Question

state and prove converse of pythagorus theorem

Answer 1

Step-by-step explanation:

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)

[Hence proved]

• The Converse of Pythagorean Theorem:

The converse of the Pythagorean Theorem is: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

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Answer 2

hi mate,

Proof: Construct another triangle, △EGF, such as AC = EG = b and BC = FG = a.

In △EGF, by Pythagoras Theorem:

EF²= EG2 + FG² = b² + a²…………(1)

In △ABC, by Pythagoras Theorem:

AB² = AC² + BC²= b² + a² …………(2)

From equation (1) and (2), we have;

EF² = AB²

EF = AB

⇒ △ ACB ≅ △EGF (By SSS postulate)

⇒ ∠G is right angle

Thus, △EGF is a right triangle.

Hence, we can say that the converse of Pythagorean theorem also holds.

Hence Proved.