Math, asked by sonali116, 1 year ago

To get 50 points... heres the question
PROOF OF PYTHAGORAS THEOREM BY DIFFERENT METHODS
(AT LEAST 5 METHODS)
PLZ ITS URGENT...

Answers

Answered by Anonymous
9

Figure in attachment

▶ We will 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 .

NOTE :

The area of triangle can be calculated by the formula :

Area = 1/2 × base × height

In the proof , c is the hypotenuse or the longest side of the triangle which is right angled .

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Answered by sk181231
0

Answer:

Refer to the attachments

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