Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. The Solutions to all the mensuration problems that have appeared in CGL lately. This is the second part of Mensuration Tricks. For part I, please click here.

This is a famous question. Just remember whenever you are forming a circle and then a square,

the side of that square is given by, a = 1.6*r (approx.), where r = radius of the circle

I have written approx. because the actual formula is 1.57*r, but it will make the calculations a bit lengthy. So, just find 1.6*r and the answer will be little less than that. Like here

Side = 1.6*84 = 134.4, so the answer is 132 cm

**Answer: (A)**

When the wire is bent in the form of a circle of radius 84cm, that means the circumference (or the length of the wire) of the circle is 2*π*84 = 44*12 cm

Now this wire forms a square of (let’s say) side ‘a’

Then, 4a (perimeter of the square) = 44*12

Hence a = 132 cm

If the given rectangular sheet of paper (length =l, breadth = b) is rolled **across its length** to form a cylinder, having a height b, then volume of cylinder = (l*l*b)/4π

If rolled **across its breadth**, then = (b*b*l)/4π

In this question the sheet is rolled along its length, so volume = (l*l*b)/4π = 12*12*5/(4* π)

Volume = 180/ π cm^{3}

**Answer: (C)**

In this question, the ratio of surface areas is given and they are asking the ratio of volumes. The word “sphere” is useless here. In such questions, just imagine area as A^{2} and volume as A^{3}. Now A^{3} is given and you have to find A^{3}. How will you do it? Simple, first take the square-root of A^{2} to convert it into A, and then take the cube of A to find A^{3}.

So for solving this question, we just have to take the square-root of 4:9. The ratio will become 2:3. Then take the cube of 2:3. Hence the answer is 8:27

**Answer: (C)**

Here again ratio of areas is given, that means A^{2} is given, and we have to find A. So 4:9 will become 2:3

**Answer: (A)**

Diameter and perimeter are directly proportional, P = D*π, where P is the perimeter and D is the diameter.

Hence a 75% increase in diameter means a 75% increase in perimeter

**Answer: (D)**

The area of base and the volume of a cone are directly proportional V = A * h/3, where V = volume and h = height of the cone

Hence a 100% increase in the area of the base would mean a 100% increase in the volume

**Answer: (B)**

A is increased by 50% hence A^{2 }(or surface area) will increase by (1.5*1.5 – 1)*100 % = 125%

*Note:* Similarly A^{3 }(or volume) will increase by (1.5*1.5*1.5 – 1)*100 % = 237.5%

**Answer: (A)**

Where ever the word “melting” is used in mensuration, it means only one thing – equate the volume

The volume of the rectangular block =** l*b*h = **21*77*24 cm^{3}

Now this volume will be equal to the volume of the sphere formed after melting the block

Volume of sphere = (4/3) * π * r^{3} = 21*77*24

Hence, r = 21 cm

**Answer: (A)**

The water rises by 5.6 cm. Take this 5.6 cm as the height of the cylindrical beaker and find its volume.

Volume of a cylinder = π*r*r*h = π * (7/2) * (7/2) * 5.6

Volume of the marbles (spherical in shape) = (4/3) * π * r^{3} = (4/3) * π * 0.7 * 0.7 * 0.7

No. of marbles dropped = Volume of beaker/Volume of a marble = 150

**Answer: (B)**

Let the radius of the big sphere be R.

Volume of a cone = (1/3) * π * R^{3} (since radius and volume are same as the radius of the sphere)

Let the radius of the smaller sphere = r

Then volume of cone = volume of smaller sphere

(1/3) * π * R^{3} = (4/3) * π * r^{3}

r : R = 1 : 2^{2/3}

Surface area of smaller sphere(s) = 4 * π * r^{2}

Surface area of larger sphere(S) = 4 * π * R^{2}

S/s = (r/R)^{2} = 1 : 2^{4/3}

**Answer: (D)**

When a cone is hollowed out from a cylinder, we get the above figure

The whole surface area of the remaining solid = Area of A + Area of B + Area of C

A = curved surface area of the cone

B = curved surface area of the cylinder

C = area of the cylindrical base

A = π * r * *l*, where *l *= slant height of the cone, which is or

Hence A = π*3*5 = 15π

B = 2πrh = 2π*3*4 = 24π

C = πr^{2} = π*3^{2} = 9π

The whole surface area of the remaining solid = 15π + 24π + 9π = 48π

**Answer: (C)**

Given, AB = 3 cm, BC = 6 cm and OF = 1 cm

Height of the cone (AC) = √(6^2 – 3^2 ) = 3√3 cm

Triangles ABC and CFO are similar (RHS similarity)

So, OC/BC = OF/AB

OC = 2 cm, therefore CG = 3 cm (OG = 1 cm)

Now, ABC and CEG are similar

GE/AB = CG/AC

So, GE =

Required volume = Volume of cone (CDE) – Volume of Sphere

= 3π – (4/3)π

= (5/3)π

**Answer: (C)**