# ML Aggarwal Triangles Exe-10.1 Class 9 ICSE Maths Solutions Ch-10

ML Aggarwal Triangles Exe-10.1 Class 9 ICSE Maths Solutions Ch-10. Step by Step Answer of Exercise-10.1 of Triangles of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Triangles Exe-10.1 Class 9 ICSE Maths Solutions Ch-10

 Board ICSE Subject Maths Class 9th Chapter-10 Triangles Topics Solution of Exe-10.1 Questions Academic Session 2024-2025

### Solution of Exe-10.1 Questions

ML Aggarwal Triangles for ICSE Class 9 Mathematics

#### Question 1. It is given that ∆ABC ≅ ∆RPQ. Is it true to say that BC = QR ? Why?

Solution: Given ∆ABC ≅ ∆RPQ

Therefore, their corresponding sides and angles are equal.

Therefore BC = PQ

Hence it is not true to say that BC = QR

#### Question 2. “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

No, it is not true statement as the angles should be included angle of there two given sides.

#### Question 3. In the given figure, AB=AC and AP=AQ. Prove that

(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.

Answer: (i) In △ APC and △AQB

AB=AC and AP=AQ [given]

From the given figure, ∠A = ∠A [common in both the triangles]

Therefore, using SAS axiom we have ∆APC ≅ ∆AQB

##### (ii) In △ APC and △AQB

AB=AC and AP=AQ [given]

From the given figure, ∠A = ∠A [common in both the triangles]

By using corresponding parts of congruent triangles concept we have

BQ = CP

##### (iii) In △ APC and △AQB

AB=AC and AP=AQ [given]

From the given figure, ∠A = ∠A [common in both the triangles]

By using corresponding parts of congruent triangles concept we have

∠APC = ∠AQB.

#### Question 4. In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that

(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.

Answer: (i) In the given figure AB = AC

P and Q are point on BA and CA produced respectively such that AP = AQ

Now we have to prove ∆APC ≅ ∆AQB

By using corresponding parts of congruent triangles concept we have

CP = BQ

∠ACP = ∠ABQ

##### (iii) ∠ACP = ∠ABQ

In ∆ APC and ∆AQB

AC = AB (Given)

AP = AQ (Given)

∠PAC =∠QAB (Vertically opposite angle)

#### Question 5. In the given figure, AD = BC and BD = AC. Prove that :

∠ADB = ∠BCA and ∠DAB = ∠CBA.

Given: in the figure, AD = BC, BD = AC

To prove :

(ii) ∠DAB = ∠CBA

Proof : in ∆ADB and ∆ACB

AB = AB (Common)

DB = AC (Given)

∠DAB = ∠CBA

#### Question 6. In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that

(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Given : in the figure ABCD is a quadrilateral

∠DAB = ∠CBA

To prove :

(i) ∆ABD = ∆BAC

(ii) ∠ABD = ∠BAC

Proof : in ∆ABD and ∆ABC

AB = AB (common)

∠DAB = ∠CBA (Given)

(i) ∆ABD ≅ ∆ABC (SAS axiom)

(ii) BD = AC

(ii) ∠ABD = ∠BAC

#### Question 7.In the given figure, AB = DC and AB || DC. Prove that AD = BC.

Given: in the given figure.

AB = DC, AB ∥ DC

To prove : AD = BC

Proof : AB ∥ DC

∠ABD = ∠CDB (Alternate angles)

In ∆ABD and ∆CDB

AB = DC

∠ABD = ∠CDB (Alternate angles)

BD = BD (common)

∆ABD ≅ CDB (SAS axiom)

#### Question 8. In the given figure. AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.

Given: in the figure, AC = AE, AB = AD

To prove : BC = DE

Proof : in ∆ABC and ∆ADE

AC = AE (given)

∠BAC = ∠DAE

∆ABC = ∆ ADE (SAS axiom)

BC = DE

#### Question 9. In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that

(i) ∆ACE ≅ ∆DBF
(ii) AE = DF.

Given : in the given figure

AB = CD

CE = BF

∠ACE = ∠DBF

To prove : (i) ∆ACE ≅ ∆DBF

AE = DE

##### (ii) AE = DF

Proof : AB = CD

AB + BC = BC + CD

AC = BD

Now in ∆ACE and ∆DBF

AC = BD (Proved)

CE = BF (Given)

∠ACE = ∠DBF (SAS axiom)

#### Question 10. In the adjoining figure, AB = AC and D is mid-point of BC. Use SSS rule of congruency to show that

(i) ∆ABD ≅ ∆ACD

(ii) AD is bisector of ∠A

(iii) AD is perpendicular to BC

We will use the SSS congruency rule to prove this.

In ∆ABD and ∆ACD,

AB=AC (given)

BD=CD (given AD is the bisector)

So, ∆ABD is congruent to ∆ACD on SSS rule.

But BDC is a straight line, so angle ADB + angle ADC = 180°

#### Question 11. Two line segment AB and CD bisect each other at O. Prove that

(i) AC = BD

(ii) ∠CAB = ∠ABD

AB and CD bisect each other at O i.e, AO=BO and CO=DO

in ΔCOA and ΔDOB
Given CO=OD,∠COA=∠BOD [ vertically opposite angles]
∴ΔCOA≅ΔBOD

##### (ii) ∠CAB=∠ABD[C.P.CT]

again
in ΔCOB and ΔAOD
CO=OD [given]
BO=AO [given]
∠COB=∠AOD [vertically opposite angles]
∴ΔCOB≅ΔAOD

##### (iii) As we know the quadrilateral whose diagonals bisect each other is a parallelogram…..

So ABCD is a parallelogram

THEN, AB || BC

#### Question 12. In the adjoining figure, find the value of and y.

as two sides are equal ,35= 2x+5

2x =30

x=15

y+5=46

y=41

—  : End of ML Aggarwal Triangles Exe-10.1 Class 9 ICSE Maths Solutions :–