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Revision Notes on Polynomials

Polynomial

A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.

Example

1. 4x³ + 3x² + x +3 is a polynomial in variable x.

2. 4x² + 3x^-1 – 4 is not a polynomial as it has negative power.

3. 3x^3/2+ 2x – 3 is not a polynomial, since it power is not an integer.

. Polynomials

Polynomials are denoted by p (x), q (x), etc.
In the above polynomials, 2x², 3y, and 2 are the terms of the polynomial.
2 and 3 are the coefficients of x² and y, respectively.
x and y are the variables.
2 is the constant term, which has no variable.

Polynomials in One Variable

If there is only one variable in the expression, then this is called the polynomial in one variable.

Example

x³ + x – 4 is polynomial in variable x and is denoted by p(x).
r² + 2 is polynomial in variable r and is denoted by p(r).

Types of polynomials on the basis of the number of terms

Types of polynomials on the basis of the number of degrees

The highest value of the power of the variable in the polynomial is the degree of the polynomial.

Zeros of a Polynomial

If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.

Example: 1

The given polynomial is p(x) = x – 4

To find the zero of the polynomial, we will equate it to zero.

x – 4 = 0

x = 4

p(4) = x – 4 = 4 – 4 = 0

This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.

So 4 is the zero of the polynomial or the root of the polynomial.

The root of the polynomial is basically the x-intercept of the polynomial.

If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.

Example: 2

Find p (1) for the polynomial p (t) = t² – t + 1

p (1) = (1)² – 1 + 1

= 1 – 1 + 1

= 1

Remainder Theorem

We know the property of division which follows in the basic division, i.e.

Dividend = (Divisor × Quotient) + Remainder

This follows the division of polynomials.

If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:

P(x) = g(x) q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x).

It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.

Let’s understand it with an example

Division of a Polynomial with a Monomial

We can see that ‘x’ is common in the above polynomial, so we can write it as

Hence, 3x²+ x + 1 and x the factors of 3x³ + x² + x.

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x² – 3x -10 by 2 + x

Step 1: Write the dividend and divisor in descending order, i.e., in the standard form. x² – 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x²/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.

Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.

Step 4: – (5x/x) = – 5

Step 5:

The remainder is zero.

Hence x² – 3x – 10 = (x + 2)(x – 5) + 0

Dividend = (Divisor × Quotient) + Remainder

The Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).

As we know,

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t – t). q (t) + r = 0

To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.

Example:

What is the remainder if a⁴ + a³ – 2a² + a + 1 is divided by a – 1.

Solution:

P(x) = a⁴ + a³ – 2a² + a + 1

To find the zero of (a – 1), we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.

Factor Theorem

The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

(y – t) is a factor of p(y), if p(t) = 0, and
P (t) = 0 if (y – t) is a factor of p (y).

Example: 1

Check whether g(x) = x – 3 is the factor of p(x) = x³– 4x²+ x + 6 using the factor theorem.

Solution:

According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.

P (3) = (3)³– 4(3)² + (3) + 6

= 27 – 36 + 3 + 6 = 0

Hence, g (x) is the factor of p (x).

Example: 2

Find the value of k, if x – 1 is a factor of p(x) = kx²– √2x + 1

Solution:

As x -1 is the factor, p(1) = 0

Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorise x² –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + bx + c = x² + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x² + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x² + 17x + 5 =6 x² + (2 + 15) x + 5

= 6 x² + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Algebraic Identities

1. (x + y)²= x²+ 2xy + y²

2. (x – y)² = x²– 2xy + y²

3. (x + y) (x – y) = x² – y²

4. (x + a) (x + b) = x² + (a + b)x + ab

5. (x + y + z)²= x²+ y²+ z²+ 2xy + 2yz + 2zx

6. (x + y)³ = x³ + y³ + 3xy(x + y) = x³+ y³ + 3x²y + 3xy²

7. (x – y)³ = x³– y³ – 3xy(x – y) = x³– y³ – 3x²y + 3xy²

8. x³+ y³ = (x + y)(x² – xy + y²)

9. x³ – y³= (x – y)(x² + xy + y²)

10. x³+ y³ + z³– 3xyz = (x + y + z)(x² + y²+ z² – xy – yz – zx) x³ + y³ + z³ = 3xyz if x + y + z = 0

Example: 2

Factorize 8x³ + 27y³ + 36x²y + 54xy²

Solution:

The given expression can be written as

= (2x)³+ (3y)³ + 3(4x²) (3y) + 3(2x) (9y²)

= (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²

= (2x + 3y)³ (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution:

4x² + y² + z² – 4xy – 2yz + 4xz = (2x)²+ (–y)² + (z)² + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]² (Using Identity V)

= (2x – y + z)² = (2x – y + z) (2x – y + z)

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May 24, 2024

STATISTICS | Study
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Statistics

Statistics is the study of collection, organization, analysis and interpretation of data.

Data

Data is a distinct piece of information in the form of fact or figures collected or represented for any specific purpose. The word data is derived from the Latin word Datum.

Collection of Data

In general, data is of two types. They are:

Primary Data

Secondary Data

Primary Data

Primary data is the data collected from any firsthand experience for an explicit use or purpose.

Secondary data

Secondary data is the data collected by any third party for a different purpose other than the user.

Presentation of Data

After collecting data it is important to present it meaningfully. There are many ways to present data.

1. Raw Data or Ungrouped Data

a. Raw Data or Ungrouped Data is the collected data without any change in its form.

Example

The marks obtained by 10 students in a Mathematics test are:

55 36 95 73 60 42 25 78 75 62

Range

Range is the difference between the highest and the lowest values of data.

For Above data: Range = 95 – 36 = 59

b. Frequency Distribution Table – Frequency Distribution Table is the data of a large number of items converted into tabular form.

Frequency is the number of times the item comes to the table.

2. Grouped Data

To present the very large number of items in the data we use a grouped distribution table.

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.

Class Size = Upper Limit – Lower Limit

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.

Exclusive or continuous Frequency Distribution

Inclusive or discontinuous Frequency Distribution – If the upper limit of a class is different from the lower limit of its next class then it is said to be an Inclusive or discontinuous Frequency Distribution.

Example

Draw the histogram of the following frequency distribution.

Daily earnings (in Rs) 700 – 749 750 – 799 800 – 849 850 – 899 900 – 949 950 – 999
No. of stores 6 9 2 7 11 5

Exclusive or continuous Frequency Distribution – If the upper limit of a class is the same as the lower limit of its next class then it is said to be exclusive or continuous Frequency Distribution

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

Graphical Representation of Data

Since a picture represents better than a thousand words, The data is presented graphically. Some of the methods of representing the data graphically are:

1. Bar Graph

2. Histogram

3. Frequency Polygon

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 the bars should also be the 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 the same.

This is the required bar graph.

2. Histogram

It is similar to Bar graph, 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

Procedure to draw the frequency polygon

First, we need to draw a histogram.

Then join the midpoint of the top of the bars to a line segment and the figure so obtained is the 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

2. Median

3. Mode

1. Mean (Average)

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 be Term

As in the above figure the no. of observations is 7 i.e. odd, so the median will beterm.

= 4^th 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.

6, 7, 10, 12, 13, 4, 8, 12

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. 4^th term is 8 and the 5^th 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: 3 Median = Mode + 2 Mean
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SURFACE AREAS AND VOLUMES | Study

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Plane figure

The figures which we can be drawn on a flat surface or that lie on a plane are called Plane Figure.

Example – Circle, Square, Rectangle etc.

Solid figures

The 3D shapes which occupy some space are called Solid Figures.

Example – Cube, Cuboid, Sphere etc. (Scroll down to continue …)

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Volume

Space occupied by any solid shape is the capacity or volume of that figure. The unit of volume is a cubic unit.

Surface Area

The area of all the faces of the solid shape is its total surface area. The unit of surface area is a square unit.

Lateral or Curved Surface Area

The surface area of the solid shape after leaving the top and bottom face of the figure is called the lateral surface of the shape. The unit of lateral surface area is a square unit.

Surface Area and Volume of a Cube

Cube is a solid shape having 6 equal square faces.

Lateral surface area of a cube	4s²
Total surface area of a cube	6s²
The volume of a cube	s³
Diagonal	√3 s, s = edge of the cube = side length of face of cube

Surface Area and Volume of a Cube

Example

What is the capacity of a cubical vessel having each side of 8 cm?

Solution

Given side = 8 cm So, Volume of the cubical vessel = l³ = (8)³ = 256 cm³.

Surface Area and volume of a Cuboid

Cuboid is a solid shape having 6 rectangular faces at a right angle.

Lateral surface area of a cuboid	2h(l + b)
Total surface area of a cuboid	2(lb + bh + lh)
Volume of a cuboid	lbh
Diagonal	l = length, b = breadth, h = height

Surface Area and volume of a Cuboid

Example

What is the surface area of a cereal box whose length, breadth and height is 20 cm, 8 cm and 30 cm respectively?

Solution

Given, length = 20 cm, breadth = 8 cm, Height = 30 cm

Total surface area of the cereal box = 2(lb + bh + lh)

= 2(20 × 8 + 8 × 30 + 20 × 30)

= 2(160 + 240 + 600)

= 2(1000) = 2000 cm².

Surface Area and Volume of a Right Circular Cylinder

If we fold a rectangular sheet with one side as its axis then it forms a cylinder. It is the curved surface of the cylinder. And if this curved surface is covered by two parallel circular bases then it forms a right circular cylinder.

Curved surface area of a Right circular cylinder	2πrh
Total surface area of a Right circular cylinder	2πr² + 2πrh = 2πr(r + h)
The volume of a Right circular cylinder	πr²h
	r = radius, h = height

Surface Area and Volume of a Right Circular Cylinder

Surface Area and Volume of a Hollow Right Circular Cylinder

If a right circular cylinder is hollow from inside then it has different curved surface and volume.

Curved surface area of a Right circular cylinder	2πh (R + r)
Total surface area of a Right circular cylinder	2πh (R + r) + 2π(R² – r²)
	R = outer radius, r = inner radius, h = height

Surface Area and Volume of a Hollow Right Circular Cylinder

Example

Find the Total surface area of a hollow cylinder whose length is 22 cm and the external radius is 7 cm with 1 cm thickness. (π = 22/7)

Solution

Given, h = 22 cm, R = 7 cm, r = 6 cm (thickness of the wall is 1 cm).

Total surface area of a hollow cylinder = 2πh(R + r) + 2π(R²– r²)

= 2(π) (22) (7+6) + 2(π)(7²– 6²)

= 572 π + 26 π = 598 π

= 1878.67 cm²

Surface Area and Volume of a Right Circular Cone

If we revolve a right-angled triangle about one of its sides by taking other as its axis then the solid shape formed is known as a Right Circular Cone.

Curved surface area of a Right Circular Cone	πrl = πr[√(h² + r²)]
Total surface area of a Right Circular Cone	πr²+ πrl = πr(r + l)
The volume of Right Circular Cone	(1/3) πr²h
	r = radius, h = height, l = slant height

Surface Area and Volume of a Right Circular Cone

Surface Area and Volume of a Sphere

A sphere is a solid shape which is completely round like a ball. It has the same curved and total surface area.

Curved or Lateral surface area of a Sphere	4πr²
Total surface area of a Sphere	4πr²
Volume of a Sphere	(4/3) πr³
	R = radius

Surface Area and Volume of a Sphere

Surface Area and Volume of a Hemisphere

If we cut the sphere in two parts then is said to be a hemisphere.

Curved or Lateral surface area of a Sphere	2πr²
Total surface area of a Sphere	3πr²
Volume of a Sphere	(2/3) πr³
	r = radius

Surface Area and Volume of a Hemisphere

Example

If we have a metal piece of cone shape with volume 523.33 cm³ and we mould it in a sphere then what will be the surface area of that sphere?

Solution

Given, volume of cone = 523.33 cm³

Volume of cone = Volume of Sphere

Volume of sphere = 100 π cm³

125 = r³

r = 5

Surface area of a sphere = 4πr²

= 314.28 cm².

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HERON’S FORMULA | Study

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Perimeter

Perimeter is defined as the outside boundary of any closed shape.

To calculate the perimeter of a given shape we need to add all the sides of the shape.

Example: The perimeter of a rectangle is the sum of its all four sides. The unit of the perimeter is the same as its length.

Perimeter of the Given rectangle = 3 + 7 + 3 + 7 cm

Perimeter of rectangle = 20 cm. (Scroll down to continue …)

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Area

Area of any closed figure is the surface enclosed by the perimeter. Unit of Area is the square of the unit of length.

Area of a triangle

The general formula to find the area of a triangle, if the height is given, is

Area of a Right Angled Triangle

To find the area of a right-angled triangle, we use the formula:

right-angled triangle, we take the two sides having the right angle, one as the base and one as height.

Example: Calculate area of a triangle of the Figure.

Data: base = 3 cm and height = 4 cm

Formula: Area of triangle = 1/2 × 3 × 4= 6 cm²

Result: Area of a triangle of the Figure is 6 cm² .

Remark: If you take base as 4 cm and height as 3 cm then also the area of the triangle will remain the same.

Area of Equilateral Triangle

Equilateral Triangle: Equilateral Triangle is defined as a triangle having three equal sides.

To calculate the area of the Equilateral Triangle ABC,

We calculate the height (altitude), AD by making the median of the triangle.

In the given example, the Height (altitude), AD touches Base of the equilateral triangle at the midpoint of BC, Say point, D.

Here the equilateral triangle ABC has three equal sides, such as:

AB = BC = AC = 10 cm.

Since, midpoint of BC divides the triangle into two right angle triangles.

The height, AD, is calculated using Pythagoras theorem.

According to Pythagoras theorem, AB² = AD² + BD²

On substituting the values we get,

(10)² = AD² + (5)²

AD² = (10)² – (5)²

AD² = 100 – 25 = 75

AD = 5√3

Now we can find the area of the triangle using the formula:

Area of triangle = 1/2 × base × height

On substituting the values we get,

Area of triangle = 1/2 × 10 × 5√3

25√3 cm²

Area of Isosceles Triangle

In the isosceles triangle also we need to find the height of the triangle then calculate the area of the triangle.

Here,

Area of a Triangle — by Heron’s Formula

The formula of the area of a triangle given by herons is called Heron’s Formula.

Heron’s Formula:

where a, b and c are the sides of the triangle and s is the semiperimeter

Generally, this formula is used when the height of the triangle is not possible to find or you can say if the triangle is a scalene triangle.

Here the sides of triangle are

AB = 12 cm

BC = 14 cm

AC = 6 cm

Application of Heron’s Formula in Finding Areas of Quadrilaterals

If we know the sides and one diagonal of the quadrilateral then we can find its area by using the Heron’s formula.

Find the area of the quadrilateral if its sides and the diagonal are given as follows.

Given, the sides of the quadrilateral

AB = 9 cm

BC = 40 cm

DC = 28 cm

AD = 15 cm

Diagonal is AC = 41 cm

Here, ∆ABC is a right angle triangle, so its area will be

Area of Quadrilateral ABCD = Area of ∆ABC + Area of ∆ADC

= 180 cm² + 126 cm²

= 306 cm²

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August 23, 2023
CIRCLES | Study
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Introduction to Circles

There are many objects in our life which are round in shape. A few examples are the clock, dart board, cartwheel, ring, Vehicle wheel, Coins, etc. (Scroll down to continue …)

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Circles

Any closed shape with all points connected at equidistant from the centre forms a Circle.

Any point which is equidistant 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 of 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 straighten it 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 a 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 = 90^o

∠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 to 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 vertices of the quadrilateral come in 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 a sum of 180º, then the quadrilateral will be cyclic.

This is the reverse of the above theorem.
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QUADRILATERALS | Study

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Quadrilateral

Any closed polygon with four sides, four angles and four vertices are called Quadrilateral. It could be regular or irregular. (Sroll down to continute …)

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Quadrilateral

Quadrilateral is a closed figure with four sides.

Characteristics of a quadrilateral

Angle Sum Property of a Quadrilateral:

Qudrilateral is a four sided closed figure.

Sum of all angles of a quadrilateral is 360°.

Types Of Quadrilaterals

Quadrilaterals are broadly classified into three categories as:

(i) Kite

(ii) Trapezium

(ii) Parallelogram

Kite:

Kite

(i) Kite has no parallel sides

(ii) Kite has a pair of equal adjacent sides.

(ii) It is not a parallelogram

Characteristics Of Kite:

Perimeter Of Square

Area Of Kite

Trapezium:

Trapezium is a quadrilateral with the following characteristics:

(i) One pair of opposite sides is parallel to each other.

(ii) The other pair of opposite sides may not be parallel to each other.

Characteristics Of Trapezium

(i) Sum of all angles of a quadrilateral is 360°.

(ii) One pair of opposite sides is parallel to each other.

(iii) The other pair of opposite sides need not be parallel to each other.

Types Of Trapezium:

Quadrilaterals are broadly classified into two categories as:

(i) Isosceles Trapezium.

(ii) Scalene Trapezium.

(i) Right Trapezium.

Isosceles Trapezium:

Isosceles Trapezium is a quadrilateral with the following characteristics:

(i) One pair of opposite sides is parallel to each other.

(ii) The other pair of opposite sides are equal.

(iii) The other pair of opposite sides need not be parallel to each other.

Isosceles Trapezium is a trapezium with the following characteristics:

(i) One pair of opposite sides is parallel to each other.

(ii) The other pair of opposite sides are equal.

(iii) The other pair of opposite sides need not be parallel to each other.

Characteristics Of Isosceles Trapezium

(i) Sum of all angles of a quadrilateral is 360°.

(ii) One pair of opposite sides is parallel to each other.

(iii) The other pair of opposite sides are equal.

(iv) The other pair of opposite sides need not be parallel to each other.

Scalene Trapezium:

Scalene trapezium: Classified by the length of the legs or the measurement of their angles.

Characteristics Of Scalene Trapezium

Right Trapezium:

Right trapezium: Has one pair of parallel sides and one pair of right angles.

Characteristics Of Right Trapezium

Perimeter Of Trapezium

Area Of Trapezium

Parallelogram:

Parallelogram is a quadrilateral with the following characteristics:

(i) Two pairs of opposite sides are parallel to each other.

(ii) Two pairs of opposite sides are equal in length.

Characteristics of a parallelogram

(i) Sum of all angles of a Parallelogram is 360°.

(ii) Two pairs of opposite sides are parallel to each other.

(ii) Two pairs of opposite sides are equal in length.

(ii) Two pairs of opposite angles are equal.

(iii) Diagonals bisect each other.

(iv) Diagonals need not be equal to each other.

(v) Diagonals divide it into two congruent triangles.

Types Of Parallelogram

Parallelograms are broadly classified into three categories as:

(i) Rectangle

(ii) Rhombus

(iii) Square

Perimeter Of Parallelogram

Area Of Parallelogram

Rectangle:

Rectangle is a quadrilateral with the following characteristics:

(i) Two pairs of opposite sides are parallel to each other.

(ii) Two pairs of opposite sides are equal in length.

(iii) All four angles are right angles. (each angle is 90 ^o).

Characteristics Of Rectangle

(i) Sum of all angles of a quadrilateral is 360°.

(ii) Two pairs of opposite sides are parallel to each other.

(ii) Two pairs of opposite sides are equal in length.

(iii) All four angles are right angles. (each angle is 90 ^o).

(iii) Diagonals bisect each other.

(iv) Diagonals are equal to each other.

(v) Diagonals of a rectangle divide it into two congruent triangles.

Conclusions:

Every Rectangle is a Parallelogram. But Every Parallelogram need not to be a Rectangle.

Condition for a rhombus to be a square:

If all four angles of a parallelogram are right angles. (each angle is 90 ^o), the parallelogram becomes a Rectangle.

Perimeter Of Rectangle

Area Of Recatangle

Rhombus:

Rhombus is a quadrilateral with the following characteristics:

(i) Two pairs of opposite sides are parallel to each other.

(ii) All four sides are equal in length.

Characteristics Of Rhombus

(i) Sum of all angles of a quadrilateral is 360°.

(ii) Two pairs of opposite sides are parallel to each other.

(ii) All four sides are equal in length.

(ii) Two pairs of opposite angles are equal.

(iii) Diagonals bisect each other.

(iv) Diagonals need not be equal to each other.

(v) Diagonals divide a Rhombus into two congruent triangles.

Conclusions:

Every Rhombus is a Parallelogram. But Every Parallelogram need not to be a Rhombus.

Condition for a rhombus to be a square:

If all the sides of a parallelogram are equal, the parallelogram becomes a Rhombus.

Perimeter Of Rhombus

Area Of Rhombus

Square:

Square is a quadrilateral with the following characteristics:

(i) Two pairs of opposite sides are parallel to each other.

(ii) All four sides are equal in length.

(iii) All four angles are right angles. (each angle is 90 ^o).

Characteristics Of Square

(i) Sum of all angles of a quadrilateral is 360°.

(ii) Two pairs of opposite sides are parallel to each other.

(iii) All four sides are equal in length.

(iv) All four angles are right angles. (each angle is 90 ^o).

(v) Diagonals bisect each other.

(vi) Diagonals need not be equal to each other.

(vii) Diagonals divide a Rhombus into two congruent triangles.

Conclusions:

Every square is a Rhombus. But Every Rhombus need not to be a square.

Condition for a rhombus to be a square:

If all the angles of a rhombus are right angles (euqal to 90^o), the rhombus becomes a square.

2. Every Square is a prallelogram. But Every prallelogram need not to be a square.

Condition for a prallelogram to be a square:

(i) If all the angles of a parallelogram are right angles (euqal to 90^o), and all the sides of a parallelogram are equal in length, the parallelogram becomes a square.

3. Every Square is a rectangle. But Every Rectangle need not to be a square.

Condition for a Rectangle to be a square:

If all the sides of a Rectangle are equal in length, the Rectangle becomes a square.

If all the sides of a parallelogram are equal, the parallelogram becomes a Rhombus.

Perimeter Of Square

Area Of Square

Important Points To Remember

The diagonals of a parallelogram are equal if and only if it is a rectangle.
If a diagonal of a parallelogram bisects one of the angles of the parallelogram then it also bisects the opposite angle.
In a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.
The angle bisectors of a parallelogram form a rectangle.

Mid Point Theorem

A line segment joining the mid points of any two sides of a triangle is parallel to the third side and length of the line segment is half of the parallel side.

Converse Of Mid Point Theorem

A line through the midpoint of a side of a triangle parallel to another side bisects the third side.

Intercept Theorem

If there are three parallel lines and the intercepts made by them on one transversal are equal then the intercepts on any other transversal are also equal.

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Angle Sum Property of a Quadrilateral

The sum of the four angles of a quadrilateral is 360°

If we draw a diagonal in the quadrilateral, it divides it into two triangles.

And we know the angle sum property of a triangle i.e. the sum of all the three angles of a triangle is 180°.

The sum of angles of ∆ADC = 180°.

The sum of angles of ∆ABC = 180°.

By adding both we get ∠A + ∠B + ∠C + ∠D = 360°

Hence, the sum of the four angles of a quadrilateral is 360°.

Example

Find ∠A and ∠D, if BC∥ AD and ∠B = 52° and ∠C = 60° in the quadrilateral ABCD.

Solution:

Given BC ∥ AD, so ∠A and ∠B are consecutive interior angles.

So ∠A + ∠B = 180° (Sum of consecutive interior angles is 180°).

∠B = 52°

∠A = 180°- 52° = 128°

∠A + ∠B + ∠C + ∠D = 360° (Sum of the four angles of a quadrilateral is 360°).

∠C = 60°

128° + 52° + 60° + ∠D = 360°

∠D = 120°

∴ ∠A = 128° and ∠D = 120 °.

Types of Quadrilaterals

Remark: A square, Rectangle and Rhombus are also a parallelogram.

Properties of a Parallelogram

Theorem 1: When we divide a parallelogram into two parts diagonally then it divides it into two congruent triangles.

∆ABD ≅ ∆CDB

Theorem 2: In a parallelogram, opposite sides will always be equal.

Theorem 3: A quadrilateral will be a parallelogram if each pair of its opposite sides will be equal.

Here, AD = BC and AB = DC

Then ABCD is a parallelogram.

Theorem 4: In a parallelogram, opposite angles are equal.

In ABCD, ∠A = ∠C and ∠B = ∠D

Theorem 5: In a quadrilateral, if each pair of opposite angles is equal, then it is said to be a parallelogram. This is the reverse of Theorem 4.

Theorem 6: The diagonals of a parallelogram bisect each other.

Here, AC and BD are the diagonals of the parallelogram ABCD.

So the bisect each other at the centre.

DE = EB and AE = EC

Theorem 7: When the diagonals of the given quadrilateral bisect each other, then it is a parallelogram.

This is the reverse of the theorem 6.

The Mid-point Theorem

1. If a line segment joins the midpoints of the two sides of the triangle then it will be parallel to the third side of the triangle.

If AB = BC and CD = DE then BD ∥ AE.

2. If a line starts from the midpoint of one line and that line is parallel to the third line then it will intersect the midpoint of the third line.

If D is the midpoint of AB and DE∥ BC then E is the midpoint of AC.

Example

Prove that C is the midpoint of BF if ABFE is a trapezium and AB ∥ EF.D is the midpoint of AE and EF∥ DC.

Solution:

Let BE cut DC at a point G.

Now in ∆AEB, D is the midpoint of AE and DG ∥ AB.

By midpoint theorem, G is the midpoint of EB.

Again in ∆BEF, G is the midpoint of BE and GC∥ EF.

So, by midpoint theorem C is the midpoint of BF.

Hence proved.

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TRIANGLES | Study

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Similar Geometric Figures:

Two geometric figures which are the same in shape, such that one is simply a copy of the other on a smaller scale or a larger scale, are called similar geometric figures.

Two geometric figures are said to be similar if and only if they have the same shape but not necessarily the same size. Two congruent geometric figures are always similar but converse may or may not be true. (Scroll down to read more …)

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Similar Polygons: Two polygons of the same number of sides are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in proportion or their corresponding sides are in the same ratio.

The same ratio of the corresponding sides is referred to as the representative fraction or the scale factor for the polygons.

Similar Triangles :

Two triangles are said to be similar,

if (i) their corresponding angles are equal and

ii) their corresponding sides are in proportion (are in the same ratio).

Basic Proportionality Theorem (or Thales Theorem) : 1

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. Or If a line is drawn parallel to one side of a triangle, intersecting the other two sides in distinct points, the other two sides are divided in the same ratio i .e.. If in ∆ABC, l∥ BC, intersecting in D and E. then

Converse of Basic Proportionality Theorem :

If a line divides any two sides of a triangle in the sameratio, the line is parallel to the third side i.e.

In ∆ABC, if l intersects AB in D and AC in E, such that:

Criteria for Similarity of Triangles:

Two triangles are said to be similar, if

(i) their corresponding angles are equal and (ii) their corresponding sides are in proportion (or are in the same ratio).

2 (i) AA or AAA Similarity Criterion : If two angles of one triangle are equal to two corresponding angles of another triangle, then the triangles are similar. If two angles of one triangle are respectively equal to the two angles of another triangle, then the third angles of the two triangles are necessarily equal, because the sum of three angles of a triangle is always 180 ⁰ .

(ii) SAS Similarity Criterion : If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the two triangles are similar. Or If two sides of a triangle are proportional to two corresponding sides of another triangle and the angles included between them are equal, then the triangles are similar.

iii) SSS Similarity Criterion : If in two triangles, sides of one triangle are proportional (or are in the same ratio) to the sides of the other triangle, then the triangles are similar. If ∆ABC~ ∆PQR by any one similarity criterion, then ∠A=∠P, ∠B=∠Q, ∠C=∠R and

i.e., A and P, B and Q, C and R are the corresponding vertices, also AB and PQ. BC and QR. CA and RP are the corresponding sides. 3 Areas of Similar Triangles: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. – The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. – The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes. – The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding angle bisectors. Pythagoras Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Converse of Pythagoras Theorem : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and similar to each other i .e ..

If in ∆ABC, ∠B=90^0 and BD ⊥ AC, then (i) ∆ADB ~ ∆ABC (ii) ∆BDC ~ ∆ABC (iii) ∆ADB ~ ∆BDC

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August 23, 2023

LINES AND ANGLES | Study

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Basic terms and Definitions

1. Point – A Point is that which has no component. It is represented by a dot.

2. Line – When we join two distinct points then we get a line. A line has no endpoints; it can be extended on both sides infinitely.

3. Line Segment Line – Segment is the part of the line which has two endpoints.

4. Ray – Ray is also a part of the line that has only one endpoint and has no end on the other side.

5. Collinear points: Points lying on the same line are called Collinear Points. Collinear and Non-collinear points

6. Non-collinear points: Points which do not lie on the same line are called Non-Collinear Points. (Scroll to continue …)

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Angles

When two rays begin from the same endpoint then they form an Angle. The two rays are the arms of the angle and the endpoint is the vertex of the angle.

Types of Angles

Complementary and Supplementary Angles

Complementary Angles are the different angles whose sum is 90°.

Complementary Angles are the different angles whose sum is 180°.

Intersecting Lines and Non-intersecting Lines

There are two ways to draw two lines-

1. The lines which cross each other from a particular point are called Intersecting Lines.

2. The lines which never cross each other at any point are called Non-intersecting Lines. These lines are called Parallel Lines and the common length between two lines is the distance between parallel lines.

Pairs of Angles Axioms

1. If a ray stands on a line, then the sum of two adjacent angles formed by that ray is 180°.

This shows that the common arm of the two angles is the ray which is standing on a line and the two adjacent angles are the linear pair of the angles. As the sum of two angles is 180° so these are supplementary angles too.

2. If the sum of two adjacent angles is 180°, then the arms which are not common of the angles form a line.

This is the reverse of the first axiom which says that the opposite is also true.

Vertically opposite Angles Theorem

When two lines intersect each other, then the vertically opposite angles so formed will be equal.

AC and BD are intersecting each other so ∠AOD = ∠BOC and ∠AOB = DOC.

Parallel Lines and a Transversal

If a line passes through two distinct lines and intersects them at distant points then this line is called Transversal Line.

Here line “l” is transversal of line m and n.

Exterior Angles – ∠1, ∠2, ∠7 and ∠8

Interior Angles – ∠3, ∠4, ∠5 and ∠6

Pairs of angles formed when a transversal intersects two lines-

1. Corresponding Angles:

∠ 1 and ∠ 5
∠ 2 and ∠ 6
∠ 4 and ∠ 8
∠ 3 and ∠ 7

2. Alternate Interior Angles:

∠ 4 and ∠ 6
∠ 3 and ∠ 5

3. Alternate Exterior Angles:

∠ 1 and ∠ 7
∠ 2 and ∠ 8

4. Interior Angles on the same side of the transversal:

∠ 4 and ∠ 5
∠ 3 and ∠ 6

Transversal Axioms

1. If a transversal intersects two parallel lines, then

Each pair of corresponding angles will be equal.
Each pair of alternate interior angles will be equal.
Each pair of interior angles on the same side of the transversal will be supplementary.

2. If a transversal intersects two lines in such a way that

Corresponding angles are equal then these two lines will be parallel to each other.
Alternate interior angles are equal then the two lines will be parallel.
Interior angles on the same side of the transversal are supplementary then the two lines will be parallel.

Lines Parallel to the Same Line

If two lines are parallel with a common line then these two lines will also be parallel to each other.

As in the above figure if AB ∥ CD and EF ∥ CD then AB ∥ EF.

Angle Sum Property of a Triangle

1. The sum of the angles of a triangle is 180º.

∠A + ∠B + ∠C = 180°

2. If we produce any side of a triangle, then the exterior angle formed is equal to the sum of the two interior opposite angles.

∠BCD = ∠BAC + ∠ABC.

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INTRODUCTION TO EUCLID’S GEOMETRY | Study

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Introduction to Euclid Geometry The necessity of geometry had been felt from ancient times in different parts of the world.

The practical problems faced by people of ancient civilization had developed this branch of mathematics.

Let us cite few examples.

With floods in the river, the demarcations of land owners on the river-side land were used to wipe out. (Scroll down to continue …)

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In order to redraw the boundaries, the idea of area was introduced, the idea of area was introduced.

The volumes of granaries could be measured by using geometry.

The existence of Egyptian pyramids indicates the use of geometry from olden times.

In Vedic period, there was a manual of geometrical construction, known as Sulbasutra’s.

Different geometrical shapes were constructed as altars to perform various Vedic rites.

The word Geometry originates from the green word ‘Geo’ (earth) and metrein (to measure) Through Geometry was developed and applied from ancient time in various part the world, it was not presented in a systematic manner.

Later in 300 BC, the Egyptian mathematician Euclid, collected all the known work and arranged it in a systematic manner.

‘Elements’ is a classic treatise in geometry which was written by Euclid.

This was the most influential book. The ‘element’ was used as a text book for several years in western Europe.

The ‘elements’ started with 28 definitions, five postulates and five common notions and systematically built the rest of plane and solid geometry.

The geometrical approach given by Euclid is known as Euclid method.

The Euclid method consists of making a small set of assumptions and then proving many other proposition from these assumptions.

The assumptions, made were obvious universal truth. The two types of assumption, made were ‘axioms’ and ‘postulates’.

Euclid’s Definitions Euclid listed 23 definitions in book 1 of the ‘elements’.

We list a few of them: 1) 2) 3) 4) 5) 6) 7) A point is that which has no part A line is a breadth less length The ends of a line are points A straight line is a line which lies evenly with the points on itself. A surface is that which has length and breadth only.

The edges of a surface are lines A plane surface is surface which lies evenly with straight lines on its self. Starting with these definitions, Euclid assumed certain assumptions, known as axioms and postulates.

Euclid’s Axioms Axioms were assumptions which were used throughout mathematics and are not specifically linked to geometry.

Few of Euclid’s axioms are:

1) Things which are equal to the same thing are equal to one another.

2) It equals are added to equals; the wholes are equal.

3) 4) 5) 6) 7) If equals are subtracted from equals, the remainders are equal.

Things which coincide one another are equal to one another.

The whole is greater than the part Things which are double of the same thing are equal to one another.

Things which are half of the same things are equal to one another.

All these axioms refer to magnitude of same kind.

Axiom – 1 can be written as follows: If x = Z and y = Z, then x = y

Axiom – 2 explains the following: If x = y, then x + Z = y + Z According to axiom – 3, If x = y, then x – Z = y – Z Axiom – 4 justifies the principle of superposition that every thing equals itself.

Axiom – 5, gives us the concept of comparison. If x is a part of y, then there is a quantity Z such that x = y + Z or x > y Note that magnitudes of the same kind can be added, subtracted or compared.

Euclid’s Postulates Euclid used the term postulate for the assumptions that were specific to geometry. Euclid’s five postulates are as follows: Postulate 1: A straight line may be drawn from any one point to any other point. Same may be stated as axiom 5.1 Given two distinct points, there is a unique line that passes through them.

Postulate 2: A terminated line can be produced indefinitely. Postulate 3: A circle can be drawn with any centre and any radius. Postulate 4: All right angles are equal to one another. Postulate 5: If a straight line falling on two straight lines makes the interior angle on the same side of it taken together less than two right angles, then two straight lines, if produced indefinitely, meet on that side on which the sum of the angles is less than two right angles. Postulates 1 to Postulates 4 are very simple and obvious and therefore they are taken a ‘self evident truths’. Postulates 5 is complex and it needs to be discussed. Suppose the line XY falls on two lines AB and CD such that ∠1 + ∠2 < 180°, then the lines AB and CD will intersect at a point. In the given figure, they intersect on left side of PQ, if both are produced. Note: In mathematics the words axiom and postulate may be used interchangeably, though they have distinct meaning according to Euclid. System of Consistent Axioms A system of axioms is said to be consistent, if it is impossible to deduce a statement from these axioms, which contradicts any of the given axioms or proposition. Proposition or Theorem The statement or results which were proved by using Euclid’s axioms and postulates are called propositions or Theorems. Theorem: Two distinct lines cannot have more than one point in common. Proof: Given: AB and CD are two lines. To prove: They intersect at one point or they do not intersect. Proof: Suppose the lines AB and CD intersect at two points P and Q. This implies the line AB passes through the points P and Q. Also the line CD passes through the points P and Q. This implies there are two lines which pass through two distinct point P and Q. But we know that one and only one line can pass through two distinct points. This axiom contradicts out assumption that two distinct lines can have more than one point in common. The lines AB and CD cannot pass through two distinct point P and Q. Equivalent Versions of Euclid’s Fifth Postulate The two different version of fifth postulate a) For every line l and for every point P not lying on l, there exist a unique line m passing through P and parallel to l. b) Two distinct intersecting lines cannot be parallel to the same line.

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August 23, 2023
LINEAR EQUATIONS IN TWO VARIABLES | Study

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Linear Equations in Two Variables | Speed Notes

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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

The equation with one variable in it is known as a Linear Equation in One Variable. (Scroll down to continue …)

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The general form for Linear Equation in One Variable is px + q = s, where p, q and s are real numbers and p ≠ 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.

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 –

Graph of the Linear Equation in One Variable

Linear Equation in Two Variables

An equation with two variables is known as a Linear Equation in Two Variables. The general form of the linear equation in two variables is

ax + by + c = 0

where a and b are coefficients and c is the constant. a ≠ 0 and b ≠ 0.

Example

6x + 2y + 5 = 0, etc.

Slope Intercept form

Generally, the linear equation in two variables is written in the slope-intercept form as this is the easiest way to find the slope of the straight line while drawing the graph of it.

The slope-intercept form is y = mx+c

Where m represents the slope of the line.

and c tells the point of intersection of the line with the y-axis.

Remark: If b = 0 i.e. if the equation is y = mx then the line will pass through the origin as the y-intercept is zero.

Solution of a Linear Equation

There is only one solution in the linear equation in one variable but there are infinitely many solutions in the linear equation in two variables.

As there are two variables, the solution will be in the form of an ordered pair, i.e. (x, y).

The pair which satisfies the equation is the solution to that particular equation.

Example:

Find the solution for the equation 2x + y = 7.

Solution:

To calculate the solution of the given equation we will take x = 0

2(0) + y = 7

y = 7

Hence, one solution is (0, 7).

To find another solution we will take y = 0

2x + 0 = 7

x = 3.5

So another solution is (3.5, 0).

Graph of a Linear Equation in Two Variables

To draw the graph of a linear equation in two variables, we need to draw a table to write the solutions of the given equation, and then plot them on the Cartesian plane.

By joining these coordinates, we get the line of that equation.

The coordinates which satisfy the given Equation lie on the line of the equation.

Every point (x, y) on the line is the solution x = a, y = b of the given Equation.

Any point, which does not lie on the line AB, is not a solution of Equation.

Example:

Draw the graph of the equation 3x + 4y = 12.

Solution:

To draw the graph of the equation 3x + 4y = 12, we need to find the solutions of the equation.

Let x = 0

3(0) + 4y = 12

y = 3

Let y = 0

3x + 4(0) = 12

x = 4

Now draw a table to write the solutions.

x 0 4

y 3 0

Now we can draw the graph easily by plotting these points on the Cartesian plane.

Linear Equation in Two Variables

Equations of Lines Parallel to the x-axis and y-axis

When we draw the graph of the linear equation in one variable then it will be a point on the number line.

x – 5 = 0

x = 5

This shows that it has only one solution i.e. x = 5, so it can be plotted on the number line.

But if we treat this equation as the linear equation in two variables then it will have infinitely many solutions and the graph will be a straight line.

x – 5 = 0 or x + (0) y – 5 = 0

This shows that this is the linear equation in two variables where the value of y is always zero. So the line will not touch the y-axis at any point.

x = 5, x = number, then the graph will be the vertical line parallel to the y-axis.

All the points on the line will be the solution of the given equation.

Equations of Lines Parallel to the x-axis and y-axis

Similarly if y = – 3, y = number then the graph will be the horizontal line parallel to the x-axis.

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August 23, 2023
COORDINATE GEOMETRY | Study

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Coordinate Geometry: Cartesian Plane | Speed Notes

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Cartesian System

A plane formed by two number lines, one horizontal

and the other vertical, such that they intersect each

other at their zeroes, and then they form a Cartesian

Plane.

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● The horizontal line is known as the x-axis and the vertical line is known

as the y-axis.

● The point where these two lines intersect each other is called the origin.

It is represented as ‘O’.

● OX and OY are the positive directions as the positive numbers lie in the

right and upward direction.

● Similarly, the left and the downward directions are the negative directions

as all the negative numbers lie there.

Quadrants of the Cartesian Plane The Cartesian plane is divided into four quadrants namely Quadrant I, Quadrant II, Quadrant III, and Quadrant IV anticlockwise from OX.

Coordinates of a Point To write the coordinates of a point we need to follow the following rules. ● Thex – coordinate of a point is marked by drawing perpendicular from the y-axis measured a length of the x-axis .It is also called the Abscissa.

They – coordinate of a point is marked by drawing a perpendicular from the x-axis measured a length of the y-axis .It is also called the Ordinate. ● While writing the coordinates of a point in the coordinate plane, the x – coordinate comes first, and then the y – coordinate. We write the coordinates in brackets. In figure, OB = CA = x coordinate (Abscissa), and CO = AB = y coordinate (Ordinate). We write the coordinate as (x, y).

Remarks:

As the origin, O has zero distance from the x-axis and the y-axis so its abscissa and ordinate are zero. Hence the coordinate of the origin is (0, 0). The relationship between the signs of the coordinates of a point and the quadrant of a point in which it lies.

Plotting a Point in the Plane if its Coordinates are Given

Steps to plot the point (2, 3) on the Cartesian plane:

● First of all, we need to draw the Cartesian plane by drawing the coordinate axes with 1 unit = 1 cm.

● To mark the x coordinates, starting from 0 moves towards the positive x-axis and counts to 2.

● To mark the y coordinate, starting from 2 moves upwards in the positive direction and counts to 3.

● Now this point is the coordinate (2, 3). Likewise, we can plot all the other points, like (-3, 1) and (-1.5,-2.5) in the figure.

Question: Are the coordinates (x, y) = (y, x)? Let x = (-4) and y = (-2) So (x, y) = (- 4, – 2) (y, x) = (- 2, – 4)

Let’s mark these coordinates on the Cartesian plane. You can see that the positions of both the points are different in the Cartesian plane. So, If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y.

Example: Plot the points (6, 4), (- 6,- 4), (- 6, 4) and (6,- 4) on the Cartesian plane.

Solution: Since both numbers 6, 4 are positive the point (6, 4) lies in the first quadrant. For x coordinate, we will move towards the right and count to 6. Then from that point go upward and count to 4. Mark that point as the coordinate (6, 4). Similarly, we can plot all the other three points.

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August 23, 2023

POLYNOMIALS | Study

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Revision Notes on Polynomials

Polynomial

A polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral powers.

Example

1. 4x³ + 3x² + x +3 is a polynomial in variable x.

2. 4x² + 3x^-1 – 4 is not a polynomial as it has negative power.

3. 3x^3/2+ 2x – 3 is not a polynomial. Polynomials

Polynomials are denoted by p (x), q (x), etc.
In the above polynomials, 2x², 3y, and 2 are the terms of the polynomial.
2 and 3 are the coefficients of x² and y, respectively.
x and y are the variables.
2 is the constant term, which has no variable.

Polynomials in One Variable

If there is only one variable in the expression, then this is called the polynomial in one variable.

Example

x³ + x – 4 is polynomial in variable x and is denoted by p(x).
r² + 2 is polynomial in variable r and is denoted by p(r).

Types of polynomials on the basis of the number of terms

Types of polynomials on the basis of the number of degrees

The highest value of the power of the variable in the polynomial is the degree of the polynomial.

Zeros of a Polynomial

If p(x) is a polynomial, then the number ‘a’ will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by equating it to zero.

Example: 1

The given polynomial is p(x) = x – 4

To find the zero of the polynomial, we will equate it to zero.

x – 4 = 0

x = 4

p(4) = x – 4 = 4 – 4 = 0

This shows that if we place 4 in place of x, we get the value of the polynomial as zero. So 4 is the zero of this polynomial. And also, we are getting the value 4 by equating the polynomial with 0.

So 4 is the zero of the polynomial or the root of the polynomial.

The root of the polynomial is basically the x-intercept of the polynomial.

If the polynomial has one root, it will intersect the x-axis at one point only, and if it has two roots, it will intersect at two points, and so on.

Example: 2

Find p (1) for the polynomial p (t) = t² – t + 1

p (1) = (1)² – 1 + 1

= 1 – 1 + 1

= 1

Remainder Theorem

We know the property of division which follows in the basic division, i.e.

Dividend = (Divisor × Quotient) + Remainder

This follows the division of polynomials.

If p(x) and g(x) are two polynomials in which the degrees of p(x) ≥ degree of g(x) and g(x) ≠ 0 are given, then we can get the q(x) and r(x) so that:

P(x) = g(x) q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x).

It says that p(x) divided by g(x), gives q(x) as a quotient and r(x) as a remainder.

Let’s understand it with an example

Division of a Polynomial with a Monomial

We can see that ‘x’ is common in the above polynomial, so we can write it as

Hence, 3x²+ x + 1 and x the factors of 3x³ + x² + x.

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x² – 3x -10 by 2 + x

Step 1: Write the dividend and divisor in descending order, i.e., in the standard form. x² – 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x²/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor by this term of the quotient and subtract it from the dividend.

Step 3: Now the remainder is our new dividend, so we will repeat the process again by dividing the dividend by the divisor.

Step 4: – (5x/x) = – 5

Step 5:

The remainder is zero.

Hence x² – 3x – 10 = (x + 2)(x – 5) + 0

Dividend = (Divisor × Quotient) + Remainder

As we know,

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t – t). q (t) + r = 0

To find the remainder or to check the multiple of the polynomial, we can use the remainder theorem.

Example:

What is the remainder if a⁴ + a³ – 2a² + a + 1 is divided by a – 1.

Solution:

P(x) = a⁴ + a³ – 2a² + a + 1

To find the zero of (a – 1), we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of the polynomial.

Factor Theorem

The factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

(y – t) is a factor of p(y), if p(t) = 0, and
P (t) = 0 if (y – t) is a factor of p (y).

Example: 1

Check whether g(x) = x – 3 is the factor of p(x) = x³– 4x²+ x + 6 using the factor theorem.

Solution:

According to the factor theorem, if x – 3 is the factor of p(x), then p(3) = 0, as the root of x – 3 is 3.

P (3) = (3)³– 4(3)² + (3) + 6

= 27 – 36 + 3 + 6 = 0

Hence, g (x) is the factor of p (x).

Example: 2

Find the value of k, if x – 1 is a factor of p(x) = kx²– √2x + 1

Solution:

As x -1 is the factor, p(1) = 0

Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorise x² –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + bx + c = x² + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x² + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x² + 17x + 5 =6 x² + (2 + 15) x + 5

= 6 x² + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Algebraic Identities

1. (x + y)²= x²+ 2xy + y²

2. (x – y)² = x²– 2xy + y²

3. (x + y) (x – y) = x² – y²

4. (x + a) (x + b) = x² + (a + b)x + ab

5. (x + y + z)²= x²+ y²+ z²+ 2xy + 2yz + 2zx

6. (x + y)³ = x³ + y³ + 3xy(x + y) = x³+ y³ + 3x²y + 3xy²

7. (x – y)³ = x³– y³ – 3xy(x – y) = x³– y³ – 3x²y + 3xy²

8. x³+ y³ = (x + y)(x² – xy + y²)

9. x³ – y³= (x – y)(x² + xy + y²)

10. x³+ y³ + z³– 3xyz = (x + y + z)(x² + y²+ z² – xy – yz – zx) x³ + y³ + z³ = 3xyz if x + y + z = 0

Example: 2

Factorize 8x³ + 27y³ + 36x²y + 54xy²

Solution:

The given expression can be written as

= (2x)³+ (3y)³ + 3(4x²) (3y) + 3(2x) (9y²)

= (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²

= (2x + 3y)³ (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution:

4x² + y² + z² – 4xy – 2yz + 4xz = (2x)²+ (–y)² + (z)² + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]² (Using Identity V)

= (2x – y + z)² = (2x – y + z) (2x – y + z)

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