# SELINA Solutions for Class 9 Maths Chapter 11 - Inequalities

## Chapter 11 - Inequalities Exercise Ex. 11

From the following figure, prove that: AB > CD.

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 a triangle PQR; QR = PR and P = 36^{o}. Which is the largest side of the triangle?

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.

If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.

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 each of the following figures, write BC, AC and CD in ascending order of their lengths.

Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.

D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.

In the following figure, BAC = 60^{o} and ABC = 65^{o}.

Prove that:

(i) CF > AF

(ii) DC > DF

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)]

In the following figure; AC = CD; BAD = 110^{o} and ACB = 74^{o}.

Prove that: BC > CD.

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)

From the following figure; prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

(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

In a quadrilateral ABCD; prove that:

(i) AB+ BC + CD > DA

(ii) AB + BC + CD + DA > 2AC

(iii) AB + BC + CD + DA > 2BD

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

In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that:

(i) BP > PA

(ii) BP > PC

(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]

P is any point inside the triangle ABC. Prove that:

BPC > BAC.

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

Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.

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)]

In the following diagram; AD = AB and AE bisects angle A. Prove that:

(i) BE = DE

(ii) ABD > C

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)]

The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB > AC, then PC > PB.

In ABC,

AB > AC,

ABC < ACB

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

In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.

Write the angles x^{o}, y^{o} and z^{o} in ascending order of their values.

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

In quadrilateral ABCD, side AB is the longest and side DC is the shortest.

Prove that:

(i) C > A

(ii) D > B.

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 triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that: ADC is greater than ADB.

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

In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that:

(i) AC > AD

(ii) AE > AC

(iii) AE > AD

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.

Given: ED = EC

Prove: AB + AD > BC.

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

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In triangle ABC, AB > AC and D is a point in side BC. Show that: AB > AD.

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