These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.2

Question 1.

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

Solution:

Question 2.

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:

We have to show that EF || QR.

If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

(i) PE = 3.9cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Ratio are not same, so EF is not parallel to QR.

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

Therefore, the ratio are same, so EF || QR.

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

Ratio are same, so EF || QR.

Question 3.

In the given figure, if LM || CB and LN || CD, Prove that \(\frac { AM }{ AB }\) = \(\frac { AN }{ AD }\)

Solution:

Given a quadrilateral ABCD, in which LM || BC and LN || CD

In ∆ABC, LM || BC

\(\frac { AM }{ MB }\) = \(\frac { AL }{ LC }\) … (i)

[Basic proportionality theorem in fig.]

In ∆ADC, LN || CD

\(\frac { AN }{ AD }\) = \(\frac { AL }{ LC }\) … (2)

[Basis proportionality theorem]

From equations (1) and (2), we have

Question 4.

In the adjoining figure DE || AC and DF || AE. Prove that \(\frac { BF }{ FE }\) = \(\frac { BE }{ EC }\)

Solution:

Given a quadrilateral ABC, in which DF || AE and DE || AC

In ∆ABE, we have

\(\frac { BD }{ AD }\) = \(\frac { BE }{ FE }\) … (i)

[Basic proportionality theorem in fig.]

In ∆ABC, we have

\(\frac { BD }{ AD }\) = \(\frac { BE }{ EC }\) … (2)

[Basis proportionality theorem]

From equations (1) and (2), we have

\(\frac { BE }{ FE }\) = \(\frac { BE }{ EC }\)

Question 5.

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

Solution:

In the triangle PQD, ED || OQ.

So applying Thales theorem, we get,

Equation (iii) shows that EF intersects PQ and PR at E and F respectively in equal ratio. So, according to the converse of the basic proportionality theorem (or Thales theorem), EF || QR. Flence Proved.

Question 6.

In the given 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 inside ∆PQR. A is any point PO and B is any point on OQ and C is | any point on OR. AB || PQ and AC || PR.

To Prove: BC || QR

Proof: In ∆POQ, AB || PQ

∴ \(\frac { OA }{ AP }\) = \(\frac { OB }{ BQ }\) …(1) [Basic proportionality theorem]

In ∆POR, AC || PR

∴\(\frac { OA }{ AP }\) = \(\frac { OC }{ OR }\) …(2) [Basic proportionality theorem]

From equations (1) and (2)

\(\frac { OB }{ BQ }\) = \(\frac { OC }{ OR }\)

But, we know that

In ∆OQR,if \(\frac { OB }{ BQ }\) = \(\frac { OC }{ OR }\)

then we can say BC || QR.

[Converse of basic proportionality theorem]

Hence Proved.

Question 7.

Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:

Given: A ∆ABC in which D is the mid-point of AB and DE || BC

To Prove : E is the mid-point of AC.

Proof : We have to prove that E is the mid-point of AC. Suppose, let E be not the mid-point of AC. Let E’ be the mid-point of AC. Join DE’.

Now. in ∆ABC, D is the mid-point of AB and E’ is the mid-point of AC. Therefore, by Theorem (6.1).

DE || BC … (1)

Also DE || BC … (2)

From (1) and (2), we find that two intersecting lines DE and DE’ are both parallel to line BC. This is a contradiction to the parallel line axiom.

So, our assumption is wrong. Hence E is the mid-point of AC.

Question 8.

Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution:

Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.

To Prove: DE || BC

Construction : Produce the line segement DE to F, such that DE = EF. Join FC.

Proof: In ΔAED and CEF, we have

AE = CE

[∵ E is the mid-point of AC]

∠AED = ∠CEF

[Vertically opposite angle]

or DE = EF

So by SAS congruence

ΔAED ≅ ΔCEF

⇒ AD = CF … (1)

[corresponding of congruent triangle]

and ∠ADE = ∠CEF … (2)

Now, D is the mid-point of AB,

AD = DB

DB = CF … (3) [From (1)]

Now, DF intersects AD and FC at D and F such that

∠ADE = ∠CFE [From (2)]

i.e., alternate interior angles are equal

∴ AD || FC

DB || CF … (4)

From (3) and (4), DBCF is a quadrilateral such that one pair of sides are equal and parallel.

∴ DBCF is a parallelogram

⇒ DF || BC and DC = BC [opposite side of parallelogram]

But, D, E, F are collinear and DE = EF

∴ DE || BC

Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the O. Show that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\)

Solution:

Question 10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) Show that ABCD is a trapezium.

Solution:

Construction: Draw OF || AB, meeting AD in F.

To Prove: Quadrilateral ABCD is a trapezium.

Proof: In ABD, we have OF || AB

Hence Proved.