Identity : For any right-angled triangle, square of the longest side (hypotenuse) is equal to the sum of squares of other two sides (opposite and adjacent).
Proof: Consider we have a square as shown in figure.
\({h}^\mathbf{2}={a}^\mathbf{2}+{b}^\mathbf{2}\); where h is Hypotenuse
a is the side opposite to the angle \(\theta \)
and b is the side adjacent to angles \(\theta.\)
The above identity has been proved by a number of Mathematicians but most commonly used proof is derived by Pythagoras. The proof is also known as ‘Pythagoras Theorem’.
Proof: Consider we have a square as shown in figure.
\(area\ of\ whole\ square={side}^2\)
\(={({a}+{b})}^\mathbf{2} \) →(1)
\(area\ of\ small\ inside\ whole\ square={side}^2\)
\(={h}^\mathbf{2}\) →(2)
\(area\ of\ each\ triangle =\frac{1}{2}\times base\times height\)
\(=\frac{1}{2}\times a\times b \)
\(=\frac{1}{2}ab\)
\(area\ of\ 4\ triangle =4\times area\ of\ each\ triangle\)
\(=4\times\frac{1}{2}ab \)
\(=\mathbf{2}{ab}\) →(3)
From equations (1), (2) and (3)
\(Area\ of\ whole\ square\ =\ area\ of\ small\ inside\ square\ +\ area\ of\ 4\ triangles\)
\(={({a}+{b})}^\mathbf{2}={h}^\mathbf{2}+\mathbf{2}{ab}\)
\(=a^2+b^2+2ab=h^2+2ab\)
Subtracting 2ab from both sides;
\(=a^2+b^2+2ab-2ab=h^2+2ab-2ab\)
\(=a^2+b^2=h^2\)
\({h}^\mathbf{2}={a}^\mathbf{2}+{b}^\mathbf{2}\)
Hence the proof.