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

(frac{mathrm{AB}}{mathrm{PQ}}=frac{mathrm{AC}}{mathrm{PR}}=frac{mathrm{AD}}{mathrm{PM}})

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.