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      Class 10 Maths

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      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        Chapter Notes – Triangles

        (1) Similar Figures : Two figures having the same shapes (and not necessarily the same size) are called similar figures.
        For Example: The given below squares are similar.

        (2) Congruent Figures : The word ‘congruent’ means equal in all aspects or the figures whose shapes and sizes are same.
        For Example: The given below squares are congruent.

         

        (3) Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
        For Example: The given below quadrilaterals ABCD and PQRS are similar.

        (4) Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
        For Example: The given below triangles ABC and DEF are similar.Image result for similar triangles

         

        (5) 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.
        Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.To Prove: AD/DB = AE/EC
        Proof:
        Firstly, join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.
        Now, area of Δ ADE = 1/2 × base × height = 1/2 AD × EN.
        So, ar(ADE) = 1/2 AD × EN, ar(BDE) = 1/2 DB × EN, ar(ADE) = 1/2 AE × DM, ar(DEC) = 1 2 EC × DM.
        Therefore, ar(ADE)/ar(BDE) = (1/2 AD × EN)/( ½ DB × EN) = AD/DB – (1)
        ar(ADE)/ar(DEC) = (1/2 AE × DM)/(1/2 EC × DM) = AE/EC – (2)
        Here, ar(BDE) = ar(DEC), since, Δ BDE and DEC are on the same base DE and between the same parallels BC and DE – (3)
        From (1), (2) and (3), we get, AD/DB = AE/EC.

        (6) Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
        For Example: If a line intersects sides AB and AC of a Δ ABC at D and E respectively and is parallel to BC.To Proove: AD/AB = AE/AC
        Proof:  Given, DE || BC.
        Now, AD/DB = AE/EC (Theorem 6.1) or, DB/AD = EC/AE.
        Adding 1 on both the sides, we get,
        (DB/AD) + 1 = (EC/AE) + 1
        AB/AD = AC/AE
        Therefore, AD/AB = AE/AC.

         

        (7) Criteria for Similarity of Triangles : For triangles, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle then they are said to be congruent triangles. The symbol ‘~’ stands for ‘is similar to’ and the symbol ‘≅’ stands for ‘is congruent to’.
        For Example: Consider two ∆ ABC and ∆ PQR as shown below:Here, ∆ ABC is congruent to ∆ PQR which is denoted as ∆ ABC ≅ ∆ PQR.
        ∆ ABC ≅ ∆ PQR means sides AB = PQ, BC = QR, CA = RP; the ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R and vertices A corresponds to P, B corresponds to Q and C corresponds to R.

        (8) Theorem: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
        Given: Two triangles ABC and DEF exists such that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F.To Proove: AB/DE = BC/EF = AC/DF.
        Proof: For the triangle DEF, we mark point P such that DP = AB and mark point Q such that DQ = AC. And join PQ.
        Hence, Δ ABC ≅ Δ DPQ.
        Therefore, ∠ B = ∠ P = ∠ E and PQ || EF.
        As per the theorem, we can say that, DP/PE = DQ/QF i.e. AB/DE = AC/DF
        Also, AB/DE = BC/EF.
        Hence, AB/DE = BC/EF = AC/DF.

        (9) Theorem: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
        For Example:
        Given: Two triangles ABC and DEF exists such that AB/DE = BC/EF = CA/FD (< 1).To Proove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F.
        Proof:
        For the triangle DEF, we mark point P such that DP = AB and mark point Q such that DQ = AC. And join PQ.
        As per the theorem, we can say that, DP/PE = DQ/QF and PQ || EF.
        Hence, ∠ P = ∠ E and ∠ Q = ∠ F.
        Also, DP/DE = DQ/DF = PQ/EF.
        Hence, DP/DE = DQ/DF = BC/EF (As per CPCT)
        On comparison, we get, BC = PQ.
        Thus, Δ ABC ≅ Δ DPQ.
        Hence, So, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F.

        (10) Theorem: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.
        For Example: Two triangles ABC and DEF exists such that AB/DE = AC/DF (< 1) and ∠ A = ∠ D.To Proove: Δ ABC ~ Δ DEF
        Proof: For the triangle DEF, we mark point P such that DP = AB and mark point Q such that DQ = AC. And join PQ.
        Now, PQ || EF and Δ ABC ≅ Δ DPQ.
        Hence, ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q.
        Therefore, Δ ABC ~ Δ DEF.

         

        (11) Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
        For Example:Two triangles ABC and PQR exists such that Δ ABC ~ Δ PQR.To Proove: ar(ABC)/ar(PQR) = (AB/PQ)2 = (BC/QR)2 = (CA/RP)2.
        Proof: Firstly, we draw altitudes AM and PN of the triangles.
        ⇒So, ar(ABC) = 1/2 BC × AM and ar(PQR) = 1/2 QR × PN
        ⇒So, ar(ABC)/ar(PQR) = (1/2 BC × AM)/( 1/2 QR × PN) = (BC × AM)/( QR × PN) – (1)
        ⇒Given, Δ ABC ~ Δ PQR, so, ∠ B = ∠ Q
        ⇒∠ M = ∠ N (As measure of both angles is of 90°)
        ⇒So, Δ ABM ~ Δ PQN (As per AA similarity criterion).
        Thus, AM/PN = AB/PQ. – (2)
        Given, Δ ABC ~ Δ PQR, so, AB/PQ = BC/QR = CA/RP – (3)
        Therefore, ar(ABC)/ar(PQR) = (AB/PQ x AM/PN) (From 1 and 3)
        ⇒                                            = (AB/PQ x AB/PQ) (From 2)
        ⇒                                            = (AB/PQ)2.
        Now, from (3), we get,
        ⇒ar(ABC)/ar(PQR) = (AB/PQ)2 = (BC/QR)2 = (CA/RP)2.

        (12) Theorem: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

        (13) Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
        For Example: A right triangle ABC right angled at B.
        To Proove: AC2 = AB2 + BC2.
        Proof:We draw BD ⊥ AC.
        As per previous theorem, we can write, Δ ADB ~ Δ ABC.
        So, AD/AB = AB/AC (Since sides are proportional) i.e. AD x AC = AB2 – (1)
        Similarly, as per previous theorem, we can write, Δ BDC ~ Δ ABC.
        So, CD/BC = BC/AC (Since sides are proportional) i.e. CD x AC = BC2 – (2)
        On adding (1) and (2), we get,
        ⇒AD x AC + CD x AC = AB2 + BC2
        ⇒
        AC (AD + CD) = AB2 + BC2
        ⇒
        AC x AC = AB2 + BC2
        ⇒
        AC2 = AB2 + BC2.

         

        (14) Theorem: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
        Given: A triangle ABC in which AC2 = AB2 + BC2.To Proove: ∠ B = 90°.
        Proof: Firstly, we construct a Δ PQR right angled at Q such that PQ = AB and QR = BC.Now, from Δ PQR, we get,
        ⇒PR2 = PQ2 + QR2 (as per Pythagoras theorem)
        ⇒PR2 = AB2 + BC2 (since PQ = AB and QR = BC) – (1)
        Given, AC2 = AB2 + BC2, so, AC = PR – (2)
        Now, for Δ ABC and Δ PQR, AB = PQ, BC = QR, AC =PR.
        Thus, Δ ABC ≅ Δ PQR (as per SSS congruence)
        Hence, ∠ B = ∠ Q (CPCT). But ∠ Q = 90° (as per construction).
        ⇒So, ∠ B = 90°.

        Prev Miscellaneous Questions
        Next NCERT Solutions – Triangles Exercise 8.1 – 8.6

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          6 Comments

        1. Priyanshu Kumar
          May 28, 2021

          How we access live tutorial classes please tell me

          • Dronstudy
            May 31, 2021

            For any information regarding live class please call us at 8287971571

        2. Rohan Yedla
          September 15, 2021

          I am not able to access any test of coordinate geometry, can you pls resolve the issue?

          • Dronstudy
            September 20, 2021

            We have fixed the issue. Please feel free to call us at 8287971571 if you face such type of issues.

        3. Ananyaa
          November 12, 2021

          Hi!
          I am a student in class 10th and ive noticed that you dont give full courses on your youtube channel. I mean, its understandable. But we request you to at least put full course of only one chapter on youtube so that we can refer to that. Only one from the book Because your videos are very good and they enrich our learning.
          Please give this feedback a chance and consider our request.

          Dronstudy lover,
          Ananyaa

        4. Nucleon IIT JEE KOTA
          December 17, 2022

          Great article, the information provided to us. Thank you

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