proof of pythagoras thoream
Answers
Answered by
1
Theorem: In right angle triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides
Given: ΔABC is a right angled triangle, right angled at B
RTP: AC² = AB² + BC²
Construction: Draw BD perpendicular to AC
Proof: ΔABD is similar to ABC
⇒ AD / AB = AB / AC
AD×AC = AB² ------------------------- (1)
Also, ΔBDC is similar to ΔABC
⇒ CD / BC = BC / AC
CD × AC = BC² ---------------------- (2)
By adding (1) and (2),
AD×AC + CD×AC = AB² + BC²
AC ( AD+CD) = AB² + BC²
AC (AC) = AB² + BC²
Therefore,
AC² = AB² + BC²
Hence, it is proved
Given: ΔABC is a right angled triangle, right angled at B
RTP: AC² = AB² + BC²
Construction: Draw BD perpendicular to AC
Proof: ΔABD is similar to ABC
⇒ AD / AB = AB / AC
AD×AC = AB² ------------------------- (1)
Also, ΔBDC is similar to ΔABC
⇒ CD / BC = BC / AC
CD × AC = BC² ---------------------- (2)
By adding (1) and (2),
AD×AC + CD×AC = AB² + BC²
AC ( AD+CD) = AB² + BC²
AC (AC) = AB² + BC²
Therefore,
AC² = AB² + BC²
Hence, it is proved
Attachments:
Answered by
2
Step-by-step explanation:
Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2.
Similar questions