Thursday, February 15th, 2018
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 45 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  √ (L^{2} + H^{2} + B^{2})  

6a^{2}  4a^{2}  a^{3}  √3a 
2Πr (r + h)  2Πrh  Πr^{2}h  No Slant height or diagonal  

Πr (r + l)  Πrl  ⅓Πr^{2}h  √(h^{2} + r^{2}) 

4Πr^{2}  4Πr^{2}  ^{4}/_{3}Πr^{3}  No Slant height or diagonal 

2Π(r₁+r₂) (r₂r₁+h)  2Πh(r₁+r₂)  Πh(r₂²r₁²)  No Slant height or diagonal 

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}) 

3Πr^{2}  2Πr^{2}  ^{2}/_{3}Πr^{3}  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 nsided 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 nsided 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: –
r/R = (Hh)/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
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)
= 14a^{2} cm^{2}
T.S.A. of three cubes = 3*6a^{2}
= 18a^{2} cm^{2}
Thus, the ratio of T.S.A of cuboid to that of cube will be
14a^{2} cm^{2}: 18a^{2} cm^{2}
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
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?
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?
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 cm^{2} then the volume of the prism is
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 cm^{3}
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 Fundas for CAT Quantitative Aptitude Preparation – Part 1
Geometry Fundas for CAT Quantitative Aptitude Preparation – Part 2
Geometry Basics for CAT – Triangle related questions and problems
All questions from CAT 2017 Quantitative Aptitude – Geometry
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a rightangled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a rightangled 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 rightangled isosceles triangle with hypotenuse BC. Let BQC be a semicircle, 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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