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.
Explore More
Similar Questions
$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
Medium
View SolutionIn quadrilateral $ACBD$,$AC = AD$ and $AB$ bisects $\angle A$ (see figure). Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?
Medium
View Solution$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
Medium
View SolutionIn the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.
Medium
View SolutionIn an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo