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Number Systems
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Introduction to Polynomials
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Exploring Algebraic Identities
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- Introduction of Constructions
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- Some Constructions of Triangles
Mensuration
Linear Equations in Two Variables
Statistics and Probability
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Introduction to Euclid’s Geometry: Axioms and Postulates
Lines and Angles
- Introduction to Lines and Angles
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- 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
Text
Examples: Sanya has a piece of land which is in the shape of a rhombus (see following Fig). She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?
Solution : Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of ∆ ABD is given by
`s = (100 + 100 + 160)/2 m = 180 m`
Therefore, area of ∆ ABD
=`sqrt (180 (180 - 100)(180 - 100)(180 - 160))`
`= sqrt (180 * 80 * 80 * 20) m^2 = 4800 m^2`
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