- Similar Figures
- |
- Similar Polygons
- |
- Basic Proportionally Theorem (or Thales Theorem)
- |
- Criteria for Similarity of Triangles
- |
- Different Criterion for similarity of the triangles
- |
- Areas of Similar Triangles
- |
- Pythagoras Theorem
- |
- Converse of Pythagoras Theorem

In this page we have *NCERT Solutions for Class 10 Mathematics:Chapter 6 Similar triangles* for
EXERCISE 6.2 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

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,

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

(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

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

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.

In the below figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD

In the given figure, LM || CB

By using basic proportionality theorem, we get,

AM/MB = AL/LC ...

Similarly, LN || CD

By using basic proportionality theorem, we get,

AN/AD = AL/LC ...

From

AM/MB = AN/AD

In the below figure DE||AC and DF||AE. Prove that BF/FE = BE/EC

In ΔABC, DE || AC

By using Basic Proportionality Theorem

BD/DA = BE/EC ...

In ΔABC, DF || AE

By using Basic Proportionality Theorem

BD/DA = BF/FE ...

From equation

BE/EC = BF/FE

In the below figure, DE||OQ and DF||OR, show that EF||QR.

In ΔPQO, DE || OQ

By using Basic Proportionality Theorem

$ \frac {PD}{DO} = \frac {PE}{EQ}$ ...

In ΔPRO, DF||OR

By using Basic Proportionality Theorem

$ \frac {PD}{DO} = \frac {PF}{FR} $ ...

From equation

$ \frac {PE}{EQ} = \frac {PF}{FR}$

By converse of Basic Proportionality Theorem

So In ΔPQR, EF || QR.

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.

In ΔOPQ, AB || PQ

By using Basic Proportionality Theorem

$\frac {OA}{AP}= \frac {OB}{BQ}$ ...

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

$ \frac {OA}{AP}= \frac {OC}{CR}...

From equation

$ \frac {OB}{BQ} = \frac {OC}{CR}$

By converse of Basic Proportionality Theorem

So In ΔOQR, BC || QR.

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).

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.

∴ AD=DB

⇒ AD/BD = 1 ...

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

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).

ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

D is the midpoint of AB (Given)

So AD=DB

Then AD/BD = 1 ...

Also, E is the mid-point of AC (Given)

So AE=EC

Then AE/EC = 1 [From equation

From equation

AD/BD = AE/EC

By converse of Basic Proportionality Theorem

Hence, DE || BC

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}$

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}$ ...

In ΔABD, we have

OE || AB (Proved above)

By using Basic Proportionality Theorem

$ \frac {DE}{EA} = \frac {DO}{BO}$ ...

From equation

$\frac {AO}{CO} = \frac {BO}{DO}$

$ \frac {AO}{BO} = \frac {CO}{DO}$

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.

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}$ ---

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}$ ...

From

$ \frac {AO}{CO} = \fac {BO}{DO}$

Rearranging the equation,we get

$ \frac {CO}{AO} = \frac {BO}{DO}$

$ \frac {DO}{OB} = \frac {CO}{AO}$ ...

From equation

$ \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

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.

We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at **[email protected]**

We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.