mathemafia Archives – Mathemafia

Class 10 – All Maths Formulas – Complete List

Sunil Kumar — Thu, 03 Jul 2025 06:49:25 +0000

Perfect for quick revision and last-minute prep!

This is an interactive, colorful and extremely engaging 💕 comprehensive ✅ list of formulas of Class 10 Mathematics. This is a one-stop 🎯 solution for the students to quickly review / recall all formulas that would be very helpful especially for last minute revision before their math exam.

Click a link displayed below to read the corresponding chapter’s formulas.

Ch 01 – Real Numbers
Ch 02 – Polynomials
Ch 03 – Pair of Linear Equations in Two Variables
Ch 04 – Quadratic Equations
Ch 05 – Arithmetic Progressions
Ch 06 – Triangles
Ch 07 – Coordinate Geometry
Ch 08 – Introduction to Trigonometry
Ch 09 – Some Applications of Trigonometry
Ch 10 – Circles
Ch 11 – Areas Related to Circles
Ch 12 – Surface Areas and Volumes
Ch 13 – Statistics
Ch 14 – Probability

Mathemafia

Creative Hub of Mathematics

👉 1400+ Math videos ✅

Join Now

Chapter 1 - Real Numbers

Square root of any prime number is irrational.

E.g. – \( \begin{aligned} &\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7} \end{aligned}\)

Click here to watch proof of √3

\( \begin{aligned} LCM (a, b) \times HCF (a, b) = a \times b \end{aligned}\)

Fundamental Theorem of Arithmetic

Every composite number can be factorised as a product of primes, and this factorisation is unique.

Click here to watch its proof

Theorem

If p is a prime and p divides a², then p divides a, where a is a positive integer.

Click here to watch its proof

Chapter 2 - Polynomials

Linear Polynomial

Degree = 1

No. of real zeros: At most 1

E.g: x

Quadratic Polynomial

Degree = 2

No. of real zeros: At most 2

E.g: x²

Cubic Polynomial

Degree = 3

No. of real zeros: At most 3

E.g: x³

Monomial

No: of terms = 1

E.g: 2x

Binomial

No: of terms = 2

E.g: 2x + 3

Trinomial

No: of terms = 3

E.g: 2x² + 3x + 2

Quadratic Polynomials

Standard form of Quadratic Polynomial

\( \begin{aligned} &ax^2+bx+c, \ \ a\ne0\\ \end{aligned}\)

Zeros of Quadratic Polynomial

If α is a zero of polynomial

\( \begin{aligned} &p(x) = ax^2+bx+c, \ \ a\ne0\\ &then \ \ p(α) = 0 \end{aligned}\)

Sum of Zeros

\( \begin{aligned} &α + β = -\frac{b}{a} \end{aligned}\)

Product of Zeros

\( \begin{aligned} &α \times β = \frac{c}{a} \end{aligned}\)

Add Graphs – Later XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Chapter 3 - Pair of Linear Equations in Two Variables

Standard form of Linear Equation in Two Variables

\( \begin{aligned} &ax+by+c =0,\\ &\text{where a and b are not both 0.} \end{aligned}\)

\( \begin{aligned} &ax+by+c =0, \ \ where \ \ a^2+b^2\ne 0 \end{aligned}\)

Standard form of Pair of Linear Equation in Two Variables

\( \begin{aligned} &a_{1}x+b_{1}y+c_{1} =0, \ \ where \ \ a_{1}^2+b_{1}^2\ne 0 \\&a_{2}x+b_{2}y+c_{2} =0, \ \ where \ \ a_{2}^2+b_{2}^2\ne 0 \end{aligned}\)

Solution of Pair of LE in 2 Variables

Intersecting Lines
➤ Consistent Pair of LE

➺ Lines Intersect at 1 point
🎯 No. of Solutions = 1

\( \begin{aligned} &\frac{a_{1}}{a_{2}} \ne \frac{b_{1}}{b_{2}} \end{aligned}\)

Coincident Lines

➤ Consistent Pair of LE
➺ Lines overlap each other

🎯 No. of Solutions = Infinite

\( \begin{aligned} &\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{aligned}\)

Parallel Lines
➤ inconsistent Pair of LE

➺ No common point

🎯 No. of Solutions = 0

\( \begin{aligned} &\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \ne \frac{c_{1}}{c_{2}} \end{aligned}\)

Click here to watch Introduction to pair of Linear equations

Chapter 4 - Quadratic Equations

Standard form of Quadratic Equation

\( \begin{aligned} &ax^2+bx+c=0, \ \ a\ne0\\ \end{aligned}\)

Roots of Quadratic Equation

Zeros of quadratic polynomial \( \begin{aligned} &ax^2+bx+c \end{aligned}\)

➤ Roots of the quadratic equation \( \begin{aligned} &ax^2+bx+c=0 \end{aligned}\)

are the same.

Sum of Roots

\( \begin{aligned} &α + β = -\frac{b}{a} \end{aligned}\)

Product of Roots

\( \begin{aligned} &α \times β = \frac{c}{a} \end{aligned}\)

Nature of Roots

Discriminant

\( \begin{aligned} &D = b^2-4ac \end{aligned}\)

D > 0 2 Distinct, Real Roots

D = 0

2 Equal, Real Roots

(coincident roots)

D < 0

No Real Roots

➤ Roots of the QE are

\( \begin{aligned} &x=\frac{-b ± \sqrt{D}}{2a} \end{aligned}\)

Click here to watch the Derivation of Quadratic Formula

Chapter 5 - Arithmetic Progression

General Form of AP

a, a + d, a + 2d, a + 3d, . . .

First Term: a

Common Difference: d

➺ d can be +ve, -ve or 0

\( \begin{aligned} &a_{2} -a_{1}=a_{3} -a_{2}=a_{4} -a_{3}=…=a_{n} -a_{n-1}=d \end{aligned}\)

nth Term of AP

\( \begin{aligned} &a_{n} = a +(n-1)d \end{aligned}\)

Sum of first n Terms

\( \begin{aligned} &S_{n} =\frac{n}{2}[2a+(n-1)d] \end{aligned}\)

Sum of first n Terms

\( \begin{aligned} &S_{n} =\frac{n}{2}(a+l) \\&\text{where }l \text{ is the last term}\end{aligned}\)

📌 Super important 🔥

Total terms in AP: 25

11th term from the end = 25 – 11 + 1 = 15th Term

If a, b, c are in AP, then

\( \begin{aligned} &b=\frac{a+c}{2} \end{aligned}\)

Chapter 6 - Triangles

Similar Figures

➺ Same shape

➺ Not necessarily the same size

🦆 🦆 🔺🔺

➺ All Circles are similar 🔵

➺ All Squares are similar 🟩

➺ All Equilateral Triangles are similar 🔺

➺ All Congruent figures are similar ✅

❌ All similar figures are not congruent

🔺 🔺 Similar Triangles 🔺🔺

Two Triangles ABC and DEF are similar if

➺ their corresponding angles are equal

\( \begin{aligned} &\angle A = \angle D, \ \ \angle B = \angle E, \ \ \angle C = \angle F\end{aligned}\)

➺ & their corresponding sides are in the same ratio (proportion).

\( \begin{aligned} &\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\end{aligned}\)

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Click here to watch its proof

Converse of Basic Proportionality Theorem

If a line divides any two sides of a triangle in the same ratio then the lines is parallel to the third side.

Click here to watch its proof

Triangles Similarity Criteria

SSS

If Corresponding Sides are in the same ratio, then the ▲s are similar.

\( \begin{aligned} &\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\end{aligned}\)

AAA

If Corresponding Angles are equal, then the ▲s are similar.

\( \begin{aligned} \angle A &= \angle D,\\ \angle B &= \angle E,\\ \ \ \angle C &= \angle F\end{aligned}\)

SAS

If one angle of a triangle is equal to one angle of another triangle and the sides including this angle are in the same ratio, then the ▲s are similar. For e.g.

\( \begin{aligned} \frac{AB}{DE}&=\frac{BC}{EF} \ \ \\ \angle B &= \angle E \end{aligned}\)

RHS

If in two right angled ▲s, hypotenuse and one side of a ▲ are proportional to hypotenuse and one side of the other ▲, then the ▲s are similar.

If two angles of one ▲ are respectively equal to the two angles of the other ▲, then the two ▲s are similar.

\( \begin{aligned} \angle A &= \angle D,\\ \angle B &= \angle E\end{aligned}\)

This case is same as AAA because 3rd angle will be equal as well.

Mathemafia

Creative Hub of Mathematics

👉 1400+ Math videos ✅

Join Now

Chapter 7 - Coordinate Geometry

Abscissa / x-coordinate

The distance of a point from the y-axis

Ordinate / y-coordinate

The distance of a point from the x-axis

Distance Formula

Distance between points A (x₁, y₁) and B (x₂, y₂)

Distance from origin

Distance of point A (x, y) from origin O (0, 0)

Section Formula

Mid-Point Formula

Chapter 8 - Introduction to Trigonometry

For Angle A

BC – Perpendicular

AB – Base

AC – Hypotenuse

For Angle C

AB – Perpendicular

BC – Base

AC – Hypotenuse

Trigonometric Ratios

\(\begin{align*}&\color{red}{sin} \ \ \color{green}{cos} \ \ \color{blue}{tan}\\ &\boxed{\color{red}{\frac{P}{H}}}\boxed{\color{green}{\frac{B}{H}}}\boxed{\color{blue}{\frac{P}{B}}}\end{align*}\)

P B P – Pandit Badri Prasad

H H B – Har Har Bole

\( \begin{aligned} sinθ = \frac{P}{H}\end{aligned}\)

\( \begin{aligned} &cosθ = \frac{B}{H}\end{aligned}\)

\( \begin{aligned} &tanθ = \frac{P}{B}\end{aligned}\)

\( \begin{aligned} &cosecθ = \frac{H}{P}\end{aligned}\)

\( \begin{aligned} &secθ = \frac{H}{B}\end{aligned}\)

\( \begin{aligned} &cotθ = \frac{B}{P}\end{aligned}\)

Imp Relationships b/w ratios

\( \begin{aligned} &sin θ= \frac{1}{cosecθ}\end{aligned}\)

\( \begin{aligned} &cos θ= \frac{1}{secθ}\end{aligned}\)

\( \begin{aligned} &tan θ= \frac{1}{cotθ}\end{aligned}\)

Trigonometric ratios of standard angles

∠A	0°	30°	45°	60°	90°
sin A	0	\(\begin{aligned} \frac {1}{2}\end{aligned} \)	\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)	\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)	1
cos A	1	\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)	\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)	\(\begin{aligned} \frac {1}{2}\end{aligned} \)	0
tan A	0	\(\begin{aligned} \frac {1}{\sqrt {3}}\end{aligned} \)	1	\(\begin{aligned} \sqrt {3} \end{aligned} \)	N.D.
cosec A	N.D.	2	\(\begin{aligned} \sqrt {2} \end{aligned} \)	\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)	1
sec A	1	\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)	\(\begin{aligned} \sqrt{2}\end{aligned} \)	2	N.D.
cot A	N.D.	\(\begin{aligned} \sqrt{3}\end{aligned} \)	1	\(\begin{aligned} \frac {1}{\sqrt{3}}\end{aligned} \)	1

N.D. – Not Defined

Trigonometric Identities

\( \begin{aligned} &sin^2θ+cos^2θ =1\end{aligned}\)

\( \begin{aligned} &1+tan^2θ =sec^2θ\end{aligned}\)

\( \begin{aligned} &1+cot^2θ =cosec^2θ\end{aligned}\)

Chapter 9 - Some Applications of Trigonometry

This chapter is about the applications of the formulas that we learnt in the previous chapter. You Students are recommended to practice different types of questions to master this chapter.

Chapter 10 - Circles

Tangent

Line that intersects the circle at only one point

Important

➺ Diameter is the longest chord

➺ Tangent is perpendicular to the radius through the point of contact

➺ A circle has infinitely many tangents

Theorem

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Click here to watch its proof

Theorem

The lengths of tangents drawn from an external point to a circle are equal.

Click here to watch its proof

Chapter 11 - Areas Related to Circles

Area of Circle = πr²

Circumference of Circle = 2πr

Area of a Sector

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{θ}{360}} \color{black}\times \color{blue}{\pi r^2}}\end{aligned}\)

Length of Arc

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{θ}{360}} \color{black}\times \color{blue}{2\pi r}}\end{aligned}\)

Area of a Segment

Area of the corresponding sector

– Area of the corresponding triangle

Area of Major Sector = Area of Circle – Area of Minor Sector

Area of Major Segment = Area of Circle – Area of Minor Segment

Chapter 12 - Surface Areas and Volumes

HS – Hemisphere
SA – Surface Area
CSA – Curved Surface Area
TSA – Total Surface Area
LSA – Lateral Surface Area

Surface Areas

TSA of a Cube

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{6} \color{blue}{a^2}}\end{aligned}\)

LSA of a Cube

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{4} \color{blue}{a^2}}\end{aligned}\)

TSA of a Cuboid

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{(lb + bh + hl)}}\end{aligned}\)

LSA of a Cuboid

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2h} \color{blue}{(l + b)}}\end{aligned}\)

SA of a Sphere

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{4} \color{blue}{\pi r^2}}\end{aligned}\)

CSA of a Solid HS

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2}}\end{aligned}\)

TSA of a Solid HS

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2} + \color{magenta}{\pi r^2}} \end{aligned}\)

TSA of a Hollow HS

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{\pi r^2} + \color{red}{2} \color{blue}{\pi r^2} } \end{aligned}\)

CSA of a Solid Cylinder

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{blue}{2 \pi r} \color{red}{h} } \end{aligned}\)

TSA of a Solid Cylinder

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{blue}{2 \pi r} \color{red}{h} + \color{red}{2}\color{blue}{\pi r^2} } \end{aligned}\)

CSA of a Hollow Cylinder

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{2} \color{blue}{ \pi r} \color{red}{h} + \color{red}{2} \color{blue}{ \pi r} \color{red}{h}} \end{aligned}\)

TSA of a Hollow Cylinder

with Base

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{ \color{red}{4} \color{blue}{\pi r} \color{red}{h} +\color{red}{2} \color{blue}{\pi r^2} } \end{aligned}\)

CSA of a Solid Cone

\(\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{blue}{\pi rl}} \end{aligned}\)

TSA of a Solid Cone

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{\pi rl} + \color{red}{\pi r^2}} \end{aligned}\)

CSA of a Hollow Cone

\(\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{red}{2} \color{blue}{\pi rl}} \end{aligned}\)

TSA of a Hollow Cone

\(\begin{aligned}&\bbox[15px, border: 3px solid green;] {\color{red}{2} \color{blue}{\pi rl}} \end{aligned}\)

Volumes

Volume of a Cube

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{a^3} } \end{aligned}\)

Volume of a Cuboid

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}l \color{black} \times \color{green}b \color{black}\times \color{red}h} \end{aligned}\)

Volume of Sphere

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{4}{3}}\color{blue}{\pi r^3}} \end{aligned}\)

Volume of HS

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{2}{3}}\color{blue}{\pi r^3}} \end{aligned}\)

Volume of Cylinder

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{blue}{\pi r^2} \color{red}{h} } \end{aligned}\)

Volume of Cone

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\color{red}{\frac{1}{3}}\color{blue}{\pi r^2} \color{red}{h} } \end{aligned}\)

Chapter 13 - Statistics

Direct Method

\(\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = \frac{\sum {f_ix_i}}{\sum{f_i}}} \end{aligned}\)

Assumed Mean Method

\(\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = a +\frac{\sum {f_id_i}}{\sum{f_i}}} \end{aligned}\)

Step Deviation Method

\(\begin{aligned} &\bbox[15px, border: 3px solid green;]{\overline{x} = a +\frac{\sum {f_iu_i}}{\sum{f_i}}\times h }\end{aligned}\)

Mode of Grouped Data

\(\begin{aligned} &\bbox[15px, border: 3px solid green;]{l +\frac{f_1-f_0}{2f_1-f_0-f_2}\times h }\end{aligned}\)

Median of Grouped Data

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{l +\frac{\frac{n}{2}-cf}{f}\times h }\end{aligned}\)

Class Marks

\(\begin{aligned}&\bbox[15px, border: 3px solid green;]{\frac{\text{Upper Class Limit}- \text{Lower Class Limit}}{2}}\end{aligned}\)

Chapter 14 - Probability

Experiment: Throwing a dice 🎲

Outcomes: { 1, 2, 3, 4, 5, 6 }

Total no. of outcomes = 6

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

➺ Outcomes are Equally Likely

Examples of Events

E1: Getting 3

E2: Getting a number > 4

E3: Getting a number < 7

E4: Getting a prime number

E5: Getting a number < 1

E – An Event 0 ≤ P(E) ≤ 1

\( \begin{aligned} &P(E) = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} \end{aligned}\)

Probability of Events (E1 – E5)

P(E1)

E1: Getting 3

Favourable Outcome: { 3 }

No. of favourable outcomes = 1

\( \begin{aligned} P(E1)=\frac{1}{6} \end{aligned}\)

P(E2)

E2: Getting a number > 4

Numbers > 4: { 5, 6 }

No. of favourable outcomes = 2

\( \begin{aligned} P(E2)=\frac{2}{6} =\frac{1}{3} \end{aligned}\)

P(E3)

E3: Getting a number < 7

Numbers < 7: { 1, 2, 3, 4, 5, 6 }

No. of favourable outcomes = 6

\( \begin{aligned} P(E3)=\frac{6}{6} =1 \end{aligned}\)

Sure Event

P(E4)

E4: Getting a prime number

Prime numbers: { 2, 3, 5 }

No. of favourable outcomes = 3

\( \begin{aligned} P(E4)=\frac{3}{6} =\frac{1}{2} \end{aligned}\)

P(E5)

E5: Getting a number < 1

Number < 1: None

No. of favourable outcomes = 0

\( \begin{aligned} P(E5)=\frac{0}{6} =0 \end{aligned}\)

Impossible Event

Sure Event 👍

E = It will rain OR it will not rain today

P(E) = 1

Impossible Event 👎

E = It will rain AND it will not rain today

P(E) = 0

Complementary Events 🎭

E = It will rain today

not E = It will not rain today

P(E) + P(not E) = 1

Equally Likely & Not Equally Likely Outcomes

Randomly take a ball out of a bag of same sized 4 red and 2 blue balls 🔴 🔴 🔴 🔴 🔵 🔵

Experiment 1

Outcomes: Red Ball 🔴, Blue Ball 🔵

Total possible outcomes = 2

P(Red Ball) = 4/6 = 2/3

P(Blue Ball) = 2/6 = 1/3

➺ Outcomes are Not Equally Likely

Experiment 2

Outcomes: A Ball of any color

(Color doesn’t matter)

Total possible outcomes = 6

P(Ball 1) = P(Ball 2) = P(Ball 3) = P(Ball 4) = P(Ball 5) = P(Ball 6) = 1/6

➺ Outcomes are Equally Likely

Mathemafia

Creative Hub of Mathematics

👉 1400+ Math videos ✅

Join Now

The post Class 10 – All Maths Formulas – Complete List appeared first on Mathemafia.

State and Prove Pythagoras Theorem

Sunil Kumar — Tue, 11 Mar 2025 11:29:49 +0000

Pythagoras Theorem

In a right angled triangle, the square of hypotenuse is equal to the sum of squares of the other two sides.

In a right angled triangle, the square of hypotenuse is equal to the sum of squares of perpendicular and base of the triangle.

To Prove

\( \begin{aligned} &BC^2 =AB^2+AC^2\end{aligned}\)

Construction

\( \begin{aligned} &\text{Draw } AD \perp BC\end{aligned}\)

Proof

Part 1

\( \begin{aligned} &In \ \ \triangle ABC \text{ and } \triangle ADB \end{aligned}\)

\( \begin{aligned} &\angle {BAC} = \angle {ADB} = 90^{\circ}\\ &\angle B = \angle B \ \ \ \ \text{(Common)} \end{aligned}\)

By AA similarity criterion

\( \begin{aligned} &\triangle ABC \sim \triangle DBA \end{aligned}\)

Very Important: We write it as △ABC is similar to △DBA. We cannot write it similar to △ADB even though △DBA and △ADB are same triangles.

Why? Click here to read it or watch the video embedded below.

\( \begin{aligned} &\Rightarrow \frac{AB}{DB} = \frac{BC}{AB}\\\\ &\Rightarrow AB^2 = BC \times DB \ \ …(1)\end{aligned}\)

Part 2

\( \begin{aligned} &In \ \ \triangle ABC \text{ and } \triangle ADC \end{aligned}\)

\( \begin{aligned} &\angle {BAC} = \angle {ADC} = 90^{\circ}\\ &\angle C = \angle C \ \ \ \ \text{(Common)} \end{aligned}\)

By AA similarity criterion

\( \begin{aligned} &\triangle ABC \sim \triangle DAC \end{aligned}\)

\( \begin{aligned} &\Rightarrow \frac{BC}{AC} = \frac{AC}{DC}\\\\ &\Rightarrow AC^2 = BC \times DC \ \ …(2)\end{aligned}\)

Part 3

Adding (1) and (2)

\( \begin{aligned} AB^2+ AC^2 &= BC \times BD + BC \times DC \\ &=BC \times (BD + DC)\\ &=BC \times BC\\ &=BC^2\\ \end{aligned}\)

Hence proved.

Watch on YouTube

Why?

\( \begin{aligned} &\text{Why }\triangle ABC \sim \triangle DBA \text{ but } \triangle ABC \not\sim \triangle ADB\end{aligned}\)

When it comes to similarity and congruence of triangles, students always find it slightly confusing but let me tell you, it is as simple as Virat Kohli hits a six 🏏 at will. All you need is to possess good mathematics skills 😜 like Kohli has (in cricket of-course).

Ok … That’s not true. You don’t need to be as good as Kohli 😜 to understand this concept. 💃🕺 Wow!!! Nacho Nacho. 💃🕺 (If this song can win oscar, one day I will get the Fields Medal. 🏅 🏆 )

Explanation

We just proved that the triangles shown below are similar. The triangles are similar when all three corresponding angles are congruent in both the triangles. In this case:

\( \begin{aligned} \angle {BAC} &= \angle {ADB} = 90^{\circ}\\ \angle B &= \angle B \ \ \ \ \text{(Common)}\\ \Rightarrow \angle {ACB} &= \angle {DAB} \end{aligned}\)

In the picture displayed above you can clearly see mapping of the corresponding angles means the mapping of the corresponding vertices of the two triangles. So while writing the similarity expression, we must write the corresponding vertices in same order on both sides as shown in Mapping Pic.

Welldone Mathemafia 👏💯 !!!

Pytha: Oh my Goras!!! I can’t believe that you have explained it so Beautifully Colorfully.

Mathemafia: Wow! You struck out Beautifully. 💔

The post State and Prove Pythagoras Theorem appeared first on Mathemafia.

Rational Root Theorem – Proof

Sunil Kumar — Sat, 15 Feb 2025 14:54:11 +0000

Rational Root Theorem

Statement: Given a polynomial

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0

with integral coefficients where a_n ≠ 0, a₀ ≠ 0. if P(x) has a rational root p/q where p and q are relatively prime integers, then p is a divisor of a₀ and q is a divisor of a_n.

Proof

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0𝑤ℎ𝑒𝑟𝑒 𝑎0,…,𝑎𝑛∈ℤ 𝑎𝑛𝑑 𝑎0,𝑎𝑛≠0.

Let p/q be a rational root of this equation where p and q are coprime integers and q ≠ 0.

⇒𝑃(𝑝𝑞)=0⇒𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎1(𝑝𝑞)+𝑎0=0

Multiply both sides by qⁿ

𝑞𝑛×[𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎2(𝑝𝑞)2+𝑎1(𝑝𝑞)+𝑎0]=𝑞𝑛×0𝑎𝑛(𝑝𝑞)𝑛𝑞𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1𝑞𝑛+𝑎𝑛−2(𝑝𝑞)𝑛−2𝑞𝑛+…….+𝑎2(𝑝𝑞)2𝑞𝑛+𝑎1(𝑝𝑞)𝑞𝑛+𝑎0𝑞𝑛=0𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Imp: We multiplied by qⁿto get rid of terms in the denominator.

Video tutorial: Proof of Rational Root Theorem

Proof of Part 1: p divides a₀

Let’s write equation (1) again

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Take a₀qⁿ to the other side of the equation.

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1= –𝑎0𝑞𝑛

Now p is common in every term on LHS. This implies:

𝑝[𝑎𝑛𝑝𝑛−1+𝑎𝑛−1𝑝𝑛−2𝑞+𝑎𝑛−2𝑝𝑛−3𝑞2+…….+𝑎2𝑝𝑞𝑛−2+𝑎1𝑞𝑛−1]= –𝑎0𝑞𝑛

Now p, q and a₀, a₁,….a_n,all are integers. Hence LHS is an integer.

⇒𝑝|𝐿𝐻𝑆⇒𝑝|𝑅𝐻𝑆⇒𝑝|𝑎0𝑞𝑛

As we know that p and q are coprime

⇒𝑝 does not divide 𝑞⇒𝑝 does not divide 𝑞𝑛

But because 𝑝|𝑎0𝑞𝑛⇒𝑝|𝑎0

Proof of Part 2: q divides a_n

Let’s go back to equation (1)

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0 …(1)

Take a_npⁿ to the other side of the equation.

𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=−𝑎𝑛𝑝𝑛

Now q is common in every term on LHS. This implies:

𝑞[𝑎𝑛−1𝑝𝑛−1+𝑎𝑛−2𝑝𝑛−2𝑞+…….+𝑎2𝑝2𝑞𝑛−3+𝑎1𝑝𝑞𝑛−2+𝑎0𝑞𝑛−1]=−𝑎𝑛𝑝𝑛

Now p, q and a₀, a₁,….a_n,all are integers. Hence LHS is an integer.

⇒𝑞|𝐿𝐻𝑆⇒𝑞|𝑅𝐻𝑆⇒𝑞|𝑎𝑛𝑝𝑛

As we know that p and q are coprime

⇒𝑞 does not divide 𝑝⇒𝑞 does not divide 𝑝𝑛

But because 𝑞|𝑎𝑛𝑝𝑛⇒𝑞|𝑎𝑛

Hope you find this proof useful.

Let’s now use this theorem to find the roots of a polynomial of degree 3.

Example of Rational Root Theorem

𝑃(𝑥)=2𝑥3–3𝑥2–2𝑥+3

Find the roots of P(x) using Rational Root Theorem.

This equation is of the form

𝑃(𝑥)=𝑎3𝑥3+𝑎2𝑥2+𝑎1𝑥+𝑎0

⇒𝑎3=2 𝑎𝑛𝑑 𝑎0=3

As per rational root theorem, if this polynomial has any rational root, say p/q, then

⇒ 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎0 , 𝑞 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎3𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 3 , 𝑞 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 2

Here are the possible values of p/q:

𝑝𝑞⇒𝑝𝑞=±1, ±3=±1, ±2=±11,±31,±12,±32=±1,±3,±12,±32

Now, if P(x) has any rational root, it has to be from the possible values of p/q written above. If P(x) = 0 for any of these values, that will be a rational root of the polynomial. The three values for which P(x) = 0 are 1, -1 and 3/2.

𝑃(𝑥)𝑃(1)𝑃(−1)𝑃(32)=2𝑥3–3𝑥2–2𝑥+3=2−3−2+3=0=−2−3+2+3=0=2×278−3×94−2×32+3=274–274−3+3=0

Rational Root Theorem is a very important theorem of Polynomials and is intended for the students learning higher mathematics and for those who are willing to dive deep into the core mathematics. For lower classes, like class 9 and class 10, this theorem comes handy when we have to find the roots of polynomials with degree 3 or more.

\(\begin{aligned} \end{aligned}\)

The post Rational Root Theorem – Proof appeared first on Mathemafia.

Class 10 – Chapter 5 – Arithmetic Progression – Last Minute Formula Revision

Sunil Kumar — Thu, 06 Feb 2025 04:39:59 +0000

⏰ Last minute Formula Revision 🔥 Arithmetic Progression, Chapter 5, Class 10

This short video covers all the formulas of chapter 5 (Arithmetic Progression, AP Series) of class 10. It is recommended to watch this before your exam to quickly recall all the formulas of this chapter.

The tutorial is available in Hinglish + 9 different languages: English, French, German, Hindi, Spanish, Indonesian, Italian, Japanese, Portuguese

The post Class 10 – Chapter 5 – Arithmetic Progression – Last Minute Formula Revision appeared first on Mathemafia.

Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision

Sunil Kumar — Sun, 02 Feb 2025 09:06:02 +0000

⏰ Last minute Formula Revision 🔥 Polynomials, Chapter 2, Class 10

This short video covers all the formulas of chapter 2 (Polynomials) of class 10. It is recommended to watch this before your exam to quickly recall all the formulas of this chapter.

The tutorial is available in Hinglish + 9 different languages: English, French, German, Hindi, Spanish, Indonesian, Italian, Japanese, Portuguese

The post Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision appeared first on Mathemafia.

Class 11 – Video Tutorials on Mathematics

Sunil Kumar — Mon, 25 Mar 2024 12:45:48 +0000

Here you will find the video tutorials for class 11 Mathematics.

Class 11 - Chapter 1 - Sets

This tutorial cover more than 50% percent of Sets, Chapter 1, of class 11 mathematics. The other half will be added once ready.

One-shot video

Class 11 - Chapter 2 - Relations (One-shot video)

This chapter has been split into 2 parts :

Relations
Functions

The reason is to cover this chapter in depth so that it should be helpful for students who are willing to appear in competitive exams as well.

Below embedded is a one-shot YouTube video that covers the Relations section in detail.

One-shot video

This is a one-shot video that covers the Functions section of chapter 2 (Relations and Functions) of class 11 in detail.

One-shot video

Class 11 - Chapter 3 - Trigonometric Functions

Trigonometry of class 11 is a very long and important topic. This chapter has been covered in multiple videos and the playlist of the same has been embedded below.

What is radian? What is the relationship between radians and degrees?

Example 1 and 2, Chapter 3, Class 11 | Trigonometric Functions | Radians to Degrees & vice versa

Example 3, Chapter 3, Class 11 | Trigonometric Functions | NCERT Maths

Example 4, Chapter 3, Class 11, Maths | Trigonometric Functions | NCERT, CBSE

Example 5, Chapter 3, Class 11, Maths | Trigonometric Functions | NCERT, CBSE

Question 1, 2 Exercise 3.1 Class 11 Maths | Chapter 3 (Trigonometric Functions) | NCERT

Question 3, Ex: 3.1, Ch 3, Class 11, NCERT Maths | Trigonometric Functions

Question 4, Ex: 3.1, Ch 3, Class 11, NCERT Maths | Trigonometric Functions

Question 5, Ex: 3.1, Ch 3, Class 11, NCERT Maths | Trigonometric Functions

Question 6 Exercise 3.1 Class 11 Maths | Chapter 3 (Trigonometric Functions) | NCERT

Question 7 Exercise 3.1 Class 11 Maths | Chapter 3 (Trigonometric Functions) | NCERT

🔥 Coordinates of a point on (1) Circle of radius r (2) Unit circle || Prove that sin^2x + cos^2x = 1

Formulas "sin(2npi + theta) = sin theta" and "sin[(2n+1)pi – theta] = sin theta"

Proof of (i) cos(- θ) = cos θ, (ii) cos(2npi + θ) = cos θ, (iii) cos(2npi – θ) = cos θ | Unit circle

sin(nπ)=0 and cos (2n+1) π/2 = 0 | Explained with unit circle

All School To College | ASTC Formula to remember the sings of trigonometric ratios in 4 quadrants

Prove that 🔥 cos (x + y) = cosx cosy – sinx siny 🔥 Full derivation by #mathemafia

Prove that 🔥 cos (x – y) = cosx cosy + sinx siny 🔥 Simple proof by #mathemafia

cos (pi/2 – x) = sin x 🔥 sin (pi/2 – x) = cos x 🔥 Solved trigonometric identities by #mathemafia

Prove that sin(x+y) = sinx cosy + cosx siny 🔥 💁👌😍 🔥 and sin(x-y) = sinx cosy – cosx siny

Prove that (i) sin(pi+x)=-sin x (ii) sin(pi-x)=sin x (iii) cos(pi+x)=-cos x (ii) cos(pi-x)=-cos x

tan(x + y)=(tan x + tan y) / (1 – tan x tan y) 🔥 & 🔥 tan(x – y)=(tan x – tan y) / (1 + tan x tan y)

cot(x + y)=(cot x cot y – 1)/(cot x + cot y) 🔥 & 🔥 cot(x – y) = (cot x cot y + 1)/(cot y – cot x)

Proof of sin 2x = 2 sinx cosx = 2 tan x /(1+tan²x) where x is not an odd multiple of pi/2 🔥

Proof of cos 2x = cos²x – sin²x = 2cos²x – 1 = 1 – 2sin²x = (1 – tan²x) /(1 + tan²x)

Proof of 🔥 tan 2x = 2 tan x /(1 – tan²x) 🔥 Class 11, Trigonometric Identities by #mathemafia

Proof of three trigonometric identities 🔥 sin 3x, cos 3x and tan 3x 🔥 by #mathemafia

Easy proof of 🔥 sin x + sin y 🔥 cos x + cos y 🔥 sin x – sin y 🔥cos x – cos y by #mathemafia

The post Class 11 – Video Tutorials on Mathematics appeared first on Mathemafia.

Class 10 | Chapter 14 | All Solutions | Probability

Sunil Syal — Fri, 17 Dec 2021 07:44:02 +0000

This playlist covers all the questions, concepts and examples of chapter 15 (Probability) of class 10, NCERT book. Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Probability – Understand Outcome and Event

Probability – What are Complementary Events

Example 1, 2, 3, 4 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 5 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 6 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 7 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 8 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 9 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 10 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 11 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Probability – Sure Event and Impossible Event

Example 12 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Example 13 – Chapter 14 – Class 10 | Probability | NCERT Maths | CBSE

Probability – What are Elementary Events

Question 1 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 2 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 3, 4, 5 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 6 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 7 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 8, 9 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 10 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 11 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 12 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 13 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 14 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 15 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 16 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 17 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 18 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 19 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 20 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 21 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 22 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 23 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 24 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

Question 25 Exercise 14.1 Chapter 14 | Class 10 | Probability | NCERT Maths | CBSE

The post Class 10 | Chapter 14 | All Solutions | Probability appeared first on Mathemafia.

Class 10 | Chapter 13 | All Solutions | Statistics (WIP)

Sunil Syal — Fri, 17 Dec 2021 07:43:48 +0000

This playlist covers all the questions, concepts and examples of chapter 14 (Statistics) of class 10, NCERT book. Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Example 1 – Chapter 13 – Class 10 | Part 1 (Finding mean by Direct Method) Statistics | NCERT Maths

Example 1 – Chapter 13 – Class 10 | Part 2 (Assumed mean and step deviation Method) Statistics

Why Virat Kohli's Average is 98.67 in T20 World Cup 2022 | How to find batting average

Example 2 – Chapter 13 – Class 10 | Statistics | NCERT Maths | CBSE

Example 3 – Chapter 13 – Class 10 | Statistics | NCERT Maths | CBSE

What are class marks in statistics? Learn with simple examples.

Example 4 & Example 5 – Chapter 13 – Class 10 | Statistics | Introduction to Mode of Grouped data

Example 6 – Chapter 13 – Class 10 | Statistics | NCERT Maths | CBSE

Best tip to quickly remember formula of MODE (measure of central tendency)

🔥🔥 Question 4 – Exercise 13.2 | Class 10 | Chapter 13 | Statistics 💁👌😍 NCERT Maths | CBSE

Class 10, MEDIAN – Ex: 13.3 || Explained and solved in LIVE class

The post Class 10 | Chapter 13 | All Solutions | Statistics (WIP) appeared first on Mathemafia.

Class 10 | Chapter 10 | All Solutions | Circles

Sunil Syal — Fri, 17 Dec 2021 07:41:58 +0000

This playlist covers all the questions, concepts and examples of chapter 10 (Circles) of class 10, NCERT book. Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Lengths of tangents from an external point are equal |Theorem 10.2| Class 10 |Circles | NCERT Maths

Example 1 – Chapter 10 – Class 10 | Circles | NCERT Maths | CBSE

Example 2 – Chapter 10 – Class 10 | Circles | NCERT Maths | CBSE

Example 3 – Chapter 10 – Class 10 | Circles | NCERT Maths | CBSE – Method 1

Example 3 – Chapter 10 – Class 10 | Circles | NCERT Maths | CBSE – Method 2 & 3

Theorem 9.7 Class 9 | Chapter 9 | Circles – Angle subtended by an arc at the centre | NCERT Maths

Equal Tangent Segments: Proving Congruence from an External Point

The post Class 10 | Chapter 10 | All Solutions | Circles appeared first on Mathemafia.

Class 10 | Chapter 8 | All Solutions | Introduction to Trigonometry

Sunil Syal — Fri, 17 Dec 2021 07:41:58 +0000

This playlist covers all the questions, concepts and examples of chapter 8 (Introduction to Trigonometry) of class 10, NCERT book. Click a thumbnail to watch the tutorial.

Watch complete chapter on YouTube

Trigonometry 💁👌😍 How to memorise values of Trigonometric ratios of some specific angles

Introduction to Trigonometry | Concepts and terms explained

Trigonometry Special 🔥 Relationship between sine and cosine of complementary angles

Example 1 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Example 2 – Chapter 8 – Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

🔥 Example 3 Chapter 8 Class 10 🔥 Introduction to Trigonometry | NCERT Maths | CBSE

Example 4 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Example 5 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Question 3, 4, 5 Exercise 8.1 | Class 10 | Chapter 8 |Introduction to Trigonometry |NCERT Maths

Trigonometry – How to derive values of all Trigonometric ratios for 45 degrees

Trigonometry – How to derive values of all Trigonometric ratios for 30, 60 degrees

Why sin 0 degree is 0 🔥 Class 10 – Chapter 8 👩‍🎓 Introduction to Trigonometry #mathemafia

Why sin 90 degree is 1 | Class 10 – Chapter 8 Introduction to Trigonometry | 90°

Example 6 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Example 7 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Example 8 – Chapter 8- Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Trigonometric Ratios of Complementary Angles

Example 9, 10, 11 – Chapter 8 – Class 10 | Introduction to Trigonometry | NCERT Maths | (Old Book)

Question 1, 2 Exercise 8.3 | Class 10 |Chapter 8 | Introduction to Trigonometry |NCERT (old Book)

Question 6, 7 Exercise 8.3 | Class 10 |Chapter 8 | Introduction to Trigonometry |NCERT (old Book)

Basic Trigonometric Identities 🔥 Super easy to remember 🔥 Amazing tip #mathemafia

Example 9, 10 – Chapter 8 – Class 10 | Introduction to Trigonometry | NCERT Maths CBSE

Example 11 – Chapter 8 – Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

Tricky Example 12 – Chapter 8 – Class 10 | Introduction to Trigonometry | NCERT Maths | CBSE

2nd method 🔥 Question 4 (v) of Ex: 8.3, class 10 🔥 Introduction to #trigonometry || #class10math

Question 7, Exercise 8.1, Class 10 | Chapter 8 | Introduction to Trigonometry | NCERT Maths | CBSE

The post Class 10 | Chapter 8 | All Solutions | Introduction to Trigonometry appeared first on Mathemafia.

mathemafia Archives – Mathemafia

Class 10 – All Maths Formulas – Complete List

Perfect for quick revision and last-minute prep!

Mathemafia

Creative Hub of Mathematics

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Quadratic Polynomials

Chapter 3 - Pair of Linear Equations in Two Variables

Solution of Pair of LE in 2 Variables

Chapter 4 - Quadratic Equations

Nature of Roots

Chapter 5 - Arithmetic Progression

Chapter 6 - Triangles

Triangles Similarity Criteria

Mathemafia

Creative Hub of Mathematics

Chapter 7 - Coordinate Geometry

Distance Formula

Section Formula

Mid-Point Formula

Chapter 8 - Introduction to Trigonometry

Trigonometric Ratios

Trigonometric ratios of standard angles

Trigonometric Identities

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Surface Areas

Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

Probability of Events (E1 – E5)

Equally Likely & Not Equally Likely Outcomes

Mathemafia

Creative Hub of Mathematics

State and Prove Pythagoras Theorem

Pythagoras Theorem

Proof

Why?

Welldone Mathemafia 👏💯 !!!

Rational Root Theorem – Proof

Rational Root Theorem

𝑃(𝑥)=2𝑥3–3𝑥2–2𝑥+3

Find the roots of P(x) using Rational Root Theorem.

Class 10 – Chapter 5 – Arithmetic Progression – Last Minute Formula Revision

⏰ Last minute Formula Revision 🔥 Arithmetic Progression, Chapter 5, Class 10

Class 10 – Chapter 2 – Polynomials – Last Minute Formula Revision

⏰ Last minute Formula Revision 🔥 Polynomials, Chapter 2, Class 10

Class 11 – Video Tutorials on Mathematics

Class 11 - Chapter 1 - Sets

One-shot video

Class 11 - Chapter 2 - Relations (One-shot video)

One-shot video

One-shot video

Class 11 - Chapter 3 - Trigonometric Functions

Class 10 | Chapter 14 | All Solutions | Probability

Class 10 | Chapter 13 | All Solutions | Statistics (WIP)

Class 10 | Chapter 10 | All Solutions | Circles

Class 10 | Chapter 8 | All Solutions | Introduction to Trigonometry