# Mensuration Formulas

In our day to day lives, we often come across several shapes and figures in different forms like books, bottles, pan, cylinders and much more and beyond. We have also heard about the area and perimeter that these objects occupy, but do we know how to calculate. The answer to this question might be either a yes or no but the basic requirement for the calculation is formulas and the process is called mensuration. In the process to ease your doubts and help you better with your studies, we have sorted out the list for your reference.

## What Is Mensuration Maths?

Mathematics is all about formulas, equations, geometry, trigonometry, numbers, arithmetic and mensuration and they are all equally different so before we talk about the formulas related to mensuration, let us understand the meaning of mensuration maths in three simple points:

• Mensuration is the part of arithmetic that manages the investigation of various geometrical shapes, their areas, and Volume.
• In the broadest sense, it is about the procedure of estimation and depends on the utilization of arithmetical equations and geometric estimations to give estimation information in regards to the width, profundity, and volume of a given item or gathering of articles.
• Estimation results acquired by the utilization of mensuration are gauges instead of real physical estimations; the computations are typically viewed as exact.

## Difference Between 2d And 3d Shapes

We have broadly discussed the meaning of mensuration above, elaborating further mensuration objects are classified into shapes which are either 2D or 3D. Furthermore let us understand the difference between the two

 2D Shapes 3D Shapes The shape which is surrounded by three or more straight lines in a plane would be referred to as a 2D shape The shape which is surrounded by a no. of surfaces or planes would be referred to as a 3D shape They  have no breadth  or height This shapes are also termed as solid shapes and have the combination of breadth and height The perimeter and area of 2D shapes could be measured. The volume of these figures could be measured like  CSA, LSA or TSA. Example: Square, Triangle, Rhombus, Trapezoid, Rectangle Example: Cube, Sphere, Cone, Hemisphere, Prism

## Important Terminologies For Mensuration:

With the meaning, a difference of the mensuration becoming more clear, let us now understand the terminologies in detail that are associated with mensuration which are explained below:

• Volume: In a 3D shape, the space included is called a Volume.

Abbreviation: V, Unit: Cm3 or m3

• Area: The area is the surface which is covered by the closed shape

Abbreviation: A, Unit: M2 or Cm2

• Perimeter: The measure of the continuous line along the boundary of the given figure is called a Perimeter.

Abbreviation: P, Unit: Cm or m

## Formula:

The formula for mensuration is based on each shape and types of shape and each of them have been explained below:

### 2D Shapes Formula:

 Shape Area (Square Units) Perimeter (Units) Rectangle l × b 2 ( l + b) Scalene Triangle √[s(s−a)(s−b)(s−c)], a+b+c Square a2 4a Circle πr2 πr2 Rhombus ½ × d1 × d2 4 x side Trapezium ½ h(a+b) a+b+c+d Parallelogram b × h 2(l+b) Equilateral Triangle (√3/4) × a2 3a Isosceles Triangle ½ × b × h 2a + b

## Mensuration Formulas for 3D Shapes

 Shape Volume  (Cubic units) Curved Surface Area (CSA) or Lateral Surface Area (LSA) (Square units) Total Surface Area (TSA) (Square units) Cuboid l × w × h — 2 (lb +bh +hl) Cylinder π r 2 h 2π r h 2πrh + 2πr2 Cube a3 S 6 a2 Hemisphere (⅔) π r3 2 π r 2 3 π r 2 Sphere (4/3) π r3 4 π r2 4 π r2 Cone (⅓) π r2 h π r l πr (r + l)

## Mensuration Problems:

We have solved a few sample problems below which might help you

Problem:1: Find the area and perimeter of a square whose side is 8cm?

Solution: Given:

Side = 8 cm

Area of a square = asquare units

Substitute the value of “a” in the formula, we get

Area of a square = 82
A = 8*8=64

Therefore, the area of a square = 64 cm2

The perimeter of a square = 4a units

P = 4 x 8 = 32

Therefore, the perimeter of a square = 32 cm.

Problem:2: Six spherical cannonballs are tightly packed into a rectangular box in one layer. Each row has two cannonballs and each column has three. What part of the box is empty?

Solution: Let the diameter of each ball be 2r.
Length of the box = 3*2r = 6r
Breadth = 2*2r = 4r
Height = 2r
Volume = 6r * 4r * 2r = 48r3
Volume of 6 balls = 6 * (4/3) * (22/7) * r3 = (176 r 3 /7)
The area of empty space = 48 r3 – (176 r 3/7)
= (160 r3/7)
The required fraction = ((160 r3/7)/48 r3)

Problem:3: Smallest side of a right-angled triangle is 13 cm less than the side of a square of perimeter 72 cm. The second-largest side of the right-angled triangle is 2 cm less than the length of the rectangle of area 112 cmÂ² and breadth 8 cm. What is the largest side of the right-angled triangle?

Solution: Side of square = 72/4 = 18 cm

Smallest side of the right-angled triangle = 18 â€“ 13 = 5 cm

Length of rectangle = 112/8 = 14 cm

Second side of the right angled triangle = 14 â€“ 2 = 12 cm

Hypotenuse of the right-angled triangle = √(25 + 144) = 13cm

We hope we were able to help ease your doubts and provide you a better understanding. For more such updates, stay tuned to AdmitKard.com / General

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