Prove that - "In a right angled triangle, the square of hypotenuse is equal to the
sum of squares of remaining two sides."
Answers
Answer:
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Answer:
Pythagoras theorem
Step-by-step explanation:
Given: 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)
⇒ XZ² = (XY² + YZ²)
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