Question

prove pythagoras theorm (class 10) - 5 marks

Answer 1

$\huge\underline\bold\red{AnswEr}$

Pythagoras Theorem Proof

Given: A right-angled triangle ABC.

To Prove- AC2 = AB2 + BC2

Construction: Draw a perpendicular BD joining AC at D.

Pythagoras Theorem Proof

Proof: First, we have to drop a perpendicular BD onto the side AC

We know, △ADB ~ △ABC

Therefore, ADAB=ABAC (Condition for similarity)

Or, AB2 = AD × AC ……………………………..……..(1)

Also, △BDC ~△ABC

Therefore, CDBC=BCAC (Condition for similarity)

Or, BC2= CD × AC ……………………………………..(2)

Adding the equations (1) and (2) we get,

AB2 + BC2 = AD × AC + CD × AC

AB2 + BC2 = AC (AD + CD)

Since, AD + CD = AC

Therefore, AC2 = AB2 + BC2

Hence, the Pythagorean theorem is proved.

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Note: Pythagorean theorem is only applicable to Right-Angled triangle

Answer 2

Proof of Pythagorean Theorem using Algebra:

Proof of Pythagorean TheoremGiven: 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)