Great Lakes - PGDM & PGPM Admissions 2026
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Geometry is one of the key areas to focus on for CAT 2025, as evident from the previous years’ exam pattern. If you have decided to prepare for CAT, you are at the right place. This article will guide you in solving some of the most common queries such as the important Geometry theorems you should learn first for CAT, the topics most frequently asked from Geometry, the frequency of Geometry questions in the exam, and how mastering this area can impact your overall percentile in CAT.
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Geometry is very important for CAT in certain ways:
Geometry constitutes a healthy weightage in CAT exam.
Geometry is interconnected with the other topics such as Mensuration, Geometry, and Coordinate geometry.
Geometry questions require conceptual clarity rather than mathematical calculations.
It creates a foundation for other topics as well.
S. No. | Year | Slot | Number of questions | Easy | Moderate | Hard |
1 | 2020 | 1 | 0 | NA | NA | NA |
2 | 2020 | 2 | 3 | 1 | 2 | - |
3 | 2020 | 3 | 0 | NA | NA | NA |
4 | 2021 | 1 | 3 | 1 | 1 | 1 |
5 | 2021 | 2 | 1 | - | 1 | - |
6 | 2021 | 3 | 3 | 1 | 1 | 1 |
7 | 2022 | 1 | 2 | 2 | - | - |
8 | 2022 | 2 | 3 | 2 | 1 | - |
9 | 2022 | 3 | 3 | 1 | 2 | - |
10 | 2023 | 1 | 2 | 1 | 1 | - |
11 | 2023 | 2 | 2 | 1 | 1 | - |
12 | 2023 | 3 | 3 | 1 | 1 | 1 |
13 | 2024 | 1 | 1 | - | 1 | - |
14 | 2024 | 2 | 2 | 1 | 1 | - |
15 | 2024 | 3 | 3 | 1 | 1 | 1 |
Topic | Repeated Concepts |
Triangles | Sine rule, BPT, Midpoint theorem, Geometrical centres, median, altitude, Pythagoras theorem, Properties of triangles, exterior angle sum property, Similarity of triangle, Congruency, Inradius, Circumradius |
Regular Polygon | Interior angles, exterior angles, relation between number of sides with interior and exterior angles |
Quadrilateral | Cyclic quadrilateral, Properties related to sum of sides |
Circles | Tangent from external point, Intersecting circles, relation between chord and tangent, perpendicular to the chord from centre, Sector and Segment |
Since Geometry is very vast topic in CAT Quantitative Aptitude, it contains several theorems and concepts to understand. You will find important theorems for CAT Geometry preparation in this article. You should have knowledge of basic concepts like
Relation between angles between two parallel lines
Angle sum properties
Exterior angle sum property
Similarity and Congruency of triangles
Properties related to quadrilaterals and Polygons
Properties of Circles, Sector, and Segment.
In this section, we are going to discuss 5 important theorems that covers almost 80% of CAT questions from geometry.
The five theorems that we are going to discuss in this article covers almost 80% of the CAT geometry questions since these theorems are most versatile.
For example: Pythagoras theorem is important for the questions related to triangle, trigonometry, quadrilateral, and circles.
The Pythagorean theorem states that in a right-angled triangle; the square of the longest side is equal to the sum of the squares of the other two sides.
The longest side in a right-angled triangle is known as Hypotenuse.
Also, if in a triangle, $a^2 + b^2 > c^2$, where '$c$' is the longest side of the triangle, the triangle is obtuse angled triangle.
if in a triangle, $a^2 + b^2 < c^2$, where '$c$' is the longest side of the triangle, the triangle is acute angled triangle.
Pythagoras theorem is directly used in a right-angled triangle:
To find the third side if two sides are given.
To check whether the given triangle is right angled triangle or not.
If one of the acute angles is given and we need to relate all the sides of an equilateral triangle.
For Example:
In a triangle ABC, right angles at A, angle B is 30o, determine the ratio of all the sides.
Solution:
For angle B, Base is AB and Perpendicular is AC.
So, $\sin 30^\circ = \frac {AC}{AB}$
⇒ $\frac 12 = \frac {AC}{AB}$
Let $AB = 2, AC = 1$, then $BC^2 = AC^2 + AB^2$
So, $AC = \sqrt {2^2 - 1^1} = \sqrt 3$
In CAT, the questions on Pythagoras theorem, are asked on
The application of Pythagoras theorem in right angled triangle to find the unknown side
The application of Pythagoras theorem in quadrilateral and circles
Q.1) In a right-angled triangle ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is:
A) 2
B) 3
C) 4
D) 4
Solution:-
Area of ∆ABC = $\frac{1}{2} \times 5 \times 12$ = 30 sq cm
So, Area of ∆ABP = $\frac {30}{1.5}$ = 20 sq cm
And, Area of ∆ABQ = $\frac {30 +20}{2}$ = 25 sq cm (Since the areas are in AP)
Now, Area of ∆ABP = $\frac{1}{2} \times 5 \times BP$ = 20 sq cm
⇒ $BP = 8$ cm
Now, Area of ∆ABQ = $\frac{1}{2} \times 5 \times BQ$ = 25 sq cm
⇒ $BQ = 10$ cm
So, PQ = $BQ-BP = 10 -8=2$ cm
Hence, the correct answer is option (1).
Q.2) In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP: PQ: QC is
A) 1:1:2
B) 1:2:4
C) 2:4:1
D) 2:4:1
Solution:-
The areas of the figures $ABP$, $APQ$, and $AQCD$ are in geometric progression.
Area of $ABP = m$.
Area of $APQ = mr$.
Area of $AQCD = mr^2$.
Also, Area of $AQCD = 4 \times$ (Area of $ABP$),
$\Rightarrow mr^2 = 4m$,
$\Rightarrow r = 2$.
So, the total area = Area of rectangle = $9 \times 6 = 54$.
So, $m + 2m + 4m = 54$,
$\Rightarrow m = \frac{54}{7}$.
Area of $ABP = m = \frac{54}{7}$,
$\Rightarrow \frac{1}{2} \times AB \times BP = \frac{54}{7}$.
$\Rightarrow BP = \frac{12}{7}$ (since $AB = 9$).
Area of $ABQ = 2m + m = \frac{162}{7}$,
$\Rightarrow \frac{1}{2} \times AB \times BQ = \frac{162}{7}$.
⇒ BQ = $\frac{36}{7}$ (Since AB = 9)
Now, PQ = BQ - BP =$\frac{36}{7}$ - $\frac{12}{7}$ = $\frac{24}{7}$
QC = BC - BQ = 6 - $\frac{36}{7}$ = $\frac{6}{7}$
Now, BP: PQ: QC = ($\frac{12}{7}$) : ($\frac{24}{7}$) : ($\frac{6}{7}$) = 2:4:1
Hence, the correct answer is option (3).
To solve the questions based on Pythagoras theorem quickly
Learn squares up to 30.
Learn Pythagorean triplets (Pythagorean triplets contain three numbers satisfying the Pythagoras theorem)
(3, 4, 5) | (6, 8, 10) | (5, 12, 13) | (8, 15, 17) |
(7, 24, 25) | (10, 24, 16) | (12, 35, 37) | (9, 40, 41) |
(15, 36, 39) | (11, 60, 61) | (20, 21, 29) | (20, 99, 101) |
(16, 63, 65) | (17, 144, 145) | (19, 180, 181) | (13, 84, 85) |
Triangles having exact similar shape but different sizes are similar triangles.
Two triangles are said to be similar if either their corresponding angles are equal or corresponding sides are in equal proportion.
Similarity Criteria | Description |
AA | If two corresponding angles are equal then the triangles are similar |
SSS | If the ratio of corresponding sides is equal then the triangles are similar. |
SAS | If the ratio of two corresponding sides is equal and the angle between these two sides are equal then the triangles are similar. |
Also,
If triangle ABC is similar to triangle PQR, then
$(\frac {AB}{PQ})^2 = (\frac {BC}{QR})^2 = (\frac {AC}{PR})^2 =\frac {ar(ABC)}{ar(PQR)} $
and $(\frac {AB}{PQ}) = (\frac {BC}{QR}) = (\frac {AC}{PR}) =\frac {ar(ABC)}{ar(PQR)} $
Application of similar triangles in the questions of height and distance can be understood with the help of following example:
A boy of height 120 cm is walking away from the base of a tower at a speed at 0.8 m/sec. If the tower is 2.8 m above the ground, find the length of the shadow of boy after 5 seconds.
From the information given in the question
The larger triangle and smaller triangle are similar as one angle of each is of 90 degrees and the other is common.
So, $\frac {280}{120} = \frac {4+x}{x}$
So, $x = 3$ metres
The Basic Proportionality Theorem (Thales’ theorem), states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
If PQ || BC, then $\frac {AP}{PB} = \frac {AQ}{QC}$
Some Important results derive from BPT:
1. $\frac {AP}{AB} = \frac {AQ}{AC}$
2. If P and Q are mid points of AB and AC respectively, then PQ = ½ BC.
In CAT, direct questions from BPT are rarely asked. They integrate BPT with the other topics like
Complex diagram of quadrilateral in which BPT will be used to relate the sides of triangles formed inside the quadrilateral to find the unknowns.
In cyclic quadrilaterals in which diagonals are intersecting.
In circles to relate chords, radius, and tangents.
In coordinate geometry where equation of two or more intersecting lines are given in a plane and we have to determine the coordinates of intersecting points or lengths of sides forming triangles to find the area etc.
Tangent is a line that touches a circle at one point only while a secant intersects a circle at two points.
Radius is perpendicular to the tangent at point of contact. Questions based on tangent and radius relations are frequently asked. An important relation is given below:
$OP$ is perpendicular to $PQ$.
So, $OQ^2 = OP^2 + PQ^2$
Perpendicular to the chord from the centre of a circle bisects the chord. An important relation is given below:
Point theorem is applied to find the length of a line segment, chord, and tangent in CAT questions.
A point outside the circle:
A tangent drawn from point B outside the circle at point A, and a secant from B intersects the circle at C and D. Then $AB^2 = BC \times BD$.
Two secants drawn from a point outside the circle:
Two secants from point C intersect the circle at A, B and D, E respectively. Then CA × CB = CD × CE.
Two intersecting chords inside the circle:
Two chords AB and CD intersecting at a point P inside a circle, then PA × PB = PC × PD.
Tangent - Secant theorem:
A tangent drawn from point P outside the circle at point C, and a secant from P intersects the circle at B and A. Then $PC^2 = PA \times PB$.
There are several formulas to find the area of a triangle but Herons formula can be used for any triangle.
If the sides of a triangle are a, b, and c, then the area of a triangle is given by
$\text {Area of a triangle} = \sqrt {s(s-a)(s-b)(s-c)}$
where s is semi perimeter of the triangle given by
$s=\frac {a+b+c}{2}$
Median is a line joining a vertex of triangle to the mid-point of the opposite side. The intersection point of all medians is known as Centroid. Centroid divides a median in the ratio 2 : 1 from vertex to the side.
There are some concepts and formulas used to determine the area of a triangle effectively.
1. For scalene triangle (three sides are known):
Use Herons formula (Discussed above)
2. For an equilateral triangle (Side is known)
Area = $\frac {\sqrt 3}{4} a^2$, where $a$ is the side of the triangle.
3. Two sides and angle between them is known
Area = $\frac{1}{2}$ × product of two sides × sine of the included angle
4. If inradius ($r$) and semi perimeter (s) is known
Area = rs
5. If circumradius ($R$) and three sides are known
Area = $\frac{abc}{4R}$, where R = circumradius
6. Universal formula to determine the area of a triangle
Area = $\frac{1}{2}$ × base × height
Herons’ formula can be used to find the area of a quadrilateral and polygon. To find the area of a quadrilateral and polygon, divide it in to triangles and apply the Herons formula to each triangle.
Example: Dimensions of quadrilateral are as shown in the figure. Find the area.
Using Pythagoras theorem, ${BD}^2 = 6^2 + 8^2$ ⇒ BD = 10
Area of ABCD = Area of ΔABD + Area of ΔBCD
Area of ΔABD = $\frac 12 \times 6 \times 8$ = 24 sq units
Find area of triangle BCD using Heron’s formula
s = (6 + 10 + 10)/2 = 13
Area of ΔBCD = $\sqrt {13(13-6)(13-10)(13-10)}$
$= 3 \sqrt {91}$ sq units
So, Area of ABCD = 24 + 3 $\sqrt {91}$ sq units
Similarly, we can determine the area of a polygon.
Q.1) From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is:
A) $\mathrm{\frac{\sqrt{3} s^2}{2}}$
B) $\mathrm{{\frac{2 s^2}{\sqrt{3}}} }$
C) $\mathrm{\frac{s^2}{2 \sqrt{3}} }$
D) $\mathrm{\frac{s^2}{2 \sqrt{3}} }$
Solution:-
Based on the question: A D, C E and B F are the three altitudes of the triangle. It has been stated that
$\mathrm{ \{\mathrm{GD}+\mathrm{GE}+\mathrm{GF}=\mathrm{s}\} }$
Now since the triangle is equilateral, let the length of each side be "a".
So the area of the triangle will be
$\mathrm{ \frac{1}{2} \times G D \times a+\frac{1}{2} \times G E \times a+\frac{1}{2} \times G F \times a=\frac{\sqrt{3}}{4} a^2 }$
Now, $\mathrm{ G D+G E+G F=\frac{\sqrt{3} a}{2} \text { or } s=\frac{\sqrt{3} a}{2} \text { or } a=\frac{2 s}{\sqrt{3}} }$
Given the area of the equilateral triangle $\mathrm{ =\frac{\sqrt{3}}{4} a^2 }$
substituting the value of 'a' from above, we get the area { in terms 's'} $\mathrm{ =\frac{s^2}{\sqrt{3}}}$
Hence, the correct answer is option (4).
Q.2) Let C1 and C2 be concentric circles such that the diameter of C1 is 2 cm longer than that of C2. If a chord of C1 has length 6 cm and is tangent to C2, then the diameter, in cm, of C1 is:
A) 10
B) 8
C) 12
D) 12
Solution:-
Now we know that the perpendicular from the centre to a chord bisects the chord. Hence at the point of intersection of tangent, the chord will be divided into two parts of 3 cm each. As you can clearly see in the diagram, a right-angled triangle is formed there.
So, $\mathrm{(r+1)^2=r^2+9 \text { or } r^2+1+2 r=r^2+9 \text { or } 2 r=8 \text { or } r=4 \mathrm{~cm}}$
Therefore, the radius of the larger circle is 5 cm and the diameter is 10 cm.
Hence, the correct answer is option (1).
Q.3) Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the midpoint of CD, then the length of AT, in cm, is:
A) $\sqrt{13}$
B) $\sqrt{14}$
C) $\sqrt{12}$
D) $\sqrt{12}$
Solution:-
Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangles forming the hexagon.
Hence, its length should be twice the side of the hexagon, in this case, 4 cm.
Now, AD divided the hexagon into two symmetrical halves.
Hence, AD bisects angle D, and hence, $\angle$ADC is 60°.
We can find out the value of AT using the cosine formula:
$\begin{aligned} & \mathrm{AT}^2=4^2+1^2-2 \times 1 \times 4 \cos 60° \\ & ⇒\mathrm{AT}^2=17-4=13 \\ & \therefore A T=\sqrt{13}\end{aligned}$
Hence, the correct answer is option (1).
Q.4) A circle of diameter 8 inches is inscribed in a triangle ABC, where $\angle A B C=90^{\circ}$. If BC = 10 inches, then the area of the triangle in square inches is:
A) 120
B) 100
C) 80
D) 80
Solution:-
Diameter = 8 inches; So, radius = 4 inches
Let p be the perpendicular and h be the hypotenuse of the triangle.
We know that inradius = $\frac{(\text { Perpendicular }+ \text { Base }- \text { Hypotenuse })}{2}$
$
⇒4=\frac{(p+10-h)}{2}
$
$
⇒\mathrm{h}-\mathrm{p}=2$
$ ⇒\mathrm{h}=\mathrm{p}+2
$
Now, $\mathrm{P}^2+100=\mathrm{h}^2$
$
\begin{aligned}
&⇒ p^2+100=(p+2)^2 \\
& ⇒p^2+100=p^2+4 p+4 \\
&⇒ 4 p=96 \\
& \therefore p=24
\end{aligned}
$
Hence, area of the $\triangle$ABC $=\frac{1}{2} \times 10 \times 24=120$
Hence, the correct answer is option (1).
Q.5) All the vertices of a rectangle lie on a circle of radius $r$. If the perimeter of the rectangle is P, then the area of the rectangle is:
A) $\frac{P^2}{16}-r^2$
B) $\frac{P^2}{8}-2 r^2$
C) $\frac{P^2}{2}-2 P r$
D) $\frac{P^2}{2}-2 P r$
Solution:-
Let the length and breadth of the rectangle be $l$ and $b$, respectively.
So, the area of the rectangle = $lb$
$\begin{aligned} &\text{Now, } l^2+b^2=4 r^2 \\ &\text{Also, } \mathrm{P}=2(l+b) \\ &⇒ \frac{P}{2}=l+b\end{aligned}$
Squaring on both sides, we get
$\begin{aligned} & \frac{P^2}{4}=l^2+b^2+2 l b \\ & ⇒\frac{P^2}{4}=4 r^2+2 l b \\ &⇒ \frac{P^2}{8}-2 r^2=l b\end{aligned}$
So, the area of the rectangle is $\frac{P^2}{8}-2 r^2$.
Hence, the correct answer is option (2).
Q.6) Suppose the medians BD and CE of a triangle ABC intersect at a point O. If the area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is:
A) 9
B) 12
C) 15
D) 15
Solution:-
Given: The medians BD and CE of a triangle ABC intersect at a point O and the area of triangle ABC is 108 sq. cm.
(Area of ABD):(Area of BDC) $=1: 1$
Therefore, the area of $A B D=54$
Area of $A D E$ : Area of $E D B=1: 1$
Therefore, the area of $A D E=27$
$\mathrm{O}$ is the centroid and it divides the medians in the ratio of 2:1
So, (Area of BEO):(Area of EOD) $=2: 1$
⇒ Area of EOD $=9$
Hence, the correct answer is option (1).
Q.7) $A B C D$ is a rectangle with sides $A B=56 \mathrm{~cm}$ and $B C=45 \mathrm{~cm}$, and $E$ is the midpoint of side $C D$. Then, the length, in cm , of radius of incircle of $\triangle A D E$ is
Solution:-
Given: $AB = 56$ cm, $BC = 45$ cm, so $AD = 45$ cm, $CD = 56$ cm.
$E$ is the midpoint of $CD \Rightarrow CE = 28$ cm.
In triangle $ADE$, use Pythagoras theorem:
$DE^2 = AD^2 + CE^2$
$⇒DE^2 = 45^2 + 28^2 = 2025 + 784 = 2809$
$⇒DE = \sqrt{2809} = 53$ cm
Using inradius formula for triangle with sides $a$, $b$, $c$:
$r = \frac{a + b - c}{2}$, where $c$ is the longest side
Here, $a = 45$, $b = 28$, $c = 53$
So, $r = \frac{45 + 28 - 53}{2} = \frac{20}{2} = 10$ cm
Hence, the correct answer is $10 \text{ cm}$.
Q.8) A quadrilateral ABCD is inscribed in a circle such that AB:CD = 2:1 and BC:AD = 5:4. If AC and BD intersect at the point E, then AE:CE equals:
A) 1:2
B) 5:8
C) 8:5
D) 8:5
Solution:-
In $\triangle AED$ and $\triangle BEC$
$\angle A D E=\angle B C E$ (Angles made by same are on the same side of the triangles are equal)
$\angle A E D=\angle B E C$ (Vertically opposite angles are equal)
$\triangle A D E \sim \triangle B C E$ (By AA similarity criteria)
So, $\frac{A D}{B C}=\frac{D E}{C E}$.............(1)
In $\triangle C E D$ and $\triangle BEA$
$\angle E D C=\angle B A E$ (Angles made by same are on the same side of the triangles are equal)
$\angle C E D=\angle B E A$ (Vertically opposite angles are equal)
$\triangle C D E \sim \triangle B A E$ (By AA similarity criteria)
So, $\frac{C D}{A B}=\frac{D E}{A E}$..................(2)
From (1) and (2), we get, $\left(\frac{A D}{B C}\right)\left(\frac{A B}{C D}\right)=\left(\frac{D E}{C E}\right)\left(\frac{A E}{C E}\right)$
So, $\left(\frac{A E}{C E}\right)=\left(\frac{4}{5}\right)\left(\frac{2}{1}\right)=\frac{8}{5}$
Hence, the correct answer is option (3).
To solve the questions on Geometry effectively, use diagrams to interpret the question, eliminate inappropriate options, learn formulas, understand the questions properly.
In MCQs, try to eliminate wrong options by using
Ratios
Using angle sum property in the questions where an unknown angle needs to determine.
Using the constraints given in the question.
Approximations and diagrams for speed
In the questions related to geometry, drawing diagrams is one of the key aspects.
To increase speed, draw an approximate and rough diagram that should make you understand the problem.
Label the key points, angles, and measurement in the diagram.
There are hundreds of formulas in Geometry. You need to memorize the key formulas of the concepts that are frequently asked in CAT quantitative aptitude.
There are some questions that can be solved using concepts easily. For such questions, avoid the formulas.
In the next section, we are going to summarize the key formulas.
Concept | Important Formulas | ||||||||||||||||||||||||
Triangle |
| ||||||||||||||||||||||||
Quadrilaterals and Polygons | Area of a quadrilateral = $\frac{1}{2} \times (\text{product of diagonals}) \times (\sin \text{ of the angle between them})$. A parallelogram circumscribed about a circle is a rhombus. Median (for a trapezium) = $\frac{1}{2} \times$ sum of the parallel sides (median is the line equidistant from the parallel sides). Properties related to diagonals
Properties of a Regular Polygon Number of diagonals in an $n$-sided regular polygon = $\frac{n(n-3)}{2}$. Measure of an interior angle of a regular polygon = $\frac{(n-2)180^\circ}{n}$. Measure of an exterior angle of a regular polygon = $\frac{360^\circ}{n}$. | ||||||||||||||||||||||||
Circles |
of the common chord.
Properties related to Cyclic Quadrilateral
Properties related to Tangents and Secants
|
Mistakes | How to Rectify |
Over-relying on memorisation vs practice | Do not ignore basic concepts, during practice try to solve questions conceptually rather than using formula. |
Ignoring Visual diagram practice | Practice by drawing approximate diagrams to solve geometry questions. |
Failing to connect theorems with real CAT questions | Include real CAT problems in your practice to understand the application of theorems in real CAT questions. |
We divide our strategy to prepare Geometry for CAT in 3 steps:
Step 1: Work on to build fundamentals:
First, understand the concepts related to angles between two parallel lines and intersecting lines such as corresponding angles, alternate interior angles, vertically opposite angles etc. Also, learn complementary angles and supplementary angles.
Understand theorems related to triangles, quadrilaterals, and Circles.
Step 2: Gain mastery in formulas and drawing diagrams:
Prepare a list of all formulas and practice questions using diagrams. This will help you in saving time and increase accuracy.
Step 3: Taking Mock tests and analyse them:
Include practice tests and CAT Mock tests in your preparation to identify your weak areas to improve. Also make a log to note down the mistakes you are making regularly. Work on the mistakes to rectify. Also, monitor your score after each test and give a direction to your preparation accordingly.
1. NCERT Mathematics Class 9 and 10
2. How to prepare quantitative Aptitude for CAT by Arun Sharma.
3. Quantitative Aptitude for CAT by Nishit K Sinha
The candidates can download the various CAT preparation resources designed by Careers360 using the links given below.
eBook Title | Download Links |
CAT 2025 Arithmetic Important Concepts and Practice Questions | |
CAT 2025 Algebra Important Concepts and Practice Questions | |
CAT 2025 Number System - Important Concepts & Practice Questions | |
CAT 2025 Exam's High Scoring Chapters and Topics | |
CAT Mock Test Series - 20 Sets, Questions with Solutions By Experts | |
Mastering CAT Exam: VARC, DILR, and Quant MCQs & Weightages | |
CAT 2025 Mastery: Chapter-wise MCQs for Success for VARC, DILR, Quant | |
CAT 2025 Quantitative Aptitude Questions with Answers | |
CAT 2025 Important Formulas | |
Past 10 years CAT Question Papers with Answers | |
CAT 2025 Quantitative Aptitude Study Material PDF - Geometry and Mensuration |
Frequently Asked Questions (FAQs)
There are few suggestions:
Draw rough and clear diagrams
Check each angle and measurements
Memorize all important formulas and use them correctly
Eliminate options
Questions on Geometry in CAT are generally moderate to tough. But with the right preparation you will be able to solve them.
You should focus mainly on
Application of Pythagoras theorem
BPT and Midpoint theorem
Similarity of triangles
Theorems related to Circles
Quadrilateral and Polygons
On Question asked by student community
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Hello dear candidate,
YES, with NEET score 344in the ST category you have high chances of getting admission in a BDS College in J&K through JKBOPEE counselling.
check the seat availability and next participate in JKBOPEE Counselling with all the necessary documents.
Thank you.
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India’s first Urban Multidisciplinary University | Case-based pedagogy | Industry-endorsed curriculum
Highest Package 27.25 LPA | Top 100 Average package 16.65 LPA | AACSB Accredited | Ranked 52 by QS International