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.
Can we have questions too
But thank you very much for answers