Text Book Solutions All Class

Q.1. What is the value of (23)2?

Solution: (23)2 = 23×2           [∵ (xm)n = xm.n]

= 26 =2 x 2 x 2 x 2 x 2 x 2 = 64

Q.2. What is the value of (625) 1/4 ?Solution:

Q.3. Which of the following are irrational numbers?

√3, √4, √5, 22/7, π, 0 

Solution:  √3, √5 and π are irrational numbers.

√3: Irrational (cannot be expressed as a fraction of two integers)

√4: Rational (equals 2, which is an integer)

√5: Irrational (cannot be expressed as a fraction of two integers)22/7: Rational (a fraction of two integers)

π: Irrational (cannot be expressed as a fraction of two integers)

0: Rational (can be expressed as 0/1)

Q.4. Which of the following are rational?

2√3, 1, 0, 3.14, π, 22/7

Solution: 1, 0, 3.14 and 22/7 are rational numbers.

2√3: Irrational (product of a rational and an irrational number)

1: Rational (an integer, can be expressed as 1/1)

0: Rational (can be expressed as 0/1)

3.14: Rational (a terminating decimal, can be expressed as 314/100)

π: Irrational (cannot be expressed as a fraction of two integers)

22/7: Rational (a fraction of two integers)

Q.5. Which of the following cube roots is not irrational?

3 √5, 3√6, 3√7, 3√8, 3√9

Solution:  3√8 is not an irrational number.

Q.6. Which three integers are equal to their own cube roots?

Solution: –1, 0 and 1

[∵ 3√-1 = -1, 3√1 = 1 and 3√0 = 0 ]

 Q.7. Is the following number rational or irrational?

0.0769230769230769230…

Solution:The given number is rational.  

The number 0.076923076923076923… is a repeating decimal.

The number is rational because it has a repeating pattern and can be expressed as a fraction of two integers.

Q.8. Is 0.666 …a “non-terminating rational” or “non-terminating recurring rational number”?

Solution: Non-terminating recurring rational number.

0.666… (where the 6 repeats indefinitely) is a repeating decimal.

It is a non-terminating recurring rational number because it has a repeating pattern and can be expressed as a fraction (specifically, 2/3)

.Q.9. Is 1.27 27 27 … non-terminating recurring or non-terminating non-recurring number?

Solution: It is a non-terminating recurring rational number.

1.272727… (where 27 repeats indefinitely) is a repeating decimal.

It is a non-terminating recurring rational number because it has a repeating pattern and can be expressed as a fraction.

Q.10. The decimal representation of √3 is 1.73205807… . Is it non-terminating nonrecurring?

Solution: YesThe decimal representation of √3 does not repeat and does not terminate.

Q.11. Is the product of a rational and an irrational number, rational?

Solution: No. It is an irrational number.

Q.12. Write the smallest non-negative integer. 

Solution: The smallest non-negative integer is ‘0’.

Q.13. Write the smallest positive integer.

Solution: The smallest positive integer is ‘1’.

Q.14. Up to how many decimal places, π and 22/7  are same?

Solution: π and 22/7 are same only up to 2 decimal places.

[∵ 22/7  = 3.142857142… and π =3.141592653589…]

Q.15. Which of the following is irrational?

Solution: √7 is irrational.

Q.16. Which of the following is the value of 0.999… expressed as p/q :

 (i) 9/10

(ii)  1

(iii)   999/1000

Solution:(ii) = 1

Q.1. Which of the following is an irrational number?

(a) 

(b) √3(c) 1/2

(d) 

Ans.An irrational number is a number that cannot be expressed as a fraction of two integers and has an infinite non-repeating decimal representation.

Let’s evaluate the options:

(a) √49/64: This is a rational number because both the numerator and denominator are perfect squares, and their square root can be expressed as a fraction of integers. √49/64 = 7/8.

(b) √3: This is an irrational number because the square root of 3 cannot be expressed as a fraction of integers, and its decimal representation goes on infinitely without repeating.

(c) 1/2: This is a rational number because it can be expressed as a fraction of integers.

(d) -√1/4: This is a rational number because √1/4 = 1/2, and the negative sign only changes the sign of the rational number.

So, the irrational number among the given options is: (b) √3

Q.2. The numberin p/q form is

(a) 267/1000

(b) 26/10

(c) 241/900

(d) 241/999

Ans. (c)

Solution:let x be the p/q form, x =

multiply both side by 100,

100 x =  …(i)

multiply both side by 10

1000 x =  ….(ii)

Subtract (ii) – (i)

1000 x – 100 x = 

900 x = 241

⇒ x = 241/900

Hence, option (c) is correct

Try yourself:

Q3: Every point on the number line represents, which of the following numbers?

A.Natural numbers

B.Irrational number

C.Rational number

D.Real numberExplanation

Ans. Every point on the number line represents

a: Real number as the number line represents all real numbers, which includes natural numbers, whole numbers, integers, rational numbers, and irrational numbers, but not imaginary numbers.

Q.4. The decimal representation of a rational number is either:

(a) Terminating or repeating

(b) Non-terminating and non-repeating

(c) Only terminating

(d) Only repeating

Ans: (a) Terminating or repeating

 A rational number is any number that can be expressed as a fraction $\frac{p}{q}$, where p and q are integers and $q\ne 0$

A terminating decimal is one that has a finite number of digits after the decimal point. 

A repeating decimal is one where a block of digits repeats infinitely.

Q.5. Insert 3 irrational number between 2.6 and 3.8

Ans. 2.6 and 3.8

Irrational numbers are non repeating non – terminating

2.61010010001…..

2.802002000200002……

3.604004000400004…….

Q.6. What is the decimal form of the following no’s.

(a) 18/11

(b) 3/26

(c) 1/17

(d) 2/13

Ans.(a) 18/11 = 1.63636363…

(b) 3/26 = 0.11538461538

(c) 1/17 = 0.05882352941

(d) 2/13 = 0.15384615384

Q.8. Simplify: 

Ans.

Q.9. Rationalise: 

Ans.

Q.10. Find the value of 

Ans.= 5+4 – 4√5 – 5 – 4 –  4√5  = -8√5

Q.11. If ,find the value of a & b.

Ans.Rationalising LHS ∴ a = 11/7 and b = 6/7

Q.12. Evaluate: 

Ans. 

Q.13. Write the value of 

Ans.= 15

Q.14. Express  in p/q form.

Ans.let x be the p/q form,so, x =  10x = 1000x = 1000x – 10x = –  990x = 15555x= 15555/990= 1037/66

Q.15. Insert five rational no’s between 3/5 and 4/5.

Ans.3/5 and 4/530/50 and 40/50∴ pick any five number between 30 and 4031/50,  32/50,  36/50,  37/50,  39/50

Introduction

Every day we come across a lot of information in the form of facts, numerical figures, tables, graphs, etc. as shown below. For example:

• Runs scored by a team,
• Profits made by a company,
• Temperatures recorded in a day of a city,
• Expenditures in various sectors by the government,
• The weather forecast, election results, and so on.

These facts or figures, which are numerical or otherwise, collected with a definite purpose are called data.
Therefore, Data is a collection of facts, such as numbers, words, measurements, observations, etc.

Suppose, the following image shows the performance of the Indian Cricket team in tests in the last 2 years against major test playing nations.

Try yourself:

What is data?

• A.A collection of facts and figures collected for a specific purpose.
• B.Information presented in the form of numerical figures, tables, and graphs.
• C.Observations and measurements recorded for analysis.
• D.All of the above.

Types of data based on the collection of facts
Qualitative data: It is descriptive data. For examples:

• Rajan is thin.
• Suman can run fast.
• The cake is orange in colour.
• She has black hair.
• He is tall.

Quantitative data: It is numerical information. For examples:

• I updated my phone 6 times in a quarter
• 83 people downloaded the latest mobile application
• She has 10 holidays in this year
• 500 people attended the seminar
• 54% of people prefer shopping online instead of going to the mall.

Graphical Representation of Data

A graphical representation is the visual display of data and its statistical results. It is more often and effective than presenting data in tabular form. Bar graphs, Histograms, and frequency polygons are different types of graphical representation, which depend on the nature of the data and the nature of statical results.

We shall study the following graphical representations in this section:-

(A) Bar graphs

(B) Histograms of uniform width, and of varying widths

(C) Frequency polygons

Also read: NCERT Solutions: Statistics (Exercise 12.1)

(A) Bar graphs

Bar graphs are the bars of uniform width that can be drawn horizontally or vertically with equal spacing between them and then the length of each bar represents the given number. Such a method of representing data is called a bar diagram or a bar graph.

We generally use bar graphs to clearly represent categorical data or any ungrouped discrete frequency observations.

Example 1: Considering the modes of transport of 30 students of class 9^th is given below:

In order to draw the bar graph for the data above, we prepare the frequency table as given below.

Now, we can represent this data using a bar graph, by following the steps as shown below:

• First, we draw two axes viz. x–axis and y–axis. Then, we decide what each axis of the graph represents. By convention, the variates being measured goes on the horizontal (x–axis) and the frequency goes on the vertical (y–axis).
• Next, decide on a numeric scale for the frequency axis. This axis represents the frequency in each category by its height. It must start at zero and include the largest frequency.
• Having decided on a range for the frequency axis we need to decide on a suitable number scale to label this axis. This should have sensible values, for example, 0, 1, 2, . . . , or 0, 10, 20 . . . , or other such values as to make sense given the data.
• Draw the axes and label them appropriately. 
• Draw a bar for each category. When drawing the bars it is essential to ensure the following:
  • the width of each bar is the same
  • the bars are separated from each other by equally sized gaps
• Range = Maximum Value – Minimum Value
• Class size =  Range/Number of classes

Using this bar graph, we can easily identify the most popular mode of transport is the metro. Bar graphs provide a simple method of quickly spotting patterns within a discrete data set.

Example 2:  In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained: 

Observe the bar graph given above and answer the following questions :

(i) How many students were born in the month of November ?

(ii) In which month were the maximum number of students born ?

Solution: (i) 4 students were born in the month of November.

(ii) The Maximum number of students were born in the month of August.

Example 3: A family with a monthly income of Rs 20,000 had planned the following expenditures per month under various heads: 

Draw a bar graph for the data above. 

Solution:

(B) Histogram

Histogram was first introduced by Karl Pearson in 1891. Bar charts have their limitations; like they cannot be used to represent continuous data. When dealing with continuous random variables different kinds of graphs are used. This type of graph is called a histogram.
At first sight, a histogram looks similar to bar charts.
However, there are two critical differences:

• The horizontal (x-axis) is a continuous scale. As a result of this, there are no gaps between the bars (unless there are no observations within a class interval).
• The height of the rectangle is only proportional to the frequency of the class if the class intervals are all equal. With histograms, it is the area of the rectangle that is proportional to its frequency.

Example 4: Consider the weights of 20 students of a class 9^th as  given below:

Now, arranging the data in ascending order.
40, 41, 42, 42, 43, 46, 46, 47, 52, 53, 53, 55, 57, 57, 58, 59, 60, 61, 62, 64.
In order to draw the histogram for the data above, we prepare the frequency table as given below.

We can represent this information using histogram, by following steps as shown below:

• Find the maximum frequency and draw the vertical (y–axis) from zero to this value.
• The range of the horizontal (x–axis) needs to include a full range of the class intervals from the frequency table.
• Draw a bar for each group in your frequency table. These should be the same width and touch each other (unless there are no data in one particular class).

Example 5: The following histogram shows the production of food grains over a period of time. 

What is the total production of food grains from 2004 to 2009?
In which periods were the production of food grains the highest and the lowest?
Solution:
The total production of food grains from 2004 to 2009 can be ascertained by adding the heights of the class intervals 2004–2007 and 2007–2009.
∴ Total production of food grains from 2004 to 2009 = 7000 tonnes + 6000 tonnes = 13000 tonnes
It is clear from the histogram that the bar corresponding to the class interval 2004–2007 is the tallest, and that corresponding to the class interval 2009–2010 is the shortest.
So, the production of food grains was the highest in the period 2004–2007 and the lowest in the period 2009–2010.

Try yourself:

Which type of data is descriptive in nature?

• A.Qualitative data
• B.Quantitative data
• C.Both A and B
• D.None of the above

(C) Frequency Polygon

It is a natural extension of the histogram. In frequency polygon rather than drawing bars, each class is represented by one point and these are joined together by straight lines. We draw frequency polygons in a similar way to drawing a histogram.

Example 6: Consider the weights of 20 students of a class 9^th as given below:

Now, arranging the data in ascending order.
40, 41, 42, 42, 43, 46, 46, 47, 52, 53, 53, 55, 57, 57, 58, 59, 60, 61, 62, 64.
In order to draw the frequency polygon for the data above, we prepare the frequency table as given below.

We can then present this information as a frequency polygon, by following the process of the steps shown below:

• Prepare a frequency table.
• Find the maximum frequency and draw the vertical (y–axis) from zero to this value.
• The range of the horizontal (x–axis) needs to include all class intervals from the frequency table.
• Draw bars for each class interval in the frequency table. These bars should be of the same width and are adjacent to each other (unless there are no data in one particular class)
• Connect the midpoints of the top side of each bar by a dotted line as shown below.

Frequency polygons can also be drawn independently without drawing histograms. For this, we require the mid-points of the class-interval. These mid-points of the class intervals are called class marks.
Class-mark = (Upper Limit + Lower Limit ) / 2

Example 7: Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table.

Draw a frequency polygon corresponding to this frequency distribution table. 

Solution:

Example 8: In a city, the weekly observations made in a study on the cost of living index are given in the following table: 

Draw a frequency polygon for the data above.

Solution:

Summary of Statistics

Introduction

Understanding Surface Area and Volume

• Surface area is like the total paper to cover a 3D object.
• Volume is the space it occupies.
• Ever wondered how we measure the wrapping needed for a gift or the space inside a box?

Wrapping Paper

• It’s how we quantify the outside and inside of shapes. So, how do we calculate these for different objects? 
• Let’s explore the answers by unravelling the mystery of surface area and delving into the roominess of volume!
• Suppose, we cut out a circle shape from a cardboard sheet.
• Then we cut many such circles identical to the first one and then pile them up in a single column. [Shown in the figure]. 
• Then the shape we obtained will be a 3-dimensional shape.
• 
• By this process, we shall obtain some solid figures such as a cuboid and cube as shown below.
• All solid objects occupy some space and have three dimensions – length, breadth, and height or depth. 
• Three-dimensional (3-d)shapes have four parts that set them apart from 2-d shapes viz. faces, vertices, edges, and volume.
• Some real-life examples which resemble solid shapes as shown below. 

Try yourself:

What is the difference between surface area and volume?

• A.Surface area is the total paper to cover a 3D object, while volume is the space it occupies.
• B.Surface area is the space inside a 3D object, while volume is the total paper to cover it.
• C.Surface area is the total space occupied by a 3D object, while volume is the space it occupies.
• D.Surface area is the space outside a 3D object, while volume is the total space occupied by it.

Surface Area of Right Circular Cones

In this chapter, when we refer to a ‘cone’, we mean a ‘right circular cone’. These cones are three-dimensional shapes with a circular base that narrows down to a single point known as the apex or vertex. A familiar example is an ice cream cone.

The curved surface of a right circular cone is called the lateral surface, and the distance from the apex to the base is known as the height. It is important to grasp the properties of right circular cones to calculate their surface area and volume, which rely on their specific geometric features.

Activity: Construction of a Cone

To grasp how a right circular cone is formed, follow these steps:

1. Cut out a right-angled triangle ABC, with the right angle at B.
2. Attach a thick string along one of the perpendicular sides, for example, AB.
3. Rotate the triangle around the string. Notice the shape that forms — a cone.
4. This shape is known as a right circular cone. Here, point A is the vertex, AB is the height, BC is the radius, and AC is the slant height of the cone.

Curved Surface Area of a Cone

Definition and Formula

The curved surface area (CSA) of a cone can be found using this formula:

• Curved Surface Area (CSA) = πrl

In this formula, r is the radius of the base, l is the slant height, and π can be approximated as 3.14 or 22/7.

Additional Information

• From the cone’s cross-section, the relationship l² = r² + h² can be derived, where h is the height of the cone.
• If the cone’s base is closed, a circular piece of paper with an area of πr² is needed, which adds to the total surface area.

Total Surface Area of a Cone = πrl + πr² = πr(l + r)

Example: Find the curved surface area of a right circular cone with a slant height of 10 cm and base radius of 7 cm.
Solution: Curved surface area = πrl = 22/7 × 7 × 10 = 220 cm²

Total Surface Area of a Cone

Definition

The total surface area of a cone is the total of its curved surface area (CSA) and the area of its base. The formulas are as follows:

Curved Surface Area (CSA)

• CSA = πrl
• where r is the radius of the base and l is the slant height of the cone.

Area of the Base

• Area of base = πr²

Total Surface Area

• Total Surface Area = CSA + Area of base = πrl + πr² = πr(l + r)

Example: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14).
Solution:

• Here, h = 16 cm and r = 12 cm.
• Using the formula l² = h² + r², we find:
• l = √(16² + 12²) = √(256 + 144) = √400 = 20 cm
• Curved surface area = πrl = 3.14 × 12 × 20 = 753.6 cm²
• Total surface area = πrl + πr² = 753.6 + 3.14 × 12 × 12 = 753.6 + 452.16 = 1205.76 cm²

Also read: Short & Long Question Answer: Surface Areas and Volume

Surface Area of Sphere

Spheres are completely round three-dimensional shapes, similar to balls or globes.

Every point on a sphere’s surface is the same distance from its centre.

Examples include:

• A basketball
• The nearly spherical shape of the Earth

Spheres do not have edges or corners.

Surface Area Calculation

The surface area of a sphere is calculated as:

Surface Area=4��Surface Area=4πr²

Here, �r is the sphere’s radius.

Example: Find the surface area of a tennis ball of radius 14 cm.

Clearly, the tennis ball is in the form of a sphere.
Here, the radius of the sphere is 14 cm.
We know that,
Surface Area of the sphere =4πr².
Therefore, the Surface Area of a tennis ball =4πr².
= 4 × 22/7 × (14 cm)².
= 4 × 22/7 × 14cm × 14 cm.
= 4 × 22 × 2cm × 14 cm.
= 2464 cm².
Hence, the Surface area of the tennis ball is 2464 cm².

Note: A sphere doesn’t have separate curved and total surface areas because its entire surface is curved. In other words, there are no flat or planar sections on a sphere. The term “surface area” for a sphere typically refers to the total surface area, which includes both the curved surface area and the area of the sphere’s base (which is also curved). 

Surface Area of Hemisphere

A hemisphere, which is half of a sphere, has a surface area calculated as 3πr².

Surface Area of Hemisphere= 3/2×4πr²=3πr²

• Imagine a sphere cut directly in half with a plane through its centre.
• This creates two equal parts, known as a hemisphere (since ‘hemi’ means ‘half’). A hemisphere consists of a curved face and a flat base.
• The curved surface area of a hemisphere is half of the sphere’s surface area, which is 1/2 of 4πr².

Curved Surface Area of a Hemisphere = 2πr²

Here, r represents the radius of the sphere from which the hemisphere is derived. The total surface area of the hemisphere, combining both faces, is calculated as 2πr² + πr². Thus, the Total Surface Area of a Hemisphere = 3πr².

Example: Calculate the curved surface area and total surface area of a hemisphere with a radius of 7 cm.

Solution: Consider a half slice of watermelon, shaped like a hemisphere with a radius of 7 cm. The curved surface area is calculated as follows:

Curved Surface Area of Hemisphere = 2πr² = 2 × (22/7) × (7 cm)² = 2 × (22/7) × 49 cm² = 308 cm².

Thus, the Curved Surface area of the watermelon slice is 308 cm².

Furthermore, the total surface area of a hemisphere is computed as:

Total Surface Area of Hemisphere = 3πr² = 3 × (22/7) × (7 cm)² = 3 × (22/7) × 49 cm² = 462 cm².

Therefore, the total surface area of the half slice of watermelon is 462 cm².

What is Volume?

Volume refers to the amount of space occupied by a three-dimensional object. It is a measure of how much “stuff” or substance an object can hold. The concept of volume is often applied to various geometric shapes, such as cubes, spheres, cylinders, and cones.

Volume of a Cone

The volume (�V) of a cone is calculated using the formula:

�=13��2ℎV= 1/3πr²h

Here, �r is the base radius, and ℎh is the height of the cone.

Example:

Given the height (ℎh) and slant height (�l) of a cone as 21 cm and 28 cm respectively, find the volume.

Solution:  Slant height (l) = 28cm; Height of cone (h) = 21cm  ;    Let radius of cone = r cm  
we know that, 

Volume of a Sphere

The volume (V) of a sphere can be understood by thinking about how much water the sphere pushes aside when it is put in water. This basic idea is very important. The volume of a sphere can be calculated using the formula:

V = 4/3 × π × r³

In this formula, r stands for the radius of the sphere.

Example: Find the volume of a sphere with a radius of 11.2 cm.
Solution:  To calculate the volume of a sphere with a radius of 11.2 cm, we can use the formula:

Try yourself:What is the formula for calculating the curved surface area of a cone?

• A.CSA = πr^2
• B.CSA = πrl
• C.CSA = 2πr
• D.CSA = 2πrl

Volume of a Hemisphere

The volume of a hemisphere is half of a sphere.

You can calculate the volume of a hemisphere using the formula:

Volume of a Hemisphere = (2/3)πr³

In this formula:

• r is the radius of the hemisphere, and
• π is a mathematical constant, roughly equal to 3.14159.

The volume of a sphere is given by Volume of a Sphere = (4/3)πr³, where r is the radius of the sphere.

Since a hemisphere is half of a sphere, its volume is half the volume of a sphere.

Example: A dome of a building is shaped like a hemisphere. The cost to whitewash it from the inside was Rs. 4989.60. If the cost to whitewash is Rs. 20 per square metre, calculate:
(i) The inside surface area of the dome
(ii) The volume of the air inside the dome
Solution: (i) The cost of whitewashing the dome from the inside is Rs. 4989.60.

The cost to whitewash 1m² is Rs. 20.

The curved surface area of the inner side of the dome is:

Curved Surface Area = Total Cost / Cost per m² = 4989.60 / 20 = 249.48 m²

(ii) Let the inner radius of the dome be r.

The curved surface area of the inner side of the dome is 249.48 m² (from (i)).

The formula for the curved surface area (CSA) of a hemisphere is:

CSA = 2πr²

So, we have:

2πr = 249.48

2 × (22/7) × r² = 249.48

r² = (249.48 × 7) / (2 × 22)

r² = 39.69

r = 6.3

Thus, the radius is 6.3 m.

The volume of air inside the given dome = Volume of hemispherical dome

Using the formula, the volume of the hemisphere is:

Volume = (2/3)πr³

= (2/3) × (22/7) × 6.3 × 6.3 × 6.3

= 523.9 (approx.)

The volume of air inside the dome is approximately 523.9 m³.

Introduction

Today, we’re diving into the fascinating world of geometry with Heron’s Formula. Have you ever wanted to calculate the area of a triangle without knowing its height? 

For instance, consider a triangular park with sides measuring 40 m, 32 m, and 24 m. If we were to use the conventional formula for area, ½ x base x height. 

We would need to know the height, which we don’t have. This is where Heron’s Formula comes in, allowing us to find the area of a triangle using just the lengths of its three sides. 

In this lesson, we will explore how to use Heron’s Formula step-by-step and apply it through engaging examples. So, get ready to unlock the secrets of triangles and discover the beauty of geometry!

Area of a Triangle — by Heron’s Formula 

Heron, a mathematician born around 10 AD, made significant contributions to applied mathematics. His works covered various mathematical and physical subjects. 

In his geometrical works, Heron derived the famous formula for the area of a triangle based on its three sides. This formula is now known as Heron’s formula or Hero’s formula

Area=�(�−�)(�−�)(�−�)

Here, �a, �b, and �c are the sides of the triangle, and �s  is the semi-perimeter i.e sum of all-side divided by 2

s= (a+b+c) /2  

�=�+�Application of Heron’s Formula

Let’s apply Heron’s formula to find the area of a triangular park with sides 40 cm, 32 cm, and 24 cm:

�=40+32+242=48

Let us take a = 40 cm, b = 24 cm, c = 32 cm,

Semi perimeter of the triangle (s) = (a + b + c)/2 

s = (40 + 32 + 24)/2 = 48 cm

�−�=48−40=8s −a = 48 − 40 = 8 cm�−�=48−24=24

s −b = 48 − 24 = 24 cm �−�=48−32=16

s −c = 48 − 32 = 16 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)]

Area=48×8×24×16=3842=384 m2

This matches the area calculated using the traditional method:

To ascertain the park’s area, the application of the formula 12×32×24½ × 32 × 24 square meters yields 384 cm². 

Try yourself:What is the formula for calculating the area of a triangle using Heron’s formula?

• A.Area = (a + b + c)/2
• B.Area = √(s(s-a)(s-b)(s-c))
• C.Area = 1/2 * base * height
• D.Area = (1/2) * a * b * sin(C)

Verification and Examples

Now, let’s verify Heron’s formula by applying it to other triangles:

Equilateral triangle (side =10 cm ) 

s= (a+b+c) /2  

=> (10+10+10) /2 

=> 30/2 =15 

=> s= 15 

replacing all values in the above area formulae we get , 

Additional Examples:

Example 1:

Given sides of triangle 8 cm, and 11 cm, and a perimeter of 32 cm, the area is calculated using Heron’s formula:

Area=16×8×5×3=30 cm2

Example 2:

How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 44 cm.

Solution: 

According to the figure,

AC = BD = 44cm, AO = 44/2 = 22cm, BO = 44/2 = 22cm

From ΔAOB,

AB² = AO² + BO²

⇒ AB² = 22² + 22²

⇒ AB² = 2 × 22²

⇒ AB = 22√2 cm

Area of square ABCD = (Side)²

= (22√2)²

= 968 cm²

Area of each triangle (I, II, III, IV) = Area of square /4

= 968 /4

= 242 cm²

To find area of lower triangle,

Let a = 20, b = 20, c = 14

s = (a + b + c)/2

⇒ s = (20 + 20 + 14)/2 = 54/2 = 27.

Area of the triangle = √[s(s-a)(s-b)(s-c)]

= √[27(27-20)(27-20)(27-14)]

= √[27×7×7×13]

= 131.14 cm²

Therefore, We get,

Area of Red = Area of IV

= 242 cm²

Area of Yellow = Area of I + Area of II

= 242 + 242

= 484 cm²

Area of Green = Area of III + Area of the lower triangle

= 242 + 131.14

= 373.14 cm²Area=125×5×45×75=15×30 m2=450 m2

Example 3:

A triangular plot has sides in the ratio 3:5:7, and its perimeter is 300 m. The area is:

Area=150×90×50×10=15003 m2

These examples illustrate Heron’s formula as a powerful tool for finding triangle areas without relying on height.

Try yourself:What is the area of a triangle with sides measuring 15 cm, 18 cm, and 24 cm?

• A.120 cmExplanation 
• Correct Answer
• B.90 cm
• C.180 cm
• D.60 c

Introduction to Circles

A circle is a unique figure; it is everywhere around us. We see the dials of clocks, buttons of shirts, coins, wheels of a vehicle, etc. All these are in the shape of a circle.

Terms related to circles
1. Chord: The chord of a circle is a straight line segment whose endpoints lie on the circle.

2. Diameter: The chord, which passes through the center of the circle is called a diameter of the circle. Diameter is the longest chord and all diameters have the same length, which is equal to two times the radius of the circle.

The length of the complete circle is called its circumference.

Try yourself:What is the theorem that states the sum of either pair of opposite angles of a cyclic quadrilateral?

• A.The perpendicular from the center of a circle to its chord
• B.Chords equidistant from the center of a circle are equal in length
• C.The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the circle
• D.The sum of either pair of opposite angles of a cyclic quadrilateral is 180°

3. Arc: The arc of a circle is a portion of the circumference of a circle.Or

A piece of a circle between two points is also called an arc.

Two points lying on the circle define two arcs: The shorter one is called a minor arc and the longer one is called a major arc.

The minor arc AB is also denoted by  and the major arc AB by  where D is some point on the arc between A and B. When A and B are ends of a diameter, then both arcs are equal and each is called a semicircle.

4. Segment: The region between a chord and either of its arc is called a segment of the circle. There are two types of segments also: which are the major segment and the minor segment.

Angle Subtended by a Chord at a Point

Theorem 1: Equal chords of circle subtend equal angles at the centre.

Given: 

A circle with centre H.

Two chords KL and JI are equal

To Prove: ∠KHL = ∠JHI
Proof:
We are given two chords KL and JI. We need to prove that ∠KHL = ∠JHI.
In triangles KHL and JHI,
HK = HJ —- radii of the same circle 
HL = HI—-  radii of the same circle 
KL = JI – given
So, ∆ KHL≅ ∆ JHI,

Thus, ∠KHL = ∠JHI  —- by CPCT. 
Hence, proved.

Theorem 2: If the angles subtended by two chords at the centre are equal, then the two chords are equal

Given: 

A circle with centre O.

∠COD = ∠AOB  are equal

To Prove: AB=CD

Proof: 

We are given two chords AB and CD. We need to prove that two chords AB and CD are equal

∠AOB = ∠COD (Vertically opposite angles ) ……………(1)

OA = OB = OC= OD (Radii of the same circle) ……………(2)

From eq. 1 and 2, we get;

∆AOB ≅ ∆COD (SAS Axiom of congruency)

Since, 

OA = OB = OC= OD

AB = CD ……… (By CPCT)

Also read: PPT: Circles

Perpendicular from the Centre to a Chord

Theorem 3: The perpendicular from the centre of a circle to a chord bisects the chord

Given:  A circle with centre O.

PQ is a chord such that OM is perpendicular to PQ

To Prove: OM bisects chord PQ i.e. PM=MQ

Constructions: Join O to Q and O to P. 
Proof: Given, in ∆QMO and ∆PMO,
∠OMP = ∠OMQ = 90° (OM ⊥ PQ) ………(1)
OP = OQ (Radii of the circle) ……….(2)
OM = OM (Common side) ………….(3)
From eq. (1), (2) and (3), we get;
∆QMO ≅ ∆PMO (R.H.S Axiom of congruency)
Hence, PM=MQ (By CPCT)

Theorem 4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

Given: 

A circle with centre O.

OM bisects chord PQ i.e. PM=MQ

To Prove: PQ is a chord such that OM is perpendicular to PQ i.e  ∠OMQ = 90°.

Constructions: Draw PQ be the chord of a circle and OM be the line from the centre that bisects the chord such that M is the mid point of the chord

Also, Join O to Q and O to P. 
Proof: 
In triangles ΔPMO and ΔQMO

PM = MQ (given, OM bisects PQ)
OP = OQ (radius of the same circle)
OM = OM (common side of both the triangles)
So, ∆PMO≅ ∆QMO
Therefore, ∠OMQ = ∠OMP —(CPCT)—–(i)
but ∠OMQ +∠OMP = 180 ° — linear pair 
Substituting equation (i) in above equation 
∠OMQ + ∠OMQ = 180 °
Therefore, 

∠OMQ = 90 °

This gives, angles ∠OMQ and ∠OMP as 90°
Hence proved 

Equal Chords and Distance from the Centre

If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal. Similar to the case of chords, equal arcs also subtend equal angles at the centre.

Theorem 5: Equal chords of a circle are equidistant (equal distance) from the centre of the circle. 

Given: 

A circle with centre O.

AB and CD are two equal chords of a circle i.e. AB=CD and OM and ON are perpendiculars to AB and CD respectively.

Constructions: Join O to B and O to D
Draw perpendicular bisector of both chords from center O (OM ⊥ AB and ON ⊥ CD) .
To Prove: OM=ON

Proof: 

Since AB = CD

BM = 1/2 AB (Perpendicular to a chord bisects it) ……..(1) 

DN = 1/2 CD (Perpendicular to a chord bisects it) ……..(2) 
Therefore, BM = DN 

In ∆OMB and ∆OND

BM = DN (proved as above) 

OB = OD (Radii of the same circle)

∠OMB = ∠OND = 90° (OM ⊥ AB and ON ⊥ CD) 

∆OMB ≅ ∆OND ( By R.H.S Axiom of Congruency) 

OM = ON ( By CPCT)
Hence proved 

Theorem 6: Chords of a circle, which are at equal distances from the centre are equal in length

Given: 

A circle with centre O.

AB and CD are at equal distance from the circle, OM and ON are perpendiculars to AB and CD respectively.

OM=ON

To Prove: AB=CD

Proof: In ∆OMB and ∆OND, 
OM = ON ………….(1) 

∠OMB = ∠OND = 90° ………..(2) 

OB = OD (Radii of the same circle) ………..(3) 

Therefore, from eq. 1, 2 and 3, we get; 

∆OMB ≅ ∆OND (By R.H.S Axiom of Congruency) 

BM = DN ( By CPCT) 

1/2 AB = 1/2 CD (Perpendicular from center bisects the chord) 

Therefore, AB = CD
Hence Proved 

Theorem 7: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given: An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle. 

To Prove :∠POQ = 2∠PAQ.

Proof:

Let’s consider three cases,

• Arc PQ is major arc.
• Arc PQ is minor arc.
• Arc PQ is semi-circle.

Let’s join AO and extend it to B.
In all three cases, ∠BOQ = ∠OAQ + ∠OQA. (Exterior angle of a triangle is equal to the sum of the two interior opposite angles).
Also in triangle ΔOAQ,
OA = OQ (Radii of Circle)
Therefore, ∠ OAQ = ∠ OQA
this gives, ∠ BOQ = 2∠OAQ —(i)
∠ BOP = 2∠OAP—-(ii)
from (i) and (ii) we get,
∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ)
∠POQ = 2 ∠PAQ
Hence Proved 
For the case (iii), where PQ is the major arc, (3) is replaced by
Reflex angle POQ = 2∠PAQ

Example 2: What is the value of ∠ABC?

Solution: According to the above theorem (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle), ∠AOC = 2 ∠ABC
Therefore, ∠ABC = 60°/2 = 30° line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e., they are concyclic).

Some other properties

• Angles in the same segment of a circle are equal.
• If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

Theorem 8: Angles in the same segment of a circle are equal. 

Given: A circle with centre O

Points P and Q on this circle subtends ∠PAQ = ∠PBQ at points A and B respectively. 

To Prove : ∠PAQ = ∠PBQ

Proof: Let P and Q be any two points on a circle to form a chord PQ, A and C any other points on the remaining part of the circle and O be the centre of the circle. Then,

∠POQ = 2∠PAQ  …… (i)

And ∠POQ = 2∠PBQ  ……. (ii) 

From above equations, we get

2∠PAQ = 2∠PBQ

Therefore, ∠PAQ = ∠PBQ
Hence Proved 

Theorem 9: If a line segment joining two points subtend equal angles at two other points lying on the same side of the line containing the line segment the four points lie on a circle.

Given: 

AB is a line segment, which subtends equal angles at two points C and D. i.e., ∠ACB = ∠ADB. 

To Prove: 

The points A, B, C and D lie on a circle.

Proof: 

Let us draw a circle through the points A, C and B.

Suppose it does not pass through the point D.

Then it will intersect AD (or extended AD) at a point, say E (or E’).

If points A,C,E and B lie on a circle,

∠ACB = ∠AEB [∴ Angles in the same segment of circle are equal]

But it is given that ∠ACB = ∠ADB

Therefore, ∠AEB = ∠ADB

This is possible only when E coincides with D. [As otherwise ∠AEB >∠ADB]

Similarly, E’ should also coincide with D. So A, B, C and D are concyclic. 

Hence Proved.

Cyclic Quadrilaterals

A quadrilateral is called cyclic if all the four vertices of it lie on a circle.

They are also called inscribed quadrilaterals.

Theorem 10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

Given: PQRS is a cyclic quadrilateral with centre O.

To Prove: 

∠PSR + ∠PQR = 180º.

∠SPQ + ∠QRS =180º.
Proof:

For chord AB, angles in same segment are equal.
∠5=∠8
Similarly, in chords BC, CD, and AD
∠1=∠6
∠2=∠4
∠7=∠3
Angle sum property of quadrilateral gives
∠A+∠B+∠C+∠D=360
∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360
2(∠1+∠2+∠7+∠8)=360
∠1+∠2+∠7+∠8=180
∠BAD+∠BCD=180
Similarly we can prove that,
∠ABC+∠ADC=180

Example 3: In the figure below, BC is the diameter of the circle, ED is a chord equal to the radius of the circle. BE and CD when extended intersect at a point F. Prove that ∠BFC = 60°.
Solution: In the figure, join AE, AD and EC. Triangle AED is an equilateral triangle.
Therefore, ∠EAD = 60°. Now, ∠ECD becomes 30°.
We know that ∠BEC = 90°.
So, by the property of exterior angles of triangle,
∠BEC = ∠ECD + ∠BFC,
90° = 30° + ∠BFC
⇒ 60° = ∠BFC
Hence, Proved.

Theorem 11: If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

Given: ABPQ is a quadrilateral, such that ∠ ABP + ∠ AQP =180 degree and ∠ QAB + ∠ QPB = 180 degree

To prove: The points A, B, P and Q lie on the circumference of a circle. 

Proof: Assume that point P does not lie on a circle drawn through points A, B and Q. 

Let the circle cut QP at point R. Join BR. ∠ QAB + ∠ QRB = 180 degree [given,sum of a pair of opposite angles of a quadrilateral is 180º]

∠ QAB + ∠ QPB = 180 degree [given]

∴ ∠ QRB = ∠ QPB 

But this cannot be true since ∠ QRB = ∠ QPB + ∠ RBP (exterior angle of the triangle) 

∴ Our assumption that the circle does not pass through P is incorrect and A, B, P and Q lie on the circumference of a circle.

∴ ABPQ is a cyclic quadrilateral.

What are  Quadrilaterals?

Quadrilaterals are shapes with four sides.
QuadrilateralsImagine the following examples:

• Square: All four sides are equal, and all angles are right angles.
• Rectangle: Opposite sides are equal, and all angles are right angles.
• Parallelogram: Opposite sides are parallel and equal in length.
• Rhombus: All four sides are equal, but angles are not necessarily right angles.
• Examples of Quadrilaterals in Daily Life: Windows, Blackboard, Study table top, Computer screens, Mobile phone screens, Pages in a book

Definition

A quadrilateral is a closed, two-dimensional figure formed by four line segments. For instance, in quadrilateral $ABCD$, the line segments are $AB$, $BC$, $CD$, and $DA$.

The term “quadrilateral” comes from Latin, where “quad” means “four” and “lateral” means “side.”

• Sides: Four segments (e.g., $AB$, $BC$, $CD$, $DA$)
• Angles: Four angles (e.g., $\angle ABC$, $\angle BCD$, $\angle CDA$, $\angle DAB$)
• Vertices: Four points (e.g., $A$, $B$, $C$, $D$)
• Diagonals: Two segments that connect opposite vertices (e.g., $AC$ and $BD$)

Properties of a Parallelogram

In geometry, a quadrilateral is a four-sided polygon with four angles and four vertices. A specific type of quadrilateral is a parallelogram, characterized by having both pairs of opposite sides parallel. Let’s explore some properties of parallelograms.

Theorem 1: A diagonal of a parallelogram divides it into two congruent triangles.

Proof: Consider parallelogram ABCD with diagonal AC. Diagonal AC divides the parallelogram into triangles ∆ABC and ∆CDA. By the alternate interior angles, we have ∠BCA = ∠DAC and ∠BAC = ∠DCA. Also, AC = AC (common side). 
Therefore, by the ASA rule, ∆ABC ≅ ∆CDA, and the diagonal AC divides the parallelogram into congruent triangles.

Example: In the figure, quadrilateral ABCD is a rectangle in which BD is diagonal. Show that ∆ ABD ≅ ∆ CDB.

Given: ABCD is a rectangle in which BD is diagonal.
To prove: ∆ ABD ≅ ∆ CDB.
Proof: Quadrilateral ABCD is a rectangle. Therefore, ABCD is also a parallelogram.
Since a diagonal of a parallelogram divides it into two congruent triangles.
Hence, ∆ ABD ≅ ∆ CDB.

Try yourself:

Which of the following shapes is NOT a quadrilateral?

• A.Square
• B.Triangle
• C.Rhombus
• D.Parallelogram

Theorem 2: In a parallelogram, opposite sides are equal.

Proof: Measure the opposite sides AB and DC of parallelogram ABCD. You will find AB = DC. 

In ΔABC and ΔCDA

AC=AC [Common/transversal]

∠BCA=∠DAC [alternate angles]

∠BAC=∠DCA [alternate angles]

ΔABC≅ΔCDA  [ASA rule]

Hence, AB=DC and AD=BC [ C.P.C.T.C]

Theorem 3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given: ABCD is a parallelogram.
To prove: AB = CD and AD = CB.
Construction: Join BD.
Proof: 
Since ABCD is a parallelogram. Therefore, AB ∥ DC and BC ∥  AD.
Now, AB ∥ DC and transversal BD intersects them at B and D respectively.
∴ ∠ ABD = ∠ CDB       ……………… (I) [Alternate interior angles]
Again, BC ∥ AD and transversal BD intersects them at B and D respectively.
∴ ∠ ADB = ∠ DBC      ……………… (II) [Alternate interior angles]
Now, in ∆ ABD and ∆ BDC, we have
∠ ABD = ∠ CDB                             [From (I)]
BD = DB                                      [Common side]
∠ ADB = ∠ DBC                              [From (II)]
Therefore, ∆ ABD ≅ ∆ CDB (By ASA-criterion of congruence)
By using corresponding parts of congruent triangles
⇒ AB = CD and AD = CB.

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

Proof: Measure the angles of a parallelogram; you will find that each pair of opposite angles is equal.

In parallelogram ABCD

AB‖CD; and AC is the transversal

Hence, ∠1=∠3….(1) (alternate interior angles)

BC‖DA; and AC is the transversal

Hence, ∠2=∠4….(2) (alternate interior angles)

Adding (1) and (2)

∠1+∠2=∠3+∠4

∠BAD=∠BCD

Similarly,

∠ADC=∠ABC

Theorem 5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Given: In the quadrilateral, the opposite angles are equal. Let the quadrilateral be $ABCD$, where $\angle A=\angle C$ and $\angle B=\angle D$.

1. Verification of Parallel Sides:
2. Since $\angle A+\angle B=180<sup\; style="box-sizing:\; border-box;\; font-family:\; Lato,\; sans-serif\; !important;\; letter-spacing:\; inherit;\; position:\; relative;\; font-size:\; 12px\; !important;\; line-height:\; 2\; !important;\; vertical-align:\; baseline;\; top:\; -0.5em;\; text-align:\; inherit;\; color:\; rgb(0,\; 0,\; 0)\; !important;\; word-break:\; break-word;">\circ </sup>$ (sum of angles in a quadrilateral equals $360<sup\; style="box-sizing:\; border-box;\; font-family:\; Lato,\; sans-serif\; !important;\; letter-spacing:\; inherit;\; position:\; relative;\; font-size:\; 12px\; !important;\; line-height:\; 2\; !important;\; vertical-align:\; baseline;\; top:\; -0.5em;\; text-align:\; inherit;\; color:\; rgb(0,\; 0,\; 0)\; !important;\; word-break:\; break-word;">\circ </sup>$, and opposite pairs add up), the adjacent angles are supplementary.
3. This implies that $AB\parallel CD$ and $AD\parallel BC$ due to the converse of the consecutive angles property.

Conclusion: If both pairs of opposite sides of a quadrilateral are parallel, it is a parallelogram.

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

Proof: Draw diagonals AC and BD of parallelogram ABCD. Measure the lengths of OA, OB, OC, and OD. You will observe that OA = OC and OB = OD, or O is the midpoint of both diagonals. 

In  ΔAOB and ΔCOD,

∠3=∠5 [alternate interior angles]

∠1=∠2  [vertically opposite angles]

AB=CD [opp. Sides of parallelogram]

ΔAOB≅ΔCOD [AAS rule]

OB=OD and OA=OC [C.P.C.T]

Hence, proved

Conversely,

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

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

Example 1: Show that each angle of a rectangle is a right angle.

Solution: Consider rectangle ABCD with ∠A = 90°. By proving that ∠B = ∠C = ∠D = 90°, it establishes that each angle of a rectangle is a right angle.

Example 2: Prove that the diagonals of a rhombus bisect each other at right angles. 

Solution: Consider rhombus ABCD. Let AC and BD be the diagonals, intersecting at point E. In triangles ABE and CDE, we observe that AE = CE (since ABCD is a rhombus). 
Similarly, BE = DE (opposite sides of a rhombus are equal). Thus, triangles ABE and CDE are congruent by the Side-Side-Side congruence criterion. Therefore, ∠AEB = ∠CED. Now, ∠AEB + ∠CED = 180° (linear pair on straight line AECD). So, each angle measures 90°. 
Hence, the diagonals AC and BD bisect each other at right angles. 

Try yourself:Which statement is true about a parallelogram?

• A.A parallelogram has all sides equal.
• B.A parallelogram has opposite sides parallel.
• C.A parallelogram has all angles equal.
• D.A parallelogram has all diagonals equal.

The Mid-point Theorem

In geometry, the Mid-point Theorem relates to the midpoints of the sides of a triangle. The theorem establishes a relationship between the line segment joining the midpoints of two sides of a triangle and the third side. 

The Midpoint Theorem of a triangle states that if you connect the midpoints of two sides of a triangle with a line segment, that line segment will be parallel to the third side of the triangle, and its length will be half the length of the third side.

Let’s explore this theorem and its converse.

Theorem 8: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Activity Observation:

• Draw a triangle and mark mid-points E and F of two sides.
• Join E and F to form line segment EF.
• Measure EF and BC. You will observe that EF = 1/2 BC.
• Measure ∠AEF and ∠ABC. You will find that ∠AEF = ∠ABC.
• Therefore, EF is parallel to BC.

Proof: Consider triangle ABC with midpoints E and F of sides AB and AC, respectively. Draw CD parallel to BA. By the ASA rule, ∆AEF ≅ ∆CDF. This implies EF = DF and BE = AE = DC. Hence, BCDE is a parallelogram, leading to EF || BC.

Theorem 9: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Proof: E is the mid-point of AB, and line l is parallel to BC with CM || BA. Prove AF = CF using the congruence of ∆AEF and ∆CDF.

Conclusion:

• The Mid-point Theorem and its converse provide a useful geometric relationship involving midpoints and parallel lines in triangles.
• These theorems are valuable tools for proving various properties and relationships within triangles.

Some Solved Examples:

 Q.1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution: Given that, AC = BD
To show that ABCD is a rectangle if the diagonals of a parallelogram are equal
To show ABCD is a rectangle, we have to prove that one of its interior angles is right-angled.
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD [SSS congruency]
∠A = ∠B [Corresponding parts of Congruent Triangles]
also, ∠A+∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Therefore, ABCD is a rectangle.
Hence Proved.

Q.2. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that, AC = BD
AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD, AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus, AC = BD [CPCT] diagonals are equal.
Now, In ΔAOB and ΔCOD, ∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
ΔAOB ≅ ΔCOD [AAS congruency]
Thus, AO = CO [CPCT].
Diagonal bisect each other.
Now, In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
ΔAOB ≅ ΔCOB [SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
Diagonals bisect each other at right angles

Q.3. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution:

Given that, ABCD is a rhombus.
AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒∠DAC = ∠BCA (Alternate interior angles)
⇒∠DCA = ∠BCA
AC bisects ∠C.
Similarly, We can prove that diagonal AC bisects ∠A.
Following the same method,
We can prove that the diagonal BD bisects ∠B and ∠D.

Introduction

A triangle, a closed figure formed by three intersecting lines, is characterized by three sides, three angles, and three vertices. 

The term ‘triangle’ is derived from ‘tri,’ signifying ‘three.’ For instance, in triangle ABC (denoted as ∆ ABC), AB, BC, and CA represent the three sides, while ∠ A, ∠ B, ∠ C are the corresponding angles, and A, B, and C are the vertices.

Congruence of Triangles

In our daily experiences, we encounter identical objects like photographs, bangles, and ATM cards, which are called congruent figures. The term ‘congruent’ implies equality in all respects, indicating figures with identical shapes and sizes. When two circles, squares, or equilateral triangles of the same dimensions overlap, we observe complete coverage, confirming them as congruent circles, congruent squares, and congruent equilateral triangles, respectively.

Real-life Applications

The study of congruence extends beyond theoretical concepts. For example:

• In ice trays, the molds for making ice are congruent, facilitating uniformity.
• The concept of congruence is applied in creating casts for identical objects, ensuring consistency.

A triangle has three sides, three angles, and three vertices.

 For example, in the triangle PQR, PQ, QR, and RP are the three sides, ∠QPR, ∠PQR, ∠PRQ are the three angles and P, Q and R, are the three vertices.

A triangle is a unique figure; it is everywhere around us. We can see the sandwiches in the shape of a triangle, traffic signals, cloth anger, set squares, etc. All these are in the shape of a triangle.

Try yourself:

What does the term ‘congruent’ imply?

• A.Equality in all respects
• B.A triangle with three sides, angles, and vertices
• C.Intersection of three lines
• D.The study of shapes and sizes

Correspondence Importance

When establishing congruence, correct correspondence between vertices is essential. For example:

• In ∆ FDE ≅ ∆ ABC, FD ↔ AB, DE ↔ BC, and EF ↔ CA.
• However, writing ∆ DEF ≅ ∆ ABC is incorrect due to improper correspondence.

Note: In congruent triangles, corresponding parts are equal. The acronym CPCT (Corresponding Parts of Congruent Triangles) is often used to denote this equality.

Symbolic Representation

Symbolically, ∆ PQR ≅ ∆ ABC is expressed as:

• PQ covers AB, QR covers BC, RP covers CA.
• ∠ P covers ∠ A, ∠ Q covers ∠ B, ∠ R covers ∠ C.
• Vertices correspond: P ↔ A, Q ↔ B, R ↔ C.

In the figures shown above, each pair is identical to each other. Such figures are called congruent (they are similar and fit over one another exactly).

Also read: NCERT Solutions: Triangles (Exercise 7.1-7.3)

Criteria for Congruence of Triangles

Axiom 1: Side-Angle-Side (SAS) congruence rule
(An axiom is a mathematical statement that is assumed to be true without proof.)
Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Given: Two triangles ABC and PQR such that AB = PQ, AC = PR and ∠BAC = ∠QPR.
To prove: ∆ ABC ≅ ∆ PQR.
Proof: This result cannot be proved with help of previously known results. So, this rule is accepted as an axiom.
Another way to check whether the given two triangles are congruent or not, we follow a practical approach.
Place ∆ABC over ∆PQR such that the side AB falls on side PQ, vertex A falls on vertex P and B on Q. Since ∠BAC = ∠QPR. Therefore, AC will fall on PR. But AC = PR and A falls on P. therefore, C will fall on R. Thus, AC coincides with PR.
Now, B falls on Q and C falls on Therefore, BC coincides with QR.
Thus, ∆ ABC when superposed on ∆ PQR, covers it exactly. Hence, by the definition of congruence, ∆ ABC ≅ ∆ PQR.

Example 1: Check whether ∆ABC and ∆PQR are congruent or not.

Solution: In ∆ ABC and ∆ PQR, we have
AB = PQ = 5 cm (Given)
∠ BAC = ∠QPR = 40o (Given)
AC = PR = 4 cm (Given)
Therefore, ∆ ABC ≅ ∆ PQR (By SAS criterion of congruence)

Example 2: In the figure below, R is the mid-point of PT and SQ. Prove that ∆ PQR ≅ ∆ TSR.

Given: PR = RT and SR = RQ.
To prove: ∆ PQR ≅ ∆ TSR.
Proof: In ∆ PQR and ∆ TSR, we have
PR = TR    (R is the mid-point of PT)
∠PRQ = ∠TRS    (Vertically opposite angles are equal)
QR = SR    (R is the mid-point of SQ)
Therefore, Δ PQR ≅ Δ TSR (By SAS-criterion of congruence)

Example 3: In the figure, it is given that PT = PU and QT = RU. Prove that ΔPTR ≅ ΔPUQ

Given: PT = PU and QT = RU.
To prove: Δ PTR ≅ Δ PUQ.
Proof: We have,
 PT = PU    ……….. (I)
And, QT = RU    ………. (II)
Adding equation (I) and (II), We get,
PT + QT = PU + RU
⇒ PQ = PR    ……….. (III)
Now, in ∆ PTR and ∆ PUQ, we have
PT = PU [Given]
⇒ ∠ TPR = ∠UPQ [Common]
⇒ PQ = PR    [From (III)]
Therefore, Δ PTR ≅ Δ PUQ [By SAS-criterion of congruence]

Theorem 1: Angle-Side-Angle (ASA) Congruence rule Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.

Given: ΔPQR and ΔMNO such that ∠PQR = ∠MNO, ∠PRQ = ∠MON and QR = NO.
To prove: Δ PQR ≅ Δ MNO.
Proof: There are three possibilities that arise.
CASE I: When PQ = MN
In this case, we have
PQ = MN
∠PQR = ∠MNO (Given)
QR = NO (Given)
Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence)
CASE II: PQ < MN
Construction: Join OS such that NS = PQ.
In Δ PQR and Δ SNO, we have
PQ = SN

∠PQR = ∠MNO    (Given)
QR = NO    (Given)
Therefore, Δ PQR ≅ Δ SNO (By SAS-criterion of congruence)
By using corresponding parts of congruent triangles
⇒ ∠PRQ = ∠SON.
But, ∠PRQ = ∠MON. (Given)
This is possible only when ray SO coincides with ray MO or S coincides with M. Therefore, PQ must be equal to MN.
∴ ∠SON = ∠MON.
Thus, in Δ PQR and Δ MNO, we have
PQ = MN
⇒ ∠PQR = ∠MNO (Given)
⇒QR = NO (Given)
Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence)
CASE III: PQ > MN.

Construction: Join SO such that NS = PQ.
In Δ PQR and Δ SNO, we have
PQ = SN
∠PQR = ∠MNO (Given)
QR = NO (Given)
Therefore, Δ PQR ≅ Δ SNO (By SAS-criterion of congruence)
By using corresponding parts of congruent triangles
⇒ ∠PRQ = ∠SON.
But, ∠PRQ = ∠MON. (Given)
This is possible only when ray SO coincides with ray MO or S coincides with M. Therefore, PQ must be equal to MN.
∴ ∠SON = ∠MON.
Thus, in Δ PQR and Δ MNO, we have
PQ = MN
⇒ ∠PQR = ∠MNO (Given)
⇒ QR = NO (Given)
Therefore, Δ PQR ≅ Δ MNO (By SAS-criterion of congruence
Hence, in all the three cases, we have Δ PQR ≅ Δ MNO.

Example 4: Check whether ∆ABC ≅ ∆PQR?

Solution: In ∆ABC and ∆PQR, we have
∠ABC = ∠PQR = 40°    (Given)
BC = QR = 5 cm    (Given)
∠BCA = ∠QRP = 60°    (Given)
Therefore, ∆ ABC ≅ ∆ PQR (By ASA-criterion of congruence)

Example 5: In the figure, ST ∥ QP and R is the mid-points of SQ, proving that R is also the mid-point of PT.

Given: ST ∥ QP and R is the mid-point of SQ.
To prove: PR = RT.
Proof: Since QP ∥ ST and transversal PT cuts them at P and T respectively.
∴ ∠RTS = ∠RPQ.    (Alternate interior angles) ………
(I) Similarly, ∠RST = ∠RQP.    (Alternate interior angles) ………
(II) Since PT and SQ intersect at R.
∴ ∠QRP = ∠SRT.    (Vertically opposite angles) ………..
(III) Thus, in Δ PQR and Δ SRT, we have
∠ PQR = ∠ RST [From (II)]
⇒ QR = RS [Given]
⇒ ∠ QRP = ∠ SRT [From (III)]
Therefore, Δ PQR ≅ Δ SRT [By ASA-criterion of congruence]
By using corresponding parts of congruent triangle.
⇒ PR = RT.
Hence, R is the mid-point of PT.

Some Properties of a Triangle

Theorem 2: Angles opposite to equal sides of an isosceles triangle are equal.
Given: ∆PQR is an isosceles triangle in which PQ = PR.

To prove: ∠ PQS = ∠PRS.
Construction: Draw the bisector PS of ∠QPR which meets QR in S.
Proof: In ∆ PQS and ∆ PRS, we have
PQ = PR    (Given)
⇒ ∠ QPS = ∠ RPS (By construction)
⇒ PS = PS (Common)
Therefore, ΔPQS ≅ ΔPRS (By SAS-criterion of congruence)
By using corresponding parts of congruent triangles
⇒ ∠PQS = ∠PRS.

Example 6: In Δ PQR, ∠ QPR = 80° and PQ = PR. Find ∠RQP and ∠PRQ.

Given that: ∠QPR = 80° and PQ = PR.
Solution: We have
PQ = PR
Since angles opposite to equal sides are equal.
⇒ ∠RQP = ∠PRQ
In Δ PQR, We have
∠QPR + ∠RQP + ∠PRQ = 180°.
⇒ ∠QPR + ∠RQP + ∠RQP = 180°. (∵ ∠ RQP = ∠PRQ)
⇒ 80° + 2 ∠RQP = 180°
⇒ 2 ∠RQP = 180° – 80°
⇒ 2 ∠RQP = 100°
⇒ ∠RQP = 100°/2
⇒ ∠RQP = 50°
Hence, ∠RQP = ∠PRQ = 50°

Example 7: In figure, PQ = PR and ∠PRS = 110°. Find ∠P.

Solution: We have,
PQ = PR
Since angles opposite to equal sides are equal.
⇒ ∠PQR = ∠PRQ.
Now, ∠PRQ + ∠PRS = 180° (∵ Linear pair of angles)
⇒ ∠PRQ+ 110° = 180°
⇒ ∠PRQ = 180° – 110°
⇒ ∠PRQ = 70°
And also, ∠PQR = 70°.
Now, ∠QPR + ∠PQR + ∠PRQ = 180°
⇒ ∠QPR+ 70° + 70° = 180° (∵ ∠PQR = ∠PRQ)
⇒ ∠QPR + 140° = 180°
⇒ ∠QPR = 180°- 140°
⇒ ∠QPR = 40°.

Theorem 3: The sides opposite to equal angles of a triangle are equal.

Given: In triangle PQR, ∠PQR = ∠PRQ.
To prove: PQ = PR.
Construction: Draw the bisector of ∠QPR and let it meet QR at S.
Proof: In ∆ PQS and ∆ PRS, we have
∠PQR = ∠PRQ    (Given)
⇒ ∠QPS = ∠RPS    (By construction)
⇒ PS = PS    (Common)
Therefore, ΔPQS ≅ ΔPRS ( By AAS-criterion of congruence)
By using corresponding parts of congruent triangles
⇒ PQ = PR.

Example 8: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have
∠ABC = ∠PQR = 40o    (Given)
∠BCA = ∠QRP = 60o    (Given)
AC = PR = 3 cm    (Given)
Therefore, ∆ ABC ≅ ∆ PQR (By AAS-criterion of congruence)

Example 9: PQ and RS are perpendiculars of equal length, to a line segment PS. Show that QR bisects PS.

Given: PQ = RS, PQ ⊥ PS and RS ⊥ PS.
To prove: OP = OS.
Proof: In triangles OPQ and ORS, we have
∠POQ = ∠SOR    [Vertically opposite angles]
∠OPQ = ∠OSR    [Each equal to 90°]
PQ = RS    [Given]
Therefore, Δ POQ ≅ ΔSOR    [By AAS-criterion of congruence]
By using corresponding parts of congruent triangles
⇒ OP = OS.
Hence, O is the mid-point of PS.

Some More Criteria for Congruence of Triangles

Theorem 4: Side-Side-Side (SSS) Congruence rule If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Given that: Two Δ ABC and Δ DEF such that AB = DE, BC = EF and AC = DF.
To prove: Δ ABC ≅ Δ DEF.
Construction: Suppose BC is the longest side. Draw EG such that
∠ FEG = ∠ ABC and EG = AB. Join GF and GD
Proof: In ΔABC and ΔGEF, we have
BC = EF    [Given]
AB = GE    [By Construction]
∠ABC = ∠GEF    [By Construction]
Therefore, Δ ABC ≅ Δ GEF [By SAS-criterion of congruence]
By using corresponding parts of congruent triangles
⇒ ∠BAC = ∠EGF and AC = GF
Now, AB = DE and AB = GE
⇒ DE = GE    ……………… (I)
Similarly, AC = DF and AC = GF
⇒ DF = GF     ……………… (II)
In Δ EGD, we have
DE = GE    [From (I)]
Since angles opposite to equal sides of an isosceles triangle are equal.
⇒ ∠EDG = ∠EGD     ……………. (III)
In Δ FGD, we have
DF = GF    [From (II)]
Since angles opposite to equal sides of an isosceles triangle are equal.
⇒ ∠FDG = ∠FGD     ……………. (IV)
From (III) and (IV), we have,
∠EDG + ∠FDG = ∠EGD + ∠FGD
⇒ ∠EDF = ∠EGF
But, ∠BAC = ∠EDF    …………….. (V)
In Δ ABC and Δ DEF, we have
AC = DF     [Given]
⇒ ∠BAC = ∠EDF    [From (V)]
And, AB = DE    [Given]
Therefore, ΔABC ≅ ΔDEF [By SAS-criterion of congruence]

Example 10: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have
BC = QR    (Given)
⇒ AB = PQ (Given)
⇒ AC = PR (Given)
Therefore, ∆ ABC ≅ ∆ PQR (By SSS-criterion of congruence)

Example 11: In the figure, it is given that PR = QS and PS = RQ. Prove that Δ SPR ≅ Δ RQS.

Proof: In triangles SPR and RQS, we have
PR = QS    [Given]
⇒ PS = QR    [Given]
⇒ RS = SR    [Given]
Therefore, Δ SPR ≅ ΔRQS [By SSS-criterion of congruence]

Theorem 5: Right angle -Hypotenuse-Side (RHS) Congruence rule If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Proof:

Consider two right triangles, $△ABC$△ABC and $△PQR$△PQR, where:

• ∠B=∠Q=90∘ (Both triangles are right-angled)
• AC is the hypotenuse of △ABC
• PR is the hypotenuse of △PQR
• AB=PQ (One side is equal)
• AC=PR (Hypotenuse is equal)

The triangles $$ABC and PQR both have right angles at $$B and Q, respectively.
The hypotenuses $$AC and $$PR are given to be equal.
The sides AB and $$PQ are also given to be equal.

Using the RHS Congruence Rule:

Since AB=PQ, and AC=PR, and both ∠B and ∠Q are right angles, the two triangles must be congruent by the RHS criterion.
Therefore, $$△ABC≅△PQR.

Example 12:  In the figure, two right triangles $$△PQR and $$△SUT are given such that $$∠PQR=∠SUT=90∘, $$PR=ST, and $$QR=UT. If $$SU is extended to $$V such that $$UV=PQ and $$VT is joined, prove that $$△PQR≅△SUT and show that $$VT=ST.

Given that: Two right triangles PQR and SUT in which ∠PQR = ∠SU = 90°, PR = ST, QR = UT.
To prove: ΔPQR ≅ ΔSUT.
Construction: Produce SU to V so that UV = PQ. Join VT.
Proof: In Δ PQR and Δ SUT, we have
PQ = VU    [By construction]
∠PQR = ∠VUT    [Each equal to 90°]
QR = UT    [Given]
Therefore, ΔPQR ≅ ΔSUT [By SAS-criterion of congruence]
By using corresponding parts of congruent triangles
⇒ ∠QPR = ∠UVT    ………….. (I)
And, PR = VT
⇒ PR = ST    ………… (II)
∴ VT = ST
Since, angles opposite to equal sides in Δ STV are equal.
⇒ ∠UST = ∠TVS    …………. (III)
From (I) and (III), we get,
∠QPR = ∠UST    ………….. (IV)
And given that, ∠RQP = ∠TUS     ………….. (V)
Adding (IV) & (V), we get,
∠QPR + ∠RQP = ∠UST + ∠TUS    …………. (VI)
∵ ∠PRQ + ∠RQP + ∠QPR = 180°
∴ ∠QPR + ∠RQP = 180 – ∠PRQ    ……….. (VII)
Similarly, ∠UST + ∠TUS = 180° – ∠STU    ………… (VIII)
From equation (VI), (VII) & (VIII), we have
180° – ∠PRQ = 180° – ∠STU
⇒ ∠PRQ= ∠STU.    …………. (IX)
Now, in Δ PQR and Δ SUT
QR = UT    [Given]
⇒ ∠PRQ = ∠STU.    [From (IX)]
And, PR = ST    [Given]
Therefore, ΔPQR ≅ ΔSUT [By SAS-criterion of congruence]

Example 13: Check whether two triangles ABC and PQR are congruent.

Solution: In ∆ ABC and ∆ PQR, we have
∠ ABC = ∠ PQR = 90°    (Given)
AC = PR    (Given)
AB = PQ    (Given)
Therefore, ∆ ABC ≅ ∆ PQR (By RHS-criterion of congruence)

Example 14: In the figure below, it is given that LM = MN, BM = MC, ML ⊥ AB and MN ⊥ AC. Prove that AB = AC.

Proof: In right-angled Δ BLM and Δ CNM, we have
BM = MC    [Given]
⇒ LM = MN    [Given]
Therefore, Δ BLM ≅ Δ CNM [By RHS-criterion of congruence]
By using corresponding parts of congruent triangles
⇒ ∠LBM = ∠NCM
Since, sides opposite to equal angles are equal.
Therefore, AB = AC.

Try yourself:

Which criterion of congruence states that two right triangles are congruent if the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle?

• A.Side-Angle-Side (SAS) criterion
• B.Angle-Side-Angle (ASA) criterion
• C.Side-Side-Side (SSS) criterion
• D.Right Angle-Hypotenuse-Side (RHS) criterion

Some more Rules

1. Angle-Angle-Angle (AAA) Rule

In ΔABC and ΔPQR, we have
∠BAC = ∠QPR    (Given)
∠ACB = ∠PRQ    (Given)
∠CBA = ∠RQP    (Given)
But, Δ ABC and Δ PQR are similar but not congruent because their sizes are different.

2. Angle-Side-Side (ASS or SSA) Rule

We have a triangle ABD. We draw AC such that AC = AD.
Now, consider the two triangles, Δ ABD and Δ ABC

Now, let us check whether we can use ASS criteria for the congruency of two triangles or not. In
Δ ABD and Δ ABC, we have
∠ABD = ∠ABC    (Given)
AB = AB    (Common)
AD = AC    (Given)
So, the corresponding Angle-Side-Side of the two triangles are equal but, these two figures are different in shape.
So, we can conclude that this method is not a universal method for proving triangles congruent.

Rules for Triangles:

Introduction to Lines and Angles

We see lines and angles all around us in different objects. In our daily life, we come across different types of angles formed between the edges of different surfaces.

An architect applies the knowledge of lines and angles for drawing a plan of a multi-storied building.
In science, lines are extensively used to represent the properties of light using the ray
diagrams. We can easily find the height of a tower or a tall building if we know the angle formed between the horizontal line and the line of sight.

Basic Terms and Definitions

• Line: It is a collection of points which has only length, no breadth or thickness and is endless in both directions. Line AB is denoted by
  
• Line segment: A portion of a line with two endpoints is called a line segment. Line segment AB is denoted by 
  
• Ray: A part of a line with one endpoint is called a ray. Ray AB is represented by 
  
• Collinear Points: If three or more points lie on the same line, they are called collinear points otherwise they are called non-collinear points.
  
• Angle: When two rays originate from the same endpoint, they form an angle. The rays making an angle are called the arms and the endpoint is called the vertex of the angle.
  

Try yourself:

Which of the following correctly defines a line?

• A.A portion of a line with two endpoints.
• B.A part of a line with one endpoint.
• C.A collection of points which has only length, no breadth or thickness and is endless in both directions.
• D.If three or more points lie on the same line, they are called collinear points.

Types of Angles

• Acute Angle is an angle whose measure is more than 0°but less than 90°
  0° < x < 90°
  
• A right angle is an angle whose measure is90°.
  y = 90°
  
• An obtuse angle is an angle whose measure is greater than 90° but less than 180°.
  90° < z < 180°
  
• A straight angle is an angle whose measure is 180°. Thus, a straight angle looks like a straight line.
  u = 180°
  
• A reflex angle is an angle whose measure is more than 180° but less than 360°
  180° < v < 360°
  

Complementary and Supplementary Angles

A pair of angles are said to be complementary if the sum of the angles is equal to 90°

∠AOB + ∠BOC =50° + 40° = 90°
∠AOB + ∠BOC =90°
∴∠AOB and ∠BOC are complementary angles.
A pair of angles are said to be supplementary if the sum of the angles is equal to 180°

∠AOB + ∠BOC = 135° + 45° = 180°
∠AOB + ∠BOC= 180°
∴ ∠AOB and ∠BOC are supplementary angles.

Example 1: If (2x − 20°) and (x + 5°) are complementary angles, find the angles.

A pair of angles is said to be complementary if the sum of the angles is equal to 90°.
If (2x − 20°)and (x + 5°) are complementary angles then their sum will be equal to 90°.
(2x − 20°) + (x + 5°) = 90°
2x − 20° + x + 5° = 90°
2x + x − 20° + 5° = 90
3x − 15° = 90° ⇒ 3x = 15° + 90°
3x = 105° ⇒ x = 105°/3 = 35°
(2x − 20°) = (2 × 35° − 20°) ⇒ 70° − 20° = 50°
(2x − 20°) = 50°
(x + 5°) = 35° + 5° = 40°
(x + 5°) = 40°

Example 2: Two supplementary are in the ratio 3: 6, find the angles.

Let the two supplementary angles be 3x and 6x,
3x + 6x = 180°

A pair of angles are said to be supplementary if the sum of the angles is equal to 180°
9x = 180°⇒ x = 180/9 = 20°
x = 20°
3x = 3 × 20 = 60°
6x = 6 × 20° = 120°

Adjacent Angles

Two angles are adjacent if
(i) they have a common vertex
(ii) they have a common arm
(iii) their non-common arms are on different sides of the common arm.

Here, ∠AOB and ∠BOC are adjacent angles because these angles have a common vertex O. Ray OB is the common arm. Rays AO and  CO are the non-common arms.
When two angles are adjacent, then their sum is always equal to the angle formed by the two non–common arms.

Therefore, ∠AOC = ∠AOB + ∠BOC
We see that ∠AOC and ∠AOB are not adjacent as their non-common arms OB and OC are on the same side of the common arm AO.

Also read: Short Answer Type Questions: Lines & Angles

Linear pair of Angles

If the non-common arms of two adjacent angles are two opposite rays, then these angles are called linear pairs of angles.

Here, OA and OB are two opposite rays and ∠AOC, ∠BOC are adjacent angles.
Therefore, ∠AOC and ∠BOC form a linear pair.

Vertically Opposite Angles

When two lines AB and CD intersect each other at point O, then there are two pairs of vertically opposite angles.

Here,
(i) ∠ AOC and ∠ BOD are vertically opposite angles.
(ii) ∠ AOD and ∠ BOC are vertically opposite angles

Intersecting and non-intersecting lines

Intersecting Lines: Two lines are said to be intersecting when the perpendicular distance between the two lines is not the same everywhere and they intersect at one point.

Here, lines AB and CD are intersecting lines.
Non-Intersecting Lines

Two lines are said to be non–intersecting when the perpendicular distance between them is the same everywhere and they do not meet.

The lengths of the common perpendiculars at different points on these lines are the same and so these lines are parallel. This equal length is called the distance between two parallel lines.
Two lines in a plane will either intersect at one point or do not intersect at all, that is they are parallel.

Pairs of Angles

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

Here, ray OC stands on line AB, then ∠AOC and ∠COB are adjacent angles.
Therefore, ∠AOC + ∠COB = 180°
In Axiom 1, it is given that a ray stands on a line and we concluded that the sum of two adjacent angles so formed is 180°. Now, if we do the reverse and take the ‘conclusion’ of Axiom 1 as ‘given’ and ‘given’ as ‘conclusion’ then it becomes:

Statement A: If the sum of two adjacent angles is 180° then a ray stands on a line.

We see that Axiom 1 and Statement A are converse of each other.
If we place a ruler along with one of the non-common arms we see that the other non-common arm also lies along the ruler.
Therefore, points A, O and B lie on the same line and ray OC stands on it.
∠ AOC + ∠COB = 115°+ 65°= 180°
Therefore, statement A is true.
The given statement can be stated in the form of an axiom as follows:

Axiom 2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
Axiom 1 and 2, together are called the linear pair axiom.
Theorem 1: If two lines intersect each other, then the vertically opposite angles are equal.

Given: Lines AB and CD intersect each other at point O. So, two pairs of vertically opposite angles are formed,
i) ∠ AOC and ∠BOD
ii) ∠ AOD and ∠COB
To prove: ∠AOC = ∠BOD and ∠AOD=∠COB
Proof: Here, ray AO stands on line CD.
∴ ∠AOC +∠AOD = 180°
(by Linear Pair axiom) ….. (1)
Similarly, ray DO stands on line AB.
∴ ∠AOD+∠BOD = 180°                (by Linear Pair axiom) ….. (2)
Using Eq 1 and 2 we get,
∠AOC +∠AOD = ∠AOD+∠BOD

Subtracting ∠AOD from both sides we get,
∠AOC +∠AOD – ∠AOD = ∠AOD+∠BOD – ∠AOD
(Euclid’s axiom 3 states that if equals are subtracted from equals, the remainders are equal)
∠AOC +∠AOD – ∠AOD = ∠BOD + ∠AOD – ∠AOD
∴ ∠ AOC = ∠ BOD
Now, we know ray AO stands on line CD.
∴ ∠AOC +∠AOD = 180°
(by Linear Pair axiom) ….. (1)
Similarly, ray CO stands on line AB.
∴ ∠AOC+∠BOC = 180°
(by Linear Pair axiom) ….. (2)
Using Eq 1 and 2 we get,
∠AOC +∠AOD = ∠AOC+∠BOC
Subtracting ∠AOC from both sides we get,
∠AOC +∠AOD – ∠AOC = ∠AOC+∠BOC – ∠AOC
(Using Euclid’s axiom 3 states which that if equals are subtracted from equals, the remainders are equal)
∠AOD + ∠AOC – ∠AOC = ∠BOC + ∠AOC- ∠AOC

∴ ∠ AOD = ∠ BOC
We see that two pairs of vertically opposite angles are equal.

Example 1: In the given figure AB and CD intersect at O. If ∠AOC+∠DOE = 60° and ∠BOD = 35°, find ∠DOE and reflex ∠AOE.

We know,
(i) ∠AOC + ∠DOE = 60°
(ii) ∠BOD = 35°
Here, lines AB and CD intersect each other at O.
So, ∠AOC = ∠BOD = 35°
(If two lines intersect each other, then the vertically opposite angles are equal)
∠AOC + ∠DOE = 60°
35° + ∠DOE = 60°
(∵ ∠AOC = 35°)
∠DOE = 60° − 35° = 25°
∠DOE = 25°

We see that COD is a straight line, that is the measure of the ∠COD is equal to 180°.
∠AOC+ ∠AOE +∠EOD = 180°
∠AOE + (∠AOC+ ∠EOD) = 180°
∠AOE + 60°= 180°
(∵ ∠AOC + ∠DOE = 60°)
∠AOE = 180° – 60°
∠AOE =120°

Reflex ∠ AOE = 360°- 120°= 240°
(reflex angle is an angle whose measure is greater than 180° but less than 360°)

Example 2: In the figure given below, lines PQ, RS and TU meet at point O. Find the value of x, hence find all the three indicated angles.

Here, lines PQ and TU intersect each other at O.
So, ∠POT = ∠UOQ = 5x (If two lines intersect each other, then the vertically opposite angles are equal)

We see that ROS is a straight line, that is the measure of the ∠ROS is equal to 180°.
∠ROP+ ∠POT +∠TOS = 180°
4x + 5x + 3x = 180°
12x = 180°
x = 180°/12
x = 15°
∠POR = 4x = 4 × 15° = 60°
∠UOQ = 5x = 5 × 15° = 75°
∠TOS = 3x = 3 × 15° = 45°

Example 3: Lines AB and CD intersect each other at O. If ∠AOD:∠AOC = 2: 4, find all the angles x, y and z .

We know,
∠AOD : ∠AOC = 2 : 4
Let, ∠AOD = 2a and ∠AOC = 4a
Ray AO stands on line CD.
∠AOD +∠AOC = 180°

(If a ray stands on a line, then the sum of two adjacent angles so formed is 1800 )
2a + 4a = 180°
6a = 180°
a = 180°/6 = 30°
∠ AOD = x = 2a = 2 × 30°= 60°
∠ AOC = z = 4a = 4 × 30°= 120°
x = y = 60° (Vertically opposite angles are equal)
So, x = 60°, y = 60° and z = 120°

Lines parallel to the same line

If two lines are parallel to the same line then they are parallel to each other. 

We draw a transversal t for the lines l, m and n, where line l is parallel to line n and line m is parallel to line n.
Now,
∠1 = ∠3(corresponding angles axiom) → Eq 1
Similarly, ∠2 = ∠3 (corresponding angles axiom) → Eq 2
Using Eq 1 and 2 we get, ∠1 = ∠2
∠1 and ∠2 are corresponding angles.
The converse of corresponding angles axiom states that if a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other
Therefore, line l is parallel to line m ( l || m )

Example 1: In the given figure, show that AB || CD.

∠BCD = ∠BCE + ∠ECD
∠BCD = 25° + 55° = 80°
We see that ∠ABC and ∠BCD are alternate interior angles.
∠ABC = ∠BCD = 80°
If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.
∴ AB ||CD
OR
∠ FEC + ∠ECD = 125° + 55° = 180°
If a transversal intersects two lines, in such a way that each pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.
∴ CD ||EF……… (1)
Now let us extend EF till it intersects BC at M and extend it till M’.
Then since CD ||EF
We can find the remaining angles.
The remaining angles are as shown:

Now,
∠ CMM’ = ∠ DCM = 80° – (Interior alternate angles)
∠ BMM’ + ∠ CMM’ = 18° – Linear Pair of angles
∠ BMM’ = 100°
∠ BMM’ + ∠MBA = 100° + 80° = 180°
If a transversal intersects two lines, in such a way that each pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.
∴ AB ||EF……… (2)
From eq (1) and (2), we get
AB ||CD

Example 2: In the figure p||q||r. From the figure, find the ratio of (a + b): (b − a).

50° + b + 30° = 180°
b + (50° + 30°) = 180°
b + 80° = 180°
b = 180° − 80°
b = 100°
We know that p||q .

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

So, 110° + a = 180°
a = 180° − 110° = 70°
a = 70°
a + b = 70° + 100° = 170°
b − a = 100° − 70° = 30°
(a + b) : ( b − a) = 170°: 30°= 17°∶ 3°

Example 3: In the figure, PQ||RS, RS||TU, and TP⊥ PQ. If ∠STU = 65°, find the value of a, b and c.

We know, PQ||RS, RS||TU, and TP⊥ PQ.
∠ STU = 65°
b + 65° = 180°
(If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary)
b = 180° − 65° = 115°
b = c = 115° (corresponding angles)
a + 65° = 90° (∵ TP⊥ PQ)
a = 90° − 65° = 25°
So,a = 25°, b = c = 115°

Angle sum property of a triangle (Old Syllabus)

We know that the sum of all the angles of a triangle is 180°.

A triangle is a plane figure formed by three intersecting lines.

• ∠A, ∠B, and ∠C are called interior angles of the triangle.
• When side BC is produced to D then we get an exterior angle,∠ACD.
• ∠BAC and ∠ABC are called its interior opposite angles.

Theorem 1: The sum of the angles of a triangle is 180°

Here, ∠1, ∠2 and ∠3 are the angles of ∆ABC.
To prove: ∠1 + ∠2 + ∠3 = 180°

Construction: We draw a line XAY parallel to BC through the vertex A. (If we draw parallel lines then we can use the properties of parallel lines)

Here, XAY is a straight line.
∴ ∠4 + ∠1 + ∠5 = 180° → Eq 1
We see that XAY is parallel to BC, so, AB and AC are transversals.
So, ∠4 = ∠2 (Alternate Angles)
∠3 = ∠5 (Alternate Angles)
Putting∠4 = ∠2, and ∠3 = ∠5 in Eq 1 we get,
∠2 + ∠1 + ∠3 = 180° → Eq 2
∠1 + ∠2 + ∠3 = 180°
Thus, the sum of three angles of a triangle is 180°.

Theorem 2: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

In ∆ABC, side BC is produced to D to form the exterior angle, ∠ACD.

To prove: ∠A + ∠B = ∠ACD
In ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angle sum property of triangle) → Eq 1
Here, BCD is a straight line.
∴ ∠ACB + ∠ACD = 180° → Eq 2
From Eq 1 and 2 we get,
∠A + ∠B + ∠ACB = ∠ACB + ∠ACD
Subtracting ∠ACB from both sides we get,
∠A + ∠B + ∠ACB – ∠ACB = ∠ACB + ∠ACD – ∠ACB
∠A + ∠B + ∠ACB – ∠ACB = ∠ACB – ∠ACB+ ∠ACD
∠A + ∠B = ∠ACD
Thus, the exterior angle is equal to the sum of the two interior opposite angles.

Example 1: The measure of the exterior angle, ∠PRS of ∆PQR is 100°.If one of the interior angles is 20°, find the measure of the other two angles of ∆PQR.

Exterior angle, ∠PRS = 100°
One of the interior angles = 20°
Let ∠Q = 20°
then the other interior opposite angle will be ∠P.
So,∠PRS = ∠Q + ∠P (exterior angle of a triangle is equal to the sum of the two interior opposite angles)
100° = 20° + ∠P
∠P = 100° − 20° = 80°
Now, ∠P + ∠Q + ∠PRQ = 180°
(Angle sum property of triangle states that the sum of the angles of a triangle is 180°)
80° + 20° + ∠PRQ = 180°
100° + ∠PRQ = 180°
∠PRQ = 180° − 100°
∠PRQ = 80°
Therefore, the measures of the three angles of ∆PQR are
∠P = 80°
∠Q = 20°
∠PRQ = 80°

Example 2: In the figure given below, PY || SZ, PS ⊥ QR, and PY bisect∠XPR. If ∠XPR = 110° then find the measure of ∠PQR and ∠PRZ.

Now, ∠XPR = 110°
∠XPY = ∠YPR (∵ PY is the bisector which bisects ∠XPR)
∴ ∠XPY = ∠YPR = 110°/2 = 55°

We know that PY is parallel to SZ.
∠YPR = ∠PRS = 55° (alternate angles)
Here, SRZ is a straight line.
∴ ∠ PRS + ∠PRZ = 180°
55°+ ∠PRZ = 180°
∠PRZ = 180°- 55°
∠PRZ = 125°

Now, ∠ XPR = ∠PRQ + ∠PQR
(exterior angle of a triangle is equal
to the sum of the two interior opposite angles)
110° = 55° + ∠PQR
∠PQR = 110° − 55° = 55°
∠PQR = 55° and ∠PRZ = 125°

Example 3: In the figure given below, PR || QS. If ∠PRS = 50° and ∠SUT = 30° then find ∠STU.

We know, PR||QS and so ∠PRQ and ∠QSU are a pair of corresponding angles.
∴ ∠PRQ = ∠QSU =50°

We see that ∠QSU is an exterior angle of the ∆ STU.
∠QSU = ∠SUT + ∠STU (exterior angle of a triangle is equal to the sum of the two interior opposite angles)
50° = 30° + ∠STU
∠STU = 50° − 30° = 20°
∠STU = 20°

Example 4: In the figure, ∠BPC = 40°, ∠ABQ = 65° and lines m and n are parallel to each other. Find x, y, and z.

∠ABQ = ∠PBC = 65° (Vertically opposite angle)
Here, x is an exterior angle of ∆PBC.
∴ x = ∠PBC + ∠BPC
x = 65° + 40° = 105°
Line m || line n.
So, ∠ABQ = ∠BQC = y = 65° (Alternate angles)
In ∆ PQR,
∠P + ∠ Q + ∠R = 180°(Angle sum property of triangle states that the sum of the angles of a triangle is 180°)
40° + y + z = 180°
40° + 65° + z = 180°
105° + z = 180°
z = 180° − 105°= 75°
x = 105°, y = 65° and z = 75°

Parallel Lines and Transversal Lines (Old Syllabus)

If two lines in the same plane do not intersect, when produced on either side, then such lines are said to be parallel to each other.

Here, lines l and m are parallel to each other.
A line that intersects two or more straight lines at distinct points is called a transversal line.

Here, line l intersects lines m and n at points A and B respectively.
We see that four angles are formed at each point A and B, namely
∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.

The pairs of angles formed, when a transversal intersects two lines are as follows:

• Corresponding Angles: The angles on the same side of a transversal are known as corresponding angles if both lies either above or below the lines.
  ∠1 & ∠5
  ∠2 & ∠6
  ∠4 & ∠8
  ∠3 & ∠7
• Alternate Interior Angles: The pairs of interior angles on opposite sides of the transversal are called alternate interior angles.
  ∠4 & ∠6
  ∠3 & ∠5
• Alternate Exterior Angles: The pairs of exterior angles on opposite sides of the transversal are called alternate exterior angles.
  ∠1 & ∠7
  ∠2 & ∠8
• Interior angles on the same side of the transversal: They are also referred to as consecutive interior angles or allied angles or co-interior angles.
  ∠4 & ∠5
  ∠3 & ∠6

Here, we see that the lines m and n are not parallel. Can you tell what will happen if line l intersects two parallel lines m and n?
When a transversal l intersects two parallel lines m and n, then the relation between angles formed are obtained as axioms and theorems.

Axiom 1: If a transversal intersects two parallel lines, then each pair of corresponding angles is equal. This axiom is known as a corresponding angle axiom.

When line l intersects two parallel lines m and n, then we see that each pair of corresponding angles is equal.
∴ ∠ 1 = ∠5, ∠2 = ∠6, ∠4 = ∠8 and ∠3 = ∠7
The converse of this axiom is as follows:
Axiom 2 (Converse of axiom 1): If a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other.
Theorem 1: If a transversal intersects two parallel lines, then each pair of alternate angles is equal.

Here, line VW intersects two parallel lines RS and TU at points A and B respectively.
The two pairs of alternate interior angles are∠SAB and ∠TBA, ∠RAB and ∠ABU
To prove:
(i) ∠SAB = ∠TBA
(ii) ∠RAB = ∠ABU
We know,
∠VAR = ∠SAB (Vertically opposite angles) → Eq 1
∠VAR = ∠TBA (Corresponding angles axiom) Eq → 2
Using Eq 1 and 2 we see that,
∠VAR = ∠SAB and∠VAR = ∠TBA. Therefore, ∠SAB = ∠TBA
Similarly,
∠SAV = ∠RAB (Vertically opposite angles) Eq → 3
∠SAV = ∠ABU (Corresponding angles axiom) Eq → 4
Using Eq 3 and 4 we get,
∠SAV = ∠RAB and ∠SAV = ∠ABU. Therefore, ∠RAB = ∠ABU
So,∠SAB = ∠TBA and ∠RAB = ∠ABU

Therefore, the pairs of alternate interior angles are equal.

Theorem 2 (Converse of theorem 1): If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.

Here, line VW intersects two parallel lines RS and TU at points A and B respectively in such a way that,∠SAB = ∠TBA and ∠RAB = ∠ABU.
To prove: RS is parallel to TU
We know,
∠SAB = ∠TBA (alternate interior angles) Eq → 1
∠SAB = ∠RAV (vertically opposite angles) Eq → 2
Using Eq 1 and 2 we see that,
∠SAB = ∠TBA and ∠SAB = ∠RAV. Therefore, ∠TBA = ∠RAV.
We know that∠TBA and ∠RAV are corresponding angles.

According to the converse of corresponding angles axiom, if a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other.
So, RS and TU are parallel lines, that are RS || TU.

Theorem 3: If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Here, line UV intersects two parallel lines AB and CD at points X and Y respectively. So, two pairs of interior angles are formed; ∠AXY and ∠CYX, ∠BXY and ∠XYD.To prove: ∠AXY + ∠CYX = 180° and ∠BXY + ∠XYD = 180°
(pair of interior angles on the same side of the transversal is supplementary)
Here, ray XY stands on line AB. Therefore, ∠AXY and ∠BXY are adjacent angles.
So, ∠AXY + ∠BXY = 180°  →  Eq 1
(linear pair axiom states that if a ray stands on a line, then the sum of two adjacent angles so formed is 180°)
Now, ∠BXY = ∠CYX (alternate interior angles)  → Eq 2
On putting ∠BXY = ∠CYX in Eq 1 we get,
∠AXY + ∠CYX = 180°
Similarly, ray XY stands on line CD. Therefore, ∠CYX and ∠XYD are adjacent angles.
So, ∠CYX + ∠XYD = 180° (linear pair axiom) Eq → 3
But, ∠CYX = ∠BXY (alternate interior angles) Eq → 4
On putting ∠CYX = ∠BXY in Eq 1 we get,
∠BXY + ∠XYD = 180°
Therefore,
∠AXY + ∠CYX = 180°
∠BXY + ∠XYD = 180°

Theorem 4 (Converse of theorem 4): If a transversal intersects two lines, in such a way that each pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

Here, line UV intersects two parallel lines AB and CD at points X and Y respectively in such a way that two pairs of interior angles on the same side of the transversal are ∠AXY and ∠CYX, ∠BXY and ∠XYD.
We know, ∠AXY + ∠CYX = 180°
∠BXY + ∠XYD = 180°
To prove: AB || CD
Here, ray XB stands on line UV. Therefore, ∠BXU and ∠BXY are adjacent angles.
So, ∠BXU + ∠BXY = 180° (linear pair axiom) → Eq 3
(linear pair axiom states that if a ray stands on a line, then the sum of two adjacent angles so formed is 180°)
It is given that,
∠BXY + ∠XYD = 180° → Eq 4
Using Eq 3 and 4 we get,
∠BXU + ∠BXY = ∠BXY + ∠XYD
On subtracting ∠BXY from both sides we get,
∠BXU + ∠BXY – ∠BXY = ∠BXY + ∠XYD – ∠BXY
∠BXU + ∠BXY – ∠BXY = ∠BXY – ∠BXY + ∠XYD
So, ∠BXU = ∠XYD
Now, ∠BXU and ∠XYD are corresponding angles.

Try yourself:According to the angle sum property of a triangle, what is the sum of the interior angles of a triangle?

• A.90º
• B.120º
• C.180º
• D.360º

According to the converse of corresponding angles axiom, if a transversal intersects two lines such that a pair of corresponding angles are equal, then the two lines are parallel to each other. So, AB || CD.

Example 1: In the figure given below, PQ ||RS, find the value of a and ∠PAB and ∠ABR.

Now, ∠PAB = ∠ABS (Alternate interior angles)
a + 220 = 2a + 10°
22° − 10° = 2a − a
a = 12°
∠PAB = a + 22°
∠PAB = 12° + 22°(∵ a = 12°)

∠PAB = 34°
We know, ∠PAB = ∠ABS = 34°
Ray AB stands on line RS. Therefore, ∠ABR and ∠ABS are adjacent angles.
So, ∠ABR + ∠ABS = 18° (linear pair axiom)
∠ABR +34° = 180°(∵ ∠PAB = 34°)
∠ABR = 180° − 34°= 126°
∠ABR = 126°

Example 2: In the figure, PQ ||RS ||TU and a ∶ b = 1 : 2, find c.

Let, a = 1x and b = 2x
Now, a and b are interior angles on the same side of the transversal.

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Therefore, a + b = 180°
x + 2x = 180° (∵ a = 3x, b = 2x)
3x = 180°
x = 180°/3 = 60°
Now, a = x = 60°
b = 2x = 2 × 60° = 120°
a = c = 60° (Alternate interior angle)
c = 60°

Example 3: In the figure PQ || RS, ∠PQX = 35° and ∠RSX = 40°, find a.

Here we draw a line m parallel to PQ through X.

Here, ∠PQX = ∠QXY = 35° (Alternate interior angle)
Again, ∠RSX = ∠YXS = 40° (Alternate interior angle)
Now, ∠QXS = ∠QXY + ∠YXS
∠QXS = 35° + 40°(∵ ∠QXY = 35° and ∠YXS = 40°)
∠QXS = 75°
∠QXS = a = 75°
Therefore, a = 75°

Introduction

The word ‘geometry’ comes from the Greek word ‘geo’, meaning the ‘earth’, and ‘metron’, meaning ‘to measure’. Thus, the word ‘geometry’ means ‘earth measurement’.

• Geometry seems to have started from the need to measure land and was explored in various ways by ancient civilisations, including Egypt.
• The Egyptians created several geometric methods and rules to calculate simple areas and to perform basic constructions. They used geometry to calculate the volumes of granaries and for building canals and pyramids.

• Excavations in the Indian subcontinent, such as in Harappa and Mohenjo-Daro, show that the Indus Valley Civilization (around 3000 BC) made extensive use of geometry. This society was highly organised, with well-planned cities featuring parallel roads and an underground drainage system.
• Thales, a Greek mathematician, is known for providing the first proof that a circle is bisected by its diameter.

• Around 300 BC, Euclid, a mathematics teacher in Alexandria, Egypt, compiled all existing knowledge into his well-known work, called ‘Elements’. He organised it into thirteen chapters, referred to as books.

Euclid’s Definitions, Axioms and Postulates

In Euclid’s era, mathematicians developed the concepts of a point, line, and plane from observing the space and solids around them. This led to a more abstract understanding of solid objects.

A solid has a shape, size, and position, and can be moved about. Therefore, a solid has three dimensions.

• Its edges are called surfaces.

• These surfaces separate different parts of space and are considered to have no thickness.
• The edges of the surfaces consist of curves or straight lines, which end in points.

Try yourself:

What is the meaning of the word ‘geometry’?

• A.The study of shapes and figures
• B.The measurement of the earth
• C.The study of ancient civilizations
• D.The development of mathematical techniques

Some of the other assumptions or definitions listed by Euclid’s are:

(i) A point is that which has no part.

(ii) A line is a breadthless length.

(iii) The ends of a line are points.

(iv) A straight line is a line which lies evenly with the points on itself.

(v) A surface is that which has length and breadth only.

(vi) The edges of a surface are lines.

(vii) A plane surface is a surface which lies evenly with the straight lines on itself.

Now, let’s look at the first definition of a point. Here, we need to explain what a ‘part’ is. If we say that a ‘part’ occupies ‘area’, then we must also explain what ‘area’ means. This leads to a never-ending series of definitions. For this reason, mathematicians decide to leave some geometric terms undefined. However, we have a basic understanding of what a point is, even if the definition doesn’t fully capture it. We often represent a point as a dot, even though a dot has some size.

Similarly, the second definition of a line talks about length and width, which are also not defined. Because of this, some terms remain undefined as a study progresses. In geometry, we consider a point, a line, and a plane (which Euclid calls a plane surface) as undefined terms. The important thing is that we can represent them in a straightforward way or describe them using ‘physical models’.

Starting from his definitions, Euclid made certain assumptions that were not meant to be proven. These assumptions are called ‘obvious universal truths’. He categorised them into two types: axioms and postulates. Common notions, often known as axioms, are used throughout mathematics and are not specifically related to geometry. For these reasons, mathematicians agree to leave some geometric terms undefined.

He divided them into two types:

Some Euclid’s axioms

(1) Things which are equal to the same thing are equal.

So,

If line segment PQ is equal to line segment RS and line segment RS is equal to line segment TU, then line segment PQ is equal to line segment TU.

(2) If equals are added to equals, the wholes are equal.
Two jars, A and B have the same quantity of sugar, which is 2 kg.

Now, we add 1 kg of sugar to both the jars, A and B.

We see that the final quantity of sugar is the same in both the jars.

(3) If equals are subtracted from equals, the remainders are equal.
We again take the example of two jars having the same quantity of sugar (2 kg).

In this case, we remove 1 kg of sugar from both the jars, A and B.

We see that the final quantity of sugar in both the jars remains the same.

(4) Things which coincide with one another are equal.
Two pages of the same book coincide with each other and hence are equal.

(5) The whole is greater than the part.

Here, we see that a whole cake is greater than a slice (part) of the cake.

(6) Things which are double of the same things are equal.
If x = 2y and z = 2y then x = z .
Suppose there are three jars of sugar, A, B, and C. Jars, A and B have an equal quantity of sugar, that is 2 kg and jar C has 1 kg of sugar.
So, we see that the quantity of sugar in jar A and B is equal to twice the quantity of sugar in jar C or we can also say that both the jars A and B, have the same quantity of sugar as their quantities are also equal to twice the quantity of sugar in jar C.

(7) Things that are halves of the same things are equal.
If x = y/2 and z = y/2 then x = z.
Suppose there are three jars of sugar, A, B, and C. Jars, A and B have an equal quantity of sugar, that is 1 kg and jar C has 2 kg of sugar.
So, we see that the quantity of sugar in jar A and B is equal to half the quantity of sugar in jar C or we can also say that both the jars A and B, have the same quantity of sugar as their quantities are also equal to half the quantity of sugar in jar C.

Example 1: Solve the equation y − 10 = 13 and state Euclid’s axioms used here.

y − 10 = 13
Adding 10 to both sides,
y − 10 + 10 = 13 + 10
y = 23
Here we have used Euclid’s axiom 2 which states that if equals are added to equals, the wholes are equals.

Example 2: In the given figure PT = QT, TR = TS, show that PR = QS. Write Euclid’s axiom to support this.

It is given that,
PT = QT Eq → 1
TR = TS Eq → 2
To show, PR = QS
Adding Eq 1 and Eq 2 we get,
PT + TR = QT + TS
PR = QS
Here, we have used Euclid’s axiom 2 which states that if equals are added to equals, the wholes are equals.

Example 3: In the given figure, we have BD = BC and AC = BC. State Euclid’s axiom to support this.

If BD = BC and AC = BC, then BD = AC
Here, we have used Euclid’s axiom 1 which states that things which are equal to the same thing are equal.

Example: Eric and David have the same weight. If both of them lose 5 kg weight, how will you compare their new weights?

Let x kg be the initial weight of both Eric and David.
On losing 5 kg, the weight of Eric and David will be (x − 5) kg.
According to Euclid’s axiom 3, if equals are subtracted from equals, the remainders are equal.
So, even after losing 5 kg, both Eric and David will have the same weight as their initial weights are also equal.

Try yourself:

Which Euclid’s axiom states that if equals are added to equals, the wholes are equal?

• A.Axiom 1
• B.Axiom 2
• C.Axiom 3
• D.Axiom 4

Euclid’s Five Postulates

Postulate 1: A straight line may be drawn from any one point to any other point

This postulate states that at least one straight line passes through two distinct points, but it does not say that there cannot be more than one such line. However, Euclid has frequently assumed that there is a unique line joining two distinct points.
This result can be stated as an axiom given below:
Axiom: Given two distinct points, there is a unique line that passes through them.
Here, we see that only one line, AB is passing through A and also through B. Now can you tell how many lines are passing through B and also through A?
We see that the same line, i.e. AB is passing through B and A. The above statement is self-evident and so, it is called an axiom.

Postulate 2: A terminated line can be produced indefinitely.

Postulate 2 states that a line segment can be extended on either side to form a line.

Postulate 3: A circle can be drawn with any center and any radius.
Here, AB = r is the radius of circle 1 and CD = R is the radius of circle 2.

Postulate 4: All right angles are equal.
Here, ∠ABC = 90°
∠XYZ = 90°
∠TUV = 90°
So, all the right angles are equal.

Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles (180°), then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
Suppose a straight line XY falls on two straight lines, PQ and RS in such a way that the interior angles ∠1 + ∠2 <180° on the left side of XY. The two lines PQ and RS, when produced indefinitely, will eventually intersect at point O on the left side of PQ.
If we see the five postulates, we notice that postulate 5 is more complex than any other postulate. The postulates 1 through 4 are so simple and obvious that these are taken as ‘self – evident truths’. As it is not possible to prove these statements, they are accepted without any proof.
Euclid used his postulates and axioms to prove other results. Then using these results, he proved some more results by applying deductive reasoning. The statements that were proved are called propositions or theorem.
Euclid deduced 465 propositions in a logical chain using his axioms, postulates, definitions, and theorems proved earlier in the chain.

Example: If S, T, and U are three points on a line, and T lies between S and U, then prove that ST + TU = SU.

We see that SU coincides with ST + TU. According to Euclid’s axiom 4, things that coincide with one another are equal.
Therefore, SU = ST + TU
Now, in this solution, it has been assumed that there is a unique line passing through two points.

Example: Prove that an equilateral triangle can be constructed on any given line segment.
(1) We first draw a line segment PQ of length 1 cm.
(2) Next, we draw a circle with point Q as center and QP as the radius. (Using Euclid’s postulate 3 – A circle can be drawn with any center and any radius)
(3) Similarly, we draw another circle with P as center and radius equal to PQ.
(4) The two circles meet at point R. Now draw the line segments, RP and RQ to form the ∆ RPQ.
To prove:∆ RPQ is an equilateral triangle
Here,
PQ = PR = 1 cm (Radii of the same circle)
PQ = QR = 1 cm (Radii of the same circle)
Using these two facts and Euclid’s axiom 1 (things which are equal to the same thing are equal) we conclude that,
PQ = PR = QR = 1 cm
Therefore, ∆ RPQ is an equilateral triangle as all the sides are equal.
Here, we see that Euclid has assumed that the two circles drawn with centers P and Q will meet each other at a point.

Theorem
Two distinct lines cannot have more than one point in common.
To prove: Lines l and m have only one point in common.
Proof: Let us suppose that the two lines intersect at two distinct points A and B which means that two lines are passing through two distinct points A and B.
According to the axiom, only one line can pass through two distinct points. Therefore, the assumption that the two lines can pass through two distinct points is wrong. So, two distinct lines cannot have more than one point in common.

Try yourself:

Which postulate states that a straight line can be extended indefinitely?

• A.Postulate 1
• B.Postulate 2
• C.Postulate 3
• D.Postulate 4

Summary

Euclid’s definition
A point is that which has no part.

A line is a breadthless length. The ends of a line are points and the 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.Euclid’s axiom
(i) Things which are equal to the same thing are equal.

Example: If  =  and  = , then  = 

(ii) If equals are added to equals, the wholes are equal.

Example: If x = y then x + z = y + z

(iii) If equals are subtracted from equals, the remainders are equal.

Example: If x = y then x − z = y − z

(iv) Things which coincide with one another are equal.

Example: Two pages of the same book

(v) The whole is greater than the part.

Example: A whole cake is greater than a slice (part) of the cake.

(vi) Things which are double of the same things are equal to one another.

Example: If x = 2y and z = 2y then x = z

(vii) Things which are halves of the same things are equal to one another.

Example: If x = y/2 and z = y/2 then x = z.

In geometry, a point, a line and a plane (or a plane surface) are considered as undefined terms.

• Euclid used the term ‘postulate’ for the assumptions that were specific to geometry.
• Common notions or axioms were assumptions used throughout mathematics and were not specifically linked to geometry.

Euclid’s postulate
Postulate 1: A straight line may be drawn from any one point to any other point.
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.
Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles (180°), then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles
Equivalent Versions of Euclid’s Fifth Postulate
‘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.

In Euclid’s era, mathematicians developed the concepts of a point, line, and plane from observing the space and solids around them. This led to a more abstract understanding of solid objects.

A solid has a shape, size, and position, and can be moved about. Therefore, a solid has three dimensions.

• Its edges are called surfaces.