NCERT Solutions Class 10 Maths Chapter 6 Triangles: Exercise 6.2
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 assists students in enhancing their problem-solving skills related to geometric shapes, particularly triangles. The exercise applies various concepts of triangle similarity along with the use of some significant theorems involving triangles. All these concepts create a solid foundation for solving several problems in Geometry. So, let's begin to discover this fascinating and significant topic of class 10 that can assist you in grasping the basics of similarity and proportionality in triangles.
1.0Download NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.2: Free PDF
2.0Introduction to Similarity of Triangles:
Let's briefly examine why and how two triangles are similar before diving into the exercise. Two triangles are similar if:
- Their corresponding angles are congruent. For instance, if two triangles, ABC and PQR, are similar, then ∠A = ∠P, ∠B = ∠Q, and ∠C = ∠R.
- The ratio of their corresponding sides is equal (that is, proportional). Like the above example, if triangles ABC and PQR are similar, then:
Once these requirements are met, the two triangles may have unequal sizes, yet they both maintain the same form, thereby forming similar triangles. The property of similarity is commonly used in various geometric proofs and exercises, including this one.
3.0Key Concepts of Exercise 6.2: Overview
Exercise 6.2 is mainly about applying the concept of similarity, especially with the help of the Basic Proportionality Theorem and its converse. These theorems are important in connecting the characteristics of proportionality to line parallelism and the ratio of the sides of similar triangles. Let's discuss these theorems and the main concepts utilised in this exercise:
Basic Proportionality Theorem (Thales' Theorem):
The theorem of basic proportionality is the most frequently applied concept when solving problems of Exercise 6.2. The theorem is such that if a line is drawn parallel to one side of a triangle, then it cuts the other two sides into segments which are proportional to each other. To better understand this theorem, consider an example: with a triangle ABC, draw a line MN parallel to the base BC of the same triangle. Then, as per the theorem,
This theorem is crucial since it allows one to find a proportionality relationship between the sides of a triangle if one side of the triangle is parallel to a line and the line cuts the other two sides of the triangle. The theorem can be proved by the use of concepts regarding the similarity of triangles.
Converse of the Basic Proportionality Theorem:
The converse of the Basic Proportionality Theorem is simply the converse of Thales' Theorem. The theorem says that if, in a triangle, a line bisects the opposite sides in an equal proportion, the line is parallel to the third side. To know this, consider the earlier example, in a triangle ABC, if MN is a line bisecting opposite sides of the triangle in equal ratio, like:
Subsequently, by this theorem, MN shall be parallel to BC. This theorem can be established through concepts related to the similarity of triangles as well as parallel line properties. This theorem holds for a great number of problems in geometry and provides a direct means of creating parallelism among triangles.
Working on Problems in Exercise:
Many questions in Exercise 6.2 can be solved using the theorems discussed above. However, this has to be attempted using the appropriate strategy to yield the required solutions. Here is how you ought to approach the problems of the exercise:
- Determine the two triangles to which Thales' theorem can be utilised effectively.
- Write down the ratio of the sides according to the needs of the questions.
- Apply the theorems, either the Basic Proportionality Theorem or its converse, as appropriate.
Start practising the problems in NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 today to solidify your understanding of these valuable mathematical concepts.
4.0NCERT Class 10 Maths Chapter 6 Triangles Exercise 6.2 : Detailed Solutions
1. In the figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solutions:
(i) In the given figure, it is given that DE || BC.
Therefore, according to the Basic Proportionality Theorem (also known as Thales' Theorem), we have:
AD/DB = AE/EC
Given AD = 1.5 cm, DB = 3 cm, and AE = 1 cm, we can substitute these values:
1.5/3 = 1/EC
Solving for EC:
EC = 3/1.5 = 2 cm
(ii) In the given figure, it is given that DE || BC.
Again, according to the Basic Proportionality Theorem:
AD/BD = AE/CE
Given BD = 7.2 cm, AE = 1.8 cm, and CE = 5.4 cm, we substitute these values:
AD/7.2 = 1.8/5.4
Solving for AD:
AD = (1.8/5.4) * 7.2 = 2.4 cm
2. E and F are points on the sides PQ and PR respectively of triangle PQR. 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 = 0.36 cm.
Solutions:
(i) In the given figure, we need to check if EF || QR.
PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 3/2 = 1.5
=> PE/EQ ≠ PF/FR
=> EF is not || QR
(ii) In figure,
PE/EQ = 4/4.5 = 8/9 and PF/FR = 8/9
=> PE/EQ = PF/FR => EF || QR
(iii) In figure,
PE/QE = 0.18/(PQ - PE) = 0.18/(1.28 - 0.18) = 0.18/1.10 = 18/110 = 9/55
PF/FR = 0.36/(PR - PF) = 0.36/(2.56 - 0.36) = 0.36/2.20 = 9/55
=> PE/QE = PF/FR
∴ EF || QR (By converse of Basic Proportionality Theorem)
3. In figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Solution:
In triangle ACB (see figure), LM || CB (Given)
=> AM/MB = AL/LC (Basic Proportionality Theorem) --- (1)
In triangle ACD (see figure), LN || CD (Given)
=> AN/ND = AL/LC (Basic Proportionality Theorem) --- (2)
From (1) and (2), we get:
AM/MB = AN/ND
=> AM/(AM + MB) = AN/(AN + ND)
=> AM/AB = AN/AD
4. In figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Solution:
In triangle ABE,
DF || AE (Given)
BD/DA = BF/FE (Basic Proportionality Theorem) ...(i)
In triangle ABC,
DE || AC (Given)
BD/DA = BE/EC (Basic Proportionality Theorem) ...(ii)
From (i) and (ii), we get
BF/FE = BE/EC.
Hence proved.
5. In figure, DE || OQ and DF || OR. Show that EF || QR.
Solution:
In the given figure, DE || OQ and DF || OR.
By the Basic Proportionality Theorem, we have:
PE/EQ = PD/DO ...(1)
and
PF/FR = PD/DO ...(2)
From equations (1) and (2), we get:
PE/EQ = PF/FR
Now, in the triangle PQR, we have proved that PE/EQ = PF/FR.
Therefore, EF || QR (By the converse of the Basic Proportionality Theorem).
6. In the figure, A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. To Show: BC || QR.
Solution:
In triangle POQ,
AB || PQ (given)
OB/BQ = OA/AP (Basic Proportionality Theorem) --- (i)
In triangle POR,
AC || PR (given)
OA/AP = OC/CR (Basic Proportionality Theorem) --- (ii)
From (i) and (ii), we get:
OB/BQ = OC/CR
Therefore, by the converse of the Basic Proportionality Theorem,
BC || QR.
7. Using the Basic proportionality theorem, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.
Solution:
In triangle ABC, D is the midpoint of AB (see figure).
i.e., AD/DB = 1
Straight line l || BC.
Line l is drawn through D and it meets AC at E.
By Basic Proportionality Theorem
AD/DB = AE/EC => AE/EC = 1 [From (1)]
=> AE = EC => 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.
Solution:
In the triangle ABC, D and E are mid points of the sides AB and AC respectively.
=> AD/DB = 1 and AE/EC = 1 (see figure)
=> AD/DB = AE/EC => DE || BC (By Converse of Basic Proportionality Theorem)
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:
We draw EOF || AB (also || CD) (see figure)
In triangle ACD, OE || CD
=> AE/ED = AO/OC ...(1)
In triangle ABD, OE || BA
=> DE/EA = DO/OB
=> AE/ED = OB/OD ...(2)
From (1) and (2)
AO/OC = OB/OD
i.e., AO/BO = CO/DO
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:
In figure, AO/BO = CO/DO
=> AO/OC = BO/OD ... (1) (given)
Through O, we draw OE || BA.
OE meets AD at E.
In triangle DAB,
EO || AB
=> DE/EA = DO/OB (by Basic Proportionality Theorem)
=> AE/ED = BO/OD ... (2)
From (1) and (2),
AO/OC = AE/ED
=> OE || CD (by converse of basic proportionality theorem)
Now, we have BA || OE and OE || CD
=> AB || CD
=> Quadrilateral ABCD is a trapezium.
5.0Benefits of NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.2
- Strengthens understanding of the Basic Proportionality Theorem (BPT)
- Essential for mastering similarity in triangles and future geometry concepts.
- Provides clear, structured solutions for better comprehension.
- Enhances logical thinking with numerical and proof-based questions.
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