# SELINA Solutions for Class 9 Maths Chapter 11 - Inequalities

## Chapter 11 - Inequalities Exercise Ex. 11

In ABC,

AB = AC[Given]

ACB = B[angles opposite to equal sides are equal]

B = 70^{0}[Given]

ACB = 70^{0} ……….(i)

Now,

ACB +ACD = 180^{0}[ BCD is a straight line]

70^{0} + ACD = 180^{0}

ACD = 110^{0} …………(ii)

In ACD,

CAD + ACD + D = 180^{0}

CAD + 110^{0} + D = 180^{0 }[From (ii)]

CAD + D = 70^{0}

But D = 40^{0 }[Given]

CAD + 40^{0}= 70^{0}

CAD = 30^{0} ………………(iii)

In ACD,

ACD = 110^{0}[From (ii)]

CAD = 30^{0}[From (iii)]

D = 40^{0 }[Given]

[Greater angle has greater side opposite to it]

Also,

AB = AC[Given]

Therefore, AB > CD.

In PQR,

QR = PR[Given]

P = Q[angles opposite to equal sides are equal]

P = 36^{0}[Given]

Q = 36^{0}

In PQR,

P + Q + R = 180^{0}

36^{0} + 36^{0} + R = 180^{0 }

R + 72^{0} = 180^{0}

R = 108^{0}

Now,

R = 108^{0}

P = 36^{0}

Q = 36^{0}

Since R is the greatest, therefore, PQ is the largest side.

The sum of any two sides of the triangle is always greater than third side of the triangle.

Third side < 13_{}+_{}8 =_{}21 cm.

The difference between any two sides of the triangle is always less than the third side of the triangle.

Third side > 13_{}-_{}8 =_{}5 cm.

Therefore, the length of the third side is between 5 cm and 9 cm, respectively.

The value of a =_{}5 cm and b_{}= 21_{}cm.

In BEC,

B + BEC + BCE = 180^{0}

B = 65^{0 }[Given]

BEC = 90^{0}[CE is perpendicular to AB]

65^{0} + 90^{0} + BCE = 180^{0}

BCE = 180^{0} - 155^{0}

BCE = 25^{0} = DCF …………(i)

In CDF,

DCF + FDC + CFD = 180^{0}

DCF = 25^{0 }[From (i)]

FDC = 90^{0}[AD is perpendicular to BC]

25^{0} + 90^{0} + CFD = 180^{0}

CFD = 180^{0} - 115^{0}

CFD = 65^{0} …………(ii)

Now, AFC + CFD = 180^{0}[AFD is a straight line]

AFC + 65^{0} = 180^{0}

AFC = 115^{0} ………(iii)

In ACE,

ACE + CEA + BAC = 180^{0}

BAC = 60^{0 }[Given]

CEA = 90^{0}[CE is perpendicular to AB]

ACE + 90^{0} + 60^{0} = 180^{0}

ACE = 180^{0} - 150^{0}

ACE = 30^{0} …………(iv)

In AFC,

AFC + ACF + FAC = 180^{0}

AFC = 115^{0 }[From (iii)]

ACF = 30^{0}[From (iv)]

115^{0} + 30^{0} + FAC = 180^{0}

FAC = 180^{0} - 145^{0}

FAC = 35^{0} …………(v)

In AFC,

FAC = 35^{0}[From (v)]

ACF = 30^{0}[From (iv)]

In CDF,

DCF = 25^{0}[From (i)]

CFD = 65^{0}[From (ii)]

ACB = 74^{0} …..(i)[Given]

ACB + ACD = 180^{0}[BCD is a straight line]

74^{0} + ACD = 180^{0}

ACD = 106^{0} ……..(ii)

In ACD,

ACD + ADC+ CAD = 180^{0}

Given that AC = CD

ADC= CAD

106^{0} + CAD + CAD = 180^{0}[From (ii)]

2CAD = 74^{0}

CAD = 37^{0} =ADC………..(iii)

Now,

BAD = 110^{0}[Given]

BAC + CAD = 110^{0}

BAC + 37^{0} = 110^{0}

BAC = 73^{0} ……..(iv)

In ABC,

B + BAC+ ACB = 180^{0}

B + 73^{0} + 74^{0} = 180^{0}[From (i) and (iv)]

B + 147^{0} = 180^{0}

B = 33^{0} ………..(v)

(i) ADC + ADB = 180^{0}[BDC is a straight line]

ADC = 90^{0}[Given]

90^{0} + ADB = 180^{0}

ADB = 90^{0} …………(i)

In ADB,

ADB = 90^{0}[From (i)]

B + BAD = 90^{0}

Therefore, B and BAD are both acute, that is less than 90^{0}.

AB > BD …….(ii)[Side opposite 90^{0 }angle is greater than

side opposite acute angle]

(ii) In ADC,

ADB = 90^{0}

C + DAC = 90^{0}

Therefore, C and DAC are both acute, that is less than 90^{0}.

AC > CD ……..(iii)[Side opposite 90^{0 }angle is greater than

side opposite acute angle]

Adding (ii) and (iii)

AB + AC > BD + CD

AB + AC > BC

Const: Join AC and BD.

(i) In ABC,

AB + BC > AC….(i)[Sum of two sides is greater than the

third side]

In ACD,

AC + CD > DA….(ii)[ Sum of two sides is greater than the

third side]

Adding (i) and (ii)

AB + BC + AC + CD > AC + DA

AB + BC + CD > AC + DA - AC

AB + BC + CD > DA …….(iii)

(ii)In ACD,

CD + DA > AC….(iv)[Sum of two sides is greater than the

third side]

Adding (i) and (iv)

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(iii) In ABD,

AB + DA > BD….(v)[Sum of two sides is greater than the

third side]

In BCD,

BC + CD > BD….(vi)[Sum of two sides is greater than the

third side]

Adding (v) and (vi)

AB + DA + BC + CD > BD + BD

AB + DA + BC + CD > 2BD

(i) In ABC,

AB = BC = CA[ABC is an equilateral triangle]

A = B = C

In ABP,

A = 60^{0}

ABP< 60^{0}

[Side opposite to greater side is greater]

(ii) In BPC,

C = 60^{0}

CBP< 60^{0}

[Side opposite to greater side is greater]

Let PBC = x and PCB = y

then,

BPC = 180^{0} - (x + y) ………(i)

Let ABP = a and ACP = b

then,

BAC = 180^{0} - (x + a) - (y + b)

BAC = 180^{0} - (x + y) - (a + b)

BAC =BPC - (a + b)

BPC = BAC + (a + b)

BPC > BAC

We know that exterior angle of a triangle is always greater than each of the interior opposite angles.

In ABD,

ADC > B ……..(i)

In ABC,

AB = AC

B = C …..(ii)

From (i) and (ii)

ADC > C

(i) In ADC,

ADC > C

AC > AD ………(iii) [side opposite to greater angle is greater]

(ii) In ABC,

AB = AC

AB > AD[ From (iii)]

Const: Join ED.

In AOB and AOD,

AB = AD[Given]

AO = AO[Common]

BAO = DAO[AO is bisector of A]

[SAS criterion]

Hence,

BO = OD………(i)[cpct]

AOB = AOD .……(ii)[cpct]

ABO = ADO ABD = ADB ………(iii)[cpct]

Now,

AOB = DOE[Vertically opposite angles]

AOD = BOE[Vertically opposite angles]

BOE = DOE ……(iv)[From (ii)]

(i) In BOE and DOE,

BO = CD[From (i)]

OE = OE[Common]

BOE = DOE[From (iv)]

[SAS criterion]

Hence, BE = DE[cpct]

(ii) In BCD,

ADB = C + CBD[Ext. angle = sum of opp. int. angles]

ADB > C

ABD > C[From (iii)]

In ABC,

AB > AC,

ABC < ACB

180^{0} -ABC > 180^{0} -ACB

Since AB is the largest side and BC is the smallest side of the triangle ABC

In the quad. ABCD,

Since AB is the longest side and DC is the shortest side.

(i) 1 > 2[AB > BC]

7 > 4[AD > DC]

1 + 7 > 2 + 4

C > A

(ii) 5 > 6[AB > AD]

3 > 8[BC > CD]

5 + 3 > 6 + 8

D > B

In ADC,

ADB = 1 + C.............(i)

In ADB,

ADC = 2 + B.................(ii)

But AC > AB[Given]

B > C

Also given, 2 = 1[AD is bisector of A]

2 + B > 1 + C …….(iii)

From (i), (ii) and (iii)

ADC > ADB

We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at right angle.

Using Pythagoras theorem in AFB,

AB^{2} = AF^{2} + BF^{2}…………..(i)

In AFD,

AD^{2} = AF^{2} + DF^{2}…………..(ii)

We know ABC is isosceles triangle and AB = AC

AC^{2} = AF^{2} + BF^{2} ……..(iii)[ From (i)]

Subtracting (ii) from (iii)

AC^{2} - AD^{2} = AF^{2} + BF^{2} - AF^{2} - DF^{2}

AC^{2} - AD^{2} = BF^{2} - DF^{2}

Let 2DF = BF

AC^{2} - AD^{2} = (2DF)^{2} - DF^{2}

AC^{2} - AD^{2} = 4DF^{2} - DF^{2}

AC^{2} = AD^{2} + 3DF^{2}

AC^{2} > AD^{2}

AC > AD

Similarly, AE > AC and AE > AD.

The sum of any two sides of the triangle is always greater than the third side of the triangle.

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