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
Draw a triangle and mark the mid-points E and F of two sides of the triangle. Join the points E and F in following fig.
Measure EF and BC. Measure ∠ AEF and ∠ ABC. EF = `1/2` BC and ∠ AEF = ∠ ABC
so, EF || BC
Theorem : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Observe following Fig. in which E and F are mid-points of AB and AC respectively and CD || BA.
∆ AEF ≅ ∆ CDF (ASA Rule)
So, EF = DF and BE = AE = DC
Therefore, BCDE is a parallelogram.
This gives EF || BC.
In this case, also note that EF = `1/2` ED = `1/2` BC.
Theorem : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
In following fig.
Observe that E is the mid-point of AB, line l is passsing through E and is parallel to BC and CM || BA.
Prove that AF = CF by using the congruence of ∆ AEF and ∆ CDF.
Shaalaa.com | Mid-point theorem
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.
In Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.
ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC =
21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.