Show that in a right-angled triangle,the hypotenuse is the longest side.
(N/A) Let us consider $\Delta ABC$ such that $\angle B = 90^\circ$.
Since the sum of angles in a triangle is $180^\circ$,we have $\angle A + \angle B + \angle C = 180^\circ$.
Substituting $\angle B = 90^\circ$,we get $\angle A + 90^\circ + \angle C = 180^\circ$,which implies $\angle A + \angle C = 90^\circ$.
Since $\angle A + \angle C = 90^\circ$ and $\angle B = 90^\circ$,it follows that $\angle B = \angle A + \angle C$.
This implies that $\angle B > \angle A$ and $\angle B > \angle C$.
In any triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle B$ (which is $AC$) is longer than the side opposite to $\angle A$ (which is $BC$) and the side opposite to $\angle C$ (which is $AB$).
Thus,$AC > BC$ and $AC > AB$.
Since $AC$ is the hypotenuse,it is the longest side of the triangle.
Since the sum of angles in a triangle is $180^\circ$,we have $\angle A + \angle B + \angle C = 180^\circ$.
Substituting $\angle B = 90^\circ$,we get $\angle A + 90^\circ + \angle C = 180^\circ$,which implies $\angle A + \angle C = 90^\circ$.
Since $\angle A + \angle C = 90^\circ$ and $\angle B = 90^\circ$,it follows that $\angle B = \angle A + \angle C$.
This implies that $\angle B > \angle A$ and $\angle B > \angle C$.
In any triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle B$ (which is $AC$) is longer than the side opposite to $\angle A$ (which is $BC$) and the side opposite to $\angle C$ (which is $AB$).
Thus,$AC > BC$ and $AC > AB$.
Since $AC$ is the hypotenuse,it is the longest side of the triangle.
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