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In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Theorem illus.png
The theorem states that in the right-angled triangle \(ABC\), \(AC^2=AB^2+BC^2\).
Proof of the theorem:
A triangle right angled at \(B\).
That is \(\angle ABC\) \(=\) \(90^{\circ}\).
To prove:
Construct a line from \(B\) to \(AC\) to intersect at \(D\) such that \(BD \perp AC\).
Theorem proof.png
Consider the triangles \(ABC\) and \(BDC\).
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of  the perpendicular are similar to the whole triangle and to each other.
By the theorem, we have \(\Delta ABC\) \(\sim\) \(\Delta BDC\).
Hence, the ratio of the corresponding sides of the triangles are equal.
That is, \(\frac{BC}{CD} = \frac{AC}{BC}\).
This implies, \(BC^{2} = AC \times CD\)        ……\((1)\)
Now consider the triangles \(ABC\) and \(ABD\).
Similarly, by the above mentioned theorem we have \(\Delta ABC\) \(\sim\) \(\Delta ABD\).
Hence, the ratio of the corresponding sides of the triangles are equal.
So, \(\frac{AB}{AD} = \frac{AC}{AB}\).
This implies, \(AB^{2} = AC \times AD\)        ……\((2)\)
Add equations \((1)\) and \((2)\) as follows:
\(BC^2 + AB^2\) \(=\) \((AC \times CD) + (AC \times AD)\)
\(=\) \(AC (CD +AD)\)
\(=\) \(AC \cdot AC\)
\(=\) \(AC^2\).
Therefore, \(AC^2 = AB^2 + BC^2\).
Hence, the proof.
In a right-angled triangle, if the measure of the hypotenuse is \(29 cm\) and one of its sides is \(21\) \(cm\) then find the length of the other side.
Let the triangle be \(ABC\) right angled at \(B\).
This implies that the side \(AC\) is the hypotenuse.
Theorem eg.png
By the Pythagorean theorem, we have \(AC^2 = AB^2 + BC^2\).
Thus, \(AB^2 = AC^2 - BC^2\).
\(\Rightarrow AB^2 = 29^2 -21^2\)
\(= 841 - 441\)
\(= 400\)
Hence, \(AB = \sqrt{400}\).
\(AB = 20\)
Therefore, the length of the other side is \(20\) \(cm\).