Question

prove the Pythagoras theorem ​

Answer 1

Answer:

The proof of Pythagorean Theorem in mathematics is very important.

In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).

In short it is written as: a2 + b2 = c2

Proof of Pythagoras Theorem

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Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.

Verification of Pythagorean Theorem

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Then, we will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the

square EFGH, each of whose side is a, so area of the square EFGH is a2.

Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF

or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c

or, b2 + c2 + 2bc = a2 + 2bc

or, b2 + c2 = a2

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)

Answer 2

hi mate,

Pythagoras Theorem Proof

Given: A right-angled triangle ABC.

To Prove- AC² = AB² + BC²

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

We know, △ADB ~ △ABC

Therefore,

AD AB

----- = -----

AB AC

(Condition for similarity)

Or, AB² = AD × AC …………………..……..(1)

Also, △BDC ~△ABC

Therefore,

CD BC

----- = -----

BC AC

(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 thoerem is proved.

i hope it helps you.