Question

state and prove Pythagoras theorm

Answer 1

Dear Student!

Here is the answer to your query.

 

Given : A right ΔABC right angled at B

To prove : AC2 = AB2 + BC2

 

Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)



⇒ AD × AC = AB2    ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)



⇒ CD × AC = BC2    ........ (2)

 

Adding (1) and (2) we get

AB2 + BC2 = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC2

∴ AC2 = AB2 + BC2
Hence Proved

Hope it helps you...
plz mark as brainliest....

 

Answer 2

Pythagoras theorm states that: The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.

The prove part has been given above.

Hope this helps..