Mensuration Made Easy With Examples

Introduction:

We usually make faces when asked to go through the topic ”MENSURATION”, the sole reason behind this is “THE FORMULAS” but wait what if we had to just understand everything and apply it in practical life ?? Read it right, today we are going to get our basic concepts cleared under this topic with few examples supporting the concepts.
Before heading to the formulas, let's know the meaning of the word MENSURATION. The word mensuration means measurement, this is a branch of mathematics which helps us in dealing with the study of plane and solid figures, their area, volume, and related parameters.

Important Definitions 

1. Area

The extent or measurement of a surface or piece of land or the total amount of space inside the boundary of flat objects is Area. The shaded region in the following image is the area of the rectangle.

2. Perimeter

The perimeter of any 2-dimensional object is the measure of the covering the inner area.
Here the dotted lines represent the boundary
If
→ l =10cm and b=4 then perimeter = 2 (l+b)

→ 2(10+4)
→ 2 x14
→ 28 cm (add the boundary measurements you get 28)

3. Circumference

When you have any curved geometric figure then their distance around the body will be called as circumference, especially for the circle.

4. Dimensional Objects

A three-dimensional shape is a solid shape that has height and depth. For example, a sphere and a cube are three-dimensional, but a circle and a square are not.

5. Curved Surface Area: 

The surfaces which are not flat, are called curved surface.
The lateral surface is the area of the vertical faces of the solid.
Curved surfaces do not include the top or bottom are, for example, CSA of the cylinder will not include the area of upper and lower circle.
Whereas the total surface area (TSA) will include both the area of both upper and bottom portion.

It becomes difficult to find area of curvy objects so with help of cylinder we can understand this: (proved by greek mathematician Archimedes)
You can fit a sphere in cylinder (with same radii and height)
Then the lateral surface of cylinder = area of sphere

Curved Surface Area of Cylinder = 2⊓rh
Curved Surface Area of sphere = 2⊓rh 
( r is same for both, we do not have h in sphere so we consider diameter to be h =2r)
Curved Surface Area = 2⊓r(2r) substituting h=2r
Curved Surface Area= 4⊓r2
Volume of sphere =4/3⊓r3 (sphere is 1/3rd of cursed surface of sphere into r)
Volume of hemisphere = 2/3 2⊓r3 (half of the volume of sphere )
Curved Surface Area of hemisphere = 4⊓r2/ 2 (half of spheres) = 2⊓r2 

NOTE: Remember friends TSA of a hemisphere is not same as CSA (don’t do this mistake of halving the TSA of the sphere because hemisphere has one side circle so you will have to add the area of the circle too:
→Total Surface Area = 2⊓r2 + ⊓r2 = 3⊓r2

CUBOID

                                               

It is 3 dimensional, has length, breadth and height.
Volume of cuboid = l x b x h
Total Surface Area = 2( l x b+ b x h + h x l)

CUBE 

                                                      

It is a form of the cuboid but all its sides all equal, best example can be a dice.same formula for cuboid and cube but in the cube all sides are same so, substitute we get

Volume = (l x b x h) suppose a is the side = (a x a x a) =a³
Total Surface Area = 2(a x a +a x a +a x a) = 2(3a2) = 6a²

Exercise with Solutions

Question1

A road roller of diameter 1.75m and length 1m has to press a ground of area 1100 SQ. meter .how many revolutions does it make?

Solution: 

We know a roller is in the shape of a cylinder, now when you roll a cylinder one revolution it makes will be equal to the curved surface area of the cylinder (you cannot take volume here because it not about how much the cylinder holds in it nor it is about the total surface of cylinder)
r=1.75/2 h=1
Area covered in one revolution= curved surface area = 2⊓rh
Total area to be pressed =1100 SQ meter
Number of revolutions= Total area to be pressed/curved surface area
= 1100 / 2⊓rh (substituting value )
Ans: -Number of revolutions = 200

Question2 

A rectangular sheet of paper of length 10 cm and breadth 24cm is rolled end to end to form a right circular cylinder of height 8 cm. find the volume of the cylinder.

Solution: 

When you roll a rectangle sheet to a cylinder shape, the base forms a circle. We need to find out what the circumference of this circle is to get the radius, which will later be utilised in volume.
Circumference = 2⊓r (here clearly they have given 10 must be the height so 24 is the circumference)
→24 = 2*22/7*r
→24*7/22*1/2=r
→r = 42/11cm
Ans:-volume = ⊓r2h = 458.18 cu.cm

Question3 

A right cylindrical vessel of 15 cm radius is filled with water. Solid spheres of diameter 6 cm are dropped one by one into it. The spheres are dropped until the water level in the vessel rises by 8 cm. then, how many solid spheres are dropped into the vessel?

Solution:

Let n balls be dropped into the cylinder
Volume of n balls will be the total increase in water level in the vessel
Given d=6 ;so r = 3 ; rise in water level h=8
→N*4/3*⊓r3 = ⊓r2h
→N*4/3 ⊓ (3)3 = ⊓ * 152 *8
Ans:-→N= 50 balls

Question4 

If a cone and a sphere have equal radii and have equal volumes, then what is the ratio between the height of the cone and diameter of the sphere?

Solution:

Let h be the height of the cone and r be the radius of the sphere as well as the radius of the base of the cone, clearly given that volume of the sphere = volume of the cone
→i.e. 4/3 ⊓r3 = 1/3 ⊓r2 h
→2r=h/2=d
Therefore the height of the cone diameter of the sphere
4r:2r
Ans:-required ratio 4:2= 2:1

Question5 

The length of a rectangular plot is 60% more than its breadth. if the difference between the length and the breadth of that rectangle is 24cm. what is the area of that rectangle?

Solution:

Let breadth = x cm then length= 160*x/100 = 8/5*x
→So 8/5*x – x = 24
→X=40
Length =64 breadth =40
Ans:-Then area will be =64*40= 2560

Question6 

A spherical ball of 6cm diameter is melted into a cone with base 12cm in diameter. find its height.

Solution:

Here diameter is given so radius will be d/2
→4/3⊓r3 = 1/3⊓R2h
→4*33 = 62 * h
Ans:-→H= 3cm

Question7 

4 containers are in the shape of a sphere of radius 7cm. find the cost of panting at Rs.2 per square metre and filling them with a liquid costing Rs.9per cubic cm.

Solution:

Surface area of the sphere = 4⊓r2
Surface area of the containers =4*22/7*72
→= 616 cm2
Cost of painting =Rs.2 * 616(surface area )= Rs.1232
For filling the containers you need to calculate the volume of the sphere using the formula 4/3⊓r3
Ans:-Cost of filling = Rs.9* volume of sphere = Rs.9*4/3*⊓*73 = Rs.12936

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