SELINA Solutions for Class 9 Maths Chapter 11 - Inequalities

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Chapter 11 - Inequalities Exercise Ex. 11

Question 1

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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 1

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

AB = AC[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB[angles opposite to equal sides are equal]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB = 700[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB = 700 ……….(i)

Now,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB +Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1800[ BCD is a straight line]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities700 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1100 …………(ii)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD + 1100 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD = 1800 [From (ii)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD = 700

But Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD = 400 [Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD + 400= 700

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 300 ………………(iii)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1100[From (ii)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 300[From (iii)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD = 400 [Given]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

[Greater angle has greater side opposite to it]

Also,

AB = AC[Given]

Therefore, AB > CD.

Question 2

In a triangle PQR; QR = PR and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesP = 36o. Which is the largest side of the triangle?

Solution 2

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesPQR,

QR = PR[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesP = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesQ[angles opposite to equal sides are equal]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesP = 360[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesQ = 360

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesPQR,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesP + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesQ + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities360 + 360 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR + 720 = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR = 1080

Now,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR = 1080

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesP = 360

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesQ = 360

Since Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesR 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 < 13Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities+Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities8 =Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities21 cm.

 

The difference between any two sides of the triangle is always less than the third side of the triangle.


Third side > 13Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities-Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities8 =Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities5 cm.



Therefore, the length of the third side is between 5 cm and 9 cm, respectively.


The value of a =Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities5 cm and bSelina Solutions Icse Class 9 Mathematics Chapter - Inequalities= 21Selina Solutions Icse Class 9 Mathematics Chapter - Inequalitiescm.

 

Question 4

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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 4

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 5

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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

Solution 5

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 6

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

Solution 6

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 7

In the following figure, Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 60o and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC = 65o.

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Prove that:

(i) CF > AF

(ii) DC > DF

Solution 7

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBEC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBEC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCE = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB = 650 [Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBEC = 900[CE is perpendicular to AB]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities650 + 900 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCE = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCE = 1800 - 1550

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCE = 250 = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDCF …………(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCDF,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDCF + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFDC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDCF = 250 [From (i)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFDC = 900[AD is perpendicular to BC]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities250 + 900 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 1800 - 1150

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 650 …………(ii)

Now, Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 1800[AFD is a straight line]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC + 650 = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC = 1150 ………(iii)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACE,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACE + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCEA + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 600 [Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCEA = 900[CE is perpendicular to AB]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACE + 900 + 600 = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACE = 1800 - 1500

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACE = 300 …………(iv)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACF + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFAC = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC = 1150 [From (iii)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACF = 300[From (iv)]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1150 + 300 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFAC = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFAC = 1800 - 1450

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFAC = 350 …………(v)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesFAC = 350[From (v)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACF = 300[From (iv)]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCDF,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDCF = 250[From (i)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCFD = 650[From (ii)]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 8

In the following figure; AC = CD; Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAD = 110o and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB = 74o.

Prove that: BC > CD.

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 8

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB = 740 …..(i)[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1800[BCD is a straight line]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities740 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD = 1060 ……..(ii)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC+ Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 1800

Given that AC = CD

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC= Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1060 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 1800[From (ii)]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities2Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 740

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 370 =Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC………..(iii)

Now,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAD = 1100[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCAD = 1100

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC + 370 = 1100

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 730 ……..(iv)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC+ Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB + 730 + 740 = 1800[From (i) and (iv)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB + 1470 = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB = 330 ………..(v)

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 9

From the following figure; prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 9

(i) Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = 1800[BDC is a straight line]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC = 900[Given]

900 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = 1800

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = 900 …………(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = 900[From (i)]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAD = 900

Therefore, Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAD are both acute, that is less than 900.

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAB > BD …….(ii)[Side opposite 900 angle is greater than

side opposite acute angle]

(ii) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = 900

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDAC = 900

Therefore, Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDAC are both acute, that is less than 900.

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAC > CD ……..(iii)[Side opposite 900 angle is greater than

side opposite acute angle]

Adding (ii) and (iii)

AB + AC > BD + CD

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAB + AC > BC

Question 10

In a quadrilateral ABCD; prove that:

(i) AB+ BC + CD > DA

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

(iii) AB + BC + CD + DA > 2BD

Solution 10

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Const: Join AC and BD.

(i) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

AB + BC > AC….(i)[Sum of two sides is greater than the

third side]

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD,

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 Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACD,

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 Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABD,

AB + DA > BD….(v)[Sum of two sides is greater than the

third side]

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCD,

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

Question 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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 11

(i) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

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

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABP,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA = 600

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABP< 600

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

[Side opposite to greater side is greater]

(ii) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC = 600

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCBP< 600

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

[Side opposite to greater side is greater]

Question 12

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

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC.

Solution 12

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Let Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesPBC = x and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesPCB = y

then,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC = 1800 - (x + y) ………(i)

Let Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABP = a and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACP = b

then,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 1800 - (x + a) - (y + b)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC = 1800 - (x + y) - (a + b)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC =Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC - (a + b)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC + (a + b)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBPC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAC

Question 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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

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

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesIn Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABD,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB ……..(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

AB = AC

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC …..(ii)

From (i) and (ii)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

(i) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAC > AD ………(iii) [side opposite to greater angle is greater]

(ii) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

AB = AC

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAB > AD[ From (iii)]

Question 14

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

(i) BE = DE

(ii) Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABD > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Solution 14

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Const: Join ED.

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOB and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOD,

AB = AD[Given]

AO = AO[Common]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBAO = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDAO[AO is bisector of Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities[SAS criterion]

Hence,

BO = OD………(i)[cpct]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOD .……(ii)[cpct]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABO = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADO Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABD = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB ………(iii)[cpct]

Now,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDOE[Vertically opposite angles]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAOD = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBOE[Vertically opposite angles]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBOE = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDOE ……(iv)[From (ii)]

 

(i) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBOE and Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDOE,

BO = CD[From (i)]

OE = OE[Common]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBOE = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesDOE[From (iv)]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities[SAS criterion]

Hence, BE = DE[cpct]

 

(ii) In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesBCD,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesCBD[Ext. angle = sum of opp. int. angles]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABD > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC[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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC,

AB > AC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC < Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1800 -Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesABC > 1800 -Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesACB

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 16

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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

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

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

Question 17

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

Prove that:

(i) Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA

(ii) Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB.

Solution 17

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In the quad. ABCD,

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

(i) Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities2[AB > BC]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities7 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities4[AD > DC]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - Inequalities1 + Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities7 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities2 + Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities4

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA

(ii) Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities5 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities6[AB > AD]

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities3 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities8[BC > CD]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - Inequalities5 + Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities3 > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities6 + Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities8

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesD > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB

Question 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: Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC is greater than Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB.

Solution 18

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB = Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC.............(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB,

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC = Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities2 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB.................(ii)

But AC > AB[Given]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC

Also given, Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities2 = Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1[AD is bisector of Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesA]

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - Inequalities2 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesB > Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities1 + Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesC …….(iii)

From (i), (ii) and (iii)

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesSelina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADC > Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesADB

Question 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 > AD

Solution 19

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

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 Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFB,

AB2 = AF2 + BF2…………..(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAFD,

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

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAC2 > AD2

Selina Solutions Icse Class 9 Mathematics Chapter - InequalitiesAC > AD

Similarly, AE > AC and AE > AD.

Question 20

Given: ED = EC

 

 

Prove: AB + AD > BC.

 

 

 Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities 

 

Solution 20

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

 

 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

 

 

 

 

 

 

 

 

 

 

Question 21

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

 

Solution 21

 

 

 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities

 

 

Selina Solutions Icse Class 9 Mathematics Chapter - Inequalities