Gujarat Board GSEB Solutions Class 10 Maths Chapter 6 Triangle Ex 6.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 6 Triangle Ex 6.2

Question 1.

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

Solution:

(i) In ∆ABC

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

⇒ \(\frac {1.5}{3}\) = \(\frac {1}{EC}\)

EC = \(\frac {3}{1.5}\)

EC = \(\frac {3}{1.5}\) = 2 cm

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:

(i) We have

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

then \(\frac {PE}{EQ}\) = \(\frac {3.9 cm}{3}\) = \(\frac {1.3}{1}\) ………(1)

and \(\frac {PF}{FR}\) = \(\frac {3.6}{2.4}\) = \(\frac {3}{2}\) \(\frac {1.5}{1}\) ……. (2)

From equation (1) and (2) we have

\(\frac {PE}{EQ}\) – \(\frac {PF}{FR}\) (by converse of BPT)

Therefore EF is not parallel to QR.

(ii) We have

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

then,

\(\frac {PE}{EQ}\) = \(\frac {4}{4.5}\) = \(\frac {40}{45}\) = \(\frac {8}{9}\) ……….(1)

and = \(\frac {PE}{RF}\) = \(\frac {8}{9}\) ……….(2)

from equation (1) and (2)

We get = (by converse of BPT)

therefore EF || QR

(iii) We have

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

\(\frac {PE}{EQ}\) = \(\frac {18}{110}\)

\(\frac {PE}{EQ}\) = \(\frac {9}{55}\) ………(1)

and \(\frac {PE}{FR}\) = \(\frac{0.36}{2.56-0.36}\)

⇒ \(\frac {PE}{FR}\) = \(\frac{0.36}{2.20}=\frac{36}{220}\)

⇒ \(\frac {PE}{FR}\) = \(\frac{9}{55}\) ………(2)

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

Therefore EF || QR (by converse of BPT)

Question 3.

In figure, if LM || CB and LN || CD, prove that (AI 2010)

\(\frac {AM}{AB}\) = \(\frac {AN}{AD}\)

Solution:

LM || BC

\(\frac {AM}{MB}\) = \(\frac {AL}{LC}\) (by BPT) ……. (1)

Now, in ∆ACD

LN || CD

Therefore = (by BPT) ……..(2)

From equation (1) and (2)

\(\frac {AM}{MB}\) = \(\frac {AN}{ND}\)

or \(\frac {MB}{AM}\) = \(\frac {ND}{AN}\) (by Invertendo)

\(\frac {MB}{AM}\) + 1 = \(\frac {ND}{AN}\) + 1 (adding 1 both sides)

\(\frac{MB + AM}{AM}\) = \(\frac{ND + AN}+{AN}\)

⇒ \(\frac{AB}{AM}\) = \(\frac{AD}{AN}\)

⇒ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\) (by Invertendo)

Question 4.

In figure, DE || AC, and DF || AE prove that

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

Solution:

In ∆ABE

DF || AE

\(\frac{AD}{BD}\) = \(\frac{FE}{BF}\) ………(1) (by BPT)

In ∆ABC

DE || AC

\(\frac{AD}{DB}\) = \(\frac{EC}{BE}\) ……..(2) (ByBPT)

From equation (1) and (2)

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

\(\frac{BF}{FE}\) = \(\frac{BE}{EC}\) (by Invertendo)

Question 5.

In figure DE || OQ and DF || OR. Show that EF || QR. (Foreign 2008)

Solution:

In ∆POQ

DE || OQ

\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ………(1) (ByBPT)

Now, in ∆PRO

DF || OR

Therefore = \(\frac{PD}{DO}\) = \(\frac{PF}{FR}\) ……..(2) (by BPT)

From eqn (1) and (2) we get

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

Therefore EF || QR (by converse of BVT)

Question 6.

In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. (CBSE 2007)

Solution:

In ∆OPQ

AB || PQ

then = \(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……(1) (by BPT)

In ∆OPR

AC || PR (given)

then = \(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……. (2) (by BPT)

From equation (1) and (2)

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

Therefore BC || QR (by converse of BPT)

Question 7.

Using theorem 6.1 (i.e. Basic proportionality theorem), prove that 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) (CBSE 2012)

Solution:

Given:

∆ABC in which, D is the midpoint of AB and

DE || BC

To prove: E is the midpoint of AC

Proof: In ∆ABC

DE || BC (given)

then \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) ………(1) (by BPT)

∵ D is the midpoint of AB then

AD = DB

Putting this value in eqn (1) we get

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

⇒ 1 = \(\frac{AE}{EC}\)

⇒ AE = EC

Hence E is the midpoint of AC.

Question 8.

Using theorem 6.2 (i.e. Converse of basic proportionality theorem), prove that the line joining the midpoints of any two sides a triangle is a parallel to the third side (Recall that you have done it in class IX)

Solution:

Given: ∆EC in which D and E are the midpoints of sides AB and AC respectively. DE is the line joining D and E.

To prove: DE || BC

Proof: D is the mid point of AB

Then AD = DB

∴ \(\frac{AD}{DB}\) = 1 …….(1)

Similarly

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

From equation (1) and (2)

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

Therefore DE || BC (by converse of BPT)

Question 9.

ABCD is a trapezium in which AB || DC and its diagonal intersect each other at the point O. Show that = \(\frac{AO}+{BO}\) = \(\frac{CO}{DO}\) (CBSE 2004)

Solution:

Given: ABCD is a trapezium in which AB || DC and diagonals of trapezium intersect each other at O.

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

Construction: Through O, draw a line 0E parallel to AB which intersect AD at E.

Proof: In ∆ADC

OE || DC

AB || DC and OE || AB then OE ||DC

then = \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) ……..(1) (By BPT)

In ∆ DBA

OE || AB

\(\frac{DE}{AE}\) = \(\frac{DO}{BO}\) (By BPT)

⇒ \(\frac{AE}{DE}\) = \(\frac{BO}{DO}\) (by Invertendo) ……(2)

From equations (1) and (2)

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

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

Question 10.

The diagonals of quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) , show that ABCD is a trapezium. (CBSE 2012)

Solution:

Given: ABCD is a quadrilateral whose diagonals intersect each other at O such that

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

To Prove: ABCD is a trapezium.

Construction: Draw a line 0E AB which intersect AD at E.

Proof: In ∆DAB

DO || DE (by construction)

then = \(\frac{DO}{BO}\) = \(\frac{DE}{AE}\) ……….(1)

But = \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) (Given)

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

from equation (1) and (2)

\(\frac{CO}{AO}\) = \(\frac{DE}{AE}\)

or \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) (by Invertendo)

∴ In ∆ADC

OE || CD (by converse of BPT)

But OE || AB (by construction)

therefore AB ||CD

Hence quadrilateral ABCD is a trapezium.