Selina Chapter 11 Inequalities ICSE Solutions Class 9 Maths

ICSE Solutions for Selina Concise Chapter 11 Inequalities Class 9 Maths

Exercise 11

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

Answer

In ΔABC, 
AB = AC [Given]
∴ ∠ACB =  ∠B [angles opposite to equal sides are equal]
 ∠B = 70°  [Given]
⇒ ∠ACB = 70° ...(i) 
Now, 
∠ACB + ∠ACD = 180° [BCD is a straight line]
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 110° ...(ii)
In ΔACD,
∠CAD + ∠ACD + ∠D = 180°
⇒ ∠CAD + 110° + ∠D = 180° [From (ii)]
⇒ ∠CAD + ∠D = 70°
But ∠D = 40°  [Given]
⇒ ∠CAD + 40° = 70°
⇒ ∠CAD = 30° ...(iii)
In ΔACD,
∠ACD = 110° [From (ii)]
∠CAD = 30°  [From (iii)]
∠D = 40° [Given]
∴∠D > ∠CAD
⇒ AC > CD [Greater angle has greater side opposite to it]
Also,
AB = AC [Given]
Therefore, AB > CD.

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

Answer

In Δ PQR, 
QR = PR  [Given]
∠P = ∠Q  [angles opposite to equal sides are equal]
∠P = 36°  [Given]

⇒∠Q = 36°
In ΔPQR,
∠P + ∠Q +∠R = 180°
⇒ 36° + 36° + ∠R = 180°

⇒ ∠R + 72° = 180°

⇒ ∠R = 108°
Now, 
∠R = 108°
∠P = 36°
∠Q = 36°
Since ∠R is the greatest, therefore, PQ is the largest side.

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.

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

Answer

In ΔABC, 
AB = AC 
⇒ ∠ABC = ∠ACB (angles opposite to equal sides are equal)
⇒ ∠ABC = ∠ACB = 67°
⇒ ∠BAC = 180° - ∠ABC - ∠ACB (angle sum property)

⇒ ∠BAC = 180° -67° - 67° = 46° 

Since ∠BAC < ∠ABC, we have 
BC < AC ...(1) 
Now, ∠ACD = 180° - ∠ACB (linear pair)

⇒ ∠ACD = 180° - 67° = 113°

Thus, in ΔACD, 
∠CAD = 180° - ∠ACD - ∠ADC

⇒ ∠CAD = 180° - 113° - 33° = 34°

Since ∠ ADC < ∠CAD, we have 
AC < CD  ...(2)
From (1) and (2), we have 
BC < AC< CD

In ΔABC, 
∠BAC < ∠ABC 
⇒BC < AC  ...(1)
Now, ∠ACB = 180° - ∠ABC - ∠BAC
⇒ ∠ACB = 180° - 73° - 47° 
⇒ ∠ACB = 60°
Now, ∠ACD = 180° - ∠ACB
⇒ ∠ACD = 180° - 60° = 120°
Now, in ΔACD, 
∠ADC = 180° - ∠ACD - ∠CAD 
⇒ ∠ADC = 180° - 120° - 31° 
⇒ ∠ADC = 29° 
Since ∠ADC < ∠CAD, we have 
AC < CD  ...(2) 
From (1) and (2), we have 
BC < AC < CD

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

Answer

∠BAC = 180° - ∠BAD = 180° - 137° = 43°

∠ABC = 180° - ∠ABE = 180° - 106° = 74°

Thus, in ΔABC,
∠ACB = 180° - ∠BAC - ∠ABC

⇒ ∠ACB = 180° - 43° - 74° = 63°

Now, ∠ABC = ∠OBC + ∠ABO
⇒ ∠ABC = 2∠OBC  (OB is bisector of ∠ABC)
⇒ 74° = 2∠OBC

⇒ ∠OBC = 37°
Similarly,
∠ACB = ∠OCB + ∠ACO
⇒ ∠ACB = 2∠OCB  (OC is bisector of ∠ACB)
⇒ 63° = 2∠OCB

⇒ ∠OCB = 31.5°

Now in ΔBOC,
∠BOC = 180° - ∠OBC - ∠OCB

⇒ ∠BOC = 180° - 37° - 31.5°

⇒ ∠BOC = 111.5°
Since, ∠BOC > ∠OBC > ∠OCB, we have 
BC > OC > OB 

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

Answer

AD > AC  (given)
⇒ ∠C > ∠ADC ...(1)
Now, ∠ADC > ∠B + ∠BAC  (Exterior Angle Property)
⇒ ∠ADC > ∠B  ...(2)
From (1) and (2), we have 
∠C >∠ADC > ∠B
⇒ ∠C > ∠B 
⇒ AB > AC

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

Prove that:

(i) CF > AF

(ii) DC > DF

Answer

In ΔBEC, 
∠B + ∠BEC + ∠BCE = 180°
∠B = 65°  [Given]
∠BEC = 90°  [CE is perpendicular to AB]
⇒ 65° + 90° + ∠BCE  = 180°
⇒ ∠BCE = 180° - 155°
⇒ ∠BCE = 25° = ∠DCF ...(i)
In ΔCDF,
∠DCF + ∠FDC + ∠CFD = 180° 
∠DCF = 25°  [From (i)]
∠FDC = 90°  [AD is perpendicular to BC]
⇒ 25° + 90° + ∠CFD  = 180°
⇒ ∠CFD = 180° - 115°
⇒ ∠CFD = 65° ...(ii)
Now, ∠AFC + ∠CFD = 180° [AFD is a straight line]
⇒ ∠AFC + 65° = 180°
⇒ ∠AFC = 115° ...(iii)
In ΔACE,
∠ACE + ∠CEA + ∠BAC = 180°
∠BAC = 60°  [Given]
∠CEA = 90°  [CE is perpendicular to AB]
⇒ ∠ACE + 90° + 60° = 180°
⇒ ∠ACE = 180° - 150°
⇒ ∠ACE = 30° ...(iv)
In ΔAFC,
∠AFC + ∠ACF + ∠FAC = 180°
∠AFC = 115° [From (iii)]
∠ACF  = 30°  [From (iv)]
⇒ 115° + 30° + ∠FAC = 180°
⇒ ∠FAC = 180° - 145°
⇒ ∠FAC = 35° ...(v)
In ΔAFC,
∠FAC = 35° [From (v)]
∠ACF  =  30°  [From (iv)]
∴∠FAC > ∠ACF
⇒CF > AF
In ΔCDF,
∠DCF = 25°  [From (i)]
∠CFD = 65°  [From (ii)]
∴∠CFD > ∠DCF
⇒ DC > DF 

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

Prove that: BC > CD

Answer

∠ACB = 74° ...(i) [Given]

∠ACB + ∠ACD = 180°  [BCD is a straight line]

⇒ 74° + ∠ACD = 180°

⇒ ∠ACD = 106° ...(ii)

In ΔACD,
∠ACD + ∠ADC + ∠CAD = 180°
Given that AC = CD
⇒ ∠ADC = ∠CAD 
⇒ 106° + ∠CAD +∠CAD = 180° [From (ii)]

⇒ 2∠CAD = 74°
⇒ ∠CAD = 37° = ∠ADC ...(iii)

Now, 
∠BAD = 110°  [Given]

∠BAC + ∠CAD = 110° 

∠BAC + 37° = 110° 

∠BAC = 73° ...(iv)

In ΔABC,
∠B + ∠BAC + ∠ACB = 180° 

⇒ ∠B + 73° + 74° = 180°  [From (i) and (iv)]

⇒ ∠B + 147° = 180°

⇒ ∠B = 33° ...(v)

∴∠BAC > ∠B  [From (iv) and (v)]
⇒BC > AC 
But, 
AC = CD  [Given]
⇒ BC > CD

9. From the following figure; prove that:

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

Answer

(i) ∠ADC + ∠ADB = 180°  [BDC is a straight line]

∠ADC = 90°  [Given]

90° + ∠ADB = 180° 

∠ADB = 90° ...(i) 

In ΔADB,

∠ADB = 90° [From (i)]
∴∠B + ∠BAD = 90° 
Therefore, ∠B and ∠BAD are both acute, that is less than 90° . 
∴AB > BD ...(ii) [Side opposite 90° angle is greater than side opposite acute angle]

(ii) In ΔADC,
∠ADB = 90° 

∠C + ∠DAC = 90° 

Therefore, ∠C and ∠DAC are both acute, that is less that 90° . 

∴AC > CD  ...(iii)  [Side opposite 90° angle is greater than side opposite acute angle]

Adding (ii) and (iii) 
AB + AC > BD + CD
⇒ AB + AC > BC

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

Construction: Join AC and BD.

(i) In ΔABC,

AB + BC > A ...(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

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

(i) In ΔABC, 
AB = BC = CA  [ABC is an equilateral triangle]
∴ ∠A = ∠B = ∠C 
∴∠A = ∠B = ∠C = 180°/3

⇒ ∠A = ∠B = ∠C = 60° 

In ΔABP, 
∠A = 60°
∠ABP < 60° 

∴∠A > ∠ABP
⇒ BP > PA  [side opposite to greater side is greater]

(ii) In ΔBPC, 
∠C = 60°
∠CBP < 60° 

∴∠C > ∠CBP
⇒ BP > PC [Side opposite to greater side is greater]

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

∠BPC > ∠BAC.

Answer

Let ∠PBC = x and ∠PCB = y 
then, 
∠BPC = 180° - (x + y)  ...(i) 

Let ∠ABP = a and ∠ACP = b
then,
∠BAC = 180° - (x + a) - (y + b)

⇒ ∠BAC = 180° - (x + y) - (a + b)

⇒ ∠BAC = ∠BPC - (a + b)

⇒ ∠BPC =  ∠BAC + (a + b)
⇒ ∠BPC > ∠BAC

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)]

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

(i) BE = DE

(ii) ∠ABD > ∠C

Answer

Construction: Join ED,
In ΔAOB and ΔAOD, 
AB = AD [Given]
AO = AO [Common]
∠BAO = ∠DAO [AO is bisector of ∠A]
∴ΔAOB ≅ ΔAOD [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 ΔBOE and ΔDOE, 

BO = CD  [From (i)]
OE = OE  [Common]
∠BOE = ∠DOE  [From (iv)]
∴ΔBOE ≅ ΔDOE  [SAS criterion]
Hence, BE = DE  [cpct]

(ii) In ΔBCD,

∠ADB = ∠C + ∠CBD  [Ext. angle = sum of opp. int. angles]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C  [From (iii)]

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

In ΔABC, 
AB > AC,
⇒ ∠ABC < ∠ACB 
∴180° - ∠ABC > 180° - ∠ACB

⇒ (180° - ∠ABC)/2 > (180° - ∠ACB)/2

⇒ 90° - (1/2)(∠ABC) > 90° - (1/2)(∠ACB)

⇒ ∠CBP > ∠BCP  [BP is bisector of ∠CBD and CP is bisector of ∠BCE]
⇒ PC > PB  [side opposite to greater angle is greater]

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°.

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

Prove that:

(i) ∠C > ∠A

(ii) ∠D > ∠B

Answer

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 > ∠B

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

In ΔADC,
∠ADB = ∠1 + ∠C ...(i)
In ΔADB, 
∠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 > ∠ADB 

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,

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

or, AC² – AD² = 4DF² – DF²

AC² = AD² + 3DF²

Hence, AC² > AD²

AC > AD

Similarly, AE > AC and AE > AD.

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.
In ΔCEB, 
CE + EB > BC
⇒ DE + EB > BC [CE = DE]
⇒ DB > BC ...(i)
In ΔADB, 
AD + AB > BD
⇒ AD + AB > BD > BC [from (i)]
⇒ AD + AB > BC

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

Answer

Given that, AB > AC 
⇒ ∠C > ∠B  ...(i) 
Also in ΔADC 
∠ADB = ∠DAC + ∠C  [Exterior angle]
⇒ ∠ADB > ∠C
⇒ ∠ADB > ∠C > ∠B  [From (i)]
⇒ ∠ADB > ∠B
⇒ AB > AD