Q.1Â Â Â Â Â In figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Sol.
(i) In fig. (i),
since DE || BC,
2 cm.
(ii) In fig. (ii),
since DE || BC,
2.4 cm.
Q.2Â Â Â Â Â Â E and F are points on the sides PQ and PB respectively of a PQR. For each of the following cases, state whether EF || QR :
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (i) PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm.
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Sol.
(i) We have,
Â Â Â Â Â PE = 3.9 cm, EQ = 4 cm,
Â Â Â Â Â PF = 3.6 cm and FR = 2.4 cm
Â Â Â Â Â Now,Â cm
Â Â Â Â Â and, cm
Â Â Â Â Â
Â Â Â Â Â EF does not divide the sides PQ and PR of PQR in the same ratio.
Â Â Â Â Â Therefore, EF is not parallel to QR.
(ii) We have, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Â Â Â Â Â Â Now,
Â Â Â Â Â Â and,
Â Â Â Â Â Â
Â Â Â Â Â Â Thus, EF divides sides PQ and PR of PQR in the same ratio.
Â Â Â Â Â Â Â Therefore, by the converse of BasicProportionality Theorem we have EF || QR.
(iii) We have, PQ = 1.28 cm, PR = 2.56 cm
Â Â Â Â Â Â Â PF = 0.18 cm and, PF = 0.36 cm
Â Â Â Â Â Â Â Therefore EQ = PQ â€“ PE = (1.28 â€“ 0.18) cm.
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.10 cm
Â Â Â Â Â Â Â Â and, ER = PR â€“ PF = (2.56 â€“ 0.36)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.20 cm
Â Â Â Â Â Â Â Â Now,
Â Â Â Â Â Â Â Â and,
Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Thus, EF divides sides PQ and PR of PQR in the same ratio.
Â Â Â Â Â Â Â Â Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR
Q.3Â Â Â Â Â Â Â In figure, if LM || CB and LN || CD, prove that .
Sol.
In ABC , we have
LM || CB [Given]
Therefore by Proportionality Theorem, we have
In , we have
LN || CD [Given]
Therefore By Basic Proportionality Theorem, we have
.............. (2)
From (1) and (2), we obtain that
Q.4Â Â Â Â Â In figure, DE || AC and DF || AE. Prove that
Sol.
In BCA, we have
DE || AC [Given]
Therefore By Basic Proportionality Theorem, we have
............ (1)
In BEA, we have
DF || AE [Given]
Therefore by Basic Proportionality Theorem, we have
............. (2)
From (1) and (2), we obtain that
Q.5 In figure, DE || OQ and DF || OR. Show that EF || QR.Â Sol.
In PQO, we have
DE || OQ [Given]
Therefore By Basic Proportionality Theorem, we have
............ (1)
In POR, we have
DF || OR [Given]
Therefore By Basic Proportionality Theorem, we have
.............. (2)
From (1) and (2), we obtain that
[By the converse of BPT]
Q.6Â Â Â Â Â 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.
Sol.
Given : O is any point within , AB || PQ and AC || PR
To prove : BC || QR
Construction : Join BC
Proof : In , we have
AB || PQ [Given]
Therefore By Basic Proportionality Theorem, we have
......... (1)
In OPR, we have
AC || PR [Given]
Therefore by Basic Proportionality Theorem, we have
............. (2)
From (1) and (2), we obtain that
Thus, in OQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by
the converse of Basic Proportionality Theorem, we have,
BC || QR
Q.7Â Â Â Â Â 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).
Sol.
Given : A ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.Â To Prove : AE = EC
Proof : Since DE || BC
Therefore by Basic Proportionality Theorem, we have
............ (1)
But, AD = DB [Because D is the mid-point of AB]
= 1
From (1), = 1 AE = EC
Hence, E is the mid-point of the third side AC.
Q.8Â Â Â Â Â 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).
Sol.
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
Since, AD = DB
Therefore = 1 and AE = EC
Therefore = 1
Therefore
i.e.
Thus, in 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.
Q.9Â Â Â Â Â ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that
Sol.
Given : A trapezium ABCD, in which AB || 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 [Construction]
Therefore by Basic Proportionality Theorem, we have
................. (1)
Again, in ABD, we have
OE || AB [Construction]
Therefore by Basic Proportionality Theorem, we have
............... (2)
From (1) and (2), we obtain that
Q.10Â Â Â The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.
Sol.
Given : A quadrilateral ABCD in which its diagonals AC and 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 in E.
Proof : In ADB, we have
OE || AB [Construction]
Therefore by Basic Proportionality Theorem, we have
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 [Construction]
Hence, AB || DC
Therefore Quadrilateral ABCD is a trapezium.
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