It is one of the easier parts, and about 6-8% of the questions in the CAT Quantitative Aptitude section come from it. There are also a number of other ability tests that have a lot of questions about this topic.
So, students who are not so good at other parts of school, like math or certain kinds of logical questions, are told to put more focus on this chapter. Even the questions in this chapter are not as hard as the ones in Geometry.
Mensuration
[Click Here for Previous Year CAT Questions]
Planes are two-dimensional, which means that they have two dimensions: length and width.
Solids have length, width, and height, which are the three dimensions.
1 km = 10 hectometre
= 1000 metre
= 1,00,000 centimetre
= 100 decametre
= 10,000 decimetre
= 10,00,000 millimetre
1 hectare = 10,000 square metre
1 are = 100 square metre
1 square hectometre = 100 square decametre
1 square decametre = 100 square metre
1 square metre = 100 square decimetre
1 square decimetre = 100 square centimetre
1 square centimetre = 100 square millimetre
Weight = Volume × density
Types of Figures and Formulae
[Click Here for Previous Year CAT Questions]
| Name | Area | Perimeter |
|---|---|---|
| Rectangle | l x b = lb | 2(l+b) |
| Square | a x a = a2 | 4a |
| Triangle | (i) \(\frac{1}{2}\) x b x h (ii) \(\sqrt {s(s-a) (s-b)(s-c)}\) | 2s = a+b+c |
| Equilateral triangle | (i) \(\frac{1}{2}\) x a x h (ii) \(\frac{\sqrt {3}}{4}\) a2 | 3a |
| Isosceles triangle | (i) \(\frac{1}{2}\) x a x h (ii) \(\frac{1}{4}\) x b x \(\sqrt {4a^2 - b^2}\) | 2a+b |
| Right angled triangle | \(\frac{1}{2}\) x b x h | b+h+d |
| Parallelogram | a x h | 2(a+b) |
| Rhombus | \(\frac{1}{2}\) x d1 x d2 | 4a |
| Trapezium | \(\frac{1}{2}\) x (a+b) x h | Sum of all sides |
| Regular Hexagon | \(\frac{3\sqrt 3}{2} a^2\) | 6a |
| Regular Octagon | 2a2 (1+ \(\sqrt 2\)) | 8a |
| Circle | \(\pi r^2\) | \(2\pi r\) |
| Semi-Circle | \(\frac{\pi r^2}{2}\) | \(\pi r + 2r\) |
| Ring or Circular path | \(\pi (R^2 - r^2)\) | Outer = \(2\pi R\) Inner = \(2\pi r\) |
| Sector of a circle | \(\pi r^2. \frac{\theta}{360^o}\) | l + 2r |
| Pathways | (l+b−w)w | 2(l+b)−4w = 2[l+b−2w] |
| Outerways | (l+ b+ 2w)2w | 2(l + b + 4w) |
| Inner path | (l + b − 2w) 2w | 2(l + b − 4w) |
Rectangles and Squares
[Click Here for Previous Year CAT Questions]
Area of a rectangle = length × breadth = l × b
Area of a square = (side) 2 = \(\frac{1}{2}\)(diagonal) 2 = \(\frac{1}{2}\) d2 = a2
Diagonal of a rectangle = \(\sqrt {l^2 + b ^2}\)
Diagonal of a square = a\(\sqrt 2\)
Perimeter of a rectangle = 2(l + b)
Perimeter of a square = 4a
Area of four walls of a room = 2 (l + b) × h
Circles
[Click Here for Previous Year CAT Questions]
Area of the Circle = πR2
Circumference of the circle = 2πR
Length of an arc = 2πR\((\frac{\theta}{360^o})\)
Area of a sector = πR2\((\frac{\theta}{360^o})\)
Area of a sector = πR2\((\frac{\theta}{360^o})\) – \(\frac{R^2}{2} sin \theta\)
Cuboid and Cube
[Click Here for Previous Year CAT Questions]
A cuboid has 6 faces, 12 edges , 8 vertices and 4 diagonals.
Volume = l × b × h
Total surface area of cuboid = 2 (lb + bh + hl)
Diagonal (d) = \(\sqrt {l^2 + b^2 + h ^2}\)
A cube has 6 equal faces, 12 equal edges, 8 vertices and 4 equal diagonals.
Volume = a3
Total surface area = 6a2
Diagonal (d) = a\(\sqrt 3\)
Euler’s Theorem → (V + F ) = (E + 2);
where-
V → number of vertices,
F → number of faces,
E → number of edges
Cylinder and Cone
[Click Here for Previous Year CAT Questions]
Volume of Cylinder = base area × height = πr 2h
Curved surface area = perimeter × height = 2πrh
Total surface of the cylinder = curved surface area + 2 (base area) = 2πrh + 2πr 2 = 2πr(h + r)
Volume of a hollow cylinder = πh (R2− r2 )
Cone
Volume (V ) = \(\frac{1}{3}\) × base area × height = \(\frac{1}{3}\)πr2h
Total surface area = πrl + πr2 = πr (l + r)
Frustum of a cone
Volume of the frustum of a cone = \(\frac{\pi}{3}\)h(r2 +Rr+R2)
Lateral surface area of the frustum of cone = πl (r + R)
Previous year CAT questions
Ques 1: A regular polygon has an even number of sides. If the product of the length of its sides and the distance between two opposite sides is 1/4th of its area, then find the number of sides it has. (CAT 2016)
- 6
- 8
- 20
- 16
Click Here for the Answer
Solution: (D)
Let 2n be the number of sides. Let S be the length of the side and P be the length of the line from the center to each side. Since there are an even number of sides, the sides next to each other will be parallel, and the distance between any two sides next to each other is 2P.
area of the polygon = 2n
S(2P) = A/4 or SP = A/8
∴ A=n(A/8)
⇒n=8 or 2n=16
Ques 2: What is the area, circumradius and inradius of a regular hexagon of side ‘a’ ? (CAT 2016)
- \(\frac{3\sqrt 3}{2} a^2, a, \frac{\sqrt 3}{2} a\)
- \(\frac{3}{2} a^2, \frac{a}{2}, \frac{\sqrt3}{2} a\)
- \(a^2, \frac{a}{2}, 2a\)
- \(\frac{\sqrt3}{2}a^2, a, \sqrt 3 a\)
Click Here for the Answer
Solution: (A)
A hexagon is nothing but 6 equilateral triangles placed around a point.
Circumradius = a
Inradius ⇒ Radius of circle inside hexagon
Inradius ≡ Altitude of each equilateral triangle r = \(\frac{\sqrt 3}{2} a\)
Area = 6 (area of equilateral triangles) = 6 x \(\frac{\sqrt 3}{4} a^2\) = \(\frac{3\sqrt 3}{2} a^2\)
Ques 3: Consider a square ABCD with mid-points E, F, G, H of AB, BC, CD and DA, respectively. Let L denote the line passing through F and H. Consider, points P and Q on L and inside ABCD, such that the angles APD and BQC both equal 120°. What is the ratio of the area of ABQCDP to the remaining area inside ABCD ? (CAT 2015)
- \(\frac{4\sqrt 2}{3}\)
- 2 + \(\sqrt 3\)
- \(\frac{10- 3 \sqrt 3}{9}\)
- \(2\sqrt 3 -1\)
Click Here for the Answer
Solution: (D)
Ques 4: Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side a in the following steps: the original square of side a is divided into n smaller squares, not necessarily of the same size, then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is __ . (CAT 2015)
- 1:1
- \(\sqrt 2 : 1\)
- \(\frac{n(4- \pi)}{4n- \pi}\)
- \(\frac{4n - \pi}{n (4- \pi)}\)
Click Here for the Answer
Solution: (A)
Ques 5: The radius of a cone is r cm and its height is h cm. The change in volume when the height is decreased by x cm is the same as the change in volume when the radius is decreased by x cm. Find the relation between x, r and h. (CAT 2014)
- \(x= \frac{2rh - r^2}{h}\)
- \(x= \frac{2rh+ r^2}{h}\)
- \(x= \frac{r^2 - 2rh}{h} \)
- x = 2r + r2
Click Here for the Answer
Solution: (A)
If the height is decreased by x cm, then
Decrease in the volume= (1 /3)[π r2 h − π r2 (h − x)]
=(1/3)πr2 x
If the radius decreased by x cm, then
Decrease in the volume= (1 / 3 ) [ π r 2h − π ( r − x ) 2 h ]
=(1/3)π[r2h−(r2−2xr+ x2)h]
=(1/3)π[2xrh− x2h]
Combining the above results, πr2 x = π [2xrh − x2h]
Canceling π and x both sides,
r2 = 2rh-xh
x = \(\frac{2rh - r^2}{h}\)
Ques 6: A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square centimetres before the cut, what is the area of smaller triangle (in cm2) formed after the cut? (CAT2013)
- 16.565
- 15.465
- 16.665
- 14.365
Click Here for the Answer
Solution: (D)
total reduction in area of triangle = (2a – \(\frac{a^2}{100}\))%
= (2 x 35 – \(\frac{35^2}{100}\))%
= (70 – \(\frac{1225}{100}\))%
=(70-12.25)%=57.75%
New triangle area will be (100 – 57.75)% of area of bigger triangle
=42.25% of 34
= \(\frac{42.25 \times 34}{100} = 14.365\ cm^2\)
Ques 7: Kunal has 64 small cubes of 1 cm3 each. He wants to arrange all of them in a cuboidal shape, such that surface area of the cuboid will be minimum. What is the diagonal of this larger cuboid? (CAT 2013)
- \(\sqrt {273}\ cm\)
- \(8 \sqrt 2 \ cm\)
- \(4 \sqrt 3 \ cm\)
- \(\sqrt 128 \ cm\)
Click Here for the Answer
Solution- (C)
For a given volume of 64 cm3 .
Cube has minimum surface area of length of edge = 4 cm
∴ Its diagonal = 4\(\sqrt3\) cm
Ques 8: A rectangle is drawn such that none of its sides has length greater than ‘a’. All lengths less than ‘a’ are equally likely. The chance that the rectangle has its diagonal greater than ‘a’ is (in terms of %) (CAT 2012)
- 29.3%
- 21.5%
- 66.66%
- 33.33%
Click Here for the Answer
Solution: (B)
required probability = \(\frac{a^2 (1- \frac{\pi}{4})}{a^2} = \frac{4- \pi}{4}\) = 0.215 or 21.5% chance
Ques 9: All rectangles with diagonal less than or equal to a will lie within the quadrant of the circle. In the figure given, OABC is a parallelogram. The area of the parallelogram is 21 sq units and the point C lies on the line x = 3. Find the coordinates of B. (CAT 2012)
- (3,10)
- (10,3)
- (10,10)
- (8,3)
Click Here for the Answer
Solution: (B)
The point C's coordinates are (3, b). This indicates that the parallelogram has a height of 3 units.
Given that the area is 21 square units, we obtain 7b=21 or b = 3.
Therefore, C's coordinates are (3, 3)
As CO = 7, the coordinates of D are (10, 3).
Ques 10: A solid sphere of radius 12 inches is melted and cast into a right circular cone whose base diameter is 2 times its slant height. If the radius of the sphere and the cone are the same, how many such cones can be made and how much material is left out? (CAT 2012)
- 4 and 1 cubic inch
- 3 and 12 cubic inches
- 4 and 0 cubic inch
- 3 and 6 cubic inches
Click Here for the Answer
Solution: (C)
As the volume of the material is the same before and after casting
Volume of the sphere = (4/3) πr3
Volume of the right circular cone = (1/3) πRH
Diameter of the base of the cone = \(\sqrt 2\)S, S = Slant Height
4R = 2S = 2(H2+R2)
2R = 2H
R = H
Volume of cone = (1/3) π.R.H
=(1/3) πr2 (r)
=(1/3) πr3
material creates exactly 4 cones of the given dimension.
Ques 11: Rekha drew a circle of radius 2 cm on a graph paper of grid 1 cm × 1 cm. She then calculated the area of the circle by adding up only the number of full unit-squares that fell within the perimeter of the circle. If the value that Rekha obtained was d sq cm less than the correct value, then find the maximum possible value of d? (CAT 2011)
- 6.28
- 7.28
- 7.56
- 8.56
Click Here for the Answer
Solution: (D)
For the value of d to be the maximum the number of full unit-squares that Rekha counts must be the minimum, which is 4 unit-squares. (i.e., 4 full unit-squares will always fall within the circle)
∴ d=πr2 −4=4π −4=4(π −1)
= 4(3.142 − 1) = 4(2.142) ≅ 8.56
Ques 12: The circumference of a cylinder is 3 feet and its height is 16 feet. An insect climbs the pole such that its motion is a spiral and one complete spiral helps it to cover 4 feet in height. Thus, when the insect reaches the top, what is the total distance covered by it? (CAT 2009)
- 16 feet
- 18 feet
- 20 feet
- 25 feet
Click Here for the Answer
Solution: (C)
For the value of d to be the maximum the number of full unit-squares that Rekha counts must be the minimum, which is 4 unit-squares. (i.e., 4 full unit-squares will always fall within the circle)
∆PQR is right angled at R.
∴ PQ = 5
Thus, for one spiral, it moves 5 feet. 4 feet height → 5 feet
So, 16 feet height → (16 × 5)/4 = 20 feet
Ques 13: A semicircle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semicircle at D. Given that, AC = 2 cm and CD = , the area of the semicircle (in sq.cm) will be __. (CAT 2006)
- 32π
- 50π
- 40.5π
- 81π
- Undeterminable
Click Here for the Answer
Solution: (B)
AC= 2cm , CD = 6 cm
∆ADB is the right triangle
In ∆ADB
(CD)2 = AC x CB
36=2 x CB
AB = 20cm
Radius of the circle =10cm
Area of the semi circle = (π/2)102 = 50 π cm2
Ques 14: P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR? (CAT 2005)
- 2r (1+\(\sqrt3\) )
- 2r (2+\(\sqrt 3\))
- r (1+\(\sqrt5\))
- 2r + \(\sqrt3\)
Click Here for the Answer
Solution: (A)
Perimeter of the quadrilateral PQSR = 2r (1+\(\sqrt 3\))
Ques 15: A square tin sheet of side 12 inches is converted into a box with open top in the following steps-The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box? (CAT 2003)
- 3
- 4
- 1
- 2
Click Here for the Answer
Solution: (D)
Volume of the box V = (12 – 2x2).x
To maximize V, \(\frac{dV}{dx} = 0\)
(12 – 2x2) + x(- 4x) = 0
x = 2
How to approach Mensuration questions in CAT
- Formulae are the very important part of this chapter.
- Students should be good with numbers and formulae to attempt all the questions in CAT.
- Practicing more questions will reduce the time taken to do a question.
Comments