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)]

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.

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 = 21 cm.

 

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,

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)

Solution 9

(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

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]

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 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,

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 14

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)]

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

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Solution 17

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 18

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 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,

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.

Solution 20

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

 

 

 

 

 

 

 

 

 

 

 

 

Solution 21