Topics
Similarity
- 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)
- Theorem of Areas of Similar Triangles
- Overview of Similarity
Pythagoras Theorem
- Pythagorean Triplet
- Property of 30°- 60°- 90° Triangle Theorem
- Property of 45°- 45°- 90° Triangle Theorem
- Theorem of Similarity in Right Angled Triangle
- Theorem of Geometric Mean
- Pythagoras Theorem
- Application of Pythagoras Theorem
- Apollonius Theorem
- Overview of Pythagoras Theorem
Circle
- Properties of Chords of a Circle
- Circles Passing Through One, Two, Three Points
- Secant and Tangent
- Tangent Theorem
- Tangent Segment Theorem
- Touching Circles
- Arc of the Circle
- Central Angle and the Measure of an Arc
- Congruence of Arcs
- Property of Sum of Measures of Arcs
- Inscribed Angle & Intercepted Arc
- Cyclic Quadrilateral
- Overview of Circle
- Theorem of Angle between Tangent and Secant
- Theorem of Internal Division of Chords
- Theorem of External Division of Chords
- Tangent Secant Segments Theorem
Geometric Constructions
- Construction of Similar Triangle
- Construction of a Tangent to a Circle at a Point on the Circle
- Construction of a Tangent to a Circle from a Point Outside the Circle
- Overview of Geometric Constructions
Co-ordinate Geometry
Trigonometry
Mensuration
Notes
We have studied Ratio and Proportion. The statement, "the numbers a and b are in the ratio m/n " is also written as, "the numbers a and b are in proportion m:n". For this concept we consider positive real numbers. We know that the lengths of line segments and area of any figure are positive real numbers. We know the formula of area of a triangle.
Area of a triangle =1/2 Base × Height
Let’s find the ratio of areas of any two triangles.
Ex. In D ABC, AD is the height and BC is the base. In D PQR, PS is the height and QR is the base.
`A(triangle ABC)/A(trianglePQR)=(1/2xxBCxxAD)/(1/2xxQRxxPS)`
`A(triangle ABC)/A(triangle PQR)=(BCxxAD)/(QRxxPS)`
Hence the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corrosponding heights. Base of a triangle is b1 and height is h1. Base of another triangle is b2 and height is h2 .
Then the ratio of their areas =`(b_1xxh_1)/(b_2xxh_2)`
Suppose some conditions are imposed on these two triangles,
Condition 1: If the heights of both triangles are equal then :
`A(triangle ABC)/A(triangle PQR)=(BCxxh)/(QRxxh)="BC"/QR"`
`A(triangle ABC)/A(triangle PQR)=b_1/b_2`
Property: The ratio of the areas of two triangles with equal heights is equal to the ratio of their corresponding bases.
Condition 2: If the bases of both triangles are equal then -
`A(triangle ABC)/A(triangle APB)=(ABxxh_1)/(ABxxh_2)`
`A(triangle ABC)/A(triangle APB)=h_1/h_2`
Property: The ratio of the areas of two triangles with equal bases is equal to the ratio of their corresponding heights.
