NCERT Solutions for Class 10 Maths Exercise 6.2 Chapter 6 Triangles – FREE PDF Download
NCERT Class 10 Maths Ch 6 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
6 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Triangles solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 6- Triangles
1. In figure (i) and (ii), DE 
BC. Find EC in (i) and AD in (ii).
BC, 
EC = 
EC = 2 cm
(ii)Since DE BC,
AD = 
EC = 2.4 cm
2. E and F are points on the sides PQ and PR respectively of a 
PQR. For each of the following cases, state whether EF QR:
(i) PE = 3.9 cm, EQ = 4 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
Now, 
= 0.97 cm
And 
= 1.2 cm
Therefore, EF does not divide the sides PQ and PR of PQR in the same ratio.
EF is not parallel to QR.
(ii)Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, 
= 
cm
And 
cm
Therefore, EF divides the sides PQ and PR of PQR in the same ratio.
EF is parallel to QR.
(iii)Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And ER = PR – PF = 2.56 – 0.36 = 2.20 cm
Now, 
= 
cm
And 
= cm
Therefore, EF divides the sides PQ and PR of PQR in the same ratio.
EF is parallel to QR.
3. In figure, if LM CB and LN CD, prove that 
ABC, LM CB 
[Basic Proportionality theorem] ……….(i)
And in ACD, LN CD
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
4. In the given figure, DE AC and DF AE. Prove that 
BCA, DE AC 
[Basic Proportionality theorem] ……….(i)
And in BEA, DF AE
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
5. In the given figure, DE OQ and DF OR. Show that EF QR.
PQO, DE OQ 
[Basic Proportionality theorem] ……….(i)
And in POR, DF OR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
EF QR [By the converse of BPT]
6. In the given figure, A, B, and C are points on OP, OQ and OR respectively such that AB PQ and AC PR. Show that BC QR.
PQR, in which AB PQ and AC PR. To prove: BC QR
Construction: Join BC.
Proof: In OPQ, AB PQ
OAAP=OBBQOAAP=OBBQ[Basic Proportionality theorem] ……….(i)
And in OPR, AC PR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
In OQR, B and C are points dividing the sides OQ and OR in the same ratio.
By the converse of Basic Proportionality theorem,
BC QR
7. Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
and the line DE is drawn parallel to BC, meeting AC at E.
To prove: AE = EC
Proof: Since DE BC
[Basic Proportionality theorem] ……….(i)
But AD = DB [Given]
[From eq. (i)]
AE = EC
Hence, E is the midpoint of the third side AC.
8. Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
sides AB and AC respectively.
To Prove: DE BC
Proof: Since D and E are the midpoints of AB and AC
respectively.
AD = DB and AE = EC
Now, AD = DB
and AE = EC
= 1
Thus, in triangle ABC, D and E are points dividing the sides AB and AC in the same ratio.
Therefore, by the converse of Basic Proportionality theorem, we have
DE BC
9. ABCD is a trapezium in which AB DC and its diagonals intersect each other at the point O. Show that 
DC and its diagonals AC and BD intersect each other at O.
To Prove: 
Construction: Through O, draw OE AB, i.e. OE DC.
Proof: In ADC, we have OE DC
[By Basic Proportionality theorem]……….(i)
Again, in ABD, we have OE AB[Construction]
[By Basic Proportionality theorem]
……….(ii)
From eq. (i) and (ii), we get
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that Show that ABCD is a trapezium.
BD intersect each other at O such that , i.e.
.
To Prove: Quadrilateral ABCD is a trapezium.
Construction: Through O, draw OE AB meeting AD at E.
Proof: In ADB, we have OE AB [By construction]
[By Basic Proportionality theorem]
= 
=
Thus in ADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore by the converse of Basic Proportionality theorem, we have
EO DC
But EO AB[By construction]
AB D
Quadrilateral ABCD is a trapezium