165-NCERT-10-6-Triangles - Ex- 6.2

NCERT

10th Mathematics

Exercise 6.2

Topic: 6 Triangles

Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.1

EXERCISE 6.2

Q1. In the following fig, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

(i) In ∆ ABC, DE ‖ BC, find EC in fig (i)

a) In ∆ ABC and ∆ ADE,

< ABC = < ADE    (corresponding angles)

< ACB = < AED    (corresponding angles)
< BAC = < DAE    (common angle)
So, ∆ ABC ~ ∆ ADE

b) Using basic proportionality theorem,

(AD)/(DB) = (AE)/(EC)

(1.5)/3 = 1/(EC)

1/2 = 1/(EC)

(EC) = 2

c) So, here EC = 2 cm.

(ii) In ∆ ABC, DE ‖ BC, find AD in fig (ii)

a) In ∆ ABC and ∆ ADE,

< ABC = < ADE    (corresponding angles)

< ACB = < AED    (corresponding angles)
< BAC = < DAE    (common angle)
So, ∆ ABC ~ ∆ ADE

b) Using basic proportionality theorem,

(AD)/(DB) = (AE)/(EC)

(AD)/(7.2) = (1.8)/(5.4)

(AD) = [(1.8)(7.2)]/(5.4)

(AD) = (1.8)[(72)/(54)]

(AD) = (1.8)[(8)/(6)]
(AD) = (18/10)[(8)/(6)]
(AD) = (3/10)[(8)]
(AD) = [(3)(8)]/10
(AD) = 24/10
(AD) = 2.4 

c) So, here AD = 2.4 cm. 

Q2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. 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 = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

1) Using basic proportionality theorem,

a) If EF ‖ QR, then we must have,

(PE)/(EQ) = (PF)/(FR) ----------- equation 1

b) LHS = (PE)/(EQ)

LHS = (3.9)/(3)

LHS = 1.3 ----------- equation 2

c) RHS = (PF)/(FR)

RHS = (3.6)/(2.4)

RHS = (36)/(24)
RHS = 3/2 

RHS = 1.5 ----------- equation 3 

2) So, from equations 2 and 3, we can say that (PE)/(EQ) ≠ (PF)/(FR).

3) So, EF is not paraller to QR.

 

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

1) Using basic proportionality theorem,

a) If EF ‖ QR, then we must have,

(PE)/(EQ) = (PF)/(FR) ----------- equation 1

b) LHS = (PE)/(EQ)

LHS = (4)/(4.5)

LHS = (40)/(45) 

LHS = 8/9 ----------- equation 2

c) RHS = (PF)/(FR)

RHS = (8)/(9)

RHS = 8/9 ----------- equation 3 

2) So, from equations 2 and 3, we can say that (PE)/(EQ) = (PF)/(FR).

3) So, EF is paraller to QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

1) First we will find EQ and FR

a) EQ = PQ - PE

EQ = 1.28 - 0.18

EQ = 1.10 ----------- equation 1

b) FR = PR - PF

FR = 2.56 - 0.36

FR = 2.20 ----------- equation 2 

2) Using basic proportionality theorem,

a) If EF ‖ QR, then we must have,

(PE)/(EQ) = (PF)/(FR) ----------- equation 3

b) LHS = (PE)/(EQ)

LHS = (0.18)/(1.10)

LHS = (18)/(110)

LHS = (9)/(55) 

LHS = 9/55 ----------- equation 4

c) RHS = (PF)/(FR)

RHS = (0.36)/(2.20)

RHS = (36)/(220)

RHS = (18)/(110)
RHS = (9)/(55) 

RHS = 9/55 ----------- equation 5 

2) So, from equations 4 and 5, we can say that (PE)/(EQ) = (PF)/(FR).

3) So, EF is paraller to QR.

Q3. In the following fig. , if LM || CB and LN || CD, prove that 

(AM)/(AB) = (AN)/(AD).

Solution:

1) In ∆ ABC, LM ‖ CB,

a) In ∆ AML and ∆ ABC,

< AML = < ABC    (corresponding angles)

< ALM = < ACB    (corresponding angles)

< MAL = < BAC    (common angle)

So, ∆ AML ~ ∆ ABC

b) Using basic proportionality theorem,

(AM)/(AB) = (AL)/(AC) ----------- equation 1

2) In ∆ ADC, LN ‖ CD,

a) In ∆ ANL and ∆ ADC,

< ANL = < ADC    (corresponding angles)

< ALN = < ACD    (corresponding angles)

< NAL = < DAC    (common angle)

So, ∆ ANL ~ ∆ ADC

b) Using basic proportionality theorem,

(AN)/(AD) = (AL)/(AC) ----------- equation 2

3) From equations 1 and 2, we have (AM)/(AB) = (AN)/(AD). Hence proved.

Q4. In the following Fig., DE || AC and DF || AE. Prove that BF/FE = BE/EC

Solution:

1) In ∆ ABC, DE ‖ AC,

a) Using basic proportionality theorem,

(BD)/(DA) = (BE)/(EC) ----------- equation 1

2) In ∆ BAE, DF ‖ AE,        

a) Using basic proportionality theorem,

(BD)/(DA) = (BF)/(FE) ----------- equation 2

3) From equations 1 and 2, we have (BE)/(EC) = (BF)/(FE). Hence proved.

Q5. In the following Fig., DE || OQ and DF || OR. Show that EF || QR.

Solution:

1) In ∆ PQO, DE ‖ OQ,

a) Using basic proportionality theorem,

(PD)/(DO) = (PE)/(EQ) ----------- equation 1

2) In ∆ POR, DF ‖ OR,        

a) Using basic proportionality theorem,

(PD)/(DO) = (PF)/(FR) ----------- equation 2

3) From equations 1 and 2, we have (PE)/(EQ) = (PF)/(FR).

4) So, EF ‖ QR. Hence proved.

Q6. In the following Fig., A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

1) In ∆ OPQ, AB ‖ PQ,

a) Using basic proportionality theorem,

(OA)/(AP) = (OB)/(BQ) ----------- equation 1

2) In ∆ OPR, AC ‖ PR,        

a) Using basic proportionality theorem,

(OA)/(AP) = (OC)/(CR) ----------- equation 2

3) From equations 1 and 2, we have (OB)/(BQ) = (OC)/(CR).

4) So, BC ‖ QR. Hence proved.

Q7. 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:

1) In ∆ ABC, point P is the midpoint of AB,

a) So we have,

(AP)/(PB) = 1 ----------- equation 1

2) In ∆ ABC, PQ ‖ BC,        

a) Using basic proportionality theorem,

(AP)/(PB) = (AQ)/(QC) ----------- equation 2

3) From equations 1 and 2, we have (AQ)/(QC) = 1.

4) Here, AQ = QC, so Q is the midpoint of AC.

5) So, line PQ bisects the third side AC of ∆ ABC. Hence proved.

Q8. 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:

1) In ∆ ABC, point P is the midpoint of AB,

a) So we have,

(AP)/(PB) = 1 ----------- equation 1

2) In ∆ ABC, point Q is the midpoint of AC,       

a) So we have,

(AQ)/(QC) = 1 ----------- equation 2

3) From equations 1 and 2, we have (AP)/(PB) = (AQ)/(QC) = 1.

4) So, PQ ‖ BC. Hence proved.

Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.

Solution:

1) In trapezium  ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.

2) Construct line PQ such that PQ ‖ AB and PQ ‖ DC.

3) In ∆ ABC, OQ ‖ AB,

a) Using basic proportionality theorem,

(BQ)/(CQ) = (AO)/(CO) ----------- equation 1

4) In ∆ BDC, OQ ‖ DC,       

a) Using basic proportionality theorem,

(BQ)/(CQ) = (BO)/(DO) ----------- equation 2

3) From equations 1 and 2, we have (AO)/(CO) = (BO)/(DO).

4) So, we have (AO)/(BO) = (CO)/(DO). Hence proved.

Q10. The diagonals of a quadrilateral ABCD intersect each other at 

the point O such that (AO)/(BO) = (CO)/(DO). Show that ABCD is a trapezium.

Solution:

1) In quadrilateral ABCD, diagonals AC and BD intersect at O.

2) Construct line PQ such that PQ ‖ AB.

3) In ∆ ABC, OQ ‖ AB,

a) Using basic proportionality theorem,

(BQ)/(CQ) = (AO)/(CO) ----------- equation 1

4) As per the problem,

(AO)/(BO) = (CO)/(DO)

(AO)/(CO) = (BO)/(DO) ----------- equation 2

5) From equations 1 and 2, we have,

(BO)/(DO) = (BQ)/(CQ) ----------- equation 3

6) So, from equation 3 and the converse of the basic proportionality theorem,

PQ ‖ AB, so AB ‖ CD.

7) Therefore quadrilateral ABCD is the trapezium.

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Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.3

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