Question

prove the pythagoras theorem using the cut and paste method

Answer 1

"Let a, b, c denote the legs and the hypotenuse of the given right triangle, and consider the two squares in the accompanying figure, each having a+b as its side. The first square is dissected into six pieces-namely, the two squares on the legs and four right triangles congruent to the given triangle. The second square is dissected into five pieces-namely, the square on the hypotenuse and four right triangles congruent to the given triangle. By subtracting equals from equals, it now follows that the square on the hypotenuse is equal to the sum of the squares on the legs" 

Consider the following figure. 

The area of the first square is given by (a+b)^2 or 4(1/2ab)+ a^2 + b^2.
The area of the second square is given by (a+b)^2 or 4(1/2ab) + c^2.
Since the squares have equal areas we can set them equal to another and subtract equals. The case (a+b)^2=(a+b)^2 is not interesting. Let's do the other case. 
4(1/2ab) + a^2 + b^2 = 4(1/2ab)+ c^2
Subtracting equals from both sides we have

concluding Pythagoras' proof.

Answer 2

Step-by-step explanation:

Pythagoras' theorem :-

→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Step-by-step explanation:

It's prove :-

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ] .

∴ △ADB ∼ △ABC [ By AA-similarity ] .

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC .

$\huge \green{ \boxed{ \sf \therefore AC^2 = AB^2 + BC^2 }}$

Hence, it is proved. ........