# NCERT Solutions for Class 10 Maths:Chapter 6 Similar triangles

In this page we have NCERT Solutions for Class 10 Mathematics:Chapter 6 Similar triangles< for EXERCISE 2 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Note: All the questions are based
Basic Proportionally 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 ΔABC , if DE || BC,

Question 1.
In below figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution
(i) In ? ABC, DE?BC
Now from Basic proportionality theorem
1.5/3 = 1/EC
EC = 3/1.5
EC = 2 = 2 cm
(ii) In ? ABC, DE?BC
Now from Basic proportionality theorem
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, PE/EQ ≠ PF/FR
So as per 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, PE/QE = PF/RF
So as per 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
∴ PE/EQ = PF/FR.
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,
From (i) and (ii), we get
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
PD/DO = PE/EQ ...(i)
In ΔPQO, DE || OQ
By using Basic Proportionality Theorem
PD/DO = PF/FR ...(ii)
From equation (i) and (ii), we get
PE/EQ = 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
OA/AP = OB/BQ ...(i)
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR ...(ii)
From equation (i) and (ii), we get
OB/BQ = 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/BD = 1 ... (i)
In ΔABC, DE || BC,
By using Basic Proportionality Theorem
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)
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
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. Show that AO/BO = 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, EO||AB
OE || DC (By Construction)
By using Basic Proportionality Theorem
AE/ED = AO/CO ...(i)
In ΔABD, we have
OE || AB (By Construction)
By using Basic Proportionality Theorem
DE/EA = DO/BO ...(ii)
From equation (i) and (ii), we get
AO/CO = BO/DO
AO/BO = CO/DO
Question 10
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = 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
AO/BO = 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
DE/EA = DO/OB ...(ii)
From (i)
AO/CO = BO/DO
CO/AO = BO/DO
DO/OB = CO/AO ...(iii)
From equation (ii) and (iii), we get
DE/EA = 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