# NCERT solutions for class 10 Maths chapter 6 Triangles (Exercise 6.2)

## Ex 6.2 Question 1.

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

Solution:

(i) In given △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using proportionality theorem]

⇒1.5/3 = 1/EC

⇒EC = 3/1.5

EC = 3×10/15 = 2 cm

(ii) In given △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using proportionality theorem]

⇒ AD/7.2 = 1.8 / 5.4

⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10

## Ex 6.2 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 = 0.63 cm

Solution:

(i)  PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Using proportionality theorem

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

PE/EQ ≠ PF/FR

∴EF is not parallel to QR.

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

Using proportionality theorem,

PE/QE = 4/4.5 = 40/45 = 8/9

PF/RF = 8/9

PE/QE = PF/RF

EF is parallel to QR.

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

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55

And, PE/FR = 0.36/2.20 = 36/220 = 9/55

PE/EQ = PF/FR

Hence, EF ∥ QR.

## Ex 6.2 Question 3.

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

Solution:

In ΔABC,  LM || CB,

AM/MB = AL/LC…………..(i)  [B.P.T]

From equation (i) and (ii),

## Ex 6.2 Question 4.

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

Solution:

In ΔABC,  DE || AC

∴BD/DA = BE/EC …………(i)    [B.P.T]

In  ΔABC, DF || AE

∴BD/DA = BF/FE ………………(ii)    [B.P.T]

From equation (i) and (ii)

BE/EC = BF/FE

## Ex 6.2 Question 5.

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

Solution:

In ΔPQO, DE || OQ

PD/DO = PE/EQ…………..(i) [B.P.T]

In ΔPQO, DE || OQ,

PD/DO = PF/FR……… (ii)  [B.P.T]

From equation (i) and (ii)

PE/EQ = PF/FR

EF || QR, in ΔPQR. [By converse of Basic Proportionality Theorem.]

## Ex 6.2 Question 6.

In the 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:

In ΔOPQ, AB || PQ

OA/AP = OB/BQ…………….(i)   [B.P.T]

In ΔOPR, AC || PR

∴ OA/AP = OC/CR……………(ii)    [B.P.T]

From equation (i) and (ii),

OB/BQ = OC/CR

In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem,]

## Ex 6.2 Question 7.

Using Basic proportionality theorem, prove that a line drawn through the mid-points 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 – In ΔABC in which D is the midpoint of AB such that AD=DB and DE || BC.

To prove – AE = AE

In ΔABC, DE || BC,

⇒ 1 = AE/EC

∴ AE = EC

Hence, DE bisect AC.

## Ex 6.2 Question 8.

Using Converse of basic proportionality theorem, 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:

In ΔABC, D and E are the midpoints of AB and AC respectively,

∴ AE=EC

⇒ AE/EC = 1

or

⇒AD/BD = AE/EC ⇒ AD/BD || AE/EC  [By converse of the Basic Proportionality Theorem]

## Ex 6.2 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:

ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O.

To prove- AO/BO = CO/DO

Construction – EO || DC

In ABD,

EO || DC

DC || AB

EO || AB

AE/ED = BO/DO………(i)    [B.P.T]

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

EO || DC

AE/ED = AO/CO ………..(i)

From equation (i) and (ii)

BO/DO = AO/CO

or

AO/BO = CO/DO

## Ex 6.2 Question 10.

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:

Given- Quadrilateral ABCD where AC and BD intersects each other at O such that,

AO/BO = CO/DO.

AO/OC = BO/OD……….(i)

In ΔDAB, EO || AB

DE/EA = DO/OB   [B.P.T]

⇒ AE/ED = BO/OD……(ii)

From wquation (i) and (ii)

AO/OC = AE/ED

∴ OE || CD [By using converse of Basic Proportionality Theorem.]

But AB || OE

∴ AB || CD

Hence, quadrilateral ABCD is a trapezium.

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