# SELINA Solutions for Class 9 Maths Chapter 11 - Inequalities

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## Chapter 11 - Inequalities Exercise Ex. 11

Solution 1

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)]

But D = 400 [Given]

In ACD,

ACD = 1100[From (ii)]

D = 400 [Given]

[Greater angle has greater side opposite to it]

Also,

AB = AC[Given]

Therefore, AB > CD.

Solution 2

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.

Solution 3

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.

Solution 4

Solution 5

Solution 6

Solution 7

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)]

Solution 8

ACB = 740 …..(i)[Given]

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

740 + ACD = 1800

ACD = 1060 ……..(ii)

In ACD,

Given that AC = CD

Now,

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)

Solution 9

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]

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]

AB + AC > BD + CD

AB + AC > BC

Solution 10

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]

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]

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]

AB + DA + BC + CD > BD + BD

AB + DA + BC + CD > 2BD

Solution 11

(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]

Solution 12

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

Solution 13

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

In ABD,

In ABC,

AB = AC

B = C …..(ii)

From (i) and (ii)

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

(ii) In ABC,

AB = AC

Solution 14

Const: Join ED.

In AOB and AOD,

AO = AO[Common]

BAO = DAO[AO is bisector of A]

[SAS criterion]

Hence,

BO = OD………(i)[cpct]

AOB = AOD .……(ii)[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]

ABD > C[From (iii)]

Solution 15

In ABC,

AB > AC,

ABC < ACB

1800 -ABC > 1800 -ACB

Solution 16

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

Solution 17

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

(i) 1 > 2[AB > BC]

1 + 7 > 2 + 4

C > A

(ii) 5 > 6[AB > AD]

3 > 8[BC > CD]

5 + 3 > 6 + 8

D > B

Solution 18

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)

Solution 19

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,

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

Similarly, AE > AC and AE > AD.

Solution 20

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

Solution 21

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