ML Aggarwal Solution Class 9 Chapter 10 Triangles Exercise 10.1

 Exercise 10.1

Question 1

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

Sol :
∆ABC ≅ ∆RPQ
∴Their corresponding sides and angles are equal






∴BC=PQ

∴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?

Sol :

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.

Sol :
Given : In figure, AB=AC, AP=AQ
To prove :
(i) ΔAPC≅ΔAQB
(ii) CP=BQ
(iii) ∠APC=∠AQB







⇒AC=AB (Given)

⇒AP=AQ (Given)

⇒∠A=∠A (Common)

(i) ∵∆APC≅∆AQB (SAS axiom)

(ii) BQ=CP (c.p.c.t.)

(iii) ∠APC=∠AQB (c.p.c.t.)


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.

Sol :
Given : In the given figure, AB=AC ,P and Q are point on BA and CA produced respectively such that AP=AQ







To prove : (i) ∆APC≅∆AQB

(ii) CP=BQ

(iii) ∠ACP=∠ABQ

Proof : In ∆APC and ∆AQB

⇒AC=AB (Given)

⇒AP=AQ (Given)

⇒∠PAC=∠QAB (Vertically opposite angle)


(i) ∴∆APC≅∆BQP (SAS axiom)

∴CP=BQ (c.pc.t)

⇒∠ACP=∠ABQ (c.p.c.t)


Question 5

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

∠ADB = ∠BCA and ∠DAB = ∠CBA.

Sol :
Given : In the figure ,AD=BC, BD=AC









To prove : 
(i) ∠ADB=∠BCA
(ii) ∠DAB=∠CBA
Proof : In ∆ADB ≅ ∆ACB
⇒AB=AB (common)
⇒AD=BC (given)
⇒DB=AC (given)
∴∆ADB≅∆ACD (SSS axiom)
∴∠ADB = ∠BCA (c.p.c.t)

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.

Sol :
Given : In the figure ABCD is a quadrilateral in which 
⇒AD=BC 
⇒∠DAB=∠CBA








To prove :

(i) ∆ABD ≅ ∆BAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Proof : In ∆ABD and  ∆ABC

⇒AB=AB (common)

⇒∠DAB=∠CBA (Given)

⇒AD=BC (Given)

(i) ∴∆ABD ≅ ∆BAC

(ii) ∴BD=AC (c.p.c.t)

(iii) ∠ABC=∠BAC (c.p.c.t)


Question 7

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

Sol :
Given : In the given figure, 
⇒AB=DC, AB||DC
To prove : AD=BC 
Proof : ∵AB||DC
∴∠ABD=∠CDB (alternate angles)

In ∆ABD=∆CDB

⇒AB=DC (Given)








⇒∠ABD=∠CDB (Alternate Angles)

⇒BD=BD (common)

∴∆ABD≅∆ADB (SAS axiom)

∴AD=BC (c.p.c.t)


Question 8

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









Sol :

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

⇒∠BAD=∠CAE

To prove : BC=DE 

Construction : Join DE.

<Diagram to be added>

Proof : In ∆ABC and ∆ADE

⇒AB=AD (given)

⇒AC=AE (given)

⇒∠BAD+∠DAC=∠DAC+∠CAE

⇒∠BAC=∠DAE

∴∆ABC≅∆ADE (SAS axiom)

∴ BC=DE (c.p.c.t)


Question 9

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

(i) ∆ACE ≅ ∆DBF

(ii) AE = DF.

Sol :
Given : In the given figure
⇒AB=CD
⇒CE=BF
⇒∠ACE=∠DBF







To prove :(i) ∆ACE ≅ ∆DBF

(ii) AE = DF.

Proof : ∵AB=CS

Adding BC to both sides

⇒AB+BC=BC+CD

⇒AC=BD 

Now in ∆ACE and ∆DBF

AC=BD (Proved)

CE=BF (Given)

∠ACE=∠DBF (Given)

(i) ∆ACE ≅ ∆DBF∴ (SAS axiom)

∴AE=DE (c.p.c.t)


Question 10

In the given 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.

Sol :
Given : In the given figure , AB=AC
D is mid point of BC
∴BD=DC












To prove :

(i) ∆ABD ≅ ∆ACD

(ii) AD is bisector of  ∠A

(iii) AD⊥BC

Proof : In ∆ABD ≅ ∆ACD

AB=AC (Given)

BD=DC (Given)

AD=AD (Common)

(i) ∴∆ABD ≅ ∆ACD

(ii) ∠BAD=∠CAD (c.p.c.t)

∴AD is the bisector of ∠A

(iii) ∠ADB=∠ADC

But ∠ADB+∠ADC=180° (linear pair)

∴∠ADB=∠ADC=90° 

∴AD⊥BC


Question 11

Two line segments AB and CD bisect each other at O. Prove that :

(i) AC = BD

(ii) ∠CAB = ∠ABD

(iii) AD || CB

(iv) AD = CB.

Sol :






In ∆AOC and ∆AOC 

OC=OD and AO=OB [AB and CD bisector each other]

Also, ∠AOC=∠BOD (vertical opposite angles)

∴ΔAOC≅ΔBOD (By S.A.S axiom of congruency)

(i) Then AC=BD (c.p.c.t)

(ii) Also ∠CAO= ∠ABD (c.pc.t)

i.e. ∠CAB=∠ABD

[∵∠CAO=∠CAB and ∠DBO=∠ABD]

(iii) We have in (ii) part

∠CAB=∠ABD

But these are Alternate angles

Hence, AD||CD

(iv) In ∆AOD and ∆BOC

OC=OD and BO=AO

(AB and CD bisect each other )

and ∠BOC=∠AOD (Vertical opposite angles)

∴∆AOD≅∆BOC [By S.A.S axiom of congruency]

Then AD=BC (c.p.c.t)

or AD=CB (Q.E.D)

 

Question 12

In each of the following diagrams, find the values of x and y.









Sol :


In ΔABD and ΔBCD,
AB=BC (given)
AD=CD (given)
BD=BD (common)

∴∆ABD≅∆BCD [By S.S.S axiom of congruency]

∴∠ABD=∠CBD (c.p.c.t)

⇒y+5=46

⇒y=46-5

⇒y=41

Also , ∠ABD=∠BDC (c.p.c.t)

⇒35=(2x+5)

⇒35=2x+5

⇒2x+5=35

⇒2x=35-5

⇒2x=30

⇒$x=\frac{30}{2}$

⇒x=15

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