**Telangana SCERT Solution Class IX (9) Math Chapter 7 Triangles Exercise 7.4**

**1.)**

△ABC is a right angled triangle at ∠B = 90°

∴ AC is the hypotenuse of △ABC.

Now,

∠A +∠B + ∠C = 180°

Or, ∠A + ∠C + 90° = 180°

Or, ∠A + ∠C =180° – 90°

Or, ∠A + ∠C = 90°

Or, ∠A + ∠C = ∠B

Clearly,

∠A and ∠C <∠B

We know,

That side opposite to the greatest angle of a triangle is the longest side.

∴ △ABC, AC is the longest side and it is also the hypotenuse

They in a right angled △ the hypotenuse is the congest side.

**2.)**

Given,

∠PBC < ∠QCB

∴ ∠A+B+∠C = 180° [∵ Sum of angles of a △ is 180°]

∠PBC is exterior angle of △ABC

∴ ∠PBC = ∠A+∠C – (i)

Similarly,

∠QCB = ∠A+∠B – (ii)

Comparing eqn. = (i) and (ii)

We see that,

∠B>∠C [∵∠PBC < ∠QCB]

∴ AC >AB [∵ Side opposite to larger angle in △ABC]

**3.)**

Given,

∠B <∠A

∠C< ∠D

∴ OB > AO [∵ Side opposite to larger angle in △AOB]

OC> OD [∵ Side opposite to larger angle in △COD

Now,

AD = AO +OD

BC = BO +OC

∴ AD<BC [∵ AO<OB and OD <OC

**4.)**

Join AC and BD

Given,

AB is smallest side

∴ In △ ABC

BC < AB

∴ ∠BAC > ∠ACB – (i) [∵ Angle opposite to longer side in △ABC]

CD is the longest side

∴ In △ACD

CD > AD

∴ ∠CAD >∠ACD – (ii) [∵ angle opposite to longer side of △ACD]

Now,

Adding eqn. (i) and (ii) We get

∠BAC + ∠CAD > ∠ACB + ∠ACD

Or, ∠A > ∠C proved.

Similarly,

In △ABD

AD >AB [∵ AB is the smallest side]

∴ ∠ABD > ∠ADB [∵ angle opposite to longer side of △ABD] – (iii)

In △BCD

CD > BC [∵ CD is the longest side]

∴ ∠CBD > ∠BDC [∵ angle opposite to the longer side of △BCD] – (iv)

Now,

Adding of eqn. (iii) and (iv) we get

∠ ABD + ∠CBD > ∠ADB + ∠BDC

Or, B > ∠D

Hence proved

**5.)**

Given,

PR > PQ

PS bisects ∠QPR

In △PQR

∠PQR > ∠PRQ [∵ PR > PQ opposite angle of longer side of △PQR]

Or, ∠PQS > ∠PRS

∠QPS = ∠RPS [∵ PS bisects ∠QPR]

∴ In △PQS

∠PQS + ∠PSQ + ∠QPS = 180° [∵Sum of angle of △ is 180°]

Or, ∠QPS = 180° – (∠PQS + ∠PSQ) – (i)

In △ PSR

∠PRS + ∠PSR + ∠RPS = 180° [∵ sum of angle of △ is 180°]

Or, ∠PSR = 180°- (∠PRS +∠RPS)

Or, ∠PSR = 180° – (∠PRS + ∠PSQ) [∵∠PSQ = ∠RPS] – (ii)

∴ Comparing eqn. (i) and (ii) we get

∠PSR > ∠QPS [∵∠PRS < ∠PQS]

Hence proved

**6.) Given,**

Two sides of a triangle are 4 cm and 6 cm

We know that

The sum of two sides of a triangle will always be greater than the third side

∴ Let,

The length of third side be x

∴ 4+X > 6 or, X > 2

4 + 6 > X or, X < 10

6 + X > 4 or, X > – 2

∴ 2 < X < 10 [∵ X > 2 so X also > – 2]

∴ all the possible solutions of X are

3, 4, 5, 6, 7, 8, 9, continue

**7.) To construct a triangle with sides 5 cm, 8cm, 1cm….**

We know that,

A triangle will always have its third side less than the sum of the other two sides

In this case,

8 > 5 +1

8 > 6

The side with 8cm is greater than the sum of other two sides thus we cannot construct a triangle of the given dimensions.