# Inequalities Concise Class-9th Selina ICSE Maths Solution

**Inequalities Concise Class-9th** Selina ICSE Maths Solution Chapter-11. We provide step by step Solutions of Exercise / lesson-11 **Inequalities ** for **ICSE** **Class-9 Concise** Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-11 A to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics .

**Inequalities Concise Class-9th** Selina ICSE Maths Solution Chapter-11

### Exercise – 11

Question 1

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

………………….

Answer

#### Question 2

In a triangle PQR; QR = PR and ∠P = 36^{o}. Which is the largest side of the triangle?

Answer

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.

Answer

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.

………………..

Answer

#### Question 5

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

………………………

Answer

#### Question 6

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

Answer

#### Question 7

n the following figure, ∠BAC = 60^{o} and ∠ABC = 65^{o}.

Prove that:

(i) CF > AF

(ii) DC > DF

Answer

#### Question 8

In the following figure; AC = CD; ∠BAD = 110^{o} and ∠ACB = 74^{o}.

Prove that: BC > CD

Answer

#### Question 9

From the following figure; prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

Answer

#### 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

Answer

Const: Join AC and BD.

(i) In ΔABC,

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

third side]

In ΔACD,

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 ΔACD,

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 ΔABD,

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

third side]

In ΔBCD,

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

………………..

Answer

#### Question 12

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

∠BPC > ∠BAC.

Answer

#### 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.

Answer

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

In ΔABD,

∠ADC > ∠B ……..(i)

In ABC,

AB = AC

∠B = ∠C …..(ii)

From (i) and (ii)

∠ADC > ∠ C

(i) In ΔADC,

∠ADC > ∠C

AC > AD ………(iii) [side opposite to greater angle is greater]

(ii) In ΔABC,

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

…………………

Answer

#### 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.

Answer

#### Question 16

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

Write the angles x^{o}, y^{o} and z^{o} in ascending order of their values.

Answer

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

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

AB > AC > BC

⇒ 180° – z° > 180° – y° > 180° – x°

⇒ – z° > -y° > – x°

⇒ z° > y° > x°.

#### Question 17

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

Prove that:

(i) ……………

(ii)………..

Answer

#### 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: ∠ADC is greater than ∠ADB.

Answer

#### 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

Answer

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,

AB^{2} = AF^{2} + BF^{2}…………..(i)

In AFD,

AD^{2} = AF^{2} + DF^{2}…………..(ii)

We know ABC is isosceles triangle and AB = AC

AC^{2} = AF^{2} + BF^{2} ……..(iii)[ From (i)]

Subtracting (ii) from (iii)

AC^{2} – AD^{2} = AF^{2} + BF^{2} – AF^{2} – DF^{2}

AC^{2} – AD^{2} = BF^{2} – DF^{2}

Let 2DF = BF

AC^{2} – AD^{2} = (2DF)^{2} – DF^{2}

or AC^{2} – AD^{2} = 4DF^{2} – DF^{2}

AC^{2} = AD^{2} + 3DF^{2}

hence AC^{2} > AD^{2}

AC > AD

Similarly, AE > AC and AE > AD.

#### Question 20

Given: ED = EC

Prove: AB + AD > BC.

…………

Answer

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

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