Topics
Compound Interest
- Compound Interest as a Repeated Simple Interest Computation with a Growing Principal
- Use of Compound Interest in Computing Amount Over a Period of 2 Or 3-years
- Use of Formula
- Finding CI from the Relation CI = A – P
Commercial Mathematics
Goods and Services Tax (G.S.T.)
Banking
Algebra
Geometry
Shares and Dividends
Symmetry
Mensuration
Linear Inequations
Quadratic Equations
- Quadratic Equations
- Method of Solving a Quadratic Equation
- Factorisation Method
- Quadratic Formula (Shreedharacharya's Rule)
- Nature of Roots of a Quadratic Equation
- Equations Reducible to Quadratic Equations
Trigonometry
Statistics
Problems on Quadratic Equations
- Method for Solving a Quadratic Word Problem
- Problems Based on Numbers
- Problems on Ages
- Problems Based on Time and Work
- Problems Based on Distance, Speed and Time
- Problems Based on Geometrical Figures
- Problems on Mensuration
- Problems on C.P. and S.P.
- Miscellaneous Problems
Ratio and Proportion
Probability
Remainder Theorem and Factor Theorem
- Function and Polynomial
- Division Algorithm for Polynomials
- Remainder Theorem
- Factor Theorem
- Applications of Factor Theorem
Matrices
Arithmetic Progression
Geometric Progression
Reflection
- Co-ordinate Geometry
- Advanced Concept of Reflection in Mathematics
- Invariant Points
- Combination of Reflections
- Using Graph Paper for Reflection
Section and Mid-Point Formulae
Equation of a Line
Similarity
Loci
- Locus
- Points Equidistant from Two Given Points
- Points Equidistant from Two Intersecting Lines
- Summary of Important Results on Locus
- Important Points on Concurrency in a Triangle
Angle and Cyclic Properties of a Circle
Tangent Properties of Circles
Constructions
Volume and Surface Area of Solids (Cylinder, Cone and Sphere)
- Mensuration of Cylinder
- Hollow Cylinder
- Mensuration of Cones
- Mensuration of a Sphere
- Hemisphere
- Conversion of Solids
- Solid Figures
- Problems on Mensuration
Trigonometrical Identities
Heights and Distances
- Angles of Elevation and Depression
- Problems based on Elevation and Depression
Graphical Representation of Statistical Data
Measures of Central Tendency (Mean, Median, Quartiles and Mode)
Probability
- Definition: Cylinder
- Properties of a Cylinder
- Activity 1
- Activity 2
CISCE: Class 10
Definition: Cylinder
A cylinder is a three-dimensional solid figure that has two identical circular bases joined by a curved surface at a particular distance from its centre, which is its height.
CISCE: Class 12
Parts of Cylinder
- Faces: A cylinder has 3 faces in total (2 flat circular faces + 1 curved face)
- Edges: A cylinder has 2 edges (one at the top and one at the bottom)
- Vertices: A cylinder has 0 vertices (as the two edges of the cylinder never meet anywhere)
CISCE: Class 12
Formula: Cylinder
Curved surface area of a cylinder = circumference of base × height
= 2πrh
Total surface area = Curved surface area + 2 (Area of cross-section)
= 2πrh + 2πr2
= 2πr(r + h)
Volume = Area of cross-section × height (or, length)
= πr2h
Maharashtra State Board: Class 6
Properties of a Cylinder
- A cylinder has one curved surface and two flat faces, which are identical.
- The two circular bases are congruent with each other.
- Its size depends on the radius of the base and the height of the curved surface.
- Unlike a cone, cube, or cuboid, a cylinder does not have any vertices. This means that a cylinder has no corners.
- The base and the top of the cylinder are identical, i.e., it has the same base — either circular or elliptical.
CISCE: Class 12
Activity 1
Making a Hollow Cylinder
Steps:
- Take a rectangular sheet of paper.
- Bring together its opposite sides (join side AB to DC).
- Tape or glue the sides together.
Observation:
- The paper takes the shape of a hollow cylinder (open at both ends).
Conclusion:
- A hollow cylinder can be formed by rolling a rectangular sheet and joining its opposite edges.
Maharashtra State Board: Class 6
Activity 2
Steps:
- Take a cylindrical tin and a rectangular sheet (one side equal to the height of the tin).
- Wrap the sheet around the tin to cover it completely.
- Mark and cut off extra paper, then unfold the sheet and lay it flat — this is the side of the cylinder.
- Take another sheet and trace two circles using the base of the tin.
- Cut out the two circles.
- Place the two circular cut-outs beside the rectangle.
Observation:
- You get one rectangle (side face) and two circles (top and bottom faces).
- This set forms the net of a closed cylinder.
Conclusion:
- A closed cylinder is made of one rectangle (curved surface) and two circles (flat circular faces).
- By joining these shapes together, we can form a 3D closed cylinder.
Example
A rectangular paper of width 14 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder. (Take `22/7` for π)
A cylinder is formed by rolling a rectangle about its width. Hence the width
of the paper becomes height and the radius of the cylinder is 20 cm.
Height of the cylinder = h = 14 cm
Radius = r = 20 cm
Volume of the cylinder = V = πr2h
= `22/7` × 20 × 20 × 14
= 17600 cm3
Hence, the volume of the cylinder is 17600 cm3.
Example
A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.
Length of the paper becomes the perimeter of the base of the cylinder and
width becomes height.
Let radius of the cylinder = r and height = h
Perimeter of the base of the cylinder = 2πr = 11
or `2 xx 22/7 xx r = 11`
Therefore,
r = `7/4` cm
Volume of the cylinder = V = πr2h
= `22/7 xx 7/4 xx 7/4 xx 4` cm3
= 38.5 cm3
Hence the volume of the cylinder is 38.5 cm3.
Example
In a building, there are 24 cylindrical pillars. For each pillar, the radius is 28 m, and the height is 4 m. Find the total cost of painting the curved surface area of the pillars at the rate of ₹ 8 per m2.
Radius of cylindrical pillar, r = 28 cm = 0.28 m, height = h = 4 m
curved surface area of a cylinder = 2πrh
curved surface area of a pillar = `2 xx 22/7 xx 0.28 xx 4` = 7.04 m2
curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m2
cost of painting an area of 1 m2 = Rs. 8
Therefore, cost of painting 1689.6 m2 = 168.96 × 8 = Rs. 1351.68.
Example
Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm2.
Let height of the cylinder = h, radius = r = 7cm
Total surface area = 2πr(h + r)
i.e., 2 × `22/7` × 7 × (7 + h) = 968
h = 15 cm
Hence, the height of the cylinder is 15 cm.
Video Tutorials
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