Question

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

Answer 1

Proof of Pythagorean Theorem using Algebra:

Proof of Pythagorean TheoremGiven: A ∆ XYZ in which ∠XYZ = 90°.

To prove: XZ2 = XY2 + YZ2
Construction: Draw YO ⊥ XZ

Proof: In ∆XOY and ∆XYZ, we have,

∠X = ∠X → common

∠XOY = ∠XYZ → each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity

⇒ XO/XY = XY/XZ

⇒ XO × XZ = XY2 ----------------- (i)
In ∆YOZ and ∆XYZ, we have,

∠Z = ∠Z → common

∠YOZ = ∠XYZ → each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity

⇒ OZ/YZ = YZ/XZ

⇒ OZ × XZ = YZ2 ----------------- (ii)

From (i) and (ii) we get,

XO × XZ + OZ × XZ = (XY2 + YZ2)

⇒ (XO + OZ) × XZ = (XY2 + YZ2)

⇒ XZ × XZ = (XY2 + YZ2)

⇒ XZ 2 = (XY2 + YZ2)


Thank You

Answer 2

Heya Mate !!!

Here's Your Answer ⏬

=> Solution ;-

To prove: XZ2 = XY2 + YZ2

=> Construction: Draw YO ⊥ XZ

=> Proof: In ∆XOY and ∆XYZ, we have,
=> ∠X = ∠X → common
=> ∠XOY = ∠XYZ → each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity

⇒ XO/XY = XY/XZ
⇒ XO × XZ = XY2 ----------------- (i)

In ∆YOZ and ∆XYZ, we have,

=> ∠Z = ∠Z → common
=> ∠YOZ = ∠XYZ → each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity

⇒ OZ/YZ = YZ/XZ
⇒ OZ × XZ = YZ2 ----------------- (ii)

From (i) and (ii) we get,

=> XO × XZ + OZ × XZ = (XY2 + YZ2)
⇒ (XO + OZ) × XZ = (XY2 + YZ2)
⇒ XZ × XZ = (XY2 + YZ2)
⇒ XZ 2 = (XY2 + YZ2)

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