Class 9 SELINA Solutions Maths Chapter 11 - Inequalities

In ABC,

AB = AC[Given]

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

B = 70⁰[Given]

ACB = 70⁰ ……….(i)

Now,

ACB +ACD = 180⁰[ BCD is a straight line]

70⁰ + ACD = 180⁰

ACD = 110⁰ …………(ii)

In ACD,

CAD + ACD + D = 180⁰

CAD + 110⁰ + D = 180⁰ [From (ii)]

CAD + D = 70⁰

But D = 40⁰ [Given]

CAD + 40⁰= 70⁰

CAD = 30⁰ ………………(iii)

In ACD,

ACD = 110⁰[From (ii)]

CAD = 30⁰[From (iii)]

D = 40⁰ [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 = 36⁰[Given]

Q = 36⁰

In PQR,

P + Q + R = 180⁰

36⁰ + 36⁰ + R = 180⁰

R + 72⁰ = 180⁰

R = 108⁰

Now,

R = 108⁰

P = 36⁰

Q = 36⁰

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= 21cm.

In BEC,

B + BEC + BCE = 180⁰

B = 65⁰ [Given]

BEC = 90⁰[CE is perpendicular to AB]

65⁰ + 90⁰ + BCE = 180⁰

BCE = 180⁰ - 155⁰

BCE = 25⁰ = DCF …………(i)

In CDF,

DCF + FDC + CFD = 180⁰

DCF = 25⁰ [From (i)]

FDC = 90⁰[AD is perpendicular to BC]

25⁰ + 90⁰ + CFD = 180⁰

CFD = 180⁰ - 115⁰

CFD = 65⁰ …………(ii)

Now, AFC + CFD = 180⁰[AFD is a straight line]

AFC + 65⁰ = 180⁰

AFC = 115⁰ ………(iii)

In ACE,

ACE + CEA + BAC = 180⁰

BAC = 60⁰ [Given]

CEA = 90⁰[CE is perpendicular to AB]

ACE + 90⁰ + 60⁰ = 180⁰

ACE = 180⁰ - 150⁰

ACE = 30⁰ …………(iv)

In AFC,

AFC + ACF + FAC = 180⁰

AFC = 115⁰ [From (iii)]

ACF = 30⁰[From (iv)]

115⁰ + 30⁰ + FAC = 180⁰

FAC = 180⁰ - 145⁰

FAC = 35⁰ …………(v)

In AFC,

FAC = 35⁰[From (v)]

ACF = 30⁰[From (iv)]

In CDF,

DCF = 25⁰[From (i)]

CFD = 65⁰[From (ii)]

ACB = 74⁰ …..(i)[Given]

ACB + ACD = 180⁰[BCD is a straight line]

74⁰ + ACD = 180⁰

ACD = 106⁰ ……..(ii)

In ACD,

ACD + ADC+ CAD = 180⁰

Given that AC = CD

ADC= CAD

106⁰ + CAD + CAD = 180⁰[From (ii)]

2CAD = 74⁰

CAD = 37⁰ =ADC………..(iii)

Now,

BAD = 110⁰[Given]

BAC + CAD = 110⁰

BAC + 37⁰ = 110⁰

BAC = 73⁰ ……..(iv)

In ABC,

B + BAC+ ACB = 180⁰

B + 73⁰ + 74⁰ = 180⁰[From (i) and (iv)]

B + 147⁰ = 180⁰

B = 33⁰ ………..(v)

(i) ADC + ADB = 180⁰[BDC is a straight line]

ADC = 90⁰[Given]

90⁰ + ADB = 180⁰

ADB = 90⁰ …………(i)

In ADB,

ADB = 90⁰[From (i)]

B + BAD = 90⁰

Therefore, B and BAD are both acute, that is less than 90⁰.

AB > BD …….(ii)[Side opposite 90⁰ angle is greater than

side opposite acute angle]

(ii) In ADC,

ADB = 90⁰

C + DAC = 90⁰

Therefore, C and DAC are both acute, that is less than 90⁰.

AC > CD ……..(iii)[Side opposite 90⁰ 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 = 60⁰

ABP< 60⁰

[Side opposite to greater side is greater]

(ii) In BPC,

C = 60⁰

CBP< 60⁰

[Side opposite to greater side is greater]

Let PBC = x and PCB = y

then,

BPC = 180⁰ - (x + y) ………(i)

Let ABP = a and ACP = b

then,

BAC = 180⁰ - (x + a) - (y + b)

BAC = 180⁰ - (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

180⁰ -ABC > 180⁰ -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,

AB² = AF² + BF²…………..(i)

In AFD,

AD² = AF² + DF²…………..(ii)

We know ABC is isosceles triangle and AB = AC

AC² = AF² + BF² ……..(iii)[ From (i)]

Subtracting (ii) from (iii)

AC² - AD² = AF² + BF² - AF² - DF²

AC² - AD² = BF² - DF²

Let 2DF = BF

AC² - AD² = (2DF)² - DF²

AC² - AD² = 4DF² - DF²

AC² = AD² + 3DF²

AC² > AD²

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.