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.