In this post, you will find the NCERT solutions for class 10 maths ex 6.2. These solutions are based on the latest curriculum of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 6.2
1. In figure (i) and (ii), DE ∥
BC. Find EC in (i) and AD in
(ii).
Solution: (i) Since DE ∥
BC,
Therefore, AD/DB = AE/EC
⇒ 1.5/3 =
1/EC
⇒ EC =
3/1.5
⇒ EC = 2 cm
(ii) Since DE ∥ BC,
Therefore, AD/DB = AE/EC
⇒
AD/7.2 = 1.8/5.4
⇒ AD = (1.8 × 7.2)/5.4
⇒ 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 = 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
Solution: (i) The
figure for the above question is as follows:
Given: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Now,
PE/EQ = 3.9/3 = 1.3 cm
And
PF/FR = 3.6/2.4 = 1.5 cm
Since, PE/EQ
≠ PF/FR
Therefore,
EF is not parallel to QR.
(ii) Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and
RF = 9 cm
Now,
PE/EQ = 4/4.5 = 8/9 cm
And
PF/FR = 8/9 cm
Since, PE/EQ
= PF/FR
Therefore, 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
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
Now,
PE/EQ = 0.18/1.10 = 18/110 = 9/55 cm
And
PF/FR = 0.36/2.20 = 36/220 = 9/55 cm
Since, PE/EQ
= PF/FR
Therefore, EF
is parallel to QR.
3. In figure, if LM ∥
CB and LN ∥
CD, prove that AM/AB = AN/AD.
Solution: In ∆ABC,
LM ∥ CB
Therefore, AM/AB = AL/AC
[From the Basic Proportionality Theorem] ………. (i)
And
in ∆ACD, LN ∥
CD
Therefore, AL/AC = AN/AD
[From the Basic
Proportionality Theorem] ………. (ii)
From
equations (i) and (ii), we have
AM/AB = AN/AD Hence proved.
4. In figure, DE ∥
AC and DF ∥
AE. Prove that BF/FE = BE/EC.
Solution: In ∆BCA, DE ∥ AC
Therefore, BE/EC = BD/DA
[From the Basic
Proportionality Theorem] ………. (i)
And
in ∆BEA, DF ∥
AE
Therefore, BF/FE = BD/DA
[From the Basic
Proportionality Theorem] ………. (ii)
From
equations (i) and (ii), we have
BF/FE
= BE/EC
Hence proved.
5. In figure, DE ∥
OQ and DF ∥
OR. Show that EF ∥
QR.
Solution: In ∆PQO, DE ∥ OQ
Therefore, PE/EQ = PD/DO
[From the Basic
Proportionality Theorem] ………. (i)
And
in ∆POR, DF ∥
OR
Therefore, PD/DO = PF/FR
[From the Basic
Proportionality Theorem] ………. (ii)
From
equations (i) and (ii), we have
PE/EQ
= PF/FR
Since in ∆PQR, PE/EQ = PF/FR
Therefore, EF
∥ QR
[By the converse of Basic Proportionality Theorem]
6. In 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: O is any point in ∆PQR, in which AB ∥
PQ and AC ∥
PR.
To prove: BC ∥ QR
Construction: Join BC.
Proof: In ∆OPQ,
AB ∥ PQ
Therefore, OA/AP = OB/BQ
[From the Basic
Proportionality Theorem] ………. (i)
And
in ∆OPR, AC ∥
PR
Therefore, OA/AP = OC/CR
[From the Basic
Proportionality Theorem] ………. (ii)
From
equations (i) and (ii), we have
OB/BQ
= OC/CR
Since, in ∆OQR, B and C are points dividing the sides OQ
and OR in the same ratio.
Therefore, AB
∥ PQ
[By the converse of Basic Proportionality Theorem]
7. Using Theorem 6.1, prove that a line drawn
through the mid-point 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: A ∆ABC,
in which D is the mid-point of side AB and the line DE is drawn parallel to BC,
meeting AC at E.
To prove: AE = EC
Proof: Since DE ∥
BC
Therefore, AD/DB = AE/EC
[From the Basic
Proportionality Theorem] ………. (i)
But
AD = DB [Given]
⇒ AD/DB = 1
⇒ AE/EC = 1 [From equation (i)]
⇒ AE = EC
Hence,
E bisects the third side AC.
8. Using Theorem 6.2, 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: A ∆ABC,
in which D and E are the mid-points of sides AB and AC respectively.
To Prove: DE ∥ BC
Proof: Since D and E are the mid-points of AB and
AC respectively.
Therefore, AD
= DB and AE = EC
Now,
AD = DB
⇒ AD/DB = 1
And
AE = EC
⇒ AE/EC = 1
Therefore, AD/DB = AE/EC = 1
⇒ AD/DB = AE/EC
Thus,
in ∆ABC, D and E are points dividing the
sides AB and AC in the same ratio.
Therefore,
DE ∥ BC
[By the 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:
Given: A trapezium
ABCD, in which AB ∥ DC and its diagonals AC and BD
intersect each other at O.
To Prove: AO/BO
= CO/DO
Construction: Through O, draw OE ∥ AB, and hence OE ∥ DC.
Proof: In ∆ADC,
we have OE ∥ DC
Therefore, AE/ED = AO/CO
[From the
Basic Proportionality Theorem] ………. (i)
Again,
in ∆ABD, we have OE ∥ AB [By
construction]
Therefore,
ED/AE = DO/BO [From the Basic Proportionality Theorem]
………. (ii)
⇒ AE/ED = BO/DO ………. (ii)
From
equations (i) and (ii), we get
AO/CO
= BO/DO
⇒ 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:
Given: A quadrilateral
ABCD, in which its diagonals AC and BD intersect each other at O such that AO/BO = CO/DO or AO/CO = BO/DO
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]
Therefore, DE/EA = DO/BO [From the Basic Proportionality Theorem]
⇒ EA/DE = BO/DO
⇒ EA/DE = BO/DO =
AO/CO [Since AO/CO = BO/DO]
⇒ EA/DE = AO/CO
Thus
in ∆ADC, E and O are points dividing
the sides AD and AC in the same ratio. Therefore, EO ∥ DC
[By the converse of Basic Proportionality Theorem]
But
EO ∥ AB
[By construction]
Therefore, AB
∥ DC
Therefore, the quadrilateral ABCD is a trapezium.