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Pythagorous Theorem Class X CBSE

Pythagoras and Thales Theorem explained

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Pythagorous Theorem Class X CBSE

1. 1. PYTHAGORAS THEOREM ANURAG X-B R.NO-54
2. 2. Pythagoras of Samos  Pythagoras (569-479 B.C) was an Ionian Greek philosopher.  He was born on the Island of Samos.  He travelled Egypt, Greece and India & returned to Samos in 530 B.C.  He made influencial contribution to philosophy and religion in late 6th century.  He is often revered as a great mathematician and scientist ; best known for the Pythagorean theorem.
3. 3. The Pythagoras Theorem  In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides BASE P E R P E N D I C U L A R A B C BC2 = AC2 + AB2
4. 4. PROOF: A C B D •We are given a right angled triangle ABC right angled at B. •We need to proved that AC2 = BC2 + AB2 • Let us draw BD ┴ AC • Now ▲ADB~ ▲ABC (as per theorem 6.7) • So, AD = AB(sides prop) AB AC
5. 5. PROOF: A C B D • Or, AD X AC = AB2 …………………… (1) • Also, ▲BDC~ ▲ABC (Theorem 6.7) • So, CD = BC(sides prop) BC AC • CD X AC = BC2 …………………… (2) • Now, let us add both Equation (1) & (2) • ADXAC + CDXAC = AB2 + BC2 • AC(AD+CD) = AB2 + BC2 • AC2= AB2 + BC2
6. 6. Thank You !
7. 7. THALES THEOREM ANURAG X-B
8. 8.  Thales(624-546 B.C) was a pre-socratic Greek philosopher and mathematician from Miletus in Asia Minor & one of the seven sages of Greece.  In Mathematics, Thales used geometry to calculate the heights of pyramids and distance of ships from the shore.
9. 9. The Thales Theorem  If a line is drawn parallel to the third side of a triangle to intersect the other two sides in distinct points, the other two sides are divided into the same ratio.  i.e. in ▲ABC, side DE ǁ BC, • Then, AD= AE DB EC A CB D E
10. 10. PROOF:  In ▲ABC, side DE ǁ BC, we need to prove that AD= AE DB EC  Let us join BE and CD, then draw DM ┴ AC & EN ┴ AB  Area of ▲ADE= 1 ADXEN 2  i.e. Ar(ADE)= 1 ADXEN 2  Ar(BDE)= 1 DBXEN 2 A CB D E N M
11. 11.  Ar(ADE)= 1 AEXDM 2  Ar(DEC)= 1 ECXDM 2  Ar(ADE) = ½ AD X EN = AD ……(1) Ar(BDE) ½ DB X EN DB  Ar(ADE) = ½ AE X DM = AE ……(2) Ar(DEC) ½ EC X DM EC  ▲BDE & ▲ DEC are on the same base DE & between the same parallel BC and DE,  So, Ar (BDE) = Ar (DEC) …….(3) • From (1), (2) & (3) AD= AE DB EC A CB D E N M
12. 12. PPREPARED BY ANURAG AND MOHIT THANK YOU !