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# Mathematics project

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### Mathematics project

1. 1. Triangles By Neeraj choithwani class 9 th c
2. 2. Triangles <ul><li>Can be classified by the number of congruent sides </li></ul><ul><li> </li></ul>
3. 3. Congruent Triangles <ul><li>Have the same SIZE and the same SHAPE and have same area </li></ul><ul><li> </li></ul>
4. 4. Scalene Triangle <ul><li>Has no congruent sides </li></ul>
5. 5. Isosceles Triangle <ul><li>Has at least two congruent sides </li></ul>
6. 6. Equilateral Triangle <ul><li>Has three congruent sides </li></ul>
7. 7. Right Triangle <ul><li>Has one right angle </li></ul>
8. 8. <ul><li>Criteria for </li></ul><ul><li>Congruency of Triangles </li></ul>
9. 9. SAS congurence Criteria <ul><li>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 </li></ul><ul><li>Both are congurent triangles </li></ul>
10. 10. ASA Congurence criteria <ul><li>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 </li></ul><ul><li>Both are congurent triangles </li></ul>
11. 11. SSS Congurence Criteria <ul><li>If a triangle include all its sides equal to the corresponding triangle’s sides the triangles can be called Congurent </li></ul>
12. 12. RHS Congurency Criteria <ul><li>If a triangle’s one right angle, one side and a hypotenese is equal to the corresponding triangles The triangles are said to be congurent. </li></ul><ul><li> </li></ul>
13. 13. Corresponding Part of congurent Triangle ( C.P.C.T. ) <ul><li>When 2 figures are congruent the corresponding parts are congruent. (angles and sides) </li></ul><ul><li>If any 2 trianglesa are congurent all their parts may be angles or line will be equal </li></ul>
14. 14. Research Work <ul><li>This research is Based on 12 th chapter “Heron’s formulae” which deals with finding the area of a triangle </li></ul><ul><li>This Research dosen’t keeps any relation </li></ul><ul><li>with opposing heron’s formulae </li></ul><ul><li>This research is a further study based on heron’s formuale </li></ul>
15. 15. Heron’s formulae <ul><li>Heron’s original formulae which is a universal formulae to find area of any triangle </li></ul><ul><li>s (s-a) (s-b) (s-c) </li></ul><ul><li>Acctually it is a generelisation on the ideas of Indian scientist “Bharamagupt-650 AD” </li></ul><ul><li>David P. Robbins in 1895 Presented this Formulae </li></ul>
16. 16. Finding area of Equilateral triangle <ul><li>3x – x / 2 3x / 2 (3x – x /2) </li></ul><ul><li>This formulae deals with Finding the area of an equilateral triangle when length of its side is given. </li></ul><ul><li>The researched formulae came up with some changes in the idea of heron’s formulae. </li></ul>
17. 17. Prooving Heron’s Formulae On an equilateral Triangle <ul><li>Given, </li></ul><ul><li>side = 4 cm </li></ul><ul><li>Solution , </li></ul><ul><li>s (s-a) (s-b) (s-c) </li></ul><ul><li>Where s = sum of all sides/2 </li></ul><ul><li>s = 4 + 4 + 4 / 2 </li></ul><ul><li>s = 12 / 2 </li></ul><ul><li>s = 6 </li></ul>4 cm 4 cm 4 cm
18. 18. <ul><li>Now, </li></ul><ul><li>s (s-a) (s-b) (s-c) </li></ul><ul><li>6 (6-4) (6-4) (6-4) </li></ul><ul><li>2 × 3 × 2 × 2 × 2 </li></ul><ul><li>4 3 </li></ul><ul><li>So, </li></ul><ul><li>Area of triangle is 4 3 </li></ul><ul><li>Hence Solved </li></ul>
19. 19. Researched Formulae <ul><li>Given, </li></ul><ul><li>side = 4 cm </li></ul><ul><li>Solution , </li></ul><ul><li>3x – x / 2 3x / 2 (3x – x /2) </li></ul><ul><li>Where x is side of Triangle </li></ul>4 cm 4 cm 4 cm
20. 20. <ul><li>( 3x – x) / 2 3x / 2 { ( 3× 4 / 2 ) - 4 } </li></ul><ul><li>(3×4 / 2) - 4 3 × 4 / 2 { ( 3× 4 / 2 ) - 4 } </li></ul><ul><li>(12 / 2) - 4 12 / 2 { ( 12 / 2 ) – 4 } </li></ul><ul><li>6-4 6 ( 6 – 4 ) </li></ul><ul><li>2 6 × 2 </li></ul><ul><li>2 2 × 3 × 2 </li></ul><ul><li>2 × 2 3 </li></ul><ul><li>4 3 </li></ul>