Exercise 6.2 – Stepwise Solutions
1. In Fig. 6.17, DE ∥ BC
(i) Find EC
Given:
AD = 1.5 cm
DB = 3 cm
AE = 1 cm
Since DE ∥ BC, by Basic Proportionality Theorem:
AD / DB = AE / EC
1.5 / 3 = 1 / EC
1/2 = 1 / EC
EC = 2 cm
Answer: EC = 2 cm
(ii) Find AD
Given:
BD = 7.2 cm
AE = 1.8 cm
EC = 5.4 cm
Since DE ∥ BC:
AD / DB = AE / EC
AD / 7.2 = 1.8 / 5.4
AD / 7.2 = 1 / 3
AD = 7.2 × 1/3
AD = 2.4 cm
Answer: AD = 2.4 cm
2. Check whether EF ∥ QR
Condition: PE/PQ = PF/PR
(i)
PE = 3.9, EQ = 3 ⇒ PQ = 6.9
PF = 3.6, FR = 2.4 ⇒ PR = 6
PE/PQ = 3.9/6.9 = 13/23
PF/PR = 3.6/6 = 3/5
Ratios not equal
EF is NOT parallel to QR
(ii)
PE = 4, QE = 4.5 ⇒ PQ = 8.5
PF = 8, RF = 9 ⇒ PR = 17
PE/PQ = 4/8.5 = 8/17
PF/PR = 8/17
Ratios equal
EF ∥ QR
(iii)
PQ = 1.28, PR = 2.56
PE = 0.18, PF = 0.36
PE/PQ = 0.18/1.28 = 9/64
PF/PR = 0.36/2.56 = 9/64
Ratios equal
EF ∥ QR
3. Prove AM/AB = AN/AD
Given: LM ∥ CB and LN ∥ CD
From LM ∥ CB:
AM / AB = AL / AC
From LN ∥ CD:
AN / AD = AL / AC
Therefore:
AM / AB = AN / AD
Proved
4. Prove BF/FE = BE/EC
Given: DE ∥ AC and DF ∥ AE
From DE ∥ AC:
BD / DA = BE / EC
From DF ∥ AE:
BD / DA = BF / FE
Therefore:
BF / FE = BE / EC
Proved
Final Answers Summary
1(i) EC = 2 cm
1(ii) AD = 2.4 cm
2(i) Not parallel
2(ii) Parallel
2(iii) Parallel
3) Proved
4) Proved
Exercise 6.2 – Stepwise Solutions (Q5 to Q10)
5. In Fig. 6.20, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.
Given:
DE ∥ OQ and DF ∥ OR
In triangle POQ, since DE ∥ OQ:
PD / PO = PE / PQ …(1)
In triangle POR, since DF ∥ OR:
PD / PO = PF / PR …(2)
From (1) and (2):
PE / PQ = PF / PR
By Converse of Basic Proportionality Theorem:
EF ∥ QR
6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.
Since AB ∥ PQ, in triangle OPQ:
OA / OP = OB / OQ …(1)
Since AC ∥ PR, in triangle OPR:
OA / OP = OC / OR …(2)
From (1) and (2):
OB / OQ = OC / OR
By Converse of Basic Proportionality Theorem:
BC ∥ QR
7. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Let DE ∥ BC in triangle ABC and D be the midpoint of AB.
By Basic Proportionality Theorem:
AD / DB = AE / EC
Since D is midpoint, AD = DB
So, AE = EC
Therefore, E is midpoint of AC.
Hence proved.
8. Prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Let D and E be midpoints of AB and AC respectively.
So, AD = DB and AE = EC
Therefore:
AD / DB = AE / EC
By Converse of Basic Proportionality Theorem:
DE ∥ BC
Hence proved.
9. ABCD is a trapezium in which AB ∥ DC and diagonals intersect at O. Show that AO/BO = CO/DO.
Since AB ∥ DC:
∠AOB = ∠COD (Vertical angles)
∠ABO = ∠CDO (Alternate interior angles)
So, triangles AOB and COD are similar.
Therefore:
AO / BO = CO / DO
Hence proved.
10. If AO/BO = CO/DO, show that ABCD is a trapezium.
Given:
AO / BO = CO / DO
So, triangles AOB and COD are similar.
Therefore:
∠ABO = ∠CDO
Hence, AB ∥ DC
Since one pair of opposite sides are parallel,
ABCD is a trapezium.
Final Answers Summary
5) EF ∥ QR
6) BC ∥ QR
7) Proved
8) Proved
9) AO/BO = CO/DO proved
10) ABCD is a trapezium