# A Complete Guide For Geometry Basics For CAT

## In this article, we are going to learn about the properties of triangles and geometry basics for CAT. Compiled by our experts at IIM Skills- online CAT coaching Contents of the article for geometry basics for CAT:

1. Theory
2. Practice questions for geometry basics for CAT
3. Syllabus for CAT

Geometry is an important topic of quantitative aptitude in not only CAT but many other competitive exams. In this post, we are going to discuss various properties of triangles and geometry basics for CAT. It is important to be strong in geometry basics for CAT since this is the basis of a good score in geometry in the Quantitative aptitude section of CAT.

### Theory in geometry basics for CAT

#### THE TRIANGLE

The triangle is a very basic and the simplest closed 2-dimensional figure. We come across triangles in every class throughout school. Every year, there was something new to learn about triangles. It makes us wonder the simplest shapes could be so complex. Triangle is a central part of the geometry portion of any class. That’s because a triangle helps understand the basics of all geometry. All types of polygons with any number of sides can be divided and represented as a group of triangles.

A triangle is a polygon having 3 sides and 3 vertices. A triangle is drawn by joining 3 points that must be non-collinear, otherwise, it would become a straight line. There are 3 internal angles whose sum total is 180 degrees, and the sum of exterior angles is 360 degrees. There are different types of triangles. They are classified on the basis of the measurement of sides and angles. There is also a comparison between two triangles like congruency, similarity, etc.

Since the triangle was such a big part of geometry, it is a measure of how well the student would be in geometry. This is also the reason why triangles and geometry are so important in Geometry basics for CAT.

Some other properties of the triangles in geometry basics for CAT are

• The sum of the length of any two sides of a triangle is always greater than the third side i.e. AB + BC > AC and the difference between the length of two sides is always less than the third side, | AB – BC | < AC.
• The value of an exterior angle is the sum of two opposite interior angles.
• The side opposite of the biggest internal angle is the longest side of the triangle and vice versa.

Classification of Triangles

Triangles can be divided into two types

Based on the length of the side

Equilateral Triangle

In the equilateral triangle, the length of all three sides is equal. Thus, all three angles are also equal i.e. 60o

Area of equilateral triangle = √3/4*(side)2

where ‘side’ is the length of the side.

Isosceles Triangle

In this triangle, the length of the two sides is equal and one is different.

Also, the angles corresponding to these sides are also equal.

AREA = ½ x base x height

Based on the measure of angle

Scalene Triangle

In a scalene triangle, all the sides measure different from each other in length and for the same reason, the angles are also varying.

Acute angle triangle

In an acute triangle, all the angles of the triangle are less than 90. An equilateral triangle is an acute triangle since all its angles are <90. An isosceles and scalene triangle can also be an acute triangle.

Obtuse Angle Triangle

In an obtuse-angled triangle, one of the angles is greater than 90. There cannot exist two obtuse angles in one triangle as the sum of all angles is 180. Therefore, in the obtuse triangle, one angle measures>90, and the other two are acute.

Right Angle triangle

A right-angled triangle is the one in which one angle measures 90 and the other two angles are acute angles and can be equal. This condition between the sides and angles of a right triangle is the basis for trigonometry.

IMPORTANT THEOREMS

These were the types of the triangle on the basis of sides and angles in geometry basics for CAT. Now we are going to learn about some formulae. These will help in the various scenarios in finding the length of sides or angles in the geometry questions.

For example, to find the area of a triangle, there is more than one method. Ways to find the area of a triangle are more than any other polygon. Let’s learn about some of them.

Heron’s Formula:

The most basic method for calculating the area of a triangle is

Area of triangle = ½ x base x height

But sometimes, when the height is not given but you know the lengths of all sides, we have a direct formula to find out the area of a triangle in that case. You do not need to do calculations to find the height before you find the area.

Let a, b, and c, be the length of the sides of the triangle then,

Area = (s*(s-a)*(s-b)*(s-c))1/2

where, s = (a+b+c)/2

‘s’ is also called the semi-perimeter because it is half the value of the perimeter.

The area through Trigonometry:

We can also use trigonometry equations to find out the area of a triangle if we have lengths of two sides and measure of one angle.

Area of triangle ABC = ½ bc x sinA = ½ ab x sinC = ½ ac x sin B

Determinant method:

The determinant method makes use of coordinate geometry to determine the area of a triangle. Therefore,

= area

where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the vertices of the triangle

There are a few other ways of determining the area of the triangle:

• Area of Triangle = abc/4r, where r is the circumradius. Circumradius is the radius of the circle that has vertices of the triangle as points on the circle
• Area of Triangle = r*s, where r = inradius and s = semi-perimeter. Inradius is the radius of the circle that is drawn with the sides of the triangle as tangents to the circle

Now we will move forward to discussing altitudes, medians, and perpendicular bisector. Many students have confusion between each of them so we will clarify all of them here. What is the definition of each and what do we call their point of intersection.

Altitudes and orthocentre

An altitude is a line segment that passes through any vertex and cuts at a right angle with the side opposite to this vertex.

Here, AD, CF, and BE are altitudes of a triangle ABC.

The orthocenter is the point of intersection of all the three altitudes of the triangle. The orthocenter may lie either inside or outside of the triangle.

Here, O is the orthocenter.

—————————–

Perpendicular Bisector and Circumcenter

A line segment that passes through any vertex of the triangle to the middle point of the opposite side and cuts it at a right angle with it is called the perpendicular bisector.

In this figure, AD, CE, and BF are perpendicular bisectors.

The circumcenter is the intersection of all three perpendicular bisectors. it is also the center of the circumcircle.

Here, G is the circumcenter.

—————————

Median and Centroid

The line segment that joins any vertex of a triangle with the mid-point of the opposite side is called the median. It divides the opposite side into two equal halves.

Here, QU, PT, and SR are medians of ABC.

The centroid is the point of intersection of the three medians of a triangle. The centroid divides the medians in a 2:1 ratio of lengths.

Here, V is the centroid.

——————————–

Angle Bisector and incentre

The line segment that bisects the angle into two angles of equal measure of the vertex from which it is drawn is called the angle bisector.

The incentre is the intersection of the three angle bisectors. It is also the center of the incircle of the triangle.

Here, I is the incentre.

————————————

Congruency and Similarity

Congruency and similarity of triangles is an important concept in geometry basics for CAT. Here are some basics about the two concepts. A comparison is necessary because some candidates confuse the two concepts.

 Congruency Similarity Definition Two triangles are said to be congruent if they have the same size and shape. All interior angles and corresponding sides measure the same. Two triangles are said to be similar when all corresponding angles of the first triangle and the second triangle are equal and the lengths of corresponding sides in both the triangles are in the same ratio. Rule 1: SAS (side angle side) Two triangles are congruent if two sides of a triangle are equal to corresponding sides of another triangle and the angle between them is also equal. Two triangles are called similar when two corresponding sides in the two triangles are in the same proportion as each other and the corresponding angles are also equal. Rule 2: SSS (Side Side Side) Two triangles are SSS congruent if all three sides in one triangle are equal to the corresponding sides in another triangle. If three sides of the first triangle are in the same proportion with the corresponding three sides of the second triangle then, they are said to be SSS similar triangles. Rule 3: AAS/ AA (Angle Angle Side) If two corresponding angles are equal and one non-included corresponding side is equal in length in both triangles, then they are said to be AAS congruent. Two triangles are said to be AA similar if two pairs of angles equal Rule 4: ASA (Angle Side Angle) This rule is only for congruency. It states that if the corresponding two angles and the included side in between them in both the triangles are equal, they are congruent. There is no ASA rule for the similarity of triangles

Some important theorems:

There are countless theorems and formulae that we studied in mathematics. Especially in geometry. Many of these theorems were connected to the triangle because the triangle is the most basic two-dimensional shape. Let us learn about a few of them.

Angle Bisector Theorem:

It states that the angle bisector of any angle inside the triangle divides the side opposite to it into the ratio of the length of the sides making the angle. To take an example, in a triangle ABC, let AD be the angle bisector of angle A and AD divide the side BC in the ratio m:n, then,

This theorem is valid for both the interior and exterior angles of the triangle. The above-given figure is for the angle bisector bisecting the interior angle. For exterior angle,

For exterior angle, in triangle ABC, let D be a point extending the line CB, such that AD be the angle bisector bisecting angle CAD of this triangle externally. Let AD divide the side BC in n:m ratio then,

The Pythagoras Theorem:

Pythagoras theorem is one of the most well-known theorems in geometry basics for CAT. It states that in the case of a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The hypotenuse is the side opposite to right angle i.e. if there’s a triangle with a and b be the length of the sides containing the right angle and c be the length of the hypotenuse then,

This theorem is one of the most widely used theorems in geometry and related fields.

Apollonius’s Theorem:

The sum of the squares of any two sides of a triangle equals twice the sum of the square of half the third side and the median of the third side. In a triangle ABC, where AM is the median of that triangle such that BM=MC; then,

Mid-point Theorem:

According to the mid-point theorem, In a triangle, the line that joins the mid-point of the two sides is parallel to the third side and is half of it. Let there be a triangle ATV; and mid-points R and S such that AR=RT and AS=SV; then,

Basic Proportionality Theorem:

Basic Proportionality theorem states that inside a triangle if a line segment is drawn parallel to one side to intersect the other two sides, the other two sides get divided in the same proportion. Let there be a triangle ABC and DE be the line parallel to BC then,

==============================================

Now that we have read the necessary theory in geometry basics for CAT, let’s look at some practice questions to get an idea of the basics. As a rule, you should do as many practice questions as your timetable allows to get a hold of the geometry basics for CAT.

QUESTION: Given sides of a triangle ABC is 6, 10, and x. Find x for which area of the triangle is maximum?

1. √19
2. 12√3
3. √136

SOLUTION:

##### We can solve this problem with Heron’s formula but there is a simpler way

We know the length of sides of a triangle is 6, 10, and x.

=> Area = 1/2 * 6 * 10 * sin∠BAC.

The area is maximum when

sin∠BAC=1;

=>∠BAC = 90◦

x = √(100+36) = √136                        (by Pythagoras theorem)

Here D is the correct answer.

You should also try to solve using the heron’s formula

=================

QUESTION: Perimeter of an △ABC is 15. All sides have integral lengths. How many triangles can be made like that?

1. 7
2. 9
3. 8
4. 4

SOLUTION:

The solution to this question is pretty simple. You just have to use the trial and error method to satisfy the condition that the sum of two sides should be greater than the third side.

Let us assume a ≤ b ≤ c.

When, a = 1, Possible sides of triangle 1, 7, 7

When, a = 2, possible sides of triangle 2, 6, 7

When, a = 3, possible sides of triangles 3, 6, 6 and 3, 5, 7

When, a = 4, possible sides of triangles 4, 4, 7 and 4, 5, 6

When, a = 5, possible sides of triangle 5, 5, 5

So a total of 7 triangles are possible.

##### Therefore, the answer is 7.

Option A is the correct answer.

==========================

QUESTION: A triangle △ABC has a length of sides as integer a, b, c such that ac = 12. How many such triangles are possible?

1. 10
2. 8
3. 9
4. 12

SOLUTION:

This is again faster to solve with the trial and error method. Let’s begin with all the possible causes of sides ab of the triangle.

ac = 12

Lengths a,c can be one of these 1, 12 or 2, 6 or 3, 4

Possible triangles with the given condition

1, 12, 12

2 , 6 , 5;      2 , 6 , 6; 2 , 6 , 7

##### Hence, the correct answer is 9 triangles.

Option C is the correct answer.

================

QUESTION: ABCDE is a regular pentagon. O is a point inside the area of pentagon such that AOB forms an equilateral triangle(all sides equal). What is the value of ∠OEA?

1. 66
2. 62
3. 54
4. 75

SOLUTION:

##### Creating a new triangle will help in understanding.

Join OE and OD.

In a regular pentagon, all 5 sides are equal, hence the internal angle opposite to the sides are also equal.

Internal angle of regular pentagon = 108°

∠BAE = ∠CDE = 108°

=> ∠BAO = 60°

=> ∠OAE = 48°

OA = OB = AB since the triangle is equilateral.

And BA = EA in a regular pentagon.

=>Triangle AEO is isosceles as AO = EA.

Let ∠OEA = ∠EOA = a

In triangle AEO,

∠OAE + 2a = 180°

48° + 2a = 180°

2a = 132°, or a = 66°

##### Therefore, the correct answer is 66°

Option A is the correct answer.

==========================

QUESTION: Consider a right-angled triangle with an inradius 2 cm and a circumradius of 7 cm. What is the area of the triangle?

1. 32 sq cm
2. 33.5 sq cm
3. 32.5 sq cm
4. 31 sq cm

SOLUTION:

Given,

r = 2 and

R = 7 (Half of length of hypotenuse)

Hypotenuse = 14

r = (a + b – h)/2

=>  2 = (a + b – 14)/2

=>  a + b – 14 = 4

=>  a + b = 18

=>  a2 + ba2 = 142

=>  a2 + (18 – a)2 = 142

=>  a2 + 324 + a2 – 36a = 196

=>  2a2 – 36a + 128 = 0

=>  a2 – 18a + 64 = 0

By finding the roots of a quadratic equation,

The 2 roots to this equation will effectively be a, (18 – a).

Product of the roots = 64.

=>  Area = 1/2 * product of roots

= 32 sq. cms

##### Therefore, the answer is 32 sq. cms

Option A is the correct answer.

===================================

QUESTION: A triangle ABC has a perimeter of 6 + 2√3 . One of the interior angles of ABC is equal to the exterior angle of a regular hexagon. Another angle of the triangle is equal to the exterior angle of a regular 12-sided polygon. Find area of the triangle.

1. 2√3
2. √3
3. √3/2
4. 3

SOLUTION:

Given, Perimeter of the triangle = 6 + 2√3

Given, one of the angles in the triangle is equal to the exterior angle of a regular hexagon ie. 60°

And the second angle is equal to the exterior angle of a regular polygon of 12 sides ie. 30°.

From this, we can calculate that the third angle in the triangle is 90°.

In a triangle with angles 60,30,90, the sides are in the ratio √3a, a, and 2a.

=>  Perimeter is sum of all sides = a(3+ √3)= 6 + 2√3 .

=>  a = (6 + 2√3)/(3+ √3) = 2.

=>  the sides of the triangle are 2√3, 2 and 4.

Area of a Right Triangle = 1/2  X Product of smaller sides = 1/2  X 2 X 2√3 = 2√3 .

##### Therefore, the answer is 2√3.

Option A is the correct answer.

### Syllabus for CAT

Syllabus for Quantitative ability

• Number Systems
• Mean median mode (statistics)
• Averages
• Percentages
• LCM and HCF
• Time and Work
• Ratio and Proportion
• Profit, Loss, and Discount
• Speed, Time, and Distance
• Quadratic Equations & Linear Equations
• Complex Numbers
• Simple and Compound Interest
• Logarithms
• Sequences and Series
• Linear equations
• Inequalities
• Probability
• Surds and Indices
• Set Theory & Function
• Permutation and Combination
• Mixtures and Alligations
• Trigonometry
• Coordinate Geometry
• Geometry
• Mensuration

Syllabus for DILR

• Tables
• Coding-Decoding
• Letter Series
• Symbol Series
• Symbol based Logic
• Number & Alphabet Analogies
• Odd one out
• Direction Sense
• Blood Relations and Family Tree
• Cryptarithmetic (Verbal Arithmetic)
• Inequalities and Conclusions – Coded Inequalities
• Data Sufficiency
• Approximation of Values
• Caselets
• Bar Graphs
• Line Charts
• Column Graphs
• Venn Diagrams
• Pie Chart
• Calendars
• Number and Letter Series
• Clocks
• Cubes
• Seating Arrangement
• Binary Logic
• Logical Matching
• Logical Sequence
• Syllogism

Syllabus for VARC

• English Usage or Grammar
• Vocabulary (Synonyms/ Antonyms)
• Fill in the blanks
• Sentence Correction
• Jumbled Paragraph
• Meaning-Usage Match
• Analogies or Reverse Analogies
• Summary Questions
• Verbal Reasoning
• Facts-Inferences-Judgments
• Categories of Passages
• Writing Styles
• Tone of Writing
• Types of Questions
• The Articles – A, An, The
• Grammar rules
• Parts of Speech in English
• Sentence Construction in English
• Punctuations
• Modifiers
• Subject-Verb Agreement
• Verbs Tenses
• Word Usage
• Verbal Reasoning
• Logical Deduction
• Statements & Assumptions
• Courses of Action
• Para jumbles (Basic Rules, Extra tips)
• Para Completion
• Sentence Exclusion
• Fact Inference Judgment
• Syllogism
• Basic Assumption And Inference
• Paragraph Summary
• Method of Reasoning and Boldfaced
• Parallel, Further Application, Evaluate
• Fallacies
• Strong Weak Arguments

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###### Author: Gaurav Rayakwar
Gaurav is a Content Writer at IIM Skills. He has a B.Tech. degree but then he switched to the creative side by doing his master's in advertising and public relations. Gaurav is also a part-time blogger and graphic designer currently living in Mumbai 