NCERT solutions for class 10 Maths
Chapter 6
Triangles
Exercise 6.2
Ex 6.2 Question 1.
In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) In given △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
(ii) In given △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using 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
Ex 6.2 Question 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:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Using proportionality theorem
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
PE/EQ ≠ PF/FR
∴EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Using proportionality theorem,
PE/QE = 4/4.5 = 40/45 = 8/9
PF/RF = 8/9
PE/QE = PF/RF
EF is parallel to QR.
(iii) 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
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55
And, PE/FR = 0.36/2.20 = 36/220 = 9/55
PE/EQ = PF/FR
Hence, EF ∥ QR.
Ex 6.2 Question 3.
In the figure, if LM || CB and LN || CD, prove that AM/MB = AN/AD
Solution:
In ΔABC, LM || CB,
AM/MB = AL/LC…………..(i) [B.P.T]
In ΔADC, LN || CD,
∴AN/AD = AL/LC…………(ii) [B.P.T]
From equation (i) and (ii),
AM/MB = AN/AD
Ex 6.2 Question 4.
In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC
Solution:
In ΔABC, DE || AC
∴BD/DA = BE/EC …………(i) [B.P.T]
In ΔABC, DF || AE
∴BD/DA = BF/FE ………………(ii) [B.P.T]
From equation (i) and (ii)
BE/EC = BF/FE
Ex 6.2 Question 5.
In the figure, DE||OQ and DF||OR, show that EF||QR.
Solution:
In ΔPQO, DE || OQ
PD/DO = PE/EQ…………..(i) [B.P.T]
In ΔPQO, DE || OQ,
PD/DO = PF/FR……… (ii) [B.P.T]
From equation (i) and (ii)
PE/EQ = PF/FR
EF || QR, in ΔPQR. [By converse of Basic Proportionality Theorem.]
Ex 6.2 Question 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:
In ΔOPQ, AB || PQ
OA/AP = OB/BQ…………….(i) [B.P.T]
In ΔOPR, AC || PR
∴ OA/AP = OC/CR……………(ii) [B.P.T]
From equation (i) and (ii),
OB/BQ = OC/CR
In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem,]
Ex 6.2 Question 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 in which D is the midpoint of AB such that AD=DB and DE || BC.
To prove – AE = AE
In ΔABC, DE || BC,
Therefore, AD/DB = AE/EC [B.P.T]
But AD=DB
⇒ AD/DB = 1
⇒ 1 = AE/EC
∴ AE = EC
Hence, DE bisect AC.
Ex 6.2 Question 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:
In ΔABC, D and E are the midpoints of AB and AC respectively,
AD=BD and AE=EC.
∴ AD=DB
⇒AD/BD = 1
∴ AE=EC
⇒ AE/EC = 1
or
⇒AD/BD = AE/EC ⇒ AD/BD || AE/EC [By converse of the Basic Proportionality Theorem]
Ex 6.2 Question 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:
ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O.
To prove- AO/BO = CO/DO
Construction – EO || DC
In ABD,
EO || DC
DC || AB
EO || AB
AE/ED = BO/DO………(i) [B.P.T]
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC,
EO || DC
AE/ED = AO/CO ………..(i)
From equation (i) and (ii)
BO/DO = AO/CO
or
AO/BO = CO/DO
Ex 6.2 Question 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 intersects each other at O such that,
AO/BO = CO/DO.
AO/OC = BO/OD……….(i)
In ΔDAB, EO || AB
DE/EA = DO/OB [B.P.T]
⇒ AE/ED = BO/OD……(ii)
From wquation (i) and (ii)
AO/OC = AE/ED
∴ OE || CD [By using converse of Basic Proportionality Theorem.]
But AB || OE
∴ AB || CD
Hence, quadrilateral ABCD is a trapezium.