Request a call back

# 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

In ΔABD,

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

### 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

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 = 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 3

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 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= 21cm.

### Solution 8

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 9

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 10

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 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]

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 12

(i) In ABC,

AB = BC = CA[ABC is an equilateral triangle]

A = B = C

In ABP,

A = 600

ABP< 600

[Side opposite to greater angle is greater]

(ii) In BPC,

C = 600

CBP< 600

[Side opposite to greater angle is greater]

### Solution 13

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 14

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 15

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)]

## Inequalities Exercise Test Yourself

### Solution 1

AB = AC (Given) (I)

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

⇒ ∠ABD = ∠ACD (II)

⇒ ∠ADC > ∠ACD [From (II)]

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

⇒ AB > AD [From (I)]

### Solution 2

⇒ ΔADC is an isosceles triangle.

⇒ ∠ADC = ∠ACD = acute angle

⇒ ∠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.

### 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)

PQ + QS + SR + RP > PS + PS

⇒ PQ + QR + RP >2PS (proved)

### Solution 4

In ABC,

AB > AC,

ABC < ACB

1800 -ABC > 1800 -ACB

### Solution 5

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

### Solution 6

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 7

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 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,

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 9

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