NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2: The NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 explain the Basic Proportionality Theorem (BPT) in a clear and simple way.
Students learn how a line parallel to one side of a triangle divides the other two sides proportionally, helping them solve problems based on ratios and geometric reasoning. These solutions are essential for board exam preparation and strengthen core concepts in class 10 Triangles exercise 6.2.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Triangles
Below are the solutions for Class 10 Maths Chapter 6 Exercise 6.2 Triangles, covering important concepts from Triangles Progressions.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 || OQSo by using 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 || PQBy using 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 mid point 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 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 mid points 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.
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)
Hence, quadrilateral ABCD is a trapezium with AB || CD.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 PDF
The Class 10 Triangles 6.2 PDF offers clear, step-by-step solutions based on the Basic Proportionality Theorem. It helps students understand concepts easily and revise quickly for exams. This PDF is a reliable resource for accurate solutions and effective practice. Download the chapter-wise PDF below to prepare confidently.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 PDF
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