Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 1.

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

Solution :

(i) we have, AD = 1.5 cm, AE = 1 cm, DB = 3 cm

DE || BC

\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

[By theorem 6.1 (BPT)]

\(\frac{1 \cdot 5}{3 \cdot 0}=\frac{1}{\mathrm{EC}}\) = 2

EC = \(\frac{3.0}{1.5}\) = 2

Hence, EC = 2 cm.

(ii) We have, DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm and DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]

\(\frac{\mathrm{AD}}{7 \cdot 2}=\frac{1 \cdot 8}{5 \cdot 4}\)

AD = \(\frac{7.2 \times 1.8}{5.4}\) = 2.4

Hence, AD = 2.4 cm.

Question 2.

E and F are points on the sides PQ and PR respectively of a APQR. 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 = 018 cm and PF = 0.36 cm.

Solution :

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

Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3.0}\) = 1.3

and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3 \cdot 6}{2 \cdot 4}\) = 1.5

∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF is not parallel to QR.

(ii) We have,

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

Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4 \cdot 5}=\frac{8}{9}\) and

\(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}\)

∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF || QR

(iii) We have,

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

EQ = PQ – PE = 1.28 – 0.18 = 11 cm,

PR = PF = 2.56 – 0.36 = 2.2 cm.

Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0 \cdot 18}{1 \cdot 1}=\frac{9}{55}\)

and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2 \cdot 2}=\frac{9}{55}\)

∵ \(\frac{P E}{E Q}=\frac{P F}{F R}\)

∴ EF || QR

Question 3.

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

Solution:

Given: In □ABCD, LM || CB and LN || CD

To Prove: \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Solution: In ∆ABC LM || CB (given)

Question 4.

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

Solution:

Given : In ∆ABC DE || AC and DF || AE

To prove: \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).

Proof: In ∆BAE,

DF || AE (given)

∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}\) ………………..(1)

[By Theorem 6.1 (BPT)]

In ∆ABC, DE || AC (given)

∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]

From(1) and (2) we get

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

Hence Proved.

Question 5.

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

Solution:

Given: In ∆POQ, DE || OQ and in POR, DF || OR

To prove: EF || QR

Proof: In ∆POQ DE || OQ (given)

∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PE}}{\mathbf{E Q}}\) ……………..(1)

[By Theorem 6.1 (BPT)]

In ∆POR DF || OR

∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}}\) ……………..(2)

[By Theorem 6.1 (BPT)]

From (1) and (2), we get

\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF | | QR [By converse of BPT]

Question 6.

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

To prove: BC || QR

Proof: In ∆OPQ

AB || PQ (given)

\(\frac{O A}{A P}=\frac{O B}{B Q}\) …………………(1)

[By theorem 6.1 (BPT)]

In ∆OPR, AC || PR (given)

\(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}\) …………….(2)

[By theorem 6.1 (BPT)]

From (1) and (2) we get

\(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)

Thus B and C are respectively points on sides OQ and OR of ∆OQR, such that \(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\).

⇒ BC || QR [By converse of BPT]

Hence Proved.

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 ∆ABC in which D is the mid-point of AB and DE || BC and meeting AC at E.

To Prove : AE = EC

Proof : Since, DE || BC

\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ……………..(1)

[By Theorem 6.1 (BPT)]

But AD = DB

[∵ D is the mid point of AB]

∴ \(\frac{\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ………………(2)

⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1

⇒ AE = EC.

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 ∆ABC in which D is the mid-point of AB and E is the mid-point of AC

To prove : DE || BC

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

AD = DB

⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = 1

and AE = EC

⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

⇒ DE || BC

[By Theorem of 6.2 (converse of BPT)]

Hence Proved.

Question 9.

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

OR

PQRS is a trapezum in which PQ || RS and its diagonals intersect each other at the point O. Prove \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\)

Solution:

Given : A trapezium ABCD in which AB || CD and its diagonals AC and BD intersect at O.

To Prove: \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

Construction: Draw OE || AB through O, which meets AD at E.

Proof : We have,

EO || AB (By construction) …………(1)

DC || AB (given) …………(2)

From (1) and (2), we get

EO || DC

∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}\) ………………..(3)

[By Theorem 6.1 (BPT)]

In ∆ DAB EO || AB

\(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}\)

∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}}\) ………………(4)

[By Theorem 6.1 (BPT)]

From (3), (4), we get

\(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}\)

⇒ \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

Hence Proved.

Question 10.

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

OR

The diagonals of a quadrilateral PQRS intersect each other at the point O, such that \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\). Show that PQRS is a trapezium.

Solution:

Given : A quadrilateral ABCD whose diagonals AC and BD intersect at a point O such that

\(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

To Prove : ABCD is a trapezium i.e. AB || DC.

Construction : Draw OE || DC which meets AD at E.

Proof: In ∆ACD,

0E || DC (By construction)

∴ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{AE}}{\mathrm{ED}}\) ……………..(1)

[By Theorem 6.1 (BPT)]

But, \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) (given)

⇒ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{DO}}\) ………………..(2)

From (1) and (2), we get

\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)

⇒ EO || AB

[By Theorem 6.2 (converse of BPT)1

But, OE || DC (By construction)

From (3) and (4), we get

AB || CD

Hence, ABCD is a trapezium. Hence Proved.