Triangles – Exercise 7.4 – (MATHEMATICS) – 9th Class

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Exercise 7.4

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

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Sol. 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.


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Sol. 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.


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Sol.    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. 


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Sol. Construction: Join B and D.
In ΔABD, AB < AD                                                [Given]

⇒       ∠ADB < ∠ABD                                               …(i)

[Longer side has greater angle opposite to it.]


In ABCD,

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.


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Sol. 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.

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Sol.  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.


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