TrianglesTrianglesBy Neeraj choithwaniBy Neeraj choithwani

class 9class 9thth c c

TrianglesTrianglesCan be classified by the number of congruent sides

Congruent Congruent TrianglesTriangles

Have the same SIZE and the same SHAPE and

have same area

Scalene TriangleScalene Triangle

Has no congruent sides

Isosceles TriangleIsosceles Triangle

Has at least two congruent sides

Equilateral TriangleEquilateral Triangle

Has three congruent sides

Right TriangleRight Triangle

Has one right angle

Criteria for

Congruency of Triangles

SAS congurence Criteria If any 2 sides and an one angle of a

triange is equal to corresponding sides and angles of other triangles then the two triangles are congurent

Both are congurent triangles

ASA Congurence criteria

If any two angles and the Included side of one triangle are equal to corresponding angle and Included side of other triangle , then the two triangles are congurent

Both are congurent triangles

SSS Congurence Criteria

If a triangle include all its sides equal to the corresponding triangle’s sides the triangles can be called Congurent

RHS Congurency Criteria

If a triangle’s one right angle, If a triangle’s one right angle, one side and a hypotenese is one side and a hypotenese is equal to the corresponding equal to the corresponding triangles The triangles are said triangles The triangles are said to be congurent.to be congurent.

Corresponding Part of congurent Triangle

( C.P.C.T. )• When 2 figures are congruent the When 2 figures are congruent the corresponding parts are congruent. corresponding parts are congruent. (angles and sides)(angles and sides)

• If any 2 trianglesa are congurent all their parts may be angles or line will be equal

Research Work

• This research is Based on 12th chapter “Heron’s formulae” which deals with finding the area of a triangle

• This Research dosen’t keeps any relation

with opposing heron’s formulae

• This research is a further study based on heron’s formuale

Heron’s formulae

• Heron’s original formulae which is a universal formulae to find area of any triangle

s (s-a) (s-b) (s-c)

• Acctually it is a generelisation on the ideas of Indian scientist “Bharamagupt-650 AD”

• David P. Robbins in 1895 Presented this Formulae

Finding area of Equilateral triangle

• 3x – x / 2 3x / 2 (3x – x /2)

• This formulae deals with Finding the area of an equilateral triangle when length of its side is given.

• The researched formulae came up with some changes in the idea of heron’s formulae.

Prooving Heron’s FormulaeOn an equilateral Triangle

Given, side = 4 cmSolution ,

s (s-a) (s-b) (s-c) Where s = sum of all sides/2 s = 4 + 4 + 4 / 2 s = 12 / 2 s = 6

Now, s (s-a) (s-b) (s-c) 6 (6-4) (6-4) (6-4) 2 × 3 × 2 × 2 × 2 4 3 So, Area of triangle is 4 3 Hence

Solved

Researched Formulae

Given, side = 4 cmSolution ,

3x – x / 2 3x / 2 (3x – x /2)

Where x is side of Triangle

(3x – x) / 2 3x / 2 { ( 3× 4 / 2 ) - 4 }

(3×4 / 2) - 4 3 × 4 / 2 { ( 3× 4 / 2 ) - 4 }

(12 / 2) - 4 12 / 2 { ( 12 / 2 ) – 4 }

6-4 6 ( 6 – 4 )

2 6 × 2

2 2 × 3 × 2

2 × 2 3 4 3