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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 – Surface Areas and Volumes

        (1)Cuboid: If l,b and h denote respectively the length, breadth and height of a cuboid, then
        (i) Total surface area of the cuboid= 2(lb+bh+lh) square units
        (ii) Volume of the cuboid = Area of the base x height = lbh cubic units
        (iii) Diagonal of the cuboid= l2+b2+h2−−−−−−−−−−√ units.
        (iv) Area of four walls of a room= 2(l+b)h sq. units.
        For Example: Two cubes each of 10cm edge are joined end to end. Find the
        (i) Surface area of the resulting cuboid
        (ii) Volume of cuboid
        (iii) Diagonal of the cuboid and
        (iv) Area of four walls of a room
        Solution: l = length of resulting cuboid = 10cm + 10cm = 20cm
        b = breadth of resulting cuboid = 10cm
        h = height of resulting cuboid = 10cm
        (i) Total Surface Area of the cuboid = 2 (lb + bh + lh) square units
        A= 2 (20 x 10 + 10 x 10 + 20 x 10) cm2
                                                               A= 1000 cm2.
        (ii) Volume of the cuboid = lbh cubic units
        V= 20 x 10 x 10 cm3
                                              V= 2000 cm3.
        (iii) Diagonal of the cuboid= 
                                                  l=
                                                  l= 24.49 cm.
        (iv) Area of four walls of a room = 2 (l + b) h
        A = 2 (20 + 10) 10
        A = 600 cm2

        (2) Cube: If the length of each edge of a cube is a units, then
        (i) Total surface area of the cube = 6a2 square units
        (ii) Volume of the cube=  a3 cubic units.
        (iii) Diagonal of the cube= 3–√a units.
        If the length of each edge of a cube is 4 cm units then find total surface area, volume of cube and diagonal of cube.
        Solution: (i) Total Surface area= 6a2 cm2
                                                       A= 6(4)2 cm2
                                                      A= 96 cm2.
        (ii) Volume of the cube = a3 cm3
                                           V= (4)3 cm3
                                           V= 64 cm3.
        (iii) Diagonal of the cube = √3a cm
                                                l= √3(4) cm
                                               l= 6.92 cm.

         

        (3) Right circular cylinder: If r and h denote respectively the radius of the base and height of a right circular cylinder, then
        (i) Area of each end= πr2
        (ii) Curved surface area of hollow cylinder= 2πrh
        (iii) Total surface area= 2πr(h+r)
        (iv) Volume=  πr2h
        For Example: The diameter of right circular cylinder is 6 cm and height is 9 cm. Find
        (i) Area of each end
        (ii) Curved surface area
        (iii) Total surface area
        (iv)Volume
        Solution:
        (i) Area of each end = π r2
                                     A = π (3)2
                                     A = 3.14 x (3)2
                                     A = 28.26 cm2.
        (ii)Curved surface area = 2 π r h
        A = 2 x 3.14 x 3 x 9
        A = 169.56 cm2.
        (iii) Total surface area = 2 π r (h + r)
        A = 2 x 3.14 x 3(9 + 3)
        A = 226.08 cm2.
        (iv)Volume = π r2 h
        V = 3.14 x (3)2x 9
        V = 254.34 cm3.

        (4) Right Circular Hollow Cylinder: If R and r denote respectively the external and internal radii of a hollow right circular cylinder, then
        (i) Area of each end= π(R2−r2)
        (ii) Curved surface area of hollow cylinder= 2π(R+r)h
        (iii) Total surface area= 2π(R+r)(R+h−r)
        (iv) Volume of material= πh(R2−r2)

        For Example: The external and internal radii of a hollow cylinder is 8cm and 6cm respectively and height is 10 cm. Find
        (i) Area of each end
        (ii) Curved surface area
        (iii) Total surface area
        (iv)Volume
        Solution:
        (i) Area of each end = π (R2 – r2)
        A= 3.14 ((8)2 – (6)2)
        A= 87.92 cm2.
        (ii) Curved surface area = 2 π h (R + r)
        A= 2 x 3.14 x 10 x (8 + 6)
        A= 879.2 cm2.
        (iii) Total surface area = 2 π (R + r) (R + h – r)
        A= 2 x 3.14 x (8 + 6) (8 + 10 – 6)
        A= 2 x 3.14 x 14 x 12
        A= 1055.04 cm2.
        (iv)Volume of material = π h (R2 – r2)
        V= 3.14 x 10 x ((8)2 – (6)2)
        V= 879.2 cm3.

        (5) Right Circular Cone: If r,h and l denote respectively the radius of base, height and slant height of a right circular cone, then
        (i) l2=r2+h2
        (ii) Curved surface area = πrl
        (iii) Total surface area= πr2+πrl
        (iv) Volume = 13πr2h
        For Example: A right circular cone is of height 8.4 cm, radius of its base is 2.1 cm. Find
        (i) Slant height
        (ii) Curved surface area
        (iii) Total surface area
        (iv)Volume
        Solution: (i) Slant height l2 = r2 + h2
                                                 l2 = (2.1)2 + (8.4)2
                                                l2 = 74.97 cm2
                                                 l = 8.66 cm.
        (ii) Curved surface area = π r l
        A= 3.14 x 2.1 x 8.66
        A= 57.10 cm2.
        (iii) Total surface area = π r (l + r)
        A= 3.14 x 2.1 (8.66 + 2.1)
        A= 70.95 cm2.
        (iv) Volume = 1/3 π r2 h
        V = 1/3 x 3.14 x (2.1)2  x 8.4
        V = 38.77 cm3.

         

        (6) Sphere: For a sphere of radius r, we have
        (i) Surface area= 4πr2
        (ii) Volume= 43πr3.
        For Example: The radius of sphere is 4 cm then find the surface area and volume.
        Solution: (i) Surface area = 4 π r2
                                               A= 4 x 3.14 x (4)2
                                               A= 200.96 cm2.
        (ii) Volume = 4/3 π r3
                       V = 4/3 x 3.14 x (4)3
                       V = 66.99 cm3.

        (7) Frustum of a right circular cone: If h is the height, l the slant height and r1 and r2 the radii of the circular bases of a frustum of a cone, then
        (i) Volume of the frustum= π3(r12+r1r2+r22)h
        (ii) Lateral surface area= π(r1+r2)l
        (iii) Total surface area= π{(r1+r2)l+r12+r22}
        (iv) Slant height of the frustum= h2+(r1−r2)2−−−−−−−−−−−−√
        (v) Height of the cone of which the frustum is a part= hr1r1−r2
        (vi) Slant height of the cone of which the frustum is a part= lr1r1−r2
        (vii) Volume of the frustum= h3{A1+A2+A1A2−−−−−√} , where Al and Az denote the areas of circular bases of the frustum.

        For Example: If 4 cm is the height, 4.12 cm is the slant height and 2.5 cm and 1.5 c, is radii of the circular bases of a frustum of cone, then find
        (i) Volume of the frustum
        (ii) Lateral surface area
        (iii) Total surface area
        (iv)Slant height of frustum
        (v) Height of the cone of which the frustum is a part
        (vi) Slant height of the cone of which the frustum is a part
        (vii) Volume of the frustum
        Solution: (i) Volume of the frustum = π/3 (r12 + r1r2 + r22) h
        V = π/3 ((2.5)2 + 2.5 x 1.5 + (1.5)2) 4
        V = π/3 x 12.25 x 4
        V = 51.28 cm3.
        (ii) Lateral surface area = π (r1 + r2) l
        A = 22/7 x (2.5 + 1.5) x 4.13
        A = 51.74 cm2.
        (iii) Total surface area = π ((r1 + r2) l + r12 + r22)
        A = 3.14 x ((2.5 + 1.5) 4.13 + (2.5)2 + (1.5)2)
        A = 78.56 cm2.
        (vi)Slant height of the cone of which the frustum is a part l = l.r1 / (r1. r2)
                                                                                                    l = (4.13 x 2.5)/(2.5 x 1.5)
                                                                                                    l = 10.325 cm.
        (v) Height of the cone of which the frustum is a part H = h.r1/(r1– r2)
        H = 4 x 2.5/(2.5 – 1.5)
        H = 10/1
        H = 10 cm.
        (vi) Slant height of the cone of which the frustum is a part l = l.r1 / (r1– r2)
                                                                                                      l= 4.13 x 2.5 / (2.5 – 1.5)
                                                                                                     l = 10.325 cm.
        (vii) Volume of the frustum = h/3 (A1 + A2 + √(A1A2))
        A1 = π r12 = 3.14 x (2.5)2 = 19.625
        A2 = π r22 = 3.14 x (1.5)2 = 7.065
        V = 4/3 (19.625 + 7.065 + √(19.625 x 7.065))
        V = 4/3 (26.69 + √138.65)
        V = 51.28 cm3.

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          6 Comments

        1. Priyanshu Kumar
          May 28, 2021

          How we access live tutorial classes please tell me

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            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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