Category: Maths Notes PPT

Calculus is known to be the branch of mathematics, that deals in the study rate of change and its application in solving equations. During the early Latin times, the idea of Calculus was derived from its original meaning “small stones” as means of computing a calculation of travelling distance or measuring and analyzing the movement of certain objects like stars from one place to another with their space covered in real-time. Apart from this, Calculus was used as a means of computation in various other fields also. 

 

For Eg: To gain a better understanding of the movement of a car about its speed at different intervals and time taken for travelling, Calculus can be used for easy computation. In other words, Calculus is a significant tool for measuring the tendency of fluctuations considering the nature of an object. 

 

Principles of Calculus

1. Limits and Infinitesimals: Infinitesimals refer to extremely small digit numbers such as values between 0 and 1. In other words, numbers that are less in value compared to a positive real number are called Infinitesimals. The concept of Calculus formulas was developed at first to compute such small values and thus, it can manipulate certain limits and principles for infinitesimals. 

2. Differential Calculus: Differential Calculus is one of the branches of calculus which is discussed further in detail. The process of deriving a derivation out of a function is termed Differentiation. Derivatives help in producing different functions and their output usually by squaring the values of the given digits. 

3. Leibniz Notation: The term Leibniz Notation is coined after the great Mathematician and German Philosopher Gottfried Wilhelm Leibniz. Leibniz Notation utilises the symbols of “dx” and “dy” to find the exact value of derivatives. The values of “dx” and “dy” are taken as assumptions to compute the further calculation of any given function. 

4. Integral Calculus: Integral Calculus is another branch of Calculus along with Differential Calculus. This particular concept is discussed more in detail further. In simple words, it is a study of the internal properties of a given function and its application in different fields. Integration refers to the idea of deriving the value of an integral. There are two types of integrals studied on a general basis namely; Indefinite Integral and Definite Integral. Indefinite integrals are denoted by “f” as they represent a number of functions that are non-constant in nature. On the other hand, definite integral functions consist of a series of functions that are constant in nature. 

5. Fundamental Theorem: The fundamental theorem of Calculus refers to the relationship between and the derivatives and their integrals. 

 

Differential calculus and integral calculus are the two major branches of calculus. Differential calculus formulas deal with the rates of change and slopes of curves. It can be used to study and examine the trend of a given graph and highlight its maximum and minimum values of curve points. The rate of change can be studied considering its quantity using Differential Calculus.  

 

For Eg: In our day-to-day life, the method of Differential Calculus can be used to scrutinize the change in rising and fall of temperature of a given place and at a given point of time. This helps to study the Weather Conditions of different places and compare them with one another later on. In the same way, the statistics related to the growth of a business entity can be compared by analyzing their profit and loss using Differential Calculus. As well, it can be used to measure the rate of change in speed of a vehicle travelling different miles or kilometers at given hours. 

 

Integral Calculus deals mainly with the accumulation of quantities and the areas under and between curves. In calculus, we use the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. These fundamentals are used by both differential and integral calculus. Unlike Differential Calculus which studies the rate of change and the number of different objects, Integral Calculus examines the nature of a particular object in the context of its speed, time-taken, and movement. It analyzes the internal properties of a change and computes advanced calculations of the area covered its length and volume.  

 

In other words, we can say that in differential calculus, an area splits up into small parts to calculate the rate of change. Whereas in integral calculus small parts are joined to calculate the area or volume and it is the method of reasoning or calculation. On one hand, Differential Calculus is used to compute the calculation of the rate of change, for instance, the velocity of the speed of a given object at a given point of time, then, on the other hand, Integral Calculus is used to study the area covered by that respective object and the time taken during the process. While Differential Calculus studies the minimum and maximum value of a given curve and highlights the curve point, Integral Calculus studies the area that has been covered under the respective curve. 

 

For calculating very small quantities, we use Calculus. Initially, the first method of doing the calculation with very small quantities was by infinitesimals. For example, an infinitesimal number could be greater than 0, but less than any number in the sequence 1, 1/2, 1/3, etc.

 

In other words, its value is less than any positive real number. According to this point of view, we can say that calculus is a collection of techniques for manipulating infinitesimals. In calculus formulas, the symbols dx and dy were taken to be infinitesimal, and the derivative formula dy/dx was simply their ratio.

 

What are the Limits?

Suppose we have a function f(x). The value, a function attains, as the variable x approaches a particular value let’s say suppose a that is., x → a is called its limit. Here, ‘a’ is some pre-assigned value. It is denoted as

lim x→a f(x) = l

• The expected value of the function f(x) shown by the points to the left of a point ‘a’ is the left-hand limit of the function at that point. It is denoted as limx→a−  f(x).

• The point to the right of the point that is ‘a’ which generally shows the value of the function is the right-hand limit of the function at that point. It can be denoted as limx→a + f(x).

Limits of functions at a point are the common and coincidence value of the left and right-hand limits.

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The value of a limit of a function f(x) at a point a that is, f(a) may vary from the value of f(x) at the point ‘a’.

 

Math Limit Formula List

 

Differential Calculus & Differential Calculus Formulas

The study of the definition, properties, and applications of the derivative of a function is known as Differential calculus. Differentiation is the process of finding the derivative. Here are some calculus differentiation formulas by which we can find a derivative of a function.

 

Differential Calculus Formulas

 

Integration Formulas

The branch of calculus where we study about integrals, accumulation of quantities, and the areas under and between curves and their properties is known as Integral Calculus. Let’s discuss some integration formulas by which we can find integral of a function. Here’s the Integration Formulas List

 

 

Questions to Be Solved

Question 1. Solve the Following Definite Integral

[int_{2}^{4}x^{2}] dx

Solution

[int_{2}^{4}x^{2}] dx

[int_{2}^{4}x^{2}] dx = [[frac{x^{3}}{3}]_{2}^{4}]

= [frac{64}{3}] – [frac{8}{3}]

= [frac {56}{3}]

= 18.66 

One of the most basic approaches of statistical analysis is the standard deviation. The Standard Deviation, abbreviated as SD and represented by the letter “, indicates how far a value has varied from the mean value. A small Standard Deviation means the results are close to the mean, whereas a big Standard Deviation means the data are widely divergent from the mean. Let’s look at how to determine the Standard Deviation of grouped and ungrouped data, as well as the random variable’s Standard Deviation.

Standard Deviation

The Standard Deviation is a statistic that indicates how much variance or dispersion there is in a group of statistics. A low Standard Deviation means that the value is close to the mean of the set (also known as the expected value), and a high Standard Deviation means that the value is spread over a wider area.

The lower case Greek letter sigma, for the population Standard Deviation, or the Latin letter s, for the sample Standard Deviation, is most usually represented in mathematical texts and equations by the lower case Greek letter sigma.

The square root of the variance is the Standard Deviation of a random variable, sample, population, data collection, or probability distribution. It is algebraically easier than the average absolute deviation, but it is less resilient in practice. The Standard Deviation has the advantage of being reported in the same unit as the data, unlike the variance.

Standard Deviation is the measure of the dispersion of data from its mean. It measures the absolute variability of a distribution. The higher is the dispersion or variability of data, the larger will be the standard deviation and the larger will be the magnitude of the deviation of value from the mean whereas the lower is the dispersion or variability of data, the lower will be the standard deviation and the lower will be the magnitude of the deviation of value from the mean. The standard deviation formula is used to find the values of a specific data that is dispersed from the mean value. It is important to observe that the value of standard deviation can never be negative.

There Are Two Types of Standard Deviation

Standard Deviation formula to calculate the value of standard deviation is given below:

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Standard Deviation Formulas For Both Sample and Population

Notations For the Sample Standard Deviation Formula and Population Standard Deviation Formula

 σ = Standard Deviation

Xi = Terms given in the data

N = Total number of terms

[overline{X}]= Mean of the data

What is Variance and Standard Deviation?

Variance –  The variance is a numerical value that represents how broadly individuals in a group may change. The variance will be larger if the individual observations change largely from the group mean and vice versa.

It is important to notice similarities between the variance of sample and variance population. They have different representations and are calculated differently. The variance of a population is represented by σ² whereas the variance of a sample is represented by s².

Standard Deviation – Standard deviation is a measure of dispersion in statistics. It gives an estimation of how individuals in data are dispersed from the mean value. Standard deviation is defined as the square root of the mean of a square of the deviation of all the values of a series derived from the arithmetic mean. It is also known as root mean square deviation.The symbol used to represent standard deviation is Greek Letter sigma (σ 2).

Variance and Standard Deviation Formula

Variance, [sigma^{2} = frac{sum_{i=1}^{n} (x_{i} – overline{x})^{2}}{n} ]

Standard Deviation, [sigma = sqrt{frac{sum_{i=1}^{n} (x_{i} – overline{x})^{2}}{n}} ] 

In the above variance and standard deviation formula:

xi   =  Data set values

[overline{x}]= Mean of the data

With the help of the variance and standard deviation formula given above, we can observe that variance is equal to the square of the standard deviation.

Mean and Standard Deviation Formula

The sample mean is the average and is calculated as the addition of all the observed outcomes from the sample divided by the total number of events. Sample mean is represented by the symbol. In Mathematical terms, sample mean formula is given as:

[overline{x} = frac{1}{n} sumlimits_{i=1}^{n} x ]

In the above sample mean formula

N is the sample size and 

X is the correspond observed values

Standard Deviation – On the other hand, standard deviation perceives the significant amount of dispersion of observations when comes up close with data. In Mathematical terms, standard dev formula is given as:

Standard Deviation, [sigma = sqrt{frac{sum_{i=1}^{n} (x_{i} – overline{x})^{2}}{n}} ] 

Standard Error of Mean Formula

The standard error of the mean is a procedure used to assess the standard deviation of a sampling distribution. It is also known as standard deviation of the mean and is represented as SEM. Generally, the population mean approximated  value is the sample mean, in a sample space. But, if we select another sample from the same population, it may obtain a different value.

Therefore, a population of the sampled means will appear to have different variance and mean values. The standard error of the mean can be determined as the standard deviation of such a sample means including all the possible samples drawn from the same population. SEM is basically an approximation of standard deviation, which has been evaluated from the sample.

The standard error of the mean formula is equal to the ratio of the standard deviation to the root of the sample size.

[SEM = frac{SD}{sqrt{N}} ]

In the above standard error of mean formula,

‘SD’ is the standard deviation 

N is the number of observations.

Variance and Standard Deviation Formula for Grouped Data 

[sigma = frac{sum f(m – mu)^{2}}{N} ]

And

[s^{2} = frac{sum f(m – overline{x})^{2}}{n – 1} ]

The calculation of standard deviation can be done by taking the square root of the variance. Hence, the standard deviation is calculated as 

Population Standard Deviation – [sigma = sqrt{sigma^{2}} ]

Sample Standard Deviation – [s = sqrt{s^{2}} ]

Here in the above variance and std deviation formula,

σ2 is the population variance, s2 is the sample variance, m is the midpoint of a class.

Standard Deviation Formula for Discrete Frequency Distribution

For the discrete frequency distribution of the type.

y : y₁, y₂, y₃,y₄

f : f₁, f₂, f₃, f₄

The formula for standard deviation becomes

[ sqrt{frac{1}{N} sumlimits_{i = 1}^{n}  f_{i}(x_{i} – overline{x})^2 }]

In the above formula, N is the total number of observations.

Standard Deviation V/S Variance

The list of standard deviation v/s variance is given below in tabulated from

Solved Examples

1. During a survey, 6 students were asked the number hours per day they give time to their studies on an average? The answers of the students are as follows: 2, 6, 5, 3, 2, 3. Calculate the standard deviation.

Solution: 

Step 1: Calculate the mean value of the given data

[ = frac{2 + 6 + 5 + 3 + 2 + 3}{6}]

[ = frac{21}{6}] 

= 3.5

Step 2: Construct a table for the above given data

Step 3 : Now, use the standard dev formula.

Sample Standard Deviation Formula – [s = sqrt{frac{sum (x_{i} – overline{x})^{2}}{n-1}} ] 

[= sqrt{frac{13.5}{6 – 1}}] 

[= sqrt{frac{13.5}{5}}] = [= sqrt{2.7}]  

= 1.643

2. A Hen lays eight eggs. The weight of each egg laid by hen is given below.

Solution:

Step 1: Let us first calculate the mean of the above data

Mean = [sum frac{X}{N} ]

[= frac{60 + 56 + 61 + 68 + 51 + 53 + 69 + 54}{8} ]

[= frac{472}{8}]

= 59

Step 2: Construct a table for the above – given data

Step 3 :  Now, use the standard dev formula

Standard Deviation Formula [= sqrt{frac{sum (x_{i} – overline{x})^{2}}{n}} ]

[= sqrt{frac{320}{8}}] = [ sqrt{40} ]

= 6.32 grams

Quiz Time

1. Find the Standard Deviation for the Given Data

    5,10,7,12,10,20,15,22,8.2

1. 6.89

2. 10.01

3. 7.26

4. 9

2. Which of the Following Is the Measure of Variability?

1. Mean

2. Median

3. Mode

4. Standard Deviation

Perfect Square Formula

When a polynomial is multiplied to itself, it results in a perfect square polynomial. 

The perfect square formula is given by the following expression:

(a + b)² = a² + 2ab + b²

How to Check Whether a Given Polynomial is a Perfect Square or Not?

To check whether a given polynomial, ax2 + bx + c is a perfect square or not, we just check the relation b2 = 4ac is satisfied or not. If the relation b2 = 4ac is satisfied, then the given polynomial is a perfect square.

Solved Examples

Q 1. Check Whether the Polynomial x2 + 6x + 8 is a Perfect Square or Not.

Solution :

Comparing the given polynomial x2 + 6x + 8 with the standard polynomial ax2 + bx + c, we get the values of a = 1, b = 6 and c = 8.

Now, check the relation b2 = 4ac is satisfied or not

LHS: b2 = 62 = 36

RHS: 4ac = 4 x 1 x 8 = 32

Here, LHS ≠ RHS. The relation is not satisfied i.e., b2 ≠ 4ac.

Therefore, the given polynomial x2 + 6x + 8 is not a perfect square.

Q 1. Check Whether the Polynomial x2 + 8x + 16 is a Perfect Square or Not.

Solution :

Comparing the given polynomial x2 + 8x + 16 with the standard polynomial ax2 + bx + c, we get the values of a = 1, b = 8 and c = 16.

Now, check the relation b2 = 4ac is satisfied or not

LHS: b2 = 82 = 64

RHS: 4ac = 4 x 1 x 16 = 64

Here, LHS = RHS. The relation b2 = 4ac is satisfied.

Therefore, the given polynomial x2 + 8x + 16 is a perfect square.

Differentiation and Integration are both quite crucial concepts in calculus which are typically used to learn the change. Calculus is not only restricted to mathematics but has a huge array of applications in various domains of science as well as the economy. Also, we may be able to spot calculus in establishing an analysis in finance as well as in the stock market. In this chapter, we will study some differentiation and integration formula with examples besides the interesting concept!

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Find below some of the basic formulas of differentiation and integration.

Differentiation and Integration Formulas

What is Differentiation in Maths?

The mathematical concept of Differentiation facilitates us to find rates of change. For example, it enables us to detect the rate of change of velocity with respect to time (called the acceleration). In addition, It also enables us to find the rate of change of x with reference to y, which graphically at y against x is the gradient of the curve. There are a set of simple rules which can be used in order to differentiate a number of functions easily.

If y = some function of x, then the derivative of y (in respect to x) is written as dy/dx.

For Example: Find out the gradient of the curve y = 2x⁴ at the point (4, 56)?

Using the formula, dy/dx = 8 x²

When x = 4, dy/dx = 8 × 7 = 56.

What is Integration in Mathematics?

Integration is a mathematical technique to find a function g(x) the derivative of which, Dg(x), is equivalent to a provided function f(x). This is denoted by the integral sign “∫,” or ∫f(x), generally termed the indefinite integral of the function. The sign dx denotes a displacement of an infinitesimal along x; therefore ∫f(x) dx becomes the aggregate sum of the product of f(x) and dx. The definite integral is written as: – “∫a^b

With ‘a’ and ‘b’ referred to the limits of integration, is equivalent (=) to g(b) − g(a), where Dg(x) = f(x).

Usefulness of Integration

Integration is a mathematical approach of adding slices to find the whole. Applying the theory of Integration, we are able to easily find areas, volumes, median points of many useful figures. That said, it is still simplest to start with finding the area under the curve of a function like one below:

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Differentiation Under the Integral Sign

Differentiation under the integral sign is an algebraic operation in calculus that is performed in order to assess certain integrals. Under a reasonably loose situation on the function being integrated, this operation enables us to swap the order of integration and differentiation. In its simplified version, called the Leibniz integral rule, differentiation under the integral sign models the ensuing equation legitimate under the formula on:

[frac{d}{dt} int^{b}_{a} f(x,t)dx = int^{b}_{a} partial_{t} f(x,t)dx]

The Usefulness of Differentiation under the Integral Sign

Many integrals that would otherwise be impracticable or need a substantially more complex approach can be solved by this method.

Differentiation under the integral sign rule can be used to assess the certain unusual definite integrals such as given below:

[I’(b) frac{d}{db} int^{1}_{0} frac{x^{b}-1}{In{x}}dx ]

[frac{d}{db} int^{1}_{0}  frac{x^b-1}{In(x)} = = int^{1}_{0} frac{partial}{partial{b}}frac{x^b -1}{In(x)} ]

[I’(b) = int^1_0 frac{In(x)x^b}{In(x)} dx]

Solved Examples For You

Example1: Differentiate the given functions

[Y = sqrt[3]{x^2}{(2x-x^2)}]

Solution1:

In the given function we would not be able to just differentiate the 1st term, and then differentiate the 2nd term and multiply the two together. This would just NOT work win the case given. However, it is still possible to perform the derivative of this function. All we are supposed to do is convert the radical into fractional exponents and multiply this across the parenthesis.

[Y = x^{frac{2}{3}}(2x-x^2) = 2x^{frac{5}{3}} -x^{frac{8}{3}}]

Here, we can differentiate the function

[Y^{1} = frac{10}{3} x^{frac{2}{3}} – frac{8}{3}x^{frac{5}{3}}]

Example2:  Find out if the given equation

[f(x) = 2x^3+ frac{300}{x^{-3}}+4 ]rising, reducing or not changing at x= −2?

Solution2:

We are already aware that the rate of change of a function is provided by the derivative of the function. Thus all we are supposed to do is to rewrite the function. Rewriting the function is required in order to deal with the 2nd term and then take the derivative.

[f(x) = 2x^3+ frac{300}{x^{-3}}+4] → [f^{1} (x) = 6x^{2} – 900^{x^{-4}} = 6x^{2} – frac{900}{x^{4}}]

Remember that had rewritten the last term in the derivative back as a fraction.

This is not something we’ve performed up to this step and is only being done here to help with the assessment in the next step. Note that it’s often simpler to do the assessment with positive exponents.

So, upon assessing the derivative we now obtain,

[f^{1} (-2) = 6(4) -frac{900}{16} = -frac{129}{4} = -32.25]

Thus, at x = −2, the derivative turns out to be negative and, hence, the function is decreasing at x = −2

Circumcentre of a triangle is a special point in the triangle at which the perpendicular bisectors of all three sides bisect. In simple words, the point of bisection of perpendicular bisectors of the sides of a triangle is what we call the circumcentre of a triangle. This is also the centre of the circle, crossing the vertices of the given triangle. That is to say, Circumcentre is also at an equal distance to all the vertices of the triangle. The triangle’s circumcentre lies inside for an acute-angled triangle, outside for an obtuse-angled triangle and at the centre of the hypotenuse for a right-angled triangle.

Perpendicular Bisector and Circumcentre of a Triangle

For better clarity and enhanced understanding of the circumcentre of a triangle, we will require to properly understand the concept of perpendicular bisector.

If we draw a perpendicular bisector of a line segment MN, then the arbitrary point P on the perpendicular bisector will be at an equal distance from the end points M, N of that line segment.

Method to Calculate The Circumcentre of a Triangle

1. Determine the midpoint of the sides provided the coordinates MN NO OM.

2. Find out the slope of a specific line.

3. By applying the slope and midpoint find out the equation of the line segment y – y₁ = m[x – x₁].

4. Calculate the equation of the other line in a similar manner.

5. Solve the 2 bisector equations by computing the interception point calculated interception point is the Circumcentre.

6. Circumcentre can be computed using the linear equations.

Properties of the Circumcentre

Applying properties of the circumcentre, we can even simply solve problems that themselves don’t involve the circumcentre directly. Because a circumcentre is a substantial structure that correlates lengths and angles, using it appropriately in a problem can be very yielding. For this reason, it is significant to know how to locate circumcentres and a proper moment to use them.

O is the circumcentre of PQR if and only if:

1. PO = QO = RO

2. QO = RO and angle ∠ POR = 2 ∠P is acute and PO are on the same side of QR.

3. QO = RO and angle ∠ POR = 2 (180 – ∠ P) when ∠P is obtuse and P, O are on opposite sides of QR.

Centroid of a Triangle

Before getting to learn the mathematical term centroid, let’s first know what we refer to by a median. In that context, a median is the line connecting the mid-points of the sides and the opposite vertices. A centroid is typically the point of bisection of the medians of the triangle and also divides the median in the ratio of 2:1.

Let’s understand the concept of the centroid of a triangle using a solved example.

Example: Identify the centroid of the triangle the coordinates of whose vertices are assigned as M(x₁, y₁), N(x₂, y₂) and O(x₃, y₃) respectively.

Solution: Now, we already know that a centroid divides the median of a triangle in the ratio 2:1. Therefore, since ‘G’ is the median, thus MG/AQ = 2/1.

Seeing that, D is tG is the Centroid of a triangle the midpoint of NO, coordinates of D are as follows:

((x₂ + x₃)/2, (y₂ + y₃)/2)

Applying the section formula, the coordinates of G will be;

{2(x₂ + x₃)/2} + 1.x₁/2 + 1, {2(y₂ + y₃)/2} + 1.y₁/2 +1]

⇒ Thus, Coordinates of G are [x₁ + x₂ + x₃/3, y₁ + y₂ + y₃/3]

If the coordinates of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃), then the coordinates of the centroid (which is usually represented by G) are given in the following manner:

[x₁ + x₂ + x₃/3, y₁ + y₂ + y₃/3].

Incentre of a Triangle

The incentre of a triangle is the point of bisection of the angle bisectors of angles of the triangle. An incentre is also referred to as the centre of the circle that touches all the sides of the triangle.

The Angle bisector typically splits the opposite sides in the ratio of remaining sides i.e. MP/PO = MN/MO = o/n.

Incentre splits the angle bisectors in the stated ratio of (n + o):a, (o + m):n and (m + n):o.

Wondering how to find the perimeter of a rhombus? Just like any polygon, the perimeter of the rhombus is the total distance around the outside, which can be simply calculated by adding up the length of each side. In the case of a rhombus, all four sides are of similar length, thus the perimeter is four times the length of aside. Or as a formula: Perimeter of rhombus = 4 a = 4 × side. Here, ‘a’ represents each side of a rhombus.

About the Perimeter of Rhombus Formula

A rhombus is a 2-dimensional (2D) geometrical figure which consists of four equal sides. Rhombus has all sides equal and its opposite angles are equivalent in measurement. Let us now talk about the rhombus formula i.e. area and perimeter of the rhombus.

The Perimeter of a Rhombus

The perimeter is the sum of the length of all 4 sides. In the rhombus all sides are equal.

Thus, the Perimeter of rhombus = 4 × side

 

So, P = 4s

 

In which,

 

S = length of a side of a rhombus

Area of Rhombus Formula

The area of a rhombus is the number of square units in the interior of the polygon. The area of a rhombus can be identified in 2 ways:

i) Multiplying the base and height as rhombus is a unique kind of parallelogram.

Area of rhombus = b × h

In which,

 

B = base of the rhombus

 

H = height of the rhombus

ii) By determining the product of the diagonal and dividing the product by 2.

Area of rhombus formula = [frac {1}{2}] × d1 × d2

 

In which,

 

d₁ × d₂ = diagonal of the rhombus

Derivation of Area of Rhombus

Let MNOP is a rhombus whose base MN = b, PN ⊥ MO, PN is a diagonal of rhombus = d₁, MO is diagonal of rhombus = d₂, and the altitude from O on MN is OZ, i.e., h.

= 2 × [frac {1}{2}] MN × OP sq units.

= 2 × [frac {1}{2}] b × h sq. units

= base × height sq. units

= 4 × [frac {1}{2}] × MZ × ZN sq. units

= 4 × [frac {1}{2}] × d₂ × [frac {1}{2}]  d₁ sq. units

 

Thus,

 

= 4 × [frac {1}{8}] d₁ × d₂ sq. units

 

= [frac {1}{2}] × d₁ × d₂

 

Hence, the area of a rhombus

 

= [frac {1}{2}] (product of diagonals) sq. units.

Fun Facts

• A rhombus consists of an inscribed circle

• In a rhombus, all sides are equal, just as a rectangle has all angles equal.

A rhombus has opposite angles equivalent to each other, while a rectangle has opposite sides equal.

Finding the Perimeter of a Rhombus when only Diagonals are known

In many questions, you will see that the length of the sides of the rhombus is not given. Instead, the question will provide you with the length of its diagonals. You can use the diagonals to find the side of the rhombus and calculate its perimeter. Here is how you can do it: 

1. Consider a rhombus PQRS with diagonals a and b, and center O.

2. Pick a triangle, say POS. Since the diagonals of a rhombus bisect each other at right angles, the length of OP will be [frac {a}{2}] and OS will be [frac {b}{2}]. 

3. Assume that the side of the rhombus is X. 

4. Using Pythagoras theorem in triangle POS, we get: 

X₂ = [frac {a} {2}^2 + frac {b} {2}^2 ], which gives us, 

X₂ = [frac {(a^2+b^2)} {4} ]

X =   [sqrt frac  {(a^2+b^2)} {2} ]   

5. Substitute the value of X in the perimeter of the rhombus formula i.e. 4X. You will get: 

4 ×  [sqrt frac  {(a^2+b^2)} {2} ]

6. 6. So, the perimeter of the rhombus will be equal to 2 ×  [sqrt {(a^2+b^2)} ].

Solved Examples

1. Evaluate the area of the rhombus MNOP having each side equal to 15 cm and one of its diagonals equal to 18 cm.

Ans:  

Given:- MNOP is a rhombus in which MN = NO = OP = PM = 15 cm

 

MO = 18 cm

 

Thus, MZ = 9 cm

 

In ∆ MZP,

 

MP² = MZ² + ZP²

 

⇒ 152 = 92 + ZP²

 

⇒ 225 = 81 + ZP²

 

⇒ 144 = ZP²

 

⇒ ZP = 12

 

Hence, NP = 2 P

 

= 2 × 12

 

= 24 cm

 

Now, to find out the area of the rhombus, we will apply the formula i.e.

 

= [frac {1}{2}] × d₁ × d₂

 

= [frac {1}{2}] × 18 × 24

 

= 216 cm²

2. Find the perimeter of a rhombus MNOP whose diagonals measure 20 cm and 24 cm respectively?

Ans: 

Given: d₁ = 20 cm

 

d₂ = 24 cm

 

MZ= [frac {20}{2}] = 10cm

 

NZ= [frac {24}{20}]= 12 cm

 

∠MZP = 90°

 

Now applying the Pythagorean Theorem, we know that

 

MN² = MZ² + NZ²

 

MN = [sqrt {(100+144)} ]

 

= 15.62 cm

 

Since, MN = NO = OP = MP,

 

Therefore, Perimeter of MNOP = 15.62 × 4 = 62.48 cm.

Properties of a Rhombus 

Identifying a rhombus is not that difficult. There are some properties of a rhombus that will help you determine whether a given figure is a rhombus or not. If a shape meets the following conditions, then it is a rhombus: 

• All the sides of a rhombus are equal. 

• The opposite sides of a rhombus are parallel to each other. 

• All the opposite angles in a rhombus will be equal. 

• The diagonals of a rhombus bisect each other at 90 degrees i.e. right angles. 

• When you add any two adjacent angles of a rhombus, the sum should be equal to 180 degrees.

Revising the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs

The Perimeter of Rhombus Formula is one of the most important formulas of Mathematics. It comes under mensuration, which is a crucial chapter of Maths. You must have a clear understanding of the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs to ensure a good score in your finals. Once you have a firm grasp of the perimeter and area of a rhombus, you will be able to solve any question related to this concept. That is why you should practice and revise the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs thoroughly. Here are some revision tips: 

• First, study the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs provided to understand the topic clearly. 

• Go through the textbook explanations of the perimeter and area of a rhombus thoroughly.

• Practice as many questions related to the perimeter and area of a rhombus as you can to become more proficient at solving mensuration problems. 

• Use different reference books and pick out the important questions based on the perimeter and area of a rhombus to understand the different types of questions that can come in your final Maths exam.

• Visit ’s e-learning platform through our website or mobile application to study the Perimeter of Rhombus Formula – Explanation, Area, Solved Examples and FAQs. Every explanation and question provided by is curated by some of the best subject matter experts to ensure accuracy and high quality. 

• Go through solved examples from the textbook or reference books to learn how to solve different kinds of questions based on the area and perimeter of a rhombus. 

• Once you have solved all the textbook questions, find out questions from previous year question papers and sample papers to understand the pattern and difficulty level of the exam.