NCERT Solutions for Class 9 Math Chapter 7.4 Triangles

---

EXERCISE 7.1 EXERCISE 7.2 EXERCISE 7.3 EXERCISE 7.5*

---

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

solution :

   Given,

   A right triangle

   Find,

   hypotenuse is the longest side

   In the Fig.,

   we know that,

   ∠A + ∠B + ∠C = 180°

   Now,

   ∠B + ∠C = 90°

   Then,

   ∠A = 90°

   ∴ ∠A = ∠B + ∠C

   ∴ ∠A > ∠B

   Similerly

   ∴ ∠A > ∠C

   ∴ BC > AC

   Similerly

   ∴ BC > AB

   ∴ BC is the longest side.

   So, hypotenuse is the longest side.

2.    In Fig. 7.48, sides AB and AC of `\▵\ `ABC are extended to points P and Q respectively. Also, ∠PBC `\<\ ` ∠QCB. Show that AC `\>\ ` AB.

solution :

   Given,

   Side AB and AC of `\▵\ `ABC are extended to points P and Q respectively. also, ∠PBC < ∠QCB.

   Find,

   AC > AB

   In the Fig.,

   ∠ABC + ∠PBC = 180°[ Linear pairs of angles ]

   ∠ABC + = 180° − ∠PBC `\-----(I)\ `

   Also,

   ∠ACB ∠QCB = 180°[ Linear pairs of angles ]

   ∠ACB = 180° − ∠QCB `\-----(II)\ `

   From (I) and (II)

   ∠ABC > ∠ACB

   Hence,

   AC > AB as sides opposite to the larger angle is always larger.

3.    In Fig. 7.49, ∠B `\<\ ` ∠A and ∠C `\<\ ` ∠D. Show that AD `\<\ ` BC.

solution :

   Given,

   ∠B < ∠A and ∠C < ∠D.

   Find,

   AD < BC

   In the Fig.,

   ∠B < ∠A[ Given ]

   AO < BO `\----- (I)\ `

   ∠C < ∠D[ Given ]

   OD < OC `\----- (II)\ `

   From (I) and (II)

   AO + OD < BO + OC

   AD ∠ BC

4.    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.

solution :

   Given,

   AB and CD are respectively the smallest and longest side of a quadrilateral ABCD.

   Find,

   ∠A > ∠C and ∠B > ∠D.

   Contruction,

   Join AC and DB.

   In `\▵\ `ABC,

   AB < BC[ Given AB is smallest line ]

   ∠ACB < ∠BAC `\-----(I)\ `[ Angle opposite to longer side is greater ]

   In `\▵\ `ACD,

   AD < DC[ Given CD is longest line ]

   ∠ACD < ∠DAC `\-----(II)\ `[ Angle opposite to longer side is greater ]

   Adding (I) and (II), we get

   ∠ACB + ∠ACD < ∠BAC + ∠DAC

   ⇒ ∠BCD < ∠BAD

   SO, ∠A > ∠C

   In `\▵\ `ABD,

   AB < AD[ Given AB is smallest line ]

   ∠ADB < ∠ABD `\-----(III)\ `[ Angle opposite to longer side is greater ]

   In `\▵\ `BCD,

   BD < DC[ Given CD is longest line ]

   ∠BDC < ∠DBC `\-----(IV)\ `[ Angle opposite to longer side is greater ]

   Adding (III) and (IV), we get

   ∠ADB + ∠BDC < ∠ABD + ∠DBC

   ⇒ ∠ADC < ∠ABC

   SO, ∠B > ∠D

5.    In Fig. 751, PR `\>\ ` PQ and Ps bisects ∠QPR. Prove that ∠PSR `\>\ ` ∠PSQ.

solution :

   Given,

   PR > PQ and PS bisect ∠QPR

   Find,

   ∠PSR > ∠PSQ

   In `\▵\ `PQR,

   PR > PQ[ Given ]

   ∠PQR > ∠PRQ `\-----(I)\ `[ Angle opposit to longer side to greater ]

   PS is the bisector of ∠QPR

   So, ∠QPS = ∠RPS `\-----(II)\ `

   In `\▵\ `PQS,

   ∠PSR = ∠PQS + ∠QPS `\-----(III)\ `[ The exterior angle of a triangle equals to the sum of opposite interior angles ]

   In `\▵\ `PSR,

   ∠PSQ = ∠PRS + ∠RPS `\-----(IV)\ `[ The exterior angle of a triangle equals to the sum of opposite interior angles ]

   Addind (I) and (II), we get

   ∠PQR + ∠QPS > ∠PRQ + ∠RPS

   From (III) and (IV),

   ∠PSR > ∠PSQ

6.    Show that of all line segments drawn from a given point not on it , the perpendicular line segment is the shortest.

solution :

   Given,

   l is a line and A is a point not lying on l. AB ⊥ l. C is any point on l other than B

   Find,

   AB < AC

   In `\▵\ `ABC,

   ∠B = 90°

   ⇒ ∠C is an acute angle.[ Angle sum property of a triangle ]

   ⇒ ∠B > ∠C

   ⇒ AC > AB[ Side opposite to greater angle is greater ]

   So, AB < AC.

---

EXERCISE 7.1 EXERCISE 7.2 EXERCISE 7.3 EXERCISE 7.5*