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

RS Aggarwal Solutions are prepared in order to make the concepts easy to learn for the students. While solving the textbook exercise problems, the students can make use of the PDF which is available. Vast research is conducted on the topics by subject experts to improve concept based approach while solving problems. The exercise wise solutions for each chapter are prepared based on the CBSE guidelines. It is helpful for the students in order to complete the whole syllabus accordingly. RS Aggarwal Solutions for Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle Exercise 9B are provided here.

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

## 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 = 50 ^{o} and âˆ B = 60^{o}. 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 = 180^{o}

By substituting the values we get

50^{o} + 60^{o} + âˆ C = 180^{o}

On further calculation

âˆ C = 180^{o} – 50^{o} – 60^{o}

By subtraction

âˆ C = 180^{o} – 110^{o}

âˆ C = 70^{o}

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 = 90 ^{o}. What is the longest side?**

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

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

**Solution:**

(i) It is given that âˆ A = 90^{o}

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

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 = 45^{o}

Based on the sum property of the triangle

âˆ A + âˆ B + âˆ C = 180^{o}

To find âˆ C

âˆ C = 180^{o} – âˆ A – âˆ B

By substituting the values in the above equation

âˆ C = 180^{o} – 45^{o} – 45^{o}

By subtraction

âˆ C = 180^{o} – 90^{o}

âˆ C = 90^{o}

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 = 100^{o} and âˆ C = 50^{o}

Based on the sum property of the triangle

âˆ A + âˆ B + âˆ C = 180^{o}

To find âˆ B

âˆ B = 180^{o} – âˆ A – âˆ C

By substituting the values in the above equation

âˆ B = 180^{o} – 100^{o} – 50^{o}

By subtraction

âˆ B = 180^{o} – 150^{o}

âˆ B = 30^{o}

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 = 70 ^{o} and âˆ B = 60^{o}, prove that **

**(i) AD > CD**

**(ii) AD > AC.**

**Solution:**

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

Based on the sum property of the triangle

âˆ A + âˆ B + âˆ C = 180^{o}

To find âˆ C

âˆ C = 180^{o} – âˆ A – âˆ B

By substituting the values in the above equation

âˆ C = 180^{o} – 70^{o} – 60^{o}

âˆ C = 180^{o} – 130^{o}

By subtraction

âˆ C = 50^{o}

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 = 70^{o} + 50^{o}

By addition

âˆ CBD = 120^{o}

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 = 180^{o}

So we get

âˆ BCD + âˆ BDC = 180^{o} – âˆ CBD

By substituting values in the above equation

âˆ BCD + âˆ BDC = 180^{o} – 120^{o}

âˆ BCD + âˆ BDC = 60^{o}

It can be written as

2 âˆ BCD = 60^{o}

By division

âˆ BCD = âˆ BDC = 30^{o}

In â–³ ACD

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

We can write it as

âˆ ACD = âˆ ACB + âˆ BCD

By substituting the values we get

âˆ ACD = 50^{o} + 30^{o}

By addition

âˆ ACD = 80^{o}

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

AD < BC

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

Consider â–³ ADB

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)

Consider â–³ ADB

We know that

DA + AB > BD â€¦.. (3)

Consider â–³ BDC

We know that

BC + CD > BD â€¦.. (4)

By adding all the equations

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)

By adding all the equations

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 = 35 ^{o}, âˆ C = 65^{o} 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 = 180^{o}

To find âˆ A

âˆ A = 180^{o} – âˆ B – âˆ C

By substituting the values

âˆ A = 180^{o} – 35^{o} – 65^{o}

By subtraction

âˆ A = 180^{o} – 100^{o}

âˆ A = 80^{o}

We know that

âˆ BAX = Â½ âˆ A

So we get

âˆ BAX = Â½ (80^{o})

By division

âˆ BAX = 40^{o}

Consider â–³ ABX

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

By sum property of a triangle

âˆ BAX + âˆ BXA + âˆ XBA = 180^{o}

To find âˆ BXA

âˆ BXA = 180^{o} – âˆ BAX – âˆ XBA

By substituting values

âˆ BXA = 180^{o} – 35^{o} – 40^{o}

By subtraction

âˆ BXA = 180^{o} – 75^{o}

âˆ BXA = 105^{o}

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 (80^{o})

By division

âˆ CAX = 40^{o}

By sum property of a triangle

âˆ AXC + âˆ CAX + âˆ CXA = 180^{o}

To find âˆ AXC

âˆ AXC = 180^{o} – âˆ CAX – âˆ CXA

By substituting values

âˆ AXC = 180^{o} – 40^{o} – 65^{o}

So we get

âˆ AXC = 180^{o} – 105^{o}

By subtraction

âˆ AXC = 75^{o}

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)

By adding both the equations

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 = 180^{o}

So we get

3 âˆ Q > 180^{o}

By division

âˆ Q > 60^{o}

So we get

âˆ Q > 2/3 (90^{o})

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)

By adding all the equations

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.

Consider â–³ ADB and â–³ ADS

We know that SD = BD

Since AD is a perpendicular we know that

âˆ ADB = âˆ ADS = 90^{o}

AD is common i.e. AD = AD

By SAS congruence criterion

â–³ ADB â‰… â–³ ADS

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.

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

- The solutions help students solve problems from Exercise 9B by gaining a better knowledge of concepts.
- Obtaining good marks in the board exams is made easier by using the solutions prepared by our subject experts.
- The answers are prepared according to the evaluation process which is followed in the CBSE board.
- The students can download the PDF and use it as a reference while solving exercise wise problems of RS Aggarwal textbook.