ICSE Class 9 Inequalities Solution New Pattern

ICSE Class 9 Inequalities Solution New Pattern By Clarify Knowledge

ICSE Class 9 Inequalities Solution New Pattern 2022

Chapter 11 - Inequalities Exercise Ex. 11

Question 1

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

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 = 180[From (ii)]

CAD + D = 700

But D = 40[Given]

CAD + 400= 700

CAD = 300 ………………(iii)

In ACD,

ACD = 1100[From (ii)]

CAD = 300[From (iii)]

D = 40[Given]

[Greater angle has greater side opposite to it]

Also,

AB = AC[Given]

Therefore, AB > CD.Question 2

In a triangle PQR; QR = PR and P = 36o. Which is the largest side of the triangle?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.Question 3

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.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.Question 4

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

Solution 4

Question 5

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

Solution 5

Question 6

D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.Solution 6

Question 7

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

Prove that:

(i) CF > AF

(ii) DC > DFSolution 7

In BEC,

B + BEC + BCE = 1800

B = 65[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 = 25[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 = 60[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 = 115[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)]

Question 8

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

Prove that: BC > CD.

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)

Question 9

From the following figure; prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

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 90angle 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 90angle is greater than

side opposite acute angle]

Adding (ii) and (iii)

AB + AC > BD + CD

AB + AC > BCQuestion 10

In a quadrilateral ABCD; prove that:

(i) AB+ BC + CD > DA

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

(iii) AB + BC + CD + DA > 2BDSolution 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 > 2BDQuestion 11

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

(i) BP > PA

(ii) BP > PC

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]Question 12

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

BPC > BAC.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 > BACQuestion 13

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.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)]Question 14

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

(i) BE = DE

(ii) ABD > C

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)]Question 15

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.Solution 15

In ABC,

AB > AC,

ABC < ACB

1800 -ABC > 1800 -ACB

Question 16

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

Write the angles xo, yo and zo in ascending order of their values.Solution 16

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

Question 17

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

Prove that:

(i) C > A

(ii) D > B.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 > BQuestion 18

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.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 > ADBQuestion 19

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 > ADSolution 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.Question 20

Given: ED = EC

Prove: AB + AD > BC.

Solution 20

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

Question 21

In triangle ABC, AB > AC and D is a point in side BC. Show that: AB > AD.Solution 21

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