Chapter 6 Triangles Exercise 6.1
Question 1: In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjC5MFfmvjdvvcyYmgHQI_nF70Dwe9XbXJXQIqddX06bt0AQssYPNOfMbF-mjtta6UD60Zd-AKDN3WQNQNaciaoF2qcwmYLLb3ETM5Y5LhBHlSTrtM9vEf0K3lvCeJwQ8iOwXR2pZYql-1B/s16000/Fig+6.17.png)
Answer:
For this question we will use Basic Proportionality theorem.
i) According to Basic Proportionality Theorem,
AD/DB = AE/EC
= 1.5/3 = 1/EC
= 1.5(EC) = 1(3)
= 1.5 EC = 3
= EC = 3/1.5
= EC = 2
Therefore, EC = 2 cm
ii) According to Basic Proportionality Theorem,
AD/DB = AE/EC
= AD/7.2 = 1.8/5.4
= AD(5.4) = 1.8(7.2)
= 5.4 AD = 12.96
= AD = 12.96/5.4
= AD = 2.4
Therefore, AD = 2.4 cm
Question 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 = 0.36 cm
Answer:
For this question we will use converse of Basic Proportionality Theorem,
i)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6woKi9Zk07AXVlReZ4GIbTlnwr_jRBEnuf70ekT9Vz3GXgCeZI9IeTc2n4UvR2Vx5Ch1Zdkojv6BYdmtD1GbNHDvOumIzj0cEFOFOp813k-BfdpmrSL3Le5TNfCjU95wsubCoUZ-wMNEb/s16000/1.png)
= 3.9/3 = 3.6/2
Since 3.9/3 ≠ 3.6/2.4. Therefore by converse of Basic Proportionality Theorem, EF is not parallel to QR.
ii)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGUfp3csrmBj_CVNDYsXNVqr5ocf9xK8pHawOrFDyhGQ5TX2qI5ykHilXVOJWbV8Jh9QRBxWcdv5IO96FPRZB05qbLAx27RBFgBo6s7VS_FiBr7dhTvGxubEDDRxJ4v8e2x5nL_MmGyo9C/s16000/2.png)
= 4/4.5 = 8/9
= 4/4.5 = 4/4.5
Since 4/4.5 = 8/9. Therefore by converse of Basic Proportionality Theorem, EF is parallel to QR.
iii)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvvjWKl3akp44a9todmwVMR4oHW1dqIBtwKquwKCpZZMflwNVegRIzZYZgOhgrU2PXNlw0zoOh-5K71ClwPEqyv0Jcxgcb4h0KmUcpBaeRzKb2HNmC0vO8tAx1H3NcXLXe7VAhK9Cij4LS/s16000/Fig.png)
= PE/PQ = PF/PR (by corollary of Basic Proportionality Theorem)
= 0.18/1.28 = 0.36/2.56
= 18/128 = 36/256 (multiplying with 100)
Simplifying both sides,
= 9/64 = 18/128
= 9/64 = 9/64
Since 0.18/1.28 = 0.36/2.56. Therefore by converse of Basic Proportionality Theorem, EF is parallel to QR.
Question 3: In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFTjlDez5ibH2KBB9DFpBxvMzdOhkd78CHfCbg8E4tCBF30TBLFJ-KsbJei9EJYERWTVNajLjFlLS6nY44RXHXwCQJFnzugUbhuKTgTLq3YBXqijLS_YOxW6w6L3YcsplX_fuaAmkwmdwB/s16000/Fig+6.18.png)
Answer:
Considering ∆ABC,
Since ML || BC,
∴ By Basic Proportionality
Theorem,
AM/AB = AL/AC …(1)
Considering ∆ADC,
Since LN || CD
∴ By Basic Proportionality
Theorem,
AN/AD = AL/AC …(2)
From (1) and (2)
AM/AB = AN/AD
Hence proved.
Question 4: In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgE84t_buQSacubi0VCBAmYBBIvLY-RYueLtKdfYwrr__jOq8PEwlrMtmd8vXJP3Mqo37KcyZFeVN1MI6wP2e-iyZ6NX5vlQ-dP0FaFjkb-HvIRb4PaqlRldbV2oIY3aARxVLBsKrBXWkkF/s16000/Fig+6.19.png)
Answer:
Considering ∆ABE
with DF || AE.
Since DF || AE
∴ By Basic Proportionality
Theorem,
BD/DA = BF/FE …(1)
Considering ∆ABC
with DE || AC.
Since DE || AC
∴ By Basic Proportionality
Theorem,
BD/DA = BE/EC …(2)
From (1) and (2)
BF/FE = BE/EC
Hence proved.
Question 5: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTxaTfo8Q2HWy1hHgttckQxqwa7t5kgseNPEOOTIY6LTO86j4k3Fo4CS9XmRapJ3PGZFbzZ22yV7uyAz2FJEhuLFZ2zLhqi58AqyZUJEJH_14Kr4TSV4W0W0zNZpoSUFfIN6YhlNMc1e0E/s16000/Fig+6.20.png)
Answer:
Considering ΔPQO,
Since ED || QO,
Therefore, by Basic Proportionality Theorem,
PE/QE = PD/DO ...(1)
Considering ΔPOR,
Since DF || OR,
Therefore, by Basic Proportionality Theorem,
PD/DO = PF/FR ...(2)
From (1) and (2),
PE/QE = PF/FR
Since, PE/QE = PF/FR, therefore, by converse of Basic Proportionality Theorem,
EF || QR.
Hence proved.
Question 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.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJvKE827F8u00nOrrWC4yY5EdxMnc_fWBNna6KWvrB28zkrOvJUH9i8aOSZ-f9Y4w2zfMDtOI3Dft_ZdetryLpaskGKUXhwbN0iyPL51Ffl6O0TeJmPnR-0CU0ooou6ydB7UuRfbv5M9TM/s16000/Fig+6.21.png)
Answer:
Considering ΔOPQ,
Since AB || PQ,
Therefore, by Basic Proportionality Theorem,
OA/AP = OB/BQ ...(1)
Considering ΔOPR,
Since AC || PR,
Therefore, by Basic Proportionality Theorem,
OA/AP = OC/CR ...(2)
From (1) and (2)
OB/BQ = OC/CR
Since OB/BQ = OC/CR, therefore by converse of Basic Proportionality theorem, BC
|| QR.
Hence proved
Question 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).
Answer:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_ulqt4j0fpreldKmatiy4A1VZjWCirX206fEWMhcfNbl25zBWpe3wP_Rn2EDRpYapYEcy1AT-vUxI4_1biGhkpPZNGyMZHIMZaVhp7un0b898d7RvoqwrcjCZMraoimMj_A-k9o5McrKf/s16000/4.5.png)
DE || BC
D is midpoint of AB
To prove:
E intersects at AC
Since DE || BC,
Therefore, by Basic Proportionality Theorem,
AD/DB = AE/EC
⇒ 1 = AE/EC (since D is midpoint of AB)
⇒ EC = AE
Since EC = AE, therefore E intersects AC and is the midpoint of AC.
Question 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).
Answer:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfX61hq6TUTbPg8ZwX4lbvhmRsv-rt_QpdW8U7odxJqNMzDfskoBerDlFSw-P9AsogHepH8qOhWV5CD8Id4fq_hpVCd1rhOzWt7Sbq4g-SGqs4VzQ6AJ3-JEzlUg3Lec9_-d-3DbpbKUGR/s16000/4.5.png)
AD/DB = AE/EC
D is midpoint of AB and E is midpoint of AC
To prove:
DE || BC
Since it is given AD/DB = AE/EC
⇒ 1 = 1 (as D and E are midpoints of AB and AC respectively)
Now,
Since AD/DB = AE/EC, therefore, by converse of Basic Proportionality Theorem, DE || BC
Hence proved.
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.
Answer:
Trapezium ABCD where AB || CD and AB & CD intersect each other at O.
To prove:
AO/BO = CO/DO
Construction:
Draw a line EO through O such that EO || DC || AB
Proof:
In ΔADC,
OE || DC (by construction)
Therefore, by Basic Proportionality Theorem,
AE/ED = AO/CO ...(1)
In ΔABD,
OE || AB (by construction)
Therefore, by Basic Proportionality Theorem,
DE/EA = DO/BO or EA/DE = BO/DO ...(2)
From (1) and (2)
AO/CO = BO/DO
= (AO)(DO) = (BO)(CO)
= AO/BO = CO/DO
Hence proved.
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.
Answer:
Quadrilateral ABCD where AC & BD intersect each other at O, such that AO/BO = CO/DO.
To prove:
ABCD is a trapezium
Construction:
Draw a line EO through O such that EO || AB, which meets AD at E.
Proof:
In Δ DAB,
EO || AB (by construction)
Therefore, by Basic Proportionality Theorem,
DE/EA = DO/OB ...(1)
Also it is given,
AO/BO = CO/DO
= (AO)(DO) = (CO)(BO)
= AO/CO = BO/DO or CO/AO = DO/BO ...(2)
From (1) and (2)
DE/EA = CO/AO
Therefore by converse of Basic Proportionality Theorem, EO is parallel to DC also EO is parallel to AB
⇒ DC || AB
Since two pair of sides (AB and CD) are parallel to each other, therefore, ABCD is a trapezium.
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