The CAT: Mensuration basic concepts
CAT Mensuration is one of the geometry topics. Aspirants should excepts at least 56 questions for mensuration. As quantitative section contain total 34 question and out of that 56 are mensuration problems. Mensuration is all most formula based topic. Clearing your mensuration concepts will enhance your accuracy level in exam. Hence, Pay attention to concepts and formula’s. As you all know geometry topic contain 10% of weightage in CAT exam.
In this article, we learn about some the important dimensions of CAT mensuration. All important aspects related to the topic will be covered. Aspirant would understand how to approach question when his concepts become clear. In addition one could easily ace mensuration just by formula’s and basic concepts clarity.
Mensuration basic concepts for CAT exam.
Mensuration is a branch of mathematics. That teaches us about the length, volume, or area. And tell us about different geometric shapes. These shapes exist 2 forms. Namely 2 dimensions or 3 dimensions. Let us known about it further. First aspirant should known the difference between these two dimensions.
Differences Between 2D and 3D dimensions.
2D dimension  3D dimension 

If a shape is surrounded by three or more straight lines in a plane, then it is a 2D dimension  If a shape is surrounded by a no. of surfaces or planes then it is a 3D dimension 
These shapes have no depth or height.  These are also called solid shapes and unlike 2D they have height or depth. 
These shapes have only two dimensions say length and breadth.  These are called Three dimensional as they have depth (or height), breadth and length. 
We can measure their area and Perimeter.  We can measure their volume, CSA, LSA or TSA. 
Important Terminologies of Mensuration for CAT exam.
Before start study formula’s aspirant should be aware of basic terminologies. Hence, Have a look on below mention table.
Terminologies  Abbreviation examiner could use.  Unit in which answer should denoted.  Explanation 
Area  A  m^{2} or cm^{2}  The area is the surface which is covered through the closed shape. 
Perimeter  P  cm or m  The measure of the continuous line along the boundary of the given figure. 
Volume  V  cm^{3} or m^{3}  The space occupied by a 3D shape is called a Volume. 
Curved Surface Area  CSA  m^{2} or cm^{2}  If there’s a curved surface, then Curved Surface Area is the total area. For Example: Sphere 
Lateral Surface area  LSA  m^{2} or cm^{2}  Lateral Surface area is the total area of all the lateral surfaces that surrounds the given figure. 
Total Surface Area  TSA  m^{2 }or cm^{2}  Total Surface area is the sum of all the curved and lateral surface areas. 
Square Unit  –  m^{2 }or cm^{2}  A Square unit is the area covered by a square of side one unit. 
Cube Unit  –  m^{3 }or cm^{3}  The volume occupied by a cube of one side one unit 
3D Shape Important Formula’s of mensuration for CAT exam.
Learn these formula’s. These formula’s would help you in solving formula based questions. Paste this table on your study table so you could revise them on regular basics. Lets have a look on below mention table:
Shapes 
Total Surface Area 
Lateral/ Curved Surface area 
Volume 
Length of Leading Diagonal/ Slant Height 
Cube 
2(LB+ BH+ HL) 
2H (L + B) 
LBH 
√ (L^{2} + H^{2} + B^{2}) 
Cuboid 
6a^{2} 
4a^{2} 
a^{3} 
√3a 
Cylinder 
2Πr (r + h) 
2Πrh 
Πr^{2}h 
No Slant height or 
Cone 
Πr (r + l) 
Πrl 
⅓Πr^{2}h 
√(h^{2} + r^{2}) 
Sphere 
4Πr^{2} 
4Πr^{2} 
^{4}/_{3}Πr^{3} 
No Slant height or 
Hollow Cylinder 
2Π(r₁+r₂) (r₂r₁+h) 
2Πh(r₁+r₂) 
Πh(r₂²r₁²) 
No Slant height or 
Frustum 
Π(R_{1} + R_{2})s + (R_{1}^{2} + R_{2}^{2}) 
Π(R_{1} + R_{2})s 
⅓Πh(R_{1}^{2} + R_{2}^{2} + R_{1}R_{2}) 
√(h^{2} + (R_{1} – R_{2})^{2}) 
Hemisphere 
3Πr^{2} 
2Πr^{2} 
^{2}/_{3}Πr^{3} 
No Slant height or 
2D Shape Important Formula’s of mensuration for CAT exam.
Learn these formula’s. These formula’s would help you in solving formula based questions. Paste this table on your study table so you could revise them on regular basics. Lets have a look on below mention table:
Area of shapes
An expression of the size of a twodimensional surface or shape in a plane is the term “area.” This is measured in square unit like cm^{2}, m^{2}, etc.

To calculate Area of Rectangle. Use this formula = Length X Breadth

To calculate Area of Triangle. Use this formula = 0.5 X Base X Height

To calculate Area of Square. Use this formula =Side X Side

To calculate Area of Circle. Use this formula =Pi X Radius X Radius

To calculate Surface Area of a Cylinder. Use this formula = 2 X Pi X Radius X (Radius + Height)

To calculate Surface Area of a Sphere. Use this formula = 4 X Pi X Radius X Radius

To calculate Surface Area of a Cube. Use this formula = 6 X Side X Side

To calculate Surface Area of a Cuboid. Use this formula = 2 X (Length X Breadth + Breadth X Height + Length X Height)
Perimeter of shapes:
A perimeter is a path that surrounds a twodimensional shape, it’s SI unit will be the meter itself.

Perimeter of Circle: 2 X Pi X Radius

Perimeter of Triangle: Side A + Side B + Side C

Perimeter of Square: 4 X Side

Perimeter of Rectangle: 2 X (Length + Breadth)
Also read: How to Prepare for CAT 2022
Some Important past year question from CAT exam.
For Example 1. A solid right circular cone of height 27 cm is cut into pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is
A) 243 B) 232 C) 256 D) 264 [CAT 2020]
Answer: If height becomes 1/3rd then radius also becomes 1/3rd
If both height and radius become 1313rd, Volume will be 1/27th
So, Top part’s volume = V
Remaining part = 26V
Totally = 27V
Given 26 V – V = 225
Hence, V = 9
For Example 2. The length, breadth, and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will (CAT 19 – Slot 1)
1. Remain the same
2. Dec. by 13.64%
3. Dec. by 15%
4. Dec. by 18.75%
5. Dec. by 30%
Answer: in the present case, let length = l = 3x, breadth = b = 2x, height = h = x
then, area of four walls = 2 (l + b) h = 2(3x + 2x) x = 10×2.
now as length gets doubled = 6x, breadth halved = x, height halved = x/2.
new area of four walls = 2 (6x + x) x/2 = 7×2.
thus there is a decrease of 30%. hence, the fifth option is the answer.
Few More questions for practice.
For Example 3. Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is
A) 3:4 B) 2:3 C) 5:6 D) 4:5 [CAT 2019]
Answer: In given case , figure can be drawn as below
Let side of equilateral triangle ABC = 3a
So side of hexagon = a
Area of triangle = (root3 /4 )*(3a)^2 = 9a^2 *(root3/4)
Area of hexagon with side a = 6*(root3 /4 )*(a)^2
Ratio of areas of hexagon to that of triangle = { 6*(root3 /4 )*(a)^2 } / { 9a^2 *(root3/4)} = 6/9 = 2/3
Thus required ratio = 2 : 3
For Example 4. A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is [CAT 2019]
A) 8464π B) 928π C) 1026(1 + π) D) 1044(4 + π)
Answer: As cylinders have the same volume and each of these has radius 3 cm. So volume of each cylinder will be equal to HCF of (405, 783 and 351) which is 27.
So volume of each cylinder = 27
No of cylinders = [ 405/27 ] + [783/27] + [351/27] = 15 + 29 + 13 =57
Using V = πr^2 h
27 = 22/7 *9 *h
So h = 3/ π cm.
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