# RS Aggarwal Solutions for Class 9 Chapter 9: Congruence of Triangles and Inequalities in a Triangle Exercise 9B

## RS Aggarwal Solutions for Class 9 Maths Exercise 9B PDF

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## Access RS Aggarwal Solutions for Class 9 Chapter 9: Congruence of Triangles and Inequalities in a Triangle Exercise 9B

1. Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

(i) 5cm, 4cm, 9cm

(ii) 8cm, 7cm, 4cm

(iii) 10cm, 5cm, 6cm

(iv) 2.5cm, 5cm, 7cm

(v) 3cm, 4cm, 8cm

Solution:

(i) No. It is not possible to construct a triangle with lengths of its sides 5cm, 4cm and 9cm because the sum of two sides is not greater than the third side i.e. 5 + 4 is not greater than 9.

(ii) Yes. It is possible to construct a triangle with lengths of its sides 8cm, 7cm and 4cm because the sum of two sides of a triangle is greater than the third side.

(iii) Yes. It is possible to construct a triangle with lengths of its sides 10cm, 5cm and 6cm because the sum of two sides of a triangle is greater than the third side.

(iv) Yes. It is possible to construct a triangle with lengths of its sides 2.5cm, 5cm and 7cm because the sum of two sides of a triangle is greater than the third side.

(v) No. It is not possible to construct a triangle with lengths of its sides 3cm, 4cm and 8cm because the sum of two sides is not greater than the third side.

2. In â–³ ABC, âˆ  A = 50o and âˆ  B = 60o. Determine the longest and the shortest sides of the triangle.

Solution:

Consider â–³ ABC

Based on the sum property we can write it as

âˆ  A + âˆ  B + âˆ  C = 180o

By substituting the values we get

50o + 60o + âˆ  C = 180o

On further calculation

âˆ  C = 180o – 50o – 60o

By subtraction

âˆ  C = 180o – 110o

âˆ  C = 70o

So we have âˆ  A < âˆ  B < âˆ  C

We get

BC < AC < AB

Therefore, the longest side of the triangle is AB and the shortest side is BC.

3.

(i) In â–³ ABC, âˆ  A = 90o. What is the longest side?

(ii) In â–³ ABC, âˆ  A = âˆ  B = 45o. Which is its longest side?

(iii) In â–³ ABC, âˆ  A = 100o and âˆ  C = 50o. Which is its shortest side?

Solution:

(i) It is given that âˆ  A = 90o

We know that in a right angled triangle the highest angle is 90o and the sum of all the angles is 180o

So we get that âˆ  A is the greatest angle in â–³ ABC

Hence, BC is the longest side which is opposite to âˆ  A

(ii) In â–³ ABC it is given that âˆ  A = âˆ  B = 45o

Based on the sum property of the triangle

âˆ  A + âˆ  B + âˆ  C = 180o

To find âˆ  C

âˆ  C = 180o – âˆ  A – âˆ  B

By substituting the values in the above equation

âˆ  C = 180o – 45o – 45o

By subtraction

âˆ  C = 180o – 90o

âˆ  C = 90o

So we get that âˆ  C is the greatest angle in â–³ ABC

Hence, AB is the longest side which is opposite to âˆ  C.

(iii) In â–³ ABC it is given that âˆ  A = 100o and âˆ  C = 50o

Based on the sum property of the triangle

âˆ  A + âˆ  B + âˆ  C = 180o

To find âˆ  B

âˆ  B = 180o – âˆ  A – âˆ  C

By substituting the values in the above equation

âˆ  B = 180o – 100o – 50o

By subtraction

âˆ  B = 180o – 150o

âˆ  B = 30o

So we get âˆ  B < âˆ  C < âˆ  A

i.e. AC < AB < BC

Hence, AC is the shortest side in â–³ ABC.

4. In â–³ ABC, side AB is produced to D such that BD = BC. If âˆ  A = 70o and âˆ  B = 60o, prove that

Solution:

In â–³ ABC it is given that âˆ  A = 70o and âˆ  B = 60o

Based on the sum property of the triangle

âˆ  A + âˆ  B + âˆ  C = 180o

To find âˆ  C

âˆ  C = 180o – âˆ  A – âˆ  B

By substituting the values in the above equation

âˆ  C = 180o – 70o – 60o

âˆ  C = 180o – 130o

By subtraction

âˆ  C = 50o

Consider â–³ BCD

We know that âˆ  CBD is the exterior angle of âˆ  ABC

So we get

âˆ  CBD = âˆ  DAC + âˆ  ACB

By substituting the values in the above equation

âˆ  CBD = 70o + 50o

âˆ  CBD = 120o

It is given that BC = BD

So we can write it as

âˆ  BCD = âˆ  BDC

Based on the sum property of the triangle

âˆ  BCD + âˆ  BDC + âˆ  CBD = 180o

So we get

âˆ  BCD + âˆ  BDC = 180o – âˆ  CBD

By substituting values in the above equation

âˆ  BCD + âˆ  BDC = 180o – 120o

âˆ  BCD + âˆ  BDC = 60o

It can be written as

2 âˆ  BCD = 60o

By division

âˆ  BCD = âˆ  BDC = 30o

In â–³ ACD

It is given that âˆ  A = 70o and âˆ  B = 60o

We can write it as

âˆ  ACD = âˆ  ACB + âˆ  BCD

By substituting the values we get

âˆ  ACD = 50o + 30o

âˆ  ACD = 80o

So we get to know that âˆ  ACD is the greatest angle and the side opposite to it i.e. AD is the longest side.

Therefore, it is proved that AD > CD

We know that âˆ  BDC is the smallest angle and the side opposite to it i.e. AC is the shortest side.

Therefore, it is proved that AD > AC.

5. In the given figure, âˆ  B < âˆ  A and âˆ  C < âˆ  D. Show that AD < BC.

Solution:

In the figure it is given that âˆ  B < âˆ  A and âˆ  C < âˆ  D

Consider triangle AOB

Since âˆ  B < âˆ  A

We get

AO < BO â€¦â€¦ (1)

Consider triangle COD

Since âˆ  C < âˆ  D

DO < CO â€¦â€¦. (2)

By adding both the equations we get

AO + DO < BO + CO

So we get

Therefore, it is proved that AD < BC.

6. AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that âˆ  A > âˆ  C and âˆ  B > âˆ  D.

Solution:

Construct two lines AC and BD in the given quadrilateral

Consider â–³ ABC

We know that BC > AB

It can be written as

âˆ  BAC > âˆ  ACB â€¦â€¦. (i)

Consider â–³ ACD

We know that CD > AD

It can be written as

âˆ  CAD > âˆ  ACD â€¦â€¦. (ii)

By adding both the equations we get

âˆ  BAC + âˆ  CAD > âˆ  ACB + âˆ  ACD

So we get

âˆ  A > âˆ  C

We know that AD > AB

It can be written as

âˆ  ABD > âˆ  ADB â€¦â€¦. (iii)

Consider â–³ BDC

We know that CD > BC

It can be written as

âˆ  CBD > âˆ  BDC â€¦â€¦. (iv)

By adding both the equations we get

âˆ  ABD + âˆ  CBD > âˆ  ADB + âˆ  BDC

So we get

âˆ  B > âˆ  D

7. In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Solution:

Consider â–³ ABC

We know that

AB + BC > AC â€¦.. (1)

Consider â–³ ACD

We know that

DA + CD > AC â€¦.. (2)

We know that

DA + AB > BD â€¦.. (3)

Consider â–³ BDC

We know that

BC + CD > BD â€¦.. (4)

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

So we get

2 (AB + BD + CD + DA) > 2 (AC + BD)

Dividing by 2 both the sides

AB + BD + CD + DA > AC + BD

Therefore, it is proved that AB + BD + CD + DA > AC + BD.

8. In a quadrilateral ABCD, show that (AB + BC + CD + DA) < 2 (BD + AC).

Solution:

Consider â–³ AOB

We know that

AO + BO > AB â€¦.. (1)

Consider â–³ BOC

We know that

BO + CO > BC â€¦.. (2)

Consider â–³ COD

We know that

CO + DO > CD â€¦.. (3)

Consider â–³ AOD

We know that

DO + AO > DA â€¦.. (4)

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

So we get

2 (AO + CO) + 2 (BO + DO) > AB + BC + CD + DA

On further calculation

2AC + 2BD > AB + BC + CD + DA

By taking 2 as common

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

So we get

AB + BC + CD + DA < 2 (AC + BD)

Therefore, it is proved that AB + BC + CD + DA < 2 (AC + BD)

9. In â–³ ABC, âˆ  B = 35o, âˆ  C = 65o and the bisector of âˆ  BAC meets BC in X. Arrange AX, BX and CX in descending order.

Solution:

Consider â–³ ABC

By sum property of a triangle

âˆ  A + âˆ  B + âˆ  C = 180o

To find âˆ  A

âˆ  A = 180o – âˆ  B – âˆ  C

By substituting the values

âˆ  A = 180o – 35o – 65o

By subtraction

âˆ  A = 180o – 100o

âˆ  A = 80o

We know that

âˆ  BAX = Â½ âˆ  A

So we get

âˆ  BAX = Â½ (80o)

By division

âˆ  BAX = 40o

Consider â–³ ABX

It is given that âˆ  B = 35o and âˆ  BAX = 40o

By sum property of a triangle

âˆ  BAX + âˆ  BXA + âˆ  XBA = 180o

To find âˆ  BXA

âˆ  BXA = 180o – âˆ  BAX – âˆ  XBA

By substituting values

âˆ  BXA = 180o – 35o – 40o

By subtraction

âˆ  BXA = 180o – 75o

âˆ  BXA = 105o

We know that âˆ  B is the smallest angle and the side opposite to it i.e. AX is the smallest side.

So we get AX < BX â€¦.. (1)

Consider â–³ AXC

âˆ  CAX = Â½ âˆ  A

So we get

âˆ  CAX = Â½ (80o)

By division

âˆ  CAX = 40o

By sum property of a triangle

âˆ  AXC + âˆ  CAX + âˆ  CXA = 180o

To find âˆ  AXC

âˆ  AXC = 180o – âˆ  CAX – âˆ  CXA

By substituting values

âˆ  AXC = 180o – 40o – 65o

So we get

âˆ  AXC = 180o – 105o

By subtraction

âˆ  AXC = 75o

So we know that âˆ  CAX is the smallest angle and the side opposite to it i.e. CX is the smallest side.

We get

CX < AX â€¦â€¦ (2)

By considering equation (1) and (2)

BX > AX > CX

Therefore, BX > AX > CX is the descending order.

10. In the given figure, PQ > PR and QS and RS are the bisectors of âˆ  Q and âˆ  R respectively. Show that SQ > SR.

Solution:

Consider â–³ PQR

It is given that PQ > PR

So we get

âˆ  PRQ > âˆ  PQR

Dividing both sides by 2 we get

Â½ âˆ  PRQ > Â½ âˆ  PQR

From the figure we get

âˆ  SRQ > âˆ  SQR

So we get

SQ > SR

Therefore, it is proved that SQ > SR.

11. D is any point on the side AC of â–³ ABC with AB = AC. Show that CD < BD.

Solution:

Consider â–³ ABC

It is given that AB = AC

So we get

âˆ  ABC = âˆ  ACB â€¦â€¦ (1)

From the figure we know that

âˆ  ABC = âˆ  ADB + âˆ  DBC

So we get

âˆ  ABC > âˆ  DBC

From equation (1)

âˆ  ACB > âˆ  DBC

i.e. âˆ  DCB > âˆ  DBC

It means that

BD > CD

So we get

CD < BD

Therefore, it is proved that CD < BD.

12. Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater 2/3 of a right angle.

Solution:

Consider â–³ PQR where PR is the longest side

So we get PR > PQ

i.e. âˆ  Q > âˆ  R â€¦.. (1)

We also know that PR > QR

i.e. âˆ  Q > âˆ  P â€¦.. (2)

âˆ  Q + âˆ  Q > âˆ  R + âˆ  P

So we get

2 âˆ  Q > âˆ  R + âˆ  P

By adding âˆ  Q on both LHS and RHS

2 âˆ  Q + âˆ  Q > âˆ  R + âˆ  P + âˆ  Q

We know that âˆ  R + âˆ  P + âˆ  Q = 180o

So we get

3 âˆ  Q > 180o

By division

âˆ  Q > 60o

So we get

âˆ  Q > 2/3 (90o)

i.e. âˆ  Q > 2/3 of a right angle

Therefore, it is proved that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater 2/3 of a right angle.

13. In the given figure, prove that

(i) CD + DA + AB > BC

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

Solution:

(i) Consider â–³ CDA

We know that CD + DA > AC â€¦.. (1)

Consider â–³ ABC

We know that AC + AB > BC â€¦.. 21)

By adding both the equations we get

CD + DA + AC + AB > AC + BC

By subtracting AC on both the sides

CD + DA + AC + AB â€“ AC > AC + BC â€“ AC

So we get

CD + DA + AB > BC

Therefore, it is proved that CD + DA + AB > BC.

(ii) Consider â–³ CDA

We know that CD + DA > AC â€¦.. (1)

Consider â–³ ABC

We know that AB + BC > AC â€¦.. (2)

By adding both the equations we get

CD + DA + AB + BC > AC + AC

So we get CD + DA + AB + BC > 2AC

Therefore, it is proved that CD + DA + AB + BC > 2AC.

14. If O is a point within â–³ ABC, show that

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > Â½ (AB + BC + CA)

Solution:

(i) It is given that O is a point within â–³ ABC

Consider â–³ ABC

We know that AB + AC > BC â€¦.. (1)

Consider â–³ OBC

We know that OB + OC > BC â€¦.. (2)

By subtracting both the equations we get

(AB + AC) â€“ (OB + OC) > BC – BC

So we get

(AB + AC) â€“ (OB + OC) > 0

AB + AC > OB + OC

Therefore, it is proved that AB + AC > OB + OC.

(ii) We know that AB + AC > OB + OC

In the same way we can write

AB + BC > OA + OC and AC + BC > OA + OB

By adding all the equations we get

AB + AC + AB + BC + AC + BC > OB + OC + OA + OC + OA + OB

So we get

2 (AB + BC + AC) > 2 (OA + OB + OC)

Dividing by 2 both sides

AB + BC + AC > OA + OB + OC

(iii) Consider â–³ OAB

We know that OA + OB > AB â€¦.. (1)

Consider â–³ OBC

We know that OB + OC > BC â€¦.. (2)

Consider â–³ OCA

OC + OA > CA â€¦â€¦ (3)

OA + OB + OB + OC + OC + OA > AB + BC + CA

So we get

2 (OA + OB + OC) > AB + BC + CA

Dividing by 2

OA + OB + OC > Â½ (AB + BC + CA)

Therefore, it is proved that OA + OB + OC > Â½ (AB + BC + CA).

15. In the given figure, AD âŠ¥ BC and CD > BD. Show that AC > AB.

Solution:

Consider point S on the line BC so that BD = SD and join AS.

We know that SD = BD

Since AD is a perpendicular we know that

By SAS congruence criterion

AB = AS (c. p. c. t)

Consider â–³ ABS

We know that AB = AS

From the figure we know that âˆ  ASB and âˆ  ABS are angles opposite to the equal sides

âˆ  ASB = âˆ  ABS â€¦. (1)

Consider â–³ ACS

From the figure we know that âˆ  ASB and âˆ  ACS are angles opposite to the equal sides

âˆ  ASB = âˆ  ACS â€¦. (2)

Considering the equations (1) and (2)

âˆ  ABS > âˆ  ACS

It can be written as

âˆ  ABC > âˆ  ACB

So we get

AC > AB

Therefore, it is proved that AC > AB.

16. In the given figure, D is a point on side BC of a â–³ ABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Solution:

Consider â–³ ABC

We know that

AB + AC > BC

It can be written as

AB + AC > BD + DC

We know that CD = DE

So we get

AB + AC > BD + DE â€¦.. (1)

Consider â–³ BED

We know that

BD + DE > BE â€¦â€¦ (2)

Considering both the equations we get

AB + AC > BE.

### Access other exercise solutions of Class 9 Maths Chapter 9: Congruence of Triangles and Inequalities in a Triangle

Exercise 9A Solutions 27 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 9 – Congruence of Triangles and Inequalities in a Triangle Exercise 9B

RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle Exercise 9B solves problems using theorems which are based on inequalities in a triangle.

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