class 10th- chapter 6- triangle - Exercise 6.2

class 10th- chapter 6- triangle - Exercise 6.2

 

 Exercise 6.2

Question 1:-

 

Solution:-

In Figure (i) and (ii) DE || BC., Find EC in (i) and AD in (iii)

(i)  Let EC= x cm

 Given :- DE || BC, therefore using Thales Theorem, we have

AD/DB = AC/EC

➱1.5/3=1/x 

➱ x=(3*1)/1.5 =2

Hence, EC=2.

(ii)            Let AD= x cm

 Given :- DE || BC, therefore

using Thales Theorem, we have

AD/DB= AC/EC

➱x / 7.2=1.8 / 5.4

➱x=(1.8 * 7.2) / 5.4=2.4

Hence, AD=2.4

 

Question:-2  

E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR. 

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:-

(i) PE/EQ= 3.9/3=1.3

and PF/FR= 3.6/2.4= 3/2=1.5

since, PE/EQ≠PF/FR

 EF∦QR

(ii)

  

 PE/EQ=4/4.5

                 =40/45=8/9

and,     PF/FR= 8/9

since, PE/EQ=PF/FR

 EF∥QR

(iii)

In△PQR, PQ=1.28 and PE=0.18

so EQ= PQ-PE=1.28-0.18=1.10

 and, PR=2.56 and PF=0.36

so, FR=PR-PF=2.56-0.36=2.20

       PE/EQ=0.18/1.10=9/55

and PF/FR=0.36/2.20=9/55

since, PE/EQ=PF/FR

        EF∥QR

 

Question:- 3

In figure, if LM ||  CB and LN ||  CD, prove that AM/AB= AN/AD 

Solution:- In ΔABC, LM || CB

 AM/AB=AL/AC [by B.P.T] ……….(i)

And in Δ ACD, LN || CD

 AN/AD=AL/AC [by B.P.T]......(ii)

From eq. (i) and (ii), we have

AM/AB= AN/AD

Hence proved.

Question 4:-

In figure, DE||AC and DF ||AE. Prove that BE/FE=BE/EC.

Solution:- 

In ΔBCA, DE || AC

 BE/EC=BD/DA [B.P.T]……….(i)

And in ΔBEA, DF || AE

BF/FE=BD/DA [B.P.T] ……….(ii)

From eq. (i) and (ii), we have

BF/FE=BE/EC

Question 5:-

 In figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:- 

In ΔPQO, DE || OQ

PE/EQ=PD/DO[B.P.T] ……….(i)

And in ΔPOR, DF || OR

PD/DO=PF/FR [B.P.T].....(ii)

From eq. (i) and (ii), we have

PE/EQ=PF/FR

EF || QR [By the converse of B.P.T]

Question 6:-

In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC||PR. Show that BC||QR.

Solution:- 

Given: O is any point in ΔPQR, in which AB || PQ and AC || PR.

To prove: BC || QR

Construction: Join BC.

Proof: In ΔOPQ, AB || PQ

OA/AP=OB/PQ  [B.P.T] ……….(i)

And in ΔOPR, AC || PR

OA/AP=OC/CR [B.P.T] ……….(ii)

From eq. (i) and (ii), we have

OB/BQ=OC/CR

In ΔOQR, B and C are points dividing the sides OQ and OR in the same ratio.

By the converse of Basic Proportionality theorem,

BC || QR

Question 7:- 

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:-

Given: A ΔABC, in which D is the mid-point of side AB

and the line DE is drawn parallel to BC, meeting AC at E.

To prove: AE = EC

Proof: Since DE || BC

AD/DB=AE/EC

       [Basic Proportionality theorem] ……….(i)

But AD = DB [Given]

AD/DB=1

AE/EC=1 [From eq. (i)]

 AE=EC

Hence, E is the mid-point of the third side AC.

Question 8:-

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution:- 

Given: A ΔABC, in which D and E are the mid-points of

sides AB and AC respectively.

To Prove: DE || BC

Proof: Since D and E are the mid-points of AB and AC

respectively.

AD = DB and AE = EC

Now, AD = DB

AD/DB=1 and AE = EC

AE/EC=1

AD/DB=AE/EC=1

AD/DB=AE/EC 

Thus, in ΔABC, D and E are points dividing the sides AB and AC in the same ratio.

Therefore, by the converse of Basic Proportionality theorem, we have

DE || BC

Question 9:- 

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO=CO/DO 

Solution:-

Given: A trapezium ABCD, in which AB || DC and its diagonals

AC and BD intersect each other at O.

To Prove: AO/BO=CO/DO 

Construction: Through O, draw OE || AB, i.e. OE|| DC.

Proof: In ΔADC, we have OE || DC

 AE/ED=AO/CO 

 [By Basic Proportionality theorem]……….(i)

Again,in ΔABD, we have OE||AB

                  [by Construction]

ED/AE=DO/BO

          [By Basic Proportionality theorem]

 AE/ED=BO/DO.......(ii)

From eq. (i) and (ii), we get

AO/CO=BO/DO 

AO/BO=CO/DO  

Question 10:- 

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO=CO/DO Such that ABCD is a trapezium.

Solution:-

Given: A quadrilateral ABCD, in which its diagonals AC and BD intersect each other at O such that AO/BO=CO/DO,i.e.

AO/CO=BO/DO.

To Prove: Quadrilateral ABCD is a trapezium.

Construction: Through O, draw OE || AB meeting AD at E.

Proof: In ΔADB, we have OE || AB [By construction]

DE/EA=OD/BO

                   [By Basic Proportionality theorem]

EA/EA=BO/DO

EA/DE= BO/DO=AO/CO

[because AO/CO=BO/DO]

EA/DE=AO/CO

Thus in ΔADC, E and O are points dividing the sides AD and AC in the same ratio. 

Therefore by the converse of Basic Proportionality theorem, we have

EO||DC

But EO||AB  [By construction]

AB||DC

 Quadrilateral ABCD is a trapezium DC

 Thank You