 60

10x = 87

Step 4: Dividing both the sides by 10

x = 8.7

Equations Reducible to the Linear Form

Sometimes there are some equations which are not linear equations but can be reduced to the linear form and then can be solved by the above methods.

Example

Solution

This is not a linear equation but can be reduced to linear form

Step 1: Multiply both the sides by (2x + 3).

Now, this is a linear equation.

Step 2: Multiply both the sides by 8.

8(x + 1) = 3(2x + 3)

8x + 8 = 6x + 9

8x – 6x = 9 – 8

2x = 1

x = 1/2

Hence the solution for the equation is x = 1/2.

Rational Numbers

 A number is called Rational if it can be expressed in the form p/q where p and q are integers (q > 0). It includes all natural, whole number and integers.

Example: 1/2, 4/3, 5/7,1 etc. 

Natural Numbers

All the positive integers from 1, 2, 3,……, ∞.

Whole Numbers

All the natural numbers including zero are called Whole Numbers.

Integers

All negative and positive numbers including zero are called Integers.

Properties of Rational Numbers

1. Closure Property

This shows that the operation of any two same types of numbers is also the same type or not.

a. Whole Numbers

If p and q are two whole numbers then

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + q will also be the whole number.	p – q will not always be a whole number.	pq will also be the whole number.	p ÷ q will not always be a whole number.
Example	6 + 0 = 6	8 – 10 = – 2	3 × 5 = 15	3 ÷ 5 = 3/5
Closed or Not	Closed	Not closed	Closed	Not closed

b. Integers

If p and q are two integers then

Operation	Addition	Subtraction	Multiplication	Division
Integers	p+q will also be an integer.	p-q will also be an integer.	pq will also be an integer.	p ÷ q will not always be an integer.
Example	– 3 + 2 = – 1	5 – 7 = – 2	– 5 × 8 = – 40	– 5 ÷ 7  = – 5/7
Closed or not	Closed	Closed	Closed	Not  closed

c. Rational Numbers

If p and q are two rational numbers then

Operation	Addition	Subtraction	Multiplication	Division
Rational Numbers	p + q will also be a rational number.	p – q will also be a rational number.	pq will also be a rational number.	p ÷ q will not always be a rational number
Example				p ÷ 0= not defined
Closed or Not	Closed	Closed	Closed	Not closed

2. Commutative Property

This shows that the position of numbers does not matter i.e. if you swap the positions of the numbers then also the result will be the same.

a. Whole Numbers

If p and q are two whole numbers then 

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + q = q + p	p – q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example	3 + 2 = 2 + 3	8 –10 ≠ 10 – 8 – 2 ≠ 2	3 × 5 = 5 × 3	3 ÷ 5 ≠ 5 ÷ 3
Commutative	yes	No	yes	No

b. Integers

If p and q are two integers then

Operation	Addition	Subtraction	Multiplication	Division
Integers	p + q = q + p	p – q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example	True	5 – 7 = – 7 – (5)	– 5 × 8 = 8 × (–5)	– 5 ÷ 7 ≠ 7 ÷ (-5)
Commutative	yes	No	yes	No

c. Rational Numbers

If p and q are two rational numbers then

Operation	Addition	Subtraction	Multiplication	Division
Rational numbers	p + q = q + p	p –q ≠ q – p	p × q = q × p	p ÷ q ≠ q ÷ p
Example
Commutative	yes	No	yes	No

3. Associative Property

This shows that the grouping of numbers does not matter i.e. we can use operations on any two numbers first and the result will be the same.

a. Whole Numbers

If p, q and r are three whole numbers then

Operation	Addition	Subtraction	Multiplication	Division
Whole number	p + (q + r) = (p + q) + r	p – (q – r) = (p – q) – r	p × (q × r) = (p × q) × r	p ÷ (q ÷ r)  ≠ (p ÷ q) ÷ r
Example	3 + (2 + 5) = (3 + 2) + 5	8 – (10 – 2) ≠ (8 -10) – 2	3 × (5 × 2) = (3 × 5) × 2	10 ÷ (5 ÷ 1) ≠ (10 ÷ 5) ÷ 1
Associative	yes	No	yes	No

b. Integers

If p, q and r are three integers then

Operation	Integers	Example	Associative
Addition	p + (q + r) = (p + q) + r	(– 6) + [(– 4)+(–5)] = [(– 6) +(– 4)] + (–5)	Yes
Subtraction	p – (q – r) = (p – q) – r	5 – (7 – 3) ≠ (5 – 7) – 3	No
Multiplication	p × (q × r) = (p × q) × r	(– 4) × [(– 8) ×(–5)] = [(– 4) × (– 8)] × (–5)	Yes
Division	p ÷ (q ÷ r) ≠ (p ÷ q) ÷ r	[(–10) ÷ 2] ÷ (–5) ≠ (–10) ÷ [2 ÷ (– 5)]	No

c. Rational Numbers

If p, q and r are three rational numbers then

Operation	Integers	Example	Associative
Addition	p + (q + r) = (p + q) + r		yes
Subtraction	p – (q – r) = (p – q) – r		No
Multiplication	p × (q × r) = (p × q) × r		yes
Division	p ÷ (q ÷ r)  ≠ (p ÷ q) ÷ r		No

The Role of Zero in Numbers (Additive Identity)

Zero is the additive identity for whole numbers, integers and rational numbers.

	Identity		Example
Whole number	a + 0 = 0 + a = a	Addition of zero to whole number	2 + 0 = 0 + 2 = 2
Integer	b + 0 = 0 + b = b	Addition of zero to an integer	False
Rational number	c + 0 = 0 + c = c	Addition of zero to a rational number	2/5 + 0 = 0 + 2/5 = 2/5

The Role of one in Numbers (Multiplicative Identity)

One is the multiplicative identity for whole numbers, integers and rational numbers.

	Identity		Example
Whole number	a ×1 = a	Multiplication of one to the whole number	5 × 1 = 5
Integer	b × 1= b	Multiplication of one to an integer	– 5 × 1 = – 5
Rational Number	c × 1= c	Multiplication of one to a rational number

Negative of a Number (Additive Inverse)

	Identity		Example
Whole number	a +(- a) = 0	Where a is a  whole number	5 + (-5) = 0
Integer	b +(- b) = 0	Where b is an integer	True
Rational number	c + (-c) = 0	Where c is a rational number

Reciprocal (Multiplicative Inverse)

The multiplicative inverse of any rational number

Example

The reciprocal of 4/5 is 5/4.

Distributivity of Multiplication over Addition and Subtraction for Rational Numbers

This shows that for all rational numbers p, q and r

1. p(q + r) = pq + pr

2. p(q – r) = pq – pr

Example

Check the distributive property of the three rational numbers 4/7,-( 2)/3 and 1/2.

Solution

Let’s find the value of

This shows that

Representation of Rational Numbers on the Number Line

On the number line, we can represent the Natural numbers, whole numbers and integers as follows

Rational Numbers can be represented as follows

Rational Numbers between Two Rational Numbers

There could be n number of rational numbers between two rational numbers. There are two methods to find rational numbers between two rational numbers.

Method 1

We have to find the equivalent fraction of the given rational numbers and write the rational numbers which come in between these numbers. These numbers are the required rational numbers.

Example

Find the rational number between 1/10 and 2/10.

Solution

As we can see that there are no visible rational numbers between these two numbers. So we need to write the equivalent fraction.

2/10 = 20/100((multiply the numerator and denominator by 10)

Hence, 2/100, 3/100, 4/100……19/100 are all the rational numbers between 1/10 and 2/10.

Method 2

We have to find the mean (average) of the two given rational numbers and the mean is the required rational number.

Example

Find the rational number between 1/10 and 2/10.

Solution

To find mean we have to divide the sum of two rational numbers by 2.

3/20 is the required rational numbers and we can find more by continuing the same process with the old and the new rational number.

Remark: 1. This shows that if p and q are two rational numbers then (p + q)/2 is a rational number between p and q so that

p < (p + q)/2 < q.

2. There are infinite rational numbers between any two rational numbers.