# Class 9 SELINA Solutions Maths Chapter 11 - Inequalities

## Inequalities Exercise Ex. 11

### Solution 1(a)

Correct option: (ii) BC > AB

In DABC,

If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

### Solution 1(b)

Correct option: (iii) BD > DC

BD = AD [Given]

In ΔABD,

If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

So, in ΔADC,

### Solution 1(c)

Correct option: (iv) BD > CD

AB = AC [Given]

### Solution 1(d)

Correct option: (ii) BD < AB

In ΔABC,

Given,

And,

Now, in ΔABD,

If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

Here,

### Solution 1(e)

Correct option: (ii) AC > AB

If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

In ΔABC,

### Solution 1(f)

Correct option: (i) AB + BC + CD + DA > AC + BD

Consider quadrilateral ABCD as follows:

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

AB + BC > AC (I)

BC + CD > BD (II)

CD + DA > AC (III)

DA + AB > BD (IV)

Adding equations (I), (II), (III), (IV),

2AB + 2BC + 2CD + 2DA > 2AC + 2BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD

### Solution 2

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.

### Solution 3

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.

### Solution 4

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.

### Solution 5

### Solution 6

### Solution 7

### Solution 8

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)]

### Solution 9

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)

### Solution 10

(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

### Solution 11

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

### Solution 12

(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 angle is greater]

(ii) In BPC,

C = 60^{0}

CBP< 60^{0}

[Side opposite to greater angle is greater]

### Solution 13

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

### Solution 14

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)]

### Solution 15

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)]

## Inequalities Exercise Test Yourself

### Solution 1

AB = AC (Given) (I)

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]

⇒ ∠ABD = ∠ACD (II)

In DABD, exterior ∠ADC = ∠ABD + ∠BAD

⇒ ∠ADC > ∠ABD

⇒ ∠ADC > ∠ACD [From (II)]

Now, in any triangle, the side opposite to the largest angle is the largest.

⇒ AC > AD

⇒ AB > AD [From (I)]

### Solution 2

AD = AC (Given)

⇒ ΔADC is an isosceles triangle.

⇒ ∠ADC = ∠ACD = acute angle

Now, ∠ADC + ∠ADB =
180^{o} [Linear pair]

⇒ ∠ADB is an obtuse angle.

Therefore, in ΔABD, ∠ADB is the largest angle.

Now, in any triangle, the side opposite to the largest angle is the largest.

⇒ AB > AD

### Solution 3

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

In ΔPQS,

PQ + QS >PS (I)

In ΔPSR,

SR + RP >PS (II)

Adding equations (I) and (II),

PQ + QS + SR + RP > PS + PS

⇒ PQ + QR + RP >2PS (proved)

### Solution 4

In ABC,

AB > AC,

ABC < ACB

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

### Solution 5

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

### Solution 6

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

### Solution 7

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

### Solution 8

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.

### Solution 9

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

### Solution 10