- In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i)
(ii)
Solution:
(i)
Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
AD/DB = AE/EC
1.5/3 = 1/x
x = (3 x 1)/1.5
x = 2
EC = 2cm
(ii)
Let AD = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
AD/DB = AE/EC
x/7.2 = 1.8/5.4
x = (1.8×7.2)/(5.4)
x = 2.4 cm
AD = 2.4cm
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 = 0.63 cm
Solution:
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
Hence, PE/EQ ≠ PF/FR
Therefore, EF is not parallel to QR.
PE = 4cm, QE = 4.5 cm, PF = 8cm , RF = 9cm
PE/EQ = 4/4.5 = 8/9
PF/FR = 8/9
Hence PE/EQ = PF/FR
Therefore, EF is parallel to QR
(ii)
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
PE/PQ = 0.18/1.28 = 18/128 = 9/64
PF/PR = 0.36/2.56 = 9/54
Hence, PE/PQ = PF/PR
Therefore, EF is parallel to QR
3. In the following figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD
Solution:
In the given figure, LM || CB
By using basic proportionality theorem, we obtain
AM/AB = AL/AC —————–(1)
Similarly, LN||CD
Therefore, AN/AD = AL/AC ————(2)
From (1) and (2), we obtain
AM/AB = AN/AD
4. In the following figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC
Solution:
In ∆ABC, DE || AC
BD/DA = BE/EC (Basic proportionality theorem)
In ∆BAE, DF || AE
BD/DA = BF/EF (Basic proportionality theorem)
5. In the following figure, DE || OQ and DF || OR, show that EF || QR
Solution:
In ∆ POQ, DE || OQ
Therefore, PE/EQ = PD/DO —————–(i)
n ∆ POQ, DF || OR
PF/FR = PD/DO (Basic proportionality theorem) ———–(ii)
From (i) and (ii), we obtain
PE/EQ = PF/FR
Therefore, EF||QR (converse of basic proportionality theorem)
6. In the following 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 ∆ POQ, AB || PQ
OA/AP = OB/BQ (Basic proportionality theorem) ———–(i)
In ∆ POR, AC || PR
OA/AP = OC/CR (Basic proportionality theorem) ———–(ii)
From (i) and (ii)
OB/BQ = OC/CR
Therefore, BC||QR
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:
Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ||BC
By using proportionality theorem, we obtain,
AQ/QC = AP/PB
AQ/QC = 1/1 (P is the midpoint of AB, AP = PB)
Therefore, AQ = QC
Or, Q is the mid-point of AC.
8. Using Converse of basic proportionality theorem, 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).
Solution:
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. i.e., AP = PB and AQ = QC It can be observed that
AP/PB = 1/1
and AQ/Qc = 1/1
AP/PB = AQ/QC
Hence, by using basic proportionality theorem, we obtain
PQ||BC
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:
Draw a line EF through point O, such that EF||CD
In ∆ADC, EO||CD
By using basic proportionality theorem, we obtain
AE/ED = AO/OC ——–(1)
In ΔABD, OE||AB
So, by using basic proportionality theorem, we obtain
ED/AE = OD/BO
Therefore, AE/ED = BO/OD ——(2_
From equations (1) & (2(, we obtain,
AO/OC = BO/OD
Therefore, AO/BO = OC/OD
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:
Let us consider the following figure for the given question.
Draw a line OE || AB
In ∆ABD, OE || AB
By using basic proportionality theorem, we obtain
AE/ED = BO/OD ———–(1)
However, it is given that
AO/OC = OB/OD —————(2)
From equations (1) and (2), we obtain,
AE/ED = AO/OC
⇒ EO || DC [By the converse of basic proportionality theorem]
⇒ AB || OE || DC
⇒ AB || CD
∴ ABCD is a trapezium.
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