49.State and prove Pythagoras theoren

1.	49.State and prove Pythagoras theoren
Answer» Given:A ∆ XYZ in which ∠XYZ = 90°. To prove:XZ^2= XY^2+ YZ^2 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 = XY^2----------------- (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 = YZ^2----------------- (ii) From (i) and (ii) we get, XO × XZ + OZ × XZ = (XY^2+ YZ^2) ⇒ (XO + OZ) × XZ = (XY^2+ YZ^2) ⇒ XZ × XZ = (XY^2+ YZ^2) ⇒ XZ^2= (XY^2+ YZ^2)

Answer»

Given:A ∆ XYZ in which ∠XYZ = 90°.

To prove:XZ^2= XY^2+ YZ^2

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 = XY^2----------------- (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 = YZ^2----------------- (ii)

From (i) and (ii) we get,

XO × XZ + OZ × XZ = (XY^2+ YZ^2)

⇒ (XO + OZ) × XZ = (XY^2+ YZ^2)

⇒ XZ × XZ = (XY^2+ YZ^2)

⇒ XZ^2= (XY^2+ YZ^2)

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