Show that in a right-angled triangle,the hypo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

Vedclass · Answer

(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.

Show that in a right-angled triangle,the hypotenuse is the longest side.

Similar Questions

$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see figure). Show that $CD$ bisects $AB$.

$ABC$ is a right-angled triangle in which $\angle A = 90^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.

$E$ and $F$ are respectively the mid-points of equal sides $AB$ and $AC$ of $\Delta ABC$ (see figure). Show that $BF = CE$.

$AB$ is a line segment. $P$ and $Q$ are points on opposite sides of $AB$ such that each of them is equidistant from the points $A$ and $B$ (see figure). Show that the line $PQ$ is the perpendicular bisector of $AB$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module