#### Topics

##### Number Systems

##### Number Systems

##### Algebra

##### Polynomials

##### Linear Equations in Two Variables

##### Algebraic Expressions

##### Algebraic Identities

##### Coordinate Geometry

##### Geometry

##### Introduction to Euclid’S Geometry

##### Lines and Angles

##### Triangles

##### Quadrilaterals

- Concept of Quadrilaterals - Sides, Adjacent Sides, Opposite Sides, Angle, Adjacent Angles and Opposite Angles
- Angle Sum Property of a Quadrilateral
- Types of Quadrilaterals
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Another Condition for a Quadrilateral to Be a Parallelogram
- The Mid-point Theorem
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Property: The Opposite Sides of a Parallelogram Are of Equal Length.
- 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

##### Area

##### Circles

- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
- Angle Subtended by a Chord at a Point
- Perpendicular from the Centre to a Chord
- Circles Passing Through One, Two, Three Points
- Equal Chords and Their Distances from the Centre
- Angle Subtended by an Arc of a Circle
- Cyclic Quadrilateral

##### Constructions

##### Mensuration

##### Areas - Heron’S Formula

##### Surface Areas and Volumes

##### Statistics and Probability

##### Statistics

##### Probability

#### text

**Construction:** To construct the bisector of a given angle.

Given an angle ABC, we want to construct its bisector.**Steps of Construction :**

1. Taking B as centre and any radius, draw an arc to intersect the rays BA and BC, say at E and D respectively [see fig.(i)]

2. Next, taking D and E as centres and with the radius more than `1/2` DE, draw arcs to intersect each other, say at F.

3. Draw the ray BF [see Fig.(ii)]. This ray BF is the required bisector of the angle ABC.

Let us see how this method gives us the required angle bisector.

Join DF and EF.

In triangles BEF and BDF,

BE = BD (Radii of the same arc)

EF = DF (Arcs of equal radii)

BF = BF (Common)

Therefore, ∆BEF ≅ ∆BDF (SSS rule)

This gives ∠EBF = ∠ DBF (CPCT)

**Construction:** To construct the perpendicular bisector of a given line segment.

Given a line segment AB, we want to construct its perpendicular bisector.**Steps of Construction :**

1. Taking A and B as centres and radius more than `1/2` AB, draw arcs on both sides of the line segment AB (to intersect each other).

2. Let these arcs intersect each other at P and Q. Join PQ see following fig.

3. Let PQ intersect AB at the point M. Then line PMQ is the required perpendicular bisector of AB.

Let us see how this method gives us the perpendicular bisector of AB.

Join A and B to both P and Q to form AP, AQ, BP and BQ.

In triangles PAQ and PBQ,

AP = BP (Arcs of equal radii)

AQ = BQ (Arcs of equal radii)

PQ = PQ (Common)

Therefore, ∆ PAQ ≅ ∆ PBQ (SSS rule)

So, ∠ APM = ∠ BPM (CPCT)

Now in triangles PMA and PMB,

AP = BP (As before)

PM = PM (Common)

∠ APM = ∠ BPM (Proved above)

Therefore, ∆ PMA ≅ ∆ PMB (SAS rule)

So, AM = BM and ∠ PMA = ∠ PMB (CPCT)

As ∠ PMA + ∠ PMB = 180° (Linear pair axiom),

we get

∠ PMA = ∠ PMB = 90°.

Therefore, PM, that is, PMQ is the perpendicular bisector of AB.

**Construction:** To construct an angle of 600 at the initial point of a given ray.

Let us take a ray AB with initial point A (see following fig.(i)) We want to construct a ray AC such that ∠ CAB = 60°. One way of doing so is given below.

**Steps of Construction :**

1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.

2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.

3. Draw the ray AC passing through E [see Fig (ii)].

Then ∠ CAB is the required angle of 60°. Now, let us see how this method gives us the required angle of 60°.

Join DE.

Then, AE = AD = DE (By construction)

Therefore, ∆ EAD is an equilateral triangle and the ∠ EAD, which is the same as ∠ CAB is equal to 60°.