ICSE Class 9 Maths Chapter 11 Inequalities

11 Inequalities

11.1 Introduction

1. The sign > means, "is greater than" i.e. if 'a' is greater than 'b', we write: a > b.

2. The sign < means, "is less than" i.e. if 'a' is less than 'b', we write: a < b.

Theorem 3

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

Given: A triangle ABC in which AB > AC.

To Prove: \(\angle ACB > \angle B\)

Construction: From AB, cut AD = AC. Join C and D.

Proof:

Statement:

In \(\triangle ACD\):

1. AC = AD

2. \(\angle ACD = \angle ADC\)

In \(\triangle BDC\):

3. Ext. \(\angle ADC > \angle B\)

\(\therefore \angle ACD > \angle B\)

\(\therefore \angle ACB > \angle B\)

Reason:

By construction

Angles opposite to equal sides

Ext. angle of a \(\triangle\) is always greater than each of its interior opposite angles.

From 2 and 3.

Since, \(\angle ACD\) is a part of \(\angle ACB\), \(\therefore \angle ACB > \angle ACD > \angle B\).

Hence Proved.

Theorem 4

(Converse of theorem 3)

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

Given: A triangle ABC in which \(\angle CAB > \angle B\)

To Prove: BC > AC.

Construction: Draw \(\angle BAD = \angle B\).

Proof:

Statement:

In \(\triangle ABD\):

1. AD = BD

In \(\triangle ADC\):

2. AD + DC > AC

\(\therefore BD + DC > AC\)

\(\therefore BC > AC\)

Reason:

Sides opposite to equal angles.

Sum of any two sides of a \(\triangle\) is always greater than the third side.

Since, AD = BD

BD + DC = BC

Hence Proved.

Theorem 5

Of all the lines, that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest.

Given: A point O outside the line AB and OP perpendicular to AB.

To Prove: OP is the shortest of all the lines that can be drawn from O to AB.

Construction: Join O with any point Q in AB.

Proof:

Statement:

In right angled \(\triangle OPQ\):

1. \(\angle OPQ > \angle OQP\)

2. OQ > OP

Similarly, it can be shown that OP is smaller than any other line that can be drawn from O to AB.

\(\therefore\) OP is the shortest line drawn from O to AB.

Reason:

Right angle is the greatest angle in a right angled \(\triangle\).

Side opp. to greater angle is greater.

Hence Proved.

Corollary 1

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

For example:

In triangle ABC.

(i) AB + AC > BC,

(ii) AB + BC > AC and (iii) BC + AC > AB.

Corollary 2

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

For example:

In triangle ABC, given for corollary 1, if AB is the largest side and AC is the smallest side; then:

(i) AB - AC < BC

(ii) AB - BC < AC and (iii) BC - AC < AB.

Problem 1

In the adjoining figure, AD bisects \(\angle A\). Arrange AB, BD and DC in the descending order of their lengths.

Solution:

\(\angle BAC = 180° - (60° + 40°) = 80°\)

Since, AD bisects \(\angle A\)

\(\therefore \angle BAD = \angle CAD = \frac{80°}{2} = 40°\)

\(\angle ADB = 180° - (60° + 40°) = 80°\)

In \(\triangle ABD\), AB > AD > BD

[Side opposite to greater angle is greater]

In \(\triangle ADC\), AD = DC

[Sides opposite to equal angles are equal]

\(\therefore\) AB > DC > BD

Ans.

[From I and II]

Problem 2

In the given figure, AC is perpendicular to line PQ and BC = CD. Show that AE is greater than AB.

Solution:

In \(\triangle ABC\) and \(\triangle ADC\)

BC = CD

[Given]

\(\angle ACB = \angle ACD\)

[Each 90°]

and, AC = AC

[Common]

\(\therefore \triangle ABC \cong \triangle ADC\)

[By S.A.S.]

\(\Rightarrow AB = AD\)

[By C.P.C.T.C.]

and so, \(\angle a = \angle b\)

[Angles opp. to equal sides]

In \(\triangle ADE\), ext. \(\angle b = \angle c + \angle DAE\)

[Ext. \(\angle\) = sum of int. opp. \(\angle\)s]

\(\Rightarrow \angle b > \angle c\)

\(\Rightarrow \angle a > \angle c\)

[\(\therefore \angle a = \angle b\)]

In \(\triangle ABE\), \(\angle a > \angle c\)

\(\Rightarrow\) AE > AB

[In a \(\triangle\), side opp. to greater angle is greater]

Hence Proved.

Problem 3

In the given figure, AB = AC. Prove that AF is greater than AE.

Solution:

Since, AB = AC \(\Rightarrow \angle B = \angle C\)

[Angles opp. to equal sides are equal]

In \(\triangle FCD\), ext. \(\angle C = \angle D + \angle a\)

\(\Rightarrow \angle C > \angle a\)

But, \(\angle a = \angle b\)

\(\therefore \angle C > \angle b\)

.....I

In \(\triangle EBD\), ext. \(\angle d = \angle B + \angle D\)

\(\Rightarrow \angle d > \angle B\)

\(\Rightarrow \angle d > \angle C\)

.....II [\(\because \angle B = \angle C\)]

\(\therefore \angle d > \angle C\) and \(\angle C > \angle b \Rightarrow \angle d > \angle b\)

In \(\triangle AEF\), \(\angle d > \angle b \Rightarrow\) AF > AE

Hence Proved.

Problem 4

In the figure, given alongside, AD bisects angle BAC. Prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

Solution:

Given, AD bisects angle BAC \(\Rightarrow \angle BAD = \angle CAD\)

(i) In triangle ADC,

ext. \(\angle ADB = \angle CAD + \angle C\)

\(\Rightarrow \angle ADB = \angle BAD + \angle C\)

[\(\therefore \angle CAD = \angle BAD\)]

\(\Rightarrow \angle ADB > \angle BAD\)

In triangle ABD, \(\angle ADB > \angle BAD\)

\(\Rightarrow\) AB > BD

[Greater angle has greater side opposite to it.]

Hence Proved.

(ii) In \(\triangle ABD\),

ext. \(\angle ADC = \angle BAD + \angle B\)

\(\Rightarrow \angle ADC = \angle CAD + \angle B\)

[\(\therefore \angle BAD = \angle CAD\)]

\(\Rightarrow \angle ADC > \angle CAD\)

In \(\triangle ADC\), \(\angle ADC > \angle CAD \Rightarrow\) AC > CD

Hence Proved.

(iii) Since, AB > BD and AC > CD

\(\therefore AB + AC > BD + CD \Rightarrow\) AB + AC > BC

Hence Proved.

Problem 5

In quadrilateral ABCD; AB is the shortest side and DC is the longest side. Prove that:

(i) \(\angle B > \angle D\)

(ii) \(\angle A > \angle C\)

Solution:

Join B and D

In \(\triangle ABD\), AD > AB

[Given, AB is the shortest]

\(\therefore \angle a > \angle c\)

[Angle opposite to the greater side is greater]

In \(\triangle BCD\), CD > BC

[Given, CD is the longest side]

\(\therefore \angle b > \angle d\)

[Angle opposite to greater side is greater]

\(\therefore \angle a + \angle b > \angle c + \angle d\)

[Adding I and II]

\(\Rightarrow\) \(\angle B > \angle D\)

Hence Proved.

Similarly, by joining A and C, it can be proved that \(\angle A > \angle C\).

Hence Proved.

Problem 6

AD is a median of triangle ABC. Prove that: AB + AC > 2AD

Solution:

According to the given statement, the figure will be as drawn alongside in which AD is a median i.e. BD = CD.

Produce AD upto a point E such that AD = DE i.e. AE = 2AD.

Also join C and E.

Since, the sum of any two sides of a triangle is greater than the third side, therefore in triangle ACE

AC + CE > AE

i.e. AC + CE > 2AD

..... I

In \(\triangle ADB\) and \(\triangle CDE\)

BD = CD

[Given]

AD = DE

[By construction]

and, \(\angle ADB = \angle CDE\)

[Vertically opposite angles]

\(\therefore \triangle ADB \cong \triangle EDC\)

[By S.A.S.]

\(\Rightarrow AB = CE\)

[C.P.C.T.C.]

Substituting CB = AB in equation I, we get:

AC + AB > 2AD

i.e. AB + AC > 2AD

Hence Proved.

As proved above, if AD is a median of the triangle ABC;

AB + AC > 2AD

Similarly, if BE and CF are also the medians:

AB + BC > 2BE

and, BC + AC > 2CF

Adding these results, we get:

2AB + 2BC + 2AC > 2AD + 2BE + 2CF

\(\Rightarrow AB + BC + AC > AD + BE + CF\)

i.e. the perimeter of a triangle is greater than the sum of the lengths of its medians.

Problem 7

P is any point in the interior of a triangle ABC. Prove that: PA + PB < AC + BC

Solution:

According to the given statement, the figure will be as shown alongside.

Produce BP to meet AC at point M.

Since, the sum of any two sides of a triangle is greater than its third side.

\(\therefore\) In \(\triangle BCM\), BC + CM > BM

......... I

and, in \(\triangle APM\), AM + PM > AP

......... II

Adding I and II, we get:

BC + CM + AM + PM > BM + AP

\(\Rightarrow BC + (CM + AM) > BM - PM + AP\)

\(\Rightarrow BC + AC > PB + PA\)

i.e. PB + PA < BC + AC

Hence Proved.

Teacher's Note

Understanding inequalities in triangles helps explain why the shortest distance between two points is a straight line - think of walking along three sides of a triangle versus cutting directly across!

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