Linear Equations in Two Variables
Introduction to Euclid’S Geometry
Lines and Angles
- Concept of Quadrilaterals - Sides, Adjacent Sides, Opposite Sides, Angle, Adjacent Angles and Opposite Angles
- Angle Sum Property of a Quadrilateral
- Types of Quadrilaterals
- Another Condition for a Quadrilateral to Be a Parallelogram
- The Mid-point Theorem
- 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
- 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
Areas - Heron’S Formula
Surface Areas and Volumes
Statistics and Probability
Theorem : The perpendicular from the centre of a circle to a chord bisects the chord.
Draw a circle. Let O be its centre. Draw a chord AB. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. So MA = MB.
Theorem : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Let AB be a chord of a circle with centre O and O is joined to the mid -point M of AB.
To prove: OM⊥ AB.
Proof: Join OA and OB
In triangles OAM and OBM,
OA =OB (radii of the same circle)
AM =BM (M is the midpoint of PQ)
OM =OM (common)
Therefore, ∆OAM ≅ ∆OBM (SSS congruence rule)
This gives , ∠OMA = ∠OMB = 90°
(Corresponding parts of congruent triangles are congruent).
Shaalaa.com | Theorem : The perpendicular from the centre of a circle to a chord bisects the chord.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. if the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?
A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball o Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.
If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.
If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.
A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.