Mensuration (Area -2 & Volume-3D)

Mensuration, Solid Geometry Questions

Mensuration is an important topic in the CAT exam, and it involves the measurement of geometric figures like areas, volumes, and perimeters. Here’s an overview tailored for CAT:

Basic Concepts in Mensuration

  1. Perimeter:
    • The perimeter is the total length of the boundary of a 2D shape.
    • Common formulas:
      • Square: P=4×side
      • Rectangle: P=2×(length+breadth)
      • Circle: C=2πr  where r is the radius.
  1. Area:
    • Area refers to the space occupied by a 2D shape.
    • Common formulas:
      • Square: A=side2
      • Rectangle: A=length×breadth
      • Triangle: A=×base×height
      • Circle: A=πr2
  1. Volume:
    • Volume measures the space occupied by a 3D object.
    • Common formulas:
      • Cube: V=side3
      • Cuboid: V=length×breadth×height
      • Cylinder: V=πr2h
      • Sphere: V=  2

Advanced Concepts

  1. Surface Area:
    • The surface area is the total area of all surfaces of a 3D object.
    • Common formulas:
      • Cube: S=6×side
      • Cuboid: S=2×(length×breadth+breadth×height+height×length)
      • Cylinder: S=2πr(h+r)
  1. Frustum of a Cone:
    • A frustum is the part of a cone that remains after cutting off the top with a plane parallel to the base.
    • Volume: V=13πh(r12+r22+r1r2) where r1​ and r2​ are the radii of the two circular ends, and  is the h height.
    • Surface Area: S=π(r1+r2)×l+πr12+ where l is the slant height.

 

  1. Composite Figures:
    • In many CAT problems, you’ll encounter composite figures, which are combinations of basic shapes.
    • Approach: Break down the composite figure into simpler shapes, calculate the required measurements (like area or volume) for each, and then sum them up.

NIMBUS MANTRA:

  1. Practice Basic Shapes First: Ensure you are comfortable with the basic shapes (square, rectangle, circle, triangle) before moving on to complex shapes.
  2. Work on Visualization: Mensuration problems often require you to visualize the shape and its dimensions. Practice sketching figures to aid in visualization.
  3. Memorize Key Formulas: While understanding is crucial, having key formulas at your fingertips will save time during the exam.
  4. Use Approximation: In some cases, approximating values (like using π≈3.14\pi \approx 3.14π≈14) can speed up calculations.
  5. Practice with Previous Year Papers: CAT often includes application-based questions in Mensuration. Practicing with previous papers will give you an idea of the types of questions to expect.

Mensuration is not just about memorizing formulas but also about applying them effectively in different scenarios, especially under time constraints, as in the CAT exam.

 

  1. A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?
  1. 9
  2. 61
  3. 98
  4. 54

Correct Answer – 54. Choice (4)

Explanatory Answer

When a 5 cc cube is sliced into 1 cc cubes, we will get 5*5*5 = 125 1 cc cubes.

In each side of the larger cube, the smaller cubes on the edges will have more than one of their sides painted.
Therefore, the cubes which are not on the edge of the larger cube and that lie on the facing sides of the larger cube will have exactly one side painted.

In each face of the larger cube, there will be 5*5 = 25 cubes. Of these, there will be 16 cubes on the edge and 3*3 = 9 cubes which are not on the edge.
Therefore, there will be 9 1-cc cubes per face that will have exactly one of their sides painted.
In total, there will be 9*6 = 54 such cubes.

 

The area of a square field is 24200 sq m. How long will a lady take to cross the field diagonally at the rate of 6.6 km/hr?

(1) 3 minutes (2) 2 minutes
(3) 2.4 minutes (4) 2 minutes 40 seconds

Correct choice – (2) Correct Answer -(2 minutes)

Solution:

Let ‘a’ meters be the length of a side of the square field.

Therefore, its area = a2 square meters. — (1)

We know that the length of the diagonal ‘d’ of a square whose side is ‘a’ meters =  a –– (2)

From (1) and (2), we can deduce that the square of the diagonal = d2 = 2a2

Or d =  meters.

The time taken to cross a length of 220 meters while traveling at 6.6 kmph is given by
(converting 1 km = 1000 meters and 1 hour = 60 minutes).

= 2 minutes

  1. The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?
(1) 20 ft (2) 25 ft
(3) 750 ft (4) 900 ft

Correct Answer – (4)

Solution:

The circumference of the front wheel is 30 ft and that of the rear wheel is 36 feet.

Let the rear wheel make n revolutions. At this time, the front wheel should have made n+5 revolutions.

As both the wheels would have covered the same distance, n*36 = (n+5)*30

36n = 30n + 150

6n = 150

n = 25.

Distance covered = 25*36 = 900 ft.

  1. If the sides of a triangle measure 72, 75 and 21, what is the measure of its in radius?
(1) 37.5 (2) 24
(3) 9 (4) 15

Correct Answer – (3)

Solution:

The sides of the triangle happen to be a Pythagorean triplet. Hence, the triangle is a right angled triangle.

For a right angled triangle,

The measure of in radius = (product of perpendicular sides) / (perimeter of triangle)

= (72 * 21) / (72 + 21 +75) = 1512 / 168 = 9

 

  1. A 5 cm cube is cut into as many 1 cm cubes as possible. What is the ratio of the surface area of the larger cube to that of the sum of the surface areas of the smaller cubes?
(1) 1 : 6 (2) 1 : 5
(3) 1 : 25 (4) 1 : 125

Correct Answer – (2)

Solution:

The volume of the larger cube = 53 = 125 cm3.

The volume of each of the smaller cubes = 13 = 1 cm3. Therefore, one would get 125 smaller cubes.

The surface area of the larger cube = 6a2 = 6(52) = 6 * 25 = 150

The surface area of each of the smaller cubes = 6 (12) = 6.

Therefore, surface area of all of the 125, 1 cm3 cubes = 125 * 6 = 750.

Therefore, the required ratio = 150 : 750 = 1 : 5

 

  1. If each interior angle of a regular polygon is 150 degrees, then it is
(1) Octagon (2) Decagon
(3) Dodecagon (4) Tetrahedron

Correct Answer – (3)

Solution:

The sum of the exterior angle and interior angle of a regular polygon is 3600.

As the measure of each interior angle is 1500, the measure of each exterior angle will be 300.

The number of sides of a regular polygon is given by the following relationship n = 360/exterior angle

As the value of each exterior angle is 300, the number of sides = 12.

  1. Four horses are tethered at 4 corners of a square field of side 70 metres so that they just cannot reach one another. The area left ungrazed by the horses is:
(1) 1050 sq.m (2) 3850 sq.m
(3) 950 sq.m (4) 1075 sq.m

Correct Answer – (1)

Solution:

The length of the rope in which the horses tied should be equal to half of the side of the square plot so that they just cannot reach one another.

Therefore, the length of the rope is 35m (70/2).

The area covered by each horse should be equal to the area of sector with radius of 70/2 = 35m(length of the rope).

Total area covered by the four horses = 4* area of sector of radius 35 metres = Area of circle of radius 35m.

Area left ungrazed by the horses = Area of square field – Area covered by four horses.

= 702 – (22/7)*35*35 = 4900 – 3850 = 1050 sq.m

 

  1. A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to the side of the square?
(1) (2)
(3) (4)

Correct Answer – (1)

Solution:

Surface area of the cylinder = Surface area of the square

Here height ‘h’ of the cylinder = side of the square since it is rolled along its length.

Base radius r =

Ratio of base radius to side of square = : a =

  1. The surface area of the three coterminous faces of a cuboid are 6, 15, 10 sq.cm respectively. Find the volume of the cuboid.
(1) 30 (2) 20
(3) 40 (4) 35

Correct Answer – (1)

Solution:

If l, b, h be the dimensions of the cuboid, then volume of the cuboid = l X b X h

 

  1. If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the length of the rectangle?
(1) 17 m (2) 31 m (3) 12 m (4) 24 m

Correct Answer – (4)

Solution:

The diagonal d = 25m. and area A = 168 m2.
Let ‘l‘ be the length and ‘b‘ be the width of the rectangle.
Therefore, l2 + b2 = d2. and lb = A
We can therefore write (l + b)2 = d2 + 2A and (l – b)2 = d2 – 2A.
Substituting and solving we get, l + b = 31 and l – b = 17. Hence l = 24 and b = 7

 

  1. A 4 cm cube is cut into 1 cm cubes. What is the percentage increase in the surface area after such cutting?
(1) 4% (2) 300% (3) 75% (4) 400%

Correct Answer – (2)

Solution:

Volume of 4 cm cube = 64 cc. When it is cut into 1 cm cube, the volume of each of the cubes = 1cc
Hence, there will be 64 such cubes. Surface area of small cubes = 6 (12) = 6 sqcm.
Therefore, the surface area of 64 such cubes = 64 * 6 = 384 sqcm.
The surface area of the large cube = 6(42) = 6*16 = 96.
% increase =  = 300%

  1. A regular hexagon is inscribed in a circle of radius r cms. What is the perimeter of the regular hexagon?
(1) 3r (2) 6r (3) r (4) 9r

Correct Answer – (2)

Solution:

A regular hexagon comprises 6 equilateral triangles, each of them having one of their vertices at the center of the hexagon. The sides of the equilateral triangle are equal to the radius of the smallest circle inscribing the hexagon. Hence, each of the side of the hexagon is equal to the radius of the hexagon and the perimeter of the hexagon is 6r.

  1. If x units are added to the length of the radius of a circle, what is the number of units by which the circumference of the circle is increased?
  1. x
  2. 2
  3. x2

Correct Answer: . Choice (4)

Explanatory Answers

Let the radius of the circle be ‘r’ units.
The circumference of the circle will therefore be  units.

If the radius is increased by ‘x’ units, the new radius will be (r + x) units.

The new circumference will be
Or the circumference increases by  units.

Correct Answer (4)

 

  1. ABCD has area equal to 28. BC is parallel to AD. BA is perpendicular to AD. If BC is 6 and AD is 8, then what is CD?
  1. 4
  2. 6

Correct Answer : . Choice (4)

Explanatory Answer

The given shape is a trapezium.
Area of a trapezium =

28 =

Height = 4.

BA is perpendicular to BC and AD.

So, drop a line parallel to BA from C to meet AD at E.
CED is a right triangle with side CE measuring 4 and ED measuring 2 units.

Hence, CD, the measure of the hypotenuse =

  1. Which of the following figures has the largest area?
  1. A circle of radius
  2. An equilateral triangle whose sides each have length 4
  3. A triangle whose sides have lengths 3, 4 and 5
  1. I
  2. II
  3. III
  4. I and III
  5. II and III

Correct Answer : II. Choice (2)

Explanatory Answer

  1. Area of circle of radius
  2. Area of an equilateral triangle whose sides each have length
  3. Area of a triangle whose sides measure 3, 4 and 5. The given triangle is a right triangle.
    Hence, its area =  = 6 sq units

We need to compare these three numbers.

Area of the circle,  = 6.28 sq units

Area of the equilateral triangle,  = 6.92
Area of the right triangle = 6 sq units.

The area of the equilateral triangle is the largest amongst the 3 objects given in this question.

 

  1. The hexagon ABCDEF is regular. That means all its sides are of the same length and all its interior angles are of the same size. Each side of the hexagon is 2m. What is the area of the rectangle BCEF?
  1. 4 sq.m
  2.  sq.m
  3. 8 sq.m
  4.  sq.m
  5. 12 sq.m

Correct Answer :  sq.m. Choice (2)

Explanatory Answer

A regular hexagon comprises six equilateral triangles – each of side 2 m, the measure of the side of the regular hexagon – as shown above. The 6 triangles are numbered 1 to 6 in the figure shown above.

BX is the altitude of triangle 1 and XF is the altitude of triangle 2.
Both triangle 1 and triangle 2 are equilateral triangles.

Hence, BX = XF =
Therefore, BF, the length of the rectangle =  m

Hence, the area of the rectangle BCEF = length * width =  sq.m

 

  1. Ram a farmer, managed to grow shaped-watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of the all the water-melons would be equal irrespective of the shape.

    A customer wants to buy water-melon for making juice, for which the skin of the water-melon has to be peeled off, and therefore is a waste. Which shape should the customer buy?

  1. Cube
  2. Hemi-sphere
  3. Cuboid
  4. Cylinder
  5. Normal spherical

Correct Answer : Normal spherical. Choice (E)

Explanatory Answer

This is the kind of question that is either very easy or very difficult depending on whether you know the concept behind the question.

For a given surface area, the volume contained increases with increasing symmetry of the object. For instance, if we are to make water melons of different shapes of the same surface area, the volume will be maximum when it is made into a sphere.

The corollary is that for a given volume, the surface area will be minimum when the object is a sphere. So, the customer should opt for spherical shaped water melons if she has to minimize wastage.

For 2-dimensional object, for a given perimeter, the area increases with increasing symmetry.

Among different triangles of a given perimeter, an equilateral triangle has the largest area.

The area increases with increasing number of sides – i.e., for a given perimeter the area of a square will larger than that of an equilateral triangle; the area of a regular pentagon of a given perimeter will be larger than that of a square and so on.

Among different regular polygons of a given perimeter / circumference a circle has the largest area.

 

  1. The figure below has been obtained by folding a rectangle. The total area of the figure (as visible) is 144 square meters. Had the rectangle not been folded, the current overlapping part would have been a square. What would have been the total area of the original unfolded rectangle?
  1. 128 square meters
  2. 154 square meters
  3. 162 square meters
  4. 172 square meters
  5. None of the above

  Correct Answer      Choice C. The total area of the original unfolded rectangle is 162 square meters.

Explanatory Answer

Folded part as shown in the first figure is a triangle – a right triangle.

The two perpendicular sides of the right triangle measure 6m each. So, the triangle is a right isosceles triangle.

When unfolded the folded area becomes a square as shown in the following figure.

The side of the square will be the width of the larger rectangle and is therefore, 6m.

Area of the square = 6 * 6 = 36 sq.m

When folded, only the area of the right triangle gets counted.
However, when unfolded the area of square gets counted.
The square comprises two congruent right triangles.

In essence, when folded only half a square is counted. When unfolded the entire square gets counted.

The area of the rectangle when unfolded = area of the rectangle when folded + area of half a square.
So area after unfolding= 144 + 18 = 162 sq.m.

The correct answer is Choice C.

 

  1. A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?
  1. 9%
  2. 16%
  3. 25%
  4. 50%
  5. None of the above

  Correct Answer      Choice D. The percentage change in the flat surface area is 50%.

Using information about volumes

Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.

Let their volumes be V, v1, v2 respectively.

Cylinder is melted and recast into cone 1 & cone 2

So V = v1 + v2.

Ratio of volumes of the two cones v1 : v2 is 3 : 4.
If v1 is 3x, v2 will be 4x

Hence, volume of cylinder V= 7x

Compute ratio of the flat surface area

The volumes of the cylinder, cone 1 and cone 2 are πR2h, 1313 πr12h and 1313 πr22h
We know the ratio of the volumes V : v1, v2 is 7 : 3 : 4
So, πR2h : 13 πr12h : 13 πr22h is 7 : 3 : 4

Cancelling π and h, which are common to all terms, we get R2 : 13 r12 : 13 r22 = 7 : 3 : 4
Or R2 : r12 : r22 = 7 : 9 : 12.

So, if R2 is 7k, r12 will be 9k and r22 will be 12k.

Flat surface area of cylinder (sum of the areas of the two circles at the top and bottom of the cylinder) = 2 * π * R2
Flat surface area of cone 1 & 2 are: π * r12 & π * r22 respectively (areas of the circle at the bottom of each of the cones).

Ratio of the flat surface area of cylinder to that of the two cones is 2 * π * R2 : (π * r12 + π * r22)
Cancelling π on both sides of the ratio we get 2R2 : (r12 + r22)
Or 2(7k) : (9k + 12k)
Or 14k : 21k

Compute percentage change

Change in flat surface area = 21k – 14k = 7k
% change in flat surface area = 7k/14k*100 = 50%.

The correct answer is Choice D.

 

  1. A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs.100 per sq. ft., find the lowest possible cost to construct 50% of the total road.
  1. Rs.70,400
  2. Rs.125,400
  3. Rs.140,800
  4. Rs.235,400
  5. None of the above

  Correct Answer      Choice B. The lowest possible cost to construct 50% of the total road is Rs.125,400.

What is to be found?

We have to find the lowest cost to construct 50% of the total road.

Data Given

A circular road is constructed outside a square field. So, the road is in the shape of a circular ring.

If we have to determine the lowest cost of constructing the road, we have to select the smallest circle that can be constructed outside the square.

Therefore, the inner circle of the ring should circumscribe the square.

Perimeter of the square = 200 ft.

Therefore, side of the square field = 50 ft.

Compute the inner and outer radii of the circular ring

The diagonal of the square field is the diameter of the circle that circumscribes it.
Measure of the diagonal of the square of side 50 ft = 502–√2 ft.
Therefore, inner diameter of the circular road = 502–√2.
Hence, inner radius of the circular road = 252–√2 ft.
Then, outer radius = 252–√2 + 72–√2 = 322–√

Compute the area of the circular ring

The area of the circular road
= π ro2 – π ri2, where ro is the outer radius and ri is the inner radius.

= 227× {(32 2–√2)2 – (252–√2)2}
= 227×2×(32 + 25)×(32 – 25)
= 2508 sq. ft.

If per sq. ft. cost is Rs. 100, then cost of constructing the road = 2508 × 100 = Rs.2,50,800.

Cost of constructing 50% of the road = 50% of the total cost = 2508002 = Rs.1,25,400

The correct answer is Choice B.

 

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