Question

Proof of Pythagoras theorem

Answer 1

proof
take a triangle abc
take one big sqaure....in it take a small square
AREA OF WHOLE SQUARE
it is a big square....with each side having a length of a+b,so the total area is:
A=(a+b)(a+b)
AREA OF THE PIECES
now let's add up the areas of all the smaller pieces:
first,the smaller(tilted) sqaure has an area of:c^2
each of the four triangles has an area of:ab/2
so all four of them together is:4ab/2=2ab
adding up the tilted square and the 4 triangles gives:A=c^2+2ab
BOTH AREAS MUST BE EQUAL
(a+b)(a+b)=c^2+2ab
(a+b)(a+b)=c^2+2ab
a^2+2ab+b^2=c^2+2ab
subtract"2ab" from both sides
a^2+b^2=c^2
HENCE PROVED

Answer 2

HELLO !!!....


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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/AB = AB /AC

⇒ AD × AC = AB2    ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

=> CD/BC = BC/AC

⇒ 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


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HOPE IT HELPS U DEAR :))

CHEERS !!!!

# NIKKY ✌✌