NCERT Solutions For Class 10 Math Chapter – 6 Exercise – 6.2
Q1. In figures (I) and (ii) DE // BC Find EC in (I) and AD in (ii).
Solution:-
In figure (I) AD/BD = AE/EC = 1.5/3 = 1/EC = EC = 2 cm.
In figure (ii) AD/BD = AE/EC = AD/7.2 = 1.8/5.4 = AD = 2.4 cm.
Q2. 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:
- PE = 3.9 cm EQ 3 cm PF = 3.6 cm and FR = 2.4 cm
- PE = 4 cm QE = 4.5 cm PF = 8 cm and RF = 9 cm
- PQ = 1.28 cm PR = 2.56 cm PE = 0.18 cm and PF = 0.36 cm.
Solution:-
- PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 3/2 = 1.5
Here , PE/EQ ≠ PF/FR , so EF is not parallel to QR.
- PE/EQ = 4/4.5 = 8/9 and PF/RF = 8/9
Here , PE/EQ = PF/RF
So by inverse of Basic proportionally Theorem EF is parallel to QR.
- PQ/PE = 1.28/0.18 = 64/9 and PR/pf = 2.56/0.36 = 64/9
As PQ/PE = PR/PF , so EF is parallel to QR.
Q3. In figure , if LM // CB and LN // CD.
Prove that AM/AB = AN/AD.
Solution:-
In ∆ABC, LM // BC
AM/MD = AL/LC (I) ( Basic proportionally theorem )
In ∆ACD , LN // CD
AN/ND = AL/LC (II) [ Basic proportionally theorem]
From (I) and (ii) we get
AM/MB = AN/ND = MB/AM = ND/AN
MB/AM + 1 = ND/AN + 1
AB/AM = AD/AN = AM/AB= AN/AD
Hence proved.
Q4. In figure , DE // AC and DF // AE.
Prove that BF/FE = BE/EC.
Solution:-
In ∆ABC DE // AC.
BE/EC = BD/AD (I) [ Basic proportionally theorem]
In ∆ABE , DF // AE.
BD/DA = BF/FE (ii) [ Reason same as above]
From (I) and (ii) we get BE/EC = BF/FE.
Hence proved.
Q5. In figure , DE // OQ and DF //OR.
Show that EF // QR.
Solution:-
In ∆PQO , DE // OQ.
PE / EQ = PD/DO (I) [ Basic proportionally theorem]
In ∆POR , DF // OR
PD/DO = PF/FR (ii) [ Reason same as above]
From (I) and (ii) , we get
PE/EQ = PF/FR
In ∆PQR , PE/EQ = PF/FR [ From (iii)]
EF // QR. [ Using converse of Basic proportionally theorem]
Q6. 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.
Solution:-
In ∆PQO , AB // PQ
PA/AO = QB/BO (I) [ Basic proportionally theorem]
In ∆POR , AC // PR
PA/AO = RC/CO (ii) [ Reason same as above]
From (I) and (ii) , we have
QB/BO = RC/CO (iii)
In ∆QOR , QB/BO = RC/CO [ From (iii)]
BC // QR. [ Using converse of basic proportionally theorem]
Q7. Using basic proportionally theorem , prove that a line drawn through the mid – point of one side of a triangle parallel to another side bisects the third side.
Solution:-
AP/PB = AQ/QC (I) [ Basic proportionally theorem]
Also , AP = PB (ii) [Given]
From (I) and (ii) ,
1 = AQ/QC
AQ = QC = Q is mid – point of AC.
Q8. Using converse of Basic proportionally theorem , prove that the line joining the mid – points of any two sides of a triangle is parallel to the third side.
Solution:-
As P and Q are mid – points of AB and AC.
AP = PB and AQ = QC (I)
AP/PB = AQ/QC (ii)
In ∆ ABC , AP/PB = AQ/QC
PQ // BC.
Q9. ABCD is a trapezium in which AB // DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution:-
In ∆ ADB , OE // AB.
DE/EA = DO/BO (I) [ Basic proportionally theorem ]
In ∆ ADC , OE // DC
DE/EA = CO/AO (ii) [ Basic proportionally theorem ]
From (I) and (ii) we have
CO/AO = DO/BO = AO/BO = CO/DO
Hence proved.
Q10. 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.
Solution:-
In ∆DAB , OE // AB
OB/OD = AE/ED (I) [ Basic proportionally theorem]
Also OA/OC = OB/OD [ Given] = OA/OC = AE/ED [ From (I)]
In ∆ADC , OA/OC = AE/ED
OE // DC (ii) [ Converse proportionally theorem]
Also OE // AB. (iii) [ From construction]
From (ii) and (iii) we have
DC // AB
Quadrilateral ABCD is a trapezium.