Answer:
Therefore Anjeli is correct and
[tex](a+b)^{2}=c^{2}+4(\frac{1}{2}ab)[/tex] is TRUE.
Step-by-step explanation:
Given:
Label on triangle as ΔABC right angle at C such that
AB = c .....(Hypotenuse)
AC = b ....(Longer leg)
BC = a ....(Shorter leg)
To Prove:
[tex](a+b)^{2}=c^{2}+4(\frac{1}{2}ab)[/tex]
Proof:
We know, in Right angle triangle ABC by Pythagoras theorem we get,
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]
Substituting the values we get
[tex]AB^{2}=BC^{2}+AC^{2}\\\\c^{2}=a^{2}+b^{2}[/tex] ...............( 1 )
Now the Left hand side of what Anjeli wrote is
Left hand side = (a+b)²
Using identity (A+B)² = A²+ B² + 2AB we get
Left hand side = a²+ b² + 2ab
From ( 1 ) we have [tex]c^{2}=a^{2}+b^{2}[/tex]
Substituting we get
Left hand side = c² + 2ab ...........................( 2 )
Now,
Right hand side =[tex]c^{2}+4(\frac{1}{2}ab)[/tex]
Dividing 4 by 2 we get 2, hence
Right hand side =[tex]c^{2}+2ab[/tex] .........................( 3 )
Therefore,
Left hand side = Right hand side From ( 2 ) and ( 3 )
[tex](a+b)^{2}=c^{2}+4(\frac{1}{2}ab)[/tex] ..True
Therefore Anjeli is correct and
[tex](a+b)^{2}=c^{2}+4(\frac{1}{2}ab)[/tex] is TRUE.
