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.