Topics
Number Systems
Number Systems
Algebra
Introduction to Polynomials
Algebraic Expressions
Algebraic Identities
Sequences and Progressions
Coordinate Geometry
Geometry
Exploring Algebraic Identities
Area
Constructions
- Introduction of Constructions
- Geometric Constructions
- Some Constructions of Triangles
Mensuration
Linear Equations in Two Variables
Statistics and Probability
Coordinate Geometry
Probability
Introduction to Euclid’s Geometry: Axioms and Postulates
Lines and Angles
- Introduction to Lines and Angles
- Basic Terms and Definitions
- Intersecting Lines and Non-intersecting Lines
- Parallel Lines
- Concept of Pairs of Angles
- Concept of Transversal Lines
- Basic Properties of a Triangle
Triangles: Congruence Theorems
4-gons (Quadrilaterals)
- Properties of Quadrilateral
- Another Condition for a Quadrilateral to Be a Parallelogram
- Theorem of Midpoints of Two Sides of a Triangle
- Property: The Opposite Sides of a Parallelogram Are of Equal Length.
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram.
- Property: The Opposite Angles of a Parallelogram Are of Equal Measure.
- Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram.
- Property: The diagonals of a parallelogram bisect each other. (at the point of their intersection)
- Theorem : If the Diagonals of a Quadrilateral Bisect Each Other, Then It is a Parallelogram
Circles
Area and Perimeter
- Area of a Triangle by Heron's Formula
- Application of Heron’s Formula in Finding Areas of Quadrilaterals
- Geometric Interpretation of the Area of a Triangle
Surface Area and Volume
Statistics
Introduction to Probability
Notes
Two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. in following fig.
Draw CD || BA and CR || BP such that D and R lie on line AP in following fig.
From this, you obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.
Therefore, ar (ABCD) = ar (PBCR)
Now ∆ ABC ≅ ∆ CDA and ∆ PBC ≅ ∆ CRP
So, ar (ABC) = `1/2` ar (ABCD) and ar (PBC) = `1/2` ar (PBCR)
Therefore, ar (ABC) = ar (PBC)
Theorem
Theorem : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Proof: Now, suppose ABCD is a parallelogram whose one of the diagonals is AC (see Fig.).
Let AN ⊥ DC. Note that
∆ ADC ≅ ∆ CBA
So, ar (ADC) = ar (CBA)
Therefore, ar (ADC) = `1/2` ar (ABCD)
=`1/2` (DC × AN)
So, area of ∆ ADC = `1/2` × base DC × corresponding altitude AN
In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height).
The two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
