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Triangles : Exercise - 6.2 (Mathematics NCERT Class 10th)


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Q.1      In figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
2Sol.

(i) In fig. (i),
since DE || BC,
{{AD} \over {DB}} = {{AE} \over {EC}} \Rightarrow {{15} \over 3} = {1 \over {EC}}
 \Rightarrow EC = {3 \over {15}} = {{3 \times 10} \over {15}} = 2 cm.
(ii) In fig. (ii),
since DE || BC,
{{AD} \over {DB}} = {{AE} \over {EC}} \Rightarrow {{AD} \over {7.2}} = {{1.8} \over {5.4}}
 \Rightarrow AD = {{18} \over {54}} \times {{72} \over {10}} = 2.4 cm.


Q.2       E and F are points on the sides PQ and PB respectively of a \Delta PQR. For each of the following cases, state whether EF || QR :
                (i) PE = 3.9 cm, EQ = 4 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
Sol.

(i) We have,
      PE = 3.9 cm, EQ = 4 cm,
      PF = 3.6 cm and FR = 2.4 cm
      Now,  {{PE} \over {EQ}} = {{3.9} \over 4} = 0.97 cm
      and, {{PF} \over {FR}} = {{3.6} \over {2.4}} + {3 \over 2} = 1.2 cm
       \Rightarrow {{PE} \over {EQ}} = {{PF} \over {FR}}

3       \Rightarrow EF does not divide the sides PQ and PR of \Delta PQR in the same ratio.
      Therefore,
EF is not parallel to QR.

(ii) We have, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
       Now, {{PE} \over {EQ}} = {4 \over {4.5}} = {{40} \over {45}} = {8 \over 9}
       and, {{PF} \over {FR}} = {8 \over 9}
        \Rightarrow {{PE} \over {EQ}} = {{PF} \over {EQ}}
       Thus, EF divides sides PQ and PR of \Delta PQR in the same ratio.
        Therefore, by the converse of Basic
Proportionality Theorem we have EF || QR.

(iii) We have, PQ = 1.28 cm, PR = 2.56 cm
        PF = 0.18 cm and, PF = 0.36 cm
        Therefore EQ = PQ – PE = (1.28 – 0.18) cm.
                                 = 1.10 cm
         and, ER = PR – PF = (2.56 – 0.36)
                       = 2.20 cm
         Now, {{PE} \over {EQ}} = {{0.18} \over {1.10}} = {{18} \over {110}} = {9 \over {55}}
         and, {{PF} \over {FR}} = {{0.36} \over {2.20}} = {{36} \over {220}} = {9 \over {55}}
          \Rightarrow {{PE} \over {EQ}} = {{PF} \over {FR}}
         Thus, EF divides sides PQ and PR of \Delta PQR in the same ratio.
         Therefore, by the converse of Basic
Proportionality Theorem, we have EF || QR


Q.3        In figure, if LM || CB and LN || CD, prove that {{AM} \over {AB}} = {{AN} \over {AD}}.
4Sol.

In \Delta ABC , we have
LM || CB [Given]
Therefore by Proportionality Theorem, we have
{{AM} \over {AB}} = {{AL} \over {AC}}
In \Delta ACD, we have
LN || CD [Given]
Therefore By Basic Proportionality Theorem, we have
{{AL} \over {AC}} = {{AN} \over {AD}} .............. (2)
From (1) and (2), we obtain that
{{AM} \over {AB}} = {{AN} \over {AD}}


Q.4      In figure, DE || AC and DF || AE. Prove that {{BF} \over {FE}} = {{BE} \over {EC}}
5Sol.

In \Delta BCA, we have
DE || AC [Given]
Therefore By Basic Proportionality Theorem, we have
{{BE} \over {EC}} = {{BD} \over {DA}} ............ (1)
In \Delta BEA, we have
DF || AE [Given]
Therefore by Basic Proportionality Theorem, we have
{{BF} \over {FE}} = {{BD} \over {DA}} ............. (2)
From (1) and (2), we obtain that
{{BF} \over {FE}} = {{BE} \over {EC}}


Q.5 In figure, DE || OQ and DF || OR. Show that EF || QR.  6Sol.

In \Delta PQO, we have
DE || OQ [Given]
Therefore By Basic Proportionality Theorem, we have
{{PE} \over {EQ}} = {{PD} \over {DO}} ............ (1)
In \Delta POR, we have
DF || OR [Given]
Therefore By Basic Proportionality Theorem, we have
{{PD} \over {DO}} = {{PF} \over {FR}} .............. (2)
From (1) and (2), we obtain that
{{PE} \over {EQ}} = {{PF} \over {FR}}
 \Rightarrow EF||QR [By the converse of BPT]


Q.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.
7Sol.

Given : O is any point within \Delta PQR, AB || PQ and AC || PR
To prove : BC || QR
Construction : Join BC
Proof : In \Delta OPQ, we have
AB || PQ [Given]
Therefore By Basic Proportionality Theorem, we have
{{OA} \over {AP}} = {{OB} \over {BQ}} ......... (1)
In \Delta OPR, we have
AC || PR [Given]
Therefore by Basic Proportionality Theorem, we have
{{OA} \over {AP}} = {{OC} \over {CR}} ............. (2)
From (1) and (2), we obtain that
{{OB} \over {BQ}} = {{OC} \over {CR}}
Thus, in \Delta 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.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).
Sol.

Given : A \Delta ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.  8To Prove : AE = EC
Proof : Since DE || BC
Therefore by Basic Proportionality Theorem, we have
{{AD} \over {DB}} = {{AE} \over {EC}} ............ (1)
But, AD = DB [Because D is the mid-point of AB]
 \Rightarrow {{AD} \over {DB}} = 1
From (1), {{AE} \over {EC}} = 1  \Rightarrow AE = EC
Hence, E is the mid-point of the third side AC.


Q.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).
Sol.

Given : A \Delta ABC, in which D and E are the mid-points of sides AB and AC respectively.
9To 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 {{AD} \over {DB}} = 1 and AE = EC
Therefore {{AE} \over {EC}} = 1
Therefore {{AD} \over {DB}} = {{AE} \over {EC}} = 1
i.e. {{AD} \over {DB}} = {{AE} \over {EC}}
Thus, in \Delta 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.9      ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that {{AO} \over {BO}} = {{CO} \over {DO}}
Sol.

Given : A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : {{AO} \over {BO}} = {{CO} \over {DO}}
Construction : Through O, draw OE || AB i.e. OE || DC
10Proof : In \Delta ADC, we have OE || DC [Construction]
Therefore by Basic Proportionality Theorem, we have
{{AE} \over {ED}} = {{AO} \over {CO}} ................. (1)
Again, in \Delta ABD, we have
OE || AB [Construction]
Therefore by Basic Proportionality Theorem, we have
{{ED} \over {AE}} = {{DO} \over {BO}} \Rightarrow {{AE} \over {ED}} = {{BO} \over {DO}} ............... (2)
From (1) and (2), we obtain that
{{AO} \over {CO}} = {{BO} \over {DO}} \Rightarrow {{AO} \over {BO}} = {{CO} \over {DO}}


Q.10    The diagonals of a quadrilateral ABCD intersect each other at the point O such that {{AO} \over {BO}} = {{CO} \over {DO}}. 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
{{AO} \over {BO}} = {{CO} \over {DO}} i.e. {{AO} \over {CO}} = {{BO} \over {DO}}
To prove : Quadrilateral ABCD is a trapezium.
Construction : Through O draw OE || AB meeting AD in E.
Proof : In \Delta ADB, we have
OE || AB [Construction]
Therefore by Basic Proportionality Theorem, we have
{{DE} \over {EA}} = {{OD} \over {BO}}
11 \Rightarrow {{EA} \over {DE}} = {{BO} \over {DO}}
 \Rightarrow {{EA} \over {DE}} = {{BO} \over {DO}} = {{AO} \over {CO}}\left[ {Therefore{{AO} \over {CO}} = {{BO} \over {DO}}\left( {given} \right)} \right]
 \Rightarrow {{EA} \over {DE}} = {{AO} \over {CO}}
Thus, in \Delta 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.