Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let us consider a right-angled triangle ABC, right-angled at B.
In ∆ABC,
∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)
∠A + 90o + ∠C = 180°
∠A + ∠C = 90°
Hence, the other two angles have to be acute (i.e., less than 90o ).
Thus, ∠B is the largest angle in ∆ABC.
So, ∠B > ∠A and ∠B > ∠C
Therefore, AC > BC and AC > AB [Using theorem 7.7 of triangles, in any triangle, the side opposite to the larger (greater) angle is longer.]
Therefore, AC is the largest side in ∆ABC.
However, AC is the hypotenuse of ∆ABC.
Therefore, the hypotenuse is the longest side in a right-angled triangle.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 7
Video Solution:
Show that in a right angled triangle, the hypotenuse is the longest side
NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.4 Question 1
Summary:
Thus, we have proved that in a right-angled triangle, the hypotenuse is the longest side.
☛ Related Questions:
- In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
- In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
- AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.
- In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.