Pythagoras Theorem
- 1. PYTHAGORAS THEOREM (PROOF OF PYTHAGORAS THEOREM)
- 2. STATEMENT :- According to the Pythagoras theorem “In a Right angled Triangle, the square of the hypotenuse is equal to the sum of individual squares of the other two sides”. i.e (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2
- 3. Right Angled Triangle hypotneuse base perpendicular A B C Acc. To Pythgoras Theorem (Hypotneuse)^2 = (Base)^2 + (Perpendicular)^2 (AC)^2 = (AB)^2 + (BC)^2
- 4. Proof of Pythagoras Theorem A B C D x 90 - x 90 90 ABC is a right angled triangle .Draw a perpendicular BD on AC. Now , there are two more right angle triangle i.e, tri.(BCD) and tri.(BAD) Therefore, AC = AD + CD In tri.(BCD), Cos x = base hypotneuse Cos x = CD BC So, CD= BC Cosx--eq1 In tri.(BAD), Cos (90 - x) = base hypotneuse Cos (90 - x) = AD AB So, AD = AB Sinx --eq2 (cos(90-x) = sinx) Also, in tri(ABC), Sin x = perpendicular hypotneuse Sinx = AB AC Cos x = base hypotneuse Cos x = BC AC
- 5. Since, AC = AD + CD By putting value of AD and CD from eq1 and eq2 We have, AC = AB Sin x + BC Cos x Now, putting Sin x = AB and Cos x = BC in above equation, AC AC We have, AC = AB( AB ) + BC( BC ) ( AC ) ( AC) AC x AC = (AB)^2 + (BC)^2 (AC)^2 = (AB)^2 + (BC)^2
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