Table of contents 
What is Mensuration? 
Prism 
Pyramid 
CAT Mensuration Questions 
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A branch of mathematics that talks about the length, volume, or area of different geometric shapes is called Mensuration. These shapes exist in 2 dimensions or 3 dimensions.
SOLID  Total Surface Area  Lateral/ Curved Surface area  Volume  Length of Leading Diagonal/ Slant Height 
 6a^{2}  4a^{2}  a^{3}  √3a 
2(LB+ BH+ HL)  2H (L + B)  LBH  √ (L^{2} + H^{2} + B^{2})  
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.
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.
A pyramid is a polyhedron formed by connecting a polygonal base and a point called apex. The lateral surface of pyramids form triangles.
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:
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) 1:3
(2) 2:3
(3) 5:9
(4) 7:9
Correct Answer is Option (4)
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
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
(1) Equal to the volume of Y
(2) Double the volume of Y
(3) Half the volume of Y
(4) Greater than the volume of Y
Correct Answer is Option (3)
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.
Problem 3: Water is poured into an empty cylindrical tank at a 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
(1) 140 cubic feet/sec
(2) 440 cubic feet/sec
(3) 700 cubic feet/sec
(4) 2200 cubic feet/sec
Correct Answer is Option (3)
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?
(1) 215 Π cm^{2}
(2) 376 Π cm^{2}
(3) 440 Π cm^{2}
(4) 542 Π cm
Correct Answer is Option (3)
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
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
(1) 291 cm^{3}
(2) 288 cm^{3}
(3) 290 cm^{3}
(4) 286 cm^{3}
Correct Answer is Option (2)
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}
Problem 6: Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of 8 mm. The largest marble has a radius of 18 mm. What is the radius (in mm) of the middle marble?
(1) 10
(2) 11
(3) 12
(4) 15
Correct Answer is Option (3)
Problem 7: Three identical cones with base radius r are placed on their base so that each is touching the other two. The radius of the circle drawn through their vertices is
(1) Smaller than r
(2) equal to r
(3) larger than r
(4) depends on the height of the cones
Correct Answer is Option (3)
The centres of the bases of the cones from a triangle of side 2r. The circumcircle of the circle will be identical to a circle drawn through the vertices of the cones and thus, it will have a radius of 2/√3 times r. which is greater than r.
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