Mensuration Formulas List

Introduction:

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.

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 x 14
→ 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

Two-Dimensional

A two-dimensional figure is a figure that has length and width but no depth. For example, circle, rectangle, square, triangle are two-dimensional.

Three-Dimensional

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 is, for example, CSA of the cylinder will not include the area of upper and lower circle. Whereas, the total surface area (TSA) will include the area of both upper and bottom portion.

TWO-DIMENSIONAL OBJECTS

Object	Perimeter	Area
Square	Four times the length = 4l	Length squared = l2
Rectangle	Twice the sum of the length and width = 2(a + b)	Length times width = ab
Triangle	Sum of all three sides = a + b + c	Half of base times height = ½ah
Trapezium	Sum of all the sides = a + b + c + d	Half of sum of super and lower base times height =½(a + d)h
Circle	Circumference = 2πr	Area = πr2

THREE DIMENSIONAL OBJECTS

Object	Total surface area	Volume
Prism	Sum of all the area of the surfaces = 2ab + 2bc + 2ca	Base area  height = abc
Cylinder	Sum of area of upper face, lower face and curved face = 2πr2 + 2πrh	Base area  height =πr2h
Cone	Sum of area of lower face and curved face =  πr2 + πrl	One third of base area  height = 1/3πr2h

Exercise with Solutions

Question 1

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=2πrh
Total area to be pressed =1100 SQ meter
Number of revolutions= Total area to be pressed/curved surface area
=1100/2πrh (substituting value )
Ans:-Number of revolutions=200

Question 2

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 utilized in volume.
Circumference = 2πr (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=h=458.18cu.cm

Question 3

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×4/3πr³= πr²h

→ N×4/3π×(3)³=π×152×8

Ans:-→N=50 balls

Question 4

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.4/3πr³=1/3πr²h

→2r=h/2=d
Therefore the height of the cone diameter of the sphere
4r:2r
Ans:-required ratio 4:2= 2:1

Question 5

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

Question 6

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
→4/3πr³=h

→4×33=62×h

Ans:-→H=3cm

Question 7

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

Solution:

Surface area of the sphere=4πr²
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=4/3πr³
Ans:-Cost of filling = Rs.9×volume of sphere = Rs.9×4/3×π×73=Rs.12936

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