Topics
Similarity
- Properties of Ratios of Areas of Two Triangles
- Basic Proportionality Theorem
- Property of an Angle Bisector of a Triangle
- Property of Three Parallel Lines and Their Transversals
- Similarity of Triangles (Corresponding Sides & Angles)
- Relation Between the Areas of Two Triangles
- Criteria for Similarity of Triangles
- Overview of Similarity
Pythagoras Theorem
- Pythagoras Theorem
- Pythagorean Triplet
- Property of 30°- 60°- 90° Triangle Theorem
- Property of 45°- 45°- 90° Triangle Theorem
- Similarity in Right Angled Triangles
- Theorem of Geometric Mean
- Right-angled Triangles and Pythagoras Property
- Converse of Pythagoras Theorem
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Apollonius Theorem
- Overview of Pythagoras Theorem
Circle
- Circles Passing Through One, Two, Three Points
- Tangent and Secant Properties
- Secant and Tangent
- Inscribed Angle Theorem
- Intersecting Chords and Tangents
- Corollaries of Inscribed Angle Theorem
- Angle Subtended by the Arc to the Point on the Circle
- Angle Subtended by the Arc to the Centre
- Overview of Circle
Geometric Constructions
Co-ordinate Geometry
Trigonometry
- Trigonometric Ratios in Terms of Coordinates of Point
- Angles in Standard Position
- Trigonometric Ratios
- Trigonometry Ratio of Zero Degree and Negative Angles
- Trigonometric Table
- Trigonometric Identities (Square Relations)
- Angles of Elevation and Depression
- Relation Among Trigonometric Ratios
- Trigonometric Ratios of Specific Angles
Mensuration
Notes
Now let's understand on what principle the conversion of Solid from One Shape to Another depends. Let's take example of a wax candle.

This candle is of cylindrical shape with a distinct height and distinct radius. If further the candle is kept in cylinder jar and melted,

wax will melt down. And once it is allowed to cool the wax candle will change into different shape. The new candle is different from the first candle, it will be more thick and it will have less height. The amount of wax is same in both cases.
i.e Volume before molding= Volume after molding
Example 1: A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution : Volume of cone = `1/3 pi r^2h`
If r is the radius of the sphere, then its volume is `4/3 pi r^3`
Since, the volume of clay in the form of the cone and the sphere remains the same, we have
`4/3 pi r^3` = `1/3 pi r^2h`
`4/3 pi r^3` =`1/3 pi xx 6 xx 6 xx 24`
`r^3` = `3 xx 3 xx 24= 3^3 xx 2^3`
`r= 3 xx 2= 6`
Therefore, the radius of the sphere is 6 cm.
Shaalaa.com | Surface Area and Volume part 10 (Shape Conversion of solid)
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