NCERT Solutions for Class 10 Maths:Chapter 6 Similar triangles EXERCISE 6.2
Triangles EXERCISE 6.2
In this page we have NCERT Solutions for Class 10 Mathematics:Chapter 6 Similar triangles for
EXERCISE 6.2 on pages 128 and 129. Hope you like them and do not forget to like , social_share
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Note: All the questions are based Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In the $\Delta ABC$ , if DE || BC
i. $\frac {AD}{DB}= \frac {AE}{EC}$
ii. $\frac {AD}{AB}= \frac {AE}{AC}$
iii. $\frac {DB}{AB}= \frac {EC}{AC}$
Question 1.
In below figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution
(i) In ΔABC, DE||BC
Now from Basic proportionality theorem
$\frac {AD}{DB} = \frac {AE}{EC}$
$ \frac {1.5}{3} = \frac {1}{EC}$
EC = 3/1.5
EC = 2 = 2 cm
(ii) In ΔABC, DE||BC
Now from Basic proportionality theorem
$\frac {AD}{DB} = \frac {AE}{EC}$
$ \frac {AD}{7.2} = \frac {1.8}{5.4}$
AD = 1.8×7.2/5.4 = 2.4 cm
Question 2
E and F are points on the sides PQ and PR respectively of a ΔPQR. 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 = 0.63 cm Solution
In ΔPQR, E and F are two points on side PQ and PR respectively.
(i) Given
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2,4 cm
Now the Ratio PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, $\frac {PE}{EQ} \neq \frac {PF}{FR}$
So, as per converse of Basic proportionality theorem
Hence, EF is not parallel to QR.
(ii) Given
PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm
Now the ratio PE/QE = 4/4.5 = 40/45 = 8/9
And PF/RF = 8/9
So, $\frac {PE}{EQ} = \frac {PF}{FR}$
So as per converse of Basic proportionality theorem
Hence, EF is parallel to QR.
(iii) Given PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm
And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55
And, PE/FR = 0.36/2.20 = 36/220 = 9/55
so, $\frac {PE}{EQ} = \frac {PF}{FR}$
So as per converse of Basic proportionality theorem
Hence, EF is parallel to QR. Question 3
In the below figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD
Solution
In the given figure, LM || CB
By using basic proportionality theorem, we get,
AM/MB = AL/LC ... (i)
Similarly, LN || CD
By using basic proportionality theorem, we get,
AN/AD = AL/LC ... (ii)
From (i) and (ii), we get
AM/MB = AN/AD Question 4
In the below figure DE||AC and DF||AE. Prove that BF/FE = BE/EC
Solution
In ΔABC, DE || AC
By using Basic Proportionality Theorem
BD/DA = BE/EC ...(i)
In ΔABC, DF || AE
By using Basic Proportionality Theorem
BD/DA = BF/FE ...(ii)
From equation (i) and (ii), we get
BE/EC = BF/FE
Question 5
In the below figure, DE||OQ and DF||OR, show that EF||QR.
Solution
In ΔPQO, DE || OQ
By using Basic Proportionality Theorem
$ \frac {PD}{DO} = \frac {PE}{EQ}$ ...(i)
In ΔPRO, DF||OR
By using Basic Proportionality Theorem
$ \frac {PD}{DO} = \frac {PF}{FR} $ ...(ii)
From equation (i) and (ii), we get
$ \frac {PE}{EQ} = \frac {PF}{FR}$
By converse of Basic Proportionality Theorem
So In ΔPQR, EF || QR.
Question 6
In the below 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
In ΔOPQ, AB || PQ
By using Basic Proportionality Theorem
$\frac {OA}{AP}= \frac {OB}{BQ}$ ...(i)
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
$ \frac {OA}{AP}= \frac {OC}{CR}...(ii)
From equation (i) and (ii), we get
$ \frac {OB}{BQ} = \frac {OC}{CR}$
By converse of Basic Proportionality Theorem
So In ΔOQR, BC || QR.
Question 7
Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). Solution
ΔABC in which D is the mid-point of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC. To Prove: E is the mid-point of AC. Proof: D is the mid-point of AB.
∴ AD=DB
⇒ AD/BD = 1 ... (i)
In ΔABC, DE || BC,
By using Basic Proportionality Theorem
AD/DB = AE/EC
Now as AD/BD=1
AE/EC=1
So E is the mid-point of AC Question 8
Using Converse of basic proportionality theorem, 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
ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC. To Prove: DE || BC Proof:
D is the midpoint of AB (Given)
So AD=DB
Then AD/BD = 1 ... (i)
Also, E is the mid-point of AC (Given)
So AE=EC
Then AE/EC = 1 [From equation (i)]
From equation (i) and (ii), we get
AD/BD = AE/EC
By converse of Basic Proportionality Theorem
Hence, DE || BC Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Prove that $ \frac {AO}{BO} = \frac {CO}{DO}$ Solution
Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.
Through O, draw EO || DC
Now as DC || AB, So, EO||AB
In ΔADC, we have
OE || DC (By Construction)
By using Basic Proportionality Theorem
$ \frac {AE}{ED} = \frac {AO}{CO}$ ...(i)
In ΔABD, we have
OE || AB (Proved above)
By using Basic Proportionality Theorem
$ \frac {DE}{EA} = \frac {DO}{BO}$ ...(ii)
From equation (i) and (ii), we get
$\frac {AO}{CO} = \frac {BO}{DO}$
$ \frac {AO}{BO} = \frac {CO}{DO}$ 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
According to the question
Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that
$ \frac {AO}{BO} = \frac {CO}{DO}$ --- -(i)
Through O, draw line EO, where EO || AB, which meets AD at E.
In ΔDAB, we have
EO || AB
[By using Basic Proportionality Theorem
$ \frac {DE}{EA} = \frac {DO}{OB}$ ...(ii)
From (i)
$ \frac {AO}{CO} = \frac {BO}{DO}$
Rearranging the equation,we get
$ \frac {CO}{AO} = \frac {BO}{DO}$
$ \frac {DO}{OB} = \frac {CO}{AO}$ ...(iii)
From equation (ii) and (iii), we get
$ \frac {DE}{EA} = \frac {CO}{AO}$
By using converse of Basic Proportionality Theorem
Therefore, EO || DC
Now EO || AB
So AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD
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