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Q.1. In figures (i) and (ii), DE y BC. Find EC in (i) and AD in (ii).
Sol.
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ 1.5/3 = 1/EC
⇒ EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.
Q.2. E and F are points on the sides PQ and PR respectively of a ΔPQR. 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
Sol.
Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Therefore, by using Basic proportionality theorem, we get,
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)
So, we get here,
PE/EQ = PF/FR
Hence, EF is parallel to QR.
Q.3.In the figure, if LM  CB and LN  CD, prove that AM/AB = AN/AD
Sol.
In the given figure, we can see, LM  CB,
By using basic proportionality theorem, we get,
AM/AB = AL/AC……………………..(i)
Similarly, given, LN  CD and using basic proportionality theorem,
∴ AN/AD = AL/AC……………………………(ii)
From equation (i) and (ii), we get,
AM/AB = AN/AD
Hence, proved.
Q.4. In the figure, DEAC and DFAE. Prove that BF/FE = BE/EC
Sol.
In ΔABC, given as, DE  AC
Thus, by using Basic Proportionality Theorem, we get,
∴ BD/DA = BE/EC ………………………………………………(i)
In ΔABC, given as, DF  AE
Thus, by using Basic Proportionality Theorem, we get,
∴ BD/DA = BF/FE ………………………………………………(ii)
From equation (i) and (ii), we get
BE/EC = BF/FE
Hence, proved.
Q.5. In the figure, DE  OQ and DF  OR. Show that EF  QR.
Sol.
Given,
In ΔPQO, DE  OQ
So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)
Again given, in ΔPQO, DE  OQ ,
So by using Basic Proportionality Theorem,
PD/DO = PF/FR………………… (ii)
From equation (i) and (ii), we get,
PE/EQ = PF/FR
Therefore, by converse of Basic Proportionality Theorem,
EF  QR, in ΔPQR.
Q.6. In the 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 here,
In ΔOPQ, AB  PQ
By using Basic Proportionality Theorem,
OA/AP = OB/BQ…………….(i)
Also given,
In ΔOPR, AC  PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR……………(ii)
From equation (i) and (ii), we get,
OB/BQ = OC/CR
Therefore, by converse of Basic Proportionality Theorem,
In ΔOQR, BC  QR.
Q.7. Using Basic proportionality theorem, prove that a line drawn through the midpoints of one side of a triangle parallel to another side bisects the third side.
Sol.
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE  BC.
We have to prove that E is the mid point of AC.
Since, D is the midpoint of AB.
∴ AD=DB
⇒ AD/DB = 1 ………… (i)
In ΔABC, DE  BC,
By using Basic Proportionality Theorem,
Therefore, AD/DB = AE/EC
From equation (i), we can write,
⇒ 1 = AE/EC
∴ AE = EC
Hence, proved, E is the midpoint of AC.
Q.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.
Sol.
Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that,
AD=BD and AE=EC.
We have to prove that: DE  BC.
Since, D is the midpoint of AB
∴ AD=DB
⇒ AD/BD = 1…………. (i)
Also given, E is the midpoint of AC.
∴ AE=EC
⇒ AE/EC = 1
From equation (i) and (ii), we get,
AD/BD = AE/EC
By converse of Basic Proportionality Theorem,
DE  BC
Hence, proved.
Q.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.
Sol. Given, ABCD is a trapezium where AB  DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO  DC  AB
In ΔADC, we have OE  DC
Therefore, By using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE  AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒ AO/BO = CO/DO
Hence, proved.
Q.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.
Sol.
Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO  DC  AB
In ΔDAB, EO  AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/OB ………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒ DO/OB = CO/AO ………….(ii)
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, By using converse of Basic Proportionality Theorem,
EO  DC also EO  AB
⇒ AB  DC.
Hence, quadrilateral ABCD is a trapezium with AB  CD.
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