Class 10 Maths Chapter 7 Triangles Exercise 7.2 NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2 focus on the criteria for similarity of triangles, including AA (Angle-Angle), SAS (Side-Angle-Side), and SSS (Side-Side-Side). These conditions help determine when two triangles are similar without comparing all sides and angles individually, making this topic an essential part of the CBSE Class 10 syllabus. 

These NCERT Solutions are explained step by step, making it easier to understand how each criterion is applied in different problems. Practising this exercise strengthens logical reasoning and will help you confidently handle geometry proofs in exams.

NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2

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

Solution:

(i) Given, in △ ABC, DE∥BC ∴ AD/DB = AE/EC [Using Basic proportionality theorem] ⇒1.5/3 = 1/EC ⇒EC = 3/1.5 EC = 3×10/15 = 2 cm Hence, EC = 2 cm.

(ii) Given, in △ ABC, DE∥BC 

∴ AD/DB = AE/EC [Using Basic proportionality theorem] 

⇒ AD/7.2 = 1.8 / 5.4 

⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10 

⇒ AD = 2.4 

Hence, AD = 2.4 cm.

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:

Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below;

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get, 

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, we get, PE/EQ ≠ PF/FR Hence, EF is not parallel to QR.

(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm

Therefore, by using Basic proportionality theorem, we get, 

PE/QE = 4/4.5 

= 40/45 

= 8/9 

And, PF/RF = 8/9 So, we get here, PE/QE = PF/RF

Hence, EF is parallel to QR.

(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

From the figure, 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 …………. (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55 ………… (ii)

So, we get here, PE/EQ = PF/FR. Hence, EF is parallel to QR.

3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

Solution:

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get, AM/AB = AL/AC ……………………..(i)

Similarly, given LN || CD and using basic proportionality theorem, 

∴ AN/AD = AL/AC ……………………………(ii)

From equations (i) and (ii), we get AM/AB = AN/AD Hence, proved.

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

Solution:

In ΔABC, given as, DE || AC. Thus, by using Basic Proportionality Theorem, we get, ∴BD/DA = BE/EC ……………………………………………… (i)

In  ΔBAE, given as, DF || AE Thus, by using Basic Proportionality Theorem, we get, ∴BD/DA = BF/FE ……………………………………………… (ii)

From equations (i) and (ii), we get BE/EC = BF/FE. Hence, proved.

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

Solution:

Given, In ΔPQO, DE || OQ

So by using the Basic Proportionality Theorem, PD/DO = PE/EQ……………… ..(i)

Again given, in ΔPOR, DF || OR, So by using Basic Proportionality Theorem, PD/DO = PF/FR………………… (ii)

From equations (i) and (ii), we get, PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem, EF || QR, in ΔPQR.

6. In the 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:

Given here, In ΔOPQ, AB || PQ

By using the Basic Proportionality Theorem, OA/AP = OB/BQ……………. (i)

Also given, In ΔOPR, AC || PR By using Basic Proportionality Theorem ∴ OA/AP = OC/CR……………(ii) 

From equations (i) and (ii) , we get, OB/BQ = OC/CR

Therefore, by converse of Basic Proportionality Theorem, In ΔOQR, BC || QR.

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:

Given, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the midpoint of AC. Since D is the mid-point of AB. ∴ AD=DB ⇒AD/DB = 1 …………………………. (i)

In ΔABC, DE || BC, By using Basic Proportionality Theorem, Therefore, AD/DB = AE/EC From equation (i), we can write, ⇒ 1 = AE/EC 

∴ AE = EC

Hence, proved, E is the midpoint of AC.

8. Using the 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:

Given, in ΔABC, D and E are the midpoints of AB and AC respectively, such that AD=BD and AE=EC.

We have to prove that: DE || BC. 

Since, D is the midpoint of AB ∴ AD=DB ⇒AD/BD = 1……………………………….. (i)

Also given, E is the mid-point of AC. 

∴ AE=EC ⇒ AE/EC = 1

From equation (i) and (ii) , we get, AD/BD = AE/EC By converse of Basic Proportionality Theorem, DE || BC Hence, proved.

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 where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB 

In ΔADC, we have OE || DCTherefore, By using Basic Proportionality Theorem AE/ED = AO/CO  ……………..(i)

Now, In ΔABD, OE || AB 

Therefore, 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 

Hence, proved.

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:

Given, Quadrilateral ABCD where AC and BD intersect each other at O such that, AO/BO = CO/DO.

We have to prove here, ABCD is a trapezium. 

From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB In ΔDAB, EO || AB

Therefore, By using Basic Proportionality Theorem DE/EA = DO/OB ……………………(i)

Also, given, AO/BO = CO/DO 

⇒ AO/CO = BO/DO 

⇒ CO/AO = DO/BO 

⇒DO/OB = CO/AO  …………………………..(ii)

From equation (i) and (ii) , we get DE/EA = CO/AO

Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB ⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.