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Mensuration Basics and 3-Dimensional Geometry Concepts for CAT

Tuesday, February 5th, 2019

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Geometry is an integral part of mathematics and mathematicians have been studying and formulating important results to simplify for years. Since, it is immensely significant it has been part of our curriculum as soon as we join middle school, though the concept taught in early years of middle school are much simpler and basic than what is taught at high and secondary years of our schooling. Because of the inevitability of deep rooted of Geometric concepts, it forms a significant part in the course structure of various competitive exams such as CAT, XAT, SSC, BANK PO, etc. Nearly 4-5 questions are asked every year on this branch of mathematics with wide ranging difficulty level from medium to hard. Geometry is that section of math’s which deals with the deduction of properties, measurement and relationship of points, lines, angles, solids and figures in space from their defining conditions by means of certain assumed properties of space. And in this blog, we will discuss about this specific measurement function of geometry that itself forms a subdivision of it and named as Mensuration.

Mensuration deals with measurement of 3D solids in terms of Total surface area, Lateral/Curved surface area and volume. Given below is the table with all the formulas of T.S.A, C.S.A and Volume of various solids.

SOLID	Total Surface Area	Lateral/ Curved Surface area	Volume	Length of Leading Diagonal/ Slant Height
	2(LB+ BH+ HL)	2H (L + B)	LBH	√ (L2 + H2 + B2)
	6a2	4a2	a3	√3a
	2Πr (r + h)	2Πrh	Πr2h	No Slant height or diagonal
	Πr (r + l)	Πrl	⅓Πr2h	√(h2 + r2)
	4Πr2	4Πr2	4/3Πr3	No Slant height or diagonal
	2Π(r₁+r₂) (r₂-r₁+h)	2Πh(r₁+r₂)	Πh(r₂²-r₁²)	No Slant height or diagonal
	Π(R1 + R2)s + (R12 + R22)	Π(R1 + R2)s	⅓Πh(R12 + R22 + R1R2)	√(h2 + (R1 – R2)2)
	3Πr2	2Πr2	2/3Πr3	No Slant height or diagonal

 

You might be acquainted with all the solids given in the above table and have done questions in your high school on them but now let’s learn about some more solids that you might be unaware of.

PRISM

Everyone must have heard this name Prism and might now a little about it through Physics or may be some of you have don’t remember anything at all. Today, I’ll tell you geometric properties of Prism.

A Prism is a polyhedron comprising n-sided polygonal base congruent with a face parallel to it whose lateral faces are parallelogram and has a same cross section all along its length.

The number of lateral faces is equal to number of sides in the polygonal base. Thus, the base could be triangular, quadrangular, hexagonal etc.

A prism is said to be right, if the side edges are perpendicular to the ends. For example, cuboids and cubes.

MENSURATION OF PRISM

Curved Surface area of Prism: perimeter of the base x Height

Total Surface area of Prism: Lateral surface area + (2 x base area)

Volume of Prism: Base area x height

PYRAMID

With the word pyramid what first pops up in our mind are pyramids of Egypt. And you are right, pyramid is of the similar in structure with the pyramids of Egypt. But its just that pyramids of Egypt are one kind of pyramids with square base. Mathematically, a pyramid is a polyhedron formed by connecting a polygonal base and a point called apex.  The lateral surface of pyramids form triangles.

A pyramid with n-sided base has n+1 vertices, n+1 faces, and 2n edges.

A right pyramid has its apex directly above the centroid of its base.

MENSURATION OF PYRAMID

Curved surface area of Pyramid: perimeter/2 x slant height

Total surface area of Pyramid: curved surface area + area of the base

Volume of the Pyramid: Base area x ⅓ x height

 Since now we are familiar with all the solids and the formulas to calculate their surface area and volume we can move on to try solving some questions but before moving forth to that keep these important points in your mind that will assist you in solving questions in simplified manner.

Points To remember: –

• Diagrammatic representation: As we are talking about 3D here, it’s crucial that we visualize the objects specified in the question. If we are unable to diagrams from the information given in the question then the whole exercise is futile. Thus, the pictorial representation of solid is the key piece in solving the question. If we are able to clearly define the interconnectedness between different solids half of the question is done. What left is to plug in the formulas using the given and interpreted parameters and you are done!
• Symmetry: Solids such as cube, spheres etc. are symmetric and thus questions can be solved with the help of symmetry. But to make use of symmetry make sure you understand the concept properly and know it’s application as well.
• If H is the height of the complete cone from which the frustum is cut, then from similar triangles

r/R = (H-h)/h

1. If a sphere is inscribed inside the cube of side a, then the radius of the sphere will be a/2. If the sphere is circumscribed about the cube of side a then the radius of sphere will be √3a/2.
2. If the largest possible sphere is inscribed in the cylinder of radius ‘R’ and height ‘h’, its radius r will be

• r = h/2 (if 2R>h)
• r = R (if 2R<h)

 Let’s do some problems now on these solids.

Problem 1: If three cubes are placed adjacently in a row, then the ratio of the total surface area of the new cuboid to the sum of the surface area of the three cubes will be

• 1:3
• 2:3
• 5:9
• 7:9

Sol: 

 

We are given 3 cubes in the above question each with side a cm. When the cubes are placed adjacent in a row then the cuboid is formed with length 3a cm (a + a+ a) and height and breadth as a cm. Thus,

T.S.A of the cuboid = 2(3a*a + a*a + 3a*a)

= 14a² cm²

T.S.A. of three cubes = 3*6a²

= 18a² cm²

Thus, the ratio of T.S.A of cuboid to that of cube will be

14a² cm²: 18a² cm²

7 : 9

Therefore, the right option is (d).

Problem 2: X and Y are two cylinders of the same height. The base of the X has diameter that is half the diameter of the base of Y. If the height of X is doubled, the volume of X becomes

• Equal to the volume of Y
• Double the volume of Y
• Half the volume of Y
• Greater than the volume of Y

Sol: Original Solids

There are two cylinders X and Y each with height H and diameter of X is 1/2 of the diameter of Y. Thus, volume of each solid is

Volume of X = Πr²H

Volume of Y = Π(2r)² H = 4ΠrH

Now, if the height of X is doubled then,

Volume of X = Πr²2H = 2 Πr²H = ½ x 4ΠrH

Thus, the resultant cylinder has volume has volume ½ of Y. Therefore, the correct answer is (c).

Problem 3: Water is poured into an empty cylindrical tank at the constant rate for 5 minutes. After the water has been poured into the tank, the depth of the water is 7 feet. The radius of the tank is 100 feet. Which of the following is the best approximation for the rate at which the water was poured into the tank?

• 140 cubic feet/sec
• 440 cubic feet/sec
• 700 cubic feet/sec
• 2200 cubic feet/sec

Sol:

We need to find out the rate at which water is flown in tank. The volume of water flown in tank in 5 minutes = Πr²h

= (22/7 * 100² * 7)

=220000 cm²

The rate at which water is flowing in the tank is = (220000/5*60) = 733.33 ft./sec

Problem 4: A conical cavity is drilled in a circular cylinder of 15 cm height and 16 cm base diameter. The height and base diameter of the cone is same as those of cylinder. Determine total surface area of the remaining solid?

• 215 Π cm²
• 376 Π cm²
• 440 Π cm²
• 542 Π cm

Sol:

In this question, we are given a cylinder out of which a cone is scraped out and we need to find out total surface area of the remaining solid given in the above figure.

Therefore, T.S.A of the remaining solid = C.S.A of cylinder + C.S.A of the cone + area of the base of cylinder

= 2Πrh + Πrl + Πr²

= Π(2*8*15 + 8*17 + 8²)

= Π(240 + 136 + 64)

= 440 cm²

Hence, the correct answer is (c).

 

Problem 5: Base of a right prism is a rectangle the ratio of whole length and breath is 3:2. If the height of the prism is 12 cm and total surface area is 288 cm² then the volume of the prism is

• 291 cm³
• 288 cm³
• 290 cm³
• 286 cm³

Sol:

Since, the prism has rectangular base then the prism formed is cuboid with the ratio of length and breath is 3:2 and height is 12 cm.

Therefore T.S.A of cuboid = 288 cm²

288 = 2(lb + bh + hl)

288 =2(3x*2x + 2x*12+ 12*3x)

288 = 2(6x² + 24x + 36x)

144 = 6x² + 60x

x² + 10x – 24 = 0

x² + 12x – 2x – 24 = 0

(x + 12) (x – 2) = 0

x = 2

Therefore, length is 6 cm and breadth is 4 cm. Thus, volume of cuboid = lbh = 6 x 4 x 12 = 288 cm³

The above questions are some examples of the kind of problems that you might be asked in the exam. But, mensuration of 3D is a diverse topic and questions with various different combination of concepts could be asked in the exam. To get more idea you can solve all the previous year question papers, they’ll help you determine the kind of formations and concept based on this topic can be asked so do practice them.

Geometry Basics for CAT – Triangle related questions and problems

CAT Questions related to Quantitative Aptitude – Geometry

All questions from CAT Exam Quantitative Aptitude – Geometry
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Mensuration – Q1: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm
Quantitative Aptitude – Geometry – Mensuration – Q2: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically.
Quantitative Aptitude – Geometry – Mensuration – Q3: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27.
Quantitative Aptitude – Geometry – Polygon – Q1: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

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