mathematics project
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TrianglesTrianglesBy Neeraj choithwaniBy Neeraj choithwani
class 9class 9thth c c
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TrianglesTrianglesCan be classified by the number of congruent sides
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Congruent Congruent TrianglesTriangles
Have the same SIZE and the same SHAPE and
have same area
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Scalene TriangleScalene Triangle
Has no congruent sides
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Isosceles TriangleIsosceles Triangle
Has at least two congruent sides
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Equilateral TriangleEquilateral Triangle
Has three congruent sides
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Right TriangleRight Triangle
Has one right angle
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Criteria for
Congruency of Triangles
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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
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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
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SSS Congurence Criteria
If a triangle include all its sides equal to the corresponding triangle’s sides the triangles can be called Congurent
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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.
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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
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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
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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
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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.
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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
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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
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Researched Formulae
Given, side = 4 cmSolution ,
3x – x / 2 3x / 2 (3x – x /2)
Where x is side of Triangle
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(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