1. Show that in a right angled triangle, the hypotenuse is the longest side.
Triangle ABC is right angled at A.
∴ ∠A > ∠B and ∠A > ∠C.
⇒ BC > AC and BC > AB
[Because the side opposite to the greater angle is longer.]
⇒ BC is the longest side.
i.e., The hypotenuse is the longest side in an right angled triangle.
2. In figure, sides AB and AC of ΔABG are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Given: ∠PBC < ∠QCB.
⇒ 180° – ∠ABC < 180° — ACB.
⇒ – ∠ABC < – ∠ACB. [ Linear pair ]
⇒ ∠ ABC > ∠ACB [ Multiplying by (-1)]
⇒ AC > AB [ Greater angle has longer side opposite to it ]
3 . In figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
As ∠B < ∠A [Given ]
⇒ OA < OB …..(i) [Greater angle has longer side opposite to it.]
∠C < ∠D
⇒ OD < OC …….(ii)
[Greater angle has longer side opposite to it.]
⇒ OA + OD < OB + OC
[Adding corresponding sides of (i) and (ii) ]
⇒ AD < BC.
4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B > ∠D.
Join B and D.
In ΔABD, AB < AD [Given]
⇒ ∠ADB < ∠ABD …(i)
[Longer side has greater angle opposite to it.]
BC < CD [Given]
⇒ ∠BDC < ∠DBC …(ii) [Longer side has greater angle opposite to it.]
Adding corresponding sides of (i) and (ii), we get
∠ADB + ∠BDC < ∠ABD + ∠DBC
⇒ ∠ADC < ∠ABC ⇒ ∠D < ∠B
Similarly, by joining A and C, we can show ∠A > ∠C.
(Students try themselves)
5. In figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.
In ΔPQR, PR > PQ [Given]
⇒ ∠Q > ∠R …(i)
[Angle opposite to longer side is greater.]
Also, ∠QPS = ∠RPS …(ii) [ PS is bisector of ∠QPR]
⇒ ∠Q + ∠QPS > ∠R < ∠RPS …..(iii) [Adding (i) and (ii)]
Also, ∠PSR = ∠Q + ∠QPS …(iv)
[Exterior angle is equal to sum of interior opposite angles ]
and ∠PSQ = ∠R + ∠RPS …(v)
∴ ∠PSR > ∠PSQ. [From (iii), (iv) and (v)]
6. Show that of all line segments drawn from a given point not On it, the perpendicular line segment is the shortest.
AP is perpendicular line segment on line l
and AQ is any other line segment.
∠APQ > ∠AQP [ in a triangle, if an angle is 90°, other two angles are less than 90°]
⇒ AQ > AP,
Hence, any line segment AQ, from point A to the line l, is greater than the perpendicular line segment AP, from point A to the line l.
Hence, AP is the shortest segment of all the segments.