To get 100!!!!!!! points... heres the question
PROOF OF PYTHAGORAS THEOREM BY DIFFERENT METHODS
(AT LEAST 5 METHODS)
PLZ ITS URGENT...
Answers
Given : A right ΔABC right angled at B
To prove : AC2 = AB2 + BC2
Construction : Draw AD ⊥ AC
Proof : ΔABD and ΔABC
∠ADB = ∠ABC = 90°
∠BAD = ∠BAC (common)
∴ ΔADB ∼ ΔABC (by AA similarly criterion)
⇒ AD × AC = AB2 ...... (1)
Now In ΔBDC and ΔABC
∠BDC = ∠ABC = 90°
∠BCD = ∠BCA (common)
∴ ΔBDC ∼ ΔABC (by AA similarly criterion)
⇒ CD × AC = BC2 ........ (2)
Adding (1) and (2) we get
AB2 + BC2 = AD × AC + CD × AC
= AC (AD + CD)
= AC × AC = AC2
∴ AC2 = AB2 + BC2
Figure in attachment
▶ We draw a right angled triangle with sides a , b and c .
▶ We name the constructed triangle Δ ABC .
▶ We extend AB to a point E such that EA = BC .
▶ We construct DE such that DE = AB .
Naming the figure's lengths
▶ Let EA = BC = b .
▶ Let DE = BC = a .
In Δ ABC and Δ AED ,
∠ABC = ∠AED [ 90° each by construction ]
BC = EA [ By construction ]
AB = DE [ By construction ]
Δ ABC ≅ Δ ADE [ S.A.S ]
AD = AC [ c.p.c.t ]
Let AD = AC = c
Area of trapezium EDBC = 1/2 ( sum of parallel sides ) ( height )
⇒ Area = 1/2 × ( a + b )( a + b )
⇒ Area = ( a + b )²/2
∠BAC + ∠EAD
⇒ ∠BAC + ∠BCA
[ Equal ∠s due to c.p.c.t rule ]
⇒ 180° - ∠ABC
⇒ 180° - 90°
⇒ 90°
We know that :
∠DAC = 180° - ( ∠EAD + ∠BAC )
⇒ ∠DAC = 180° - 90°
⇒ ∠DAC = 90°
Hence Δ DAC is right angled .
Area of ( Δ DAC + Δ ABC + Δ AED ) = area of trapezium
⇒ 1/2 × ab + 1/2 ab + 1/2 c² = ( a + b )²/2
⇒ ab + 1/2 c² = ( a + b )²/2
⇒ 2 ab + c² = a² + b² + 2 ab
⇒ a² + b² = c²
Hence the Pythagoras theorem is proved .
