Triangles EX 6.2 Solutions
Question - 1 : - In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Answer - 1 : - 
Solution
(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.
Answer - 2 : -
(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
Question - 3 : - In the figure, if LM || CB and LN || CD, provethat AM/MB = AN/AD
Answer - 3 : - 
Question - 4 : - In the figure,DE||AC and DF||AE. Prove that BF/FE = BE/EC
Answer - 4 : - 
Question - 5 : - In the figure, DE||OQ and DF||OR, show thatEF||QR.
Answer - 5 : - 
Question - 6 : - In the figure, A, B and C are points on OP, OQand OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Answer - 6 : - 
Question - 7 : - Using Basic proportionality theorem, provethat a line drawn through the mid-points of one side of a triangle parallel toanother side bisects the third side. (Recall that you have proved it in ClassIX).
Answer - 7 : - 
Solution
Question - 8 : - Using Converse ofbasic proportionality theorem, prove that the line joining the mid-points ofany two sides of a triangle is parallel to the third side. (Recall that youhave done it in Class IX).
Answer - 8 : - 
Question - 9 : - ABCD is a trapeziumin which AB || DC and its diagonals intersect each other at the point O. Showthat AO/BO = CO/DO.
Answer - 9 : -
Given, ABCD is a trapezium where AB || DC anddiagonals 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 ProportionalityTheorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic ProportionalityTheorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii),we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.
Question - 10 : - The diagonals of aquadrilateral ABCD intersect each other at the point O such that AO/BO =CO/DO. Show that ABCD is a trapezium.
Answer - 10 : -
Given, Quadrilateral ABCD where AC and BDintersects 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 ADat E, in such a way that,
EO || DC || AB
In ΔDAB, EO || AB
Therefore, By using Basic ProportionalityTheorem
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 BasicProportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium withAB || CD.