Class 10 - Mathematics
Triangles - Exercise 6.2
Top Block 1
Exercise 6.2
Question : 1:In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
Answer :
(i) Let EC = x cm
Given that DE || BC, therefore using Thales theorem, we get
AD/DB = AC/EC
⇒ 1.5/3 = 1/x
⇒ x = 3/1.5
⇒ x = 3/(15/10)
⇒ x = (3 * 10)/15
⇒ x = 30/15
⇒ x = 2
Hence, EC = 2 cm
(ii) Let AD = x cm
Given that DE || BC, therefore using Thales theorem, we get
AD/DB = AC/EC
⇒ x/7.2 = 1.8/5.4
⇒ x/7.2 = 1/3
⇒ x = 7.2/3
⇒ x = 2.4
Hence, AD = 2.4 cm
Question : 2:E and F are points on the sides PQ and PR respectively of a Δ 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 :
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm,
therefore
Mddle block 1
PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5 cm
Since PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
PF/FR = 8/9 cm
Since PE/EQ = PF/FR
Hence, according to converse of Thales theorem, EF || QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm, therefore
PE/PQ = 0.18/1.28 = 18/128 = 9/64 cm
PE/PR = 0.36/2.56 = 36/256 = 9/64 cm
Hence, according to converse of Thales theorem, EF || QR.
Question : 3:In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Answer :
Given, in triangle ABC, LM || CB, therefore
According to Thales theorem, we have
AM/AB = AL/AC ………………1
Similarly, in triangle ADC, LN || CD, therefore
According to Thales theorem, we have
AN/AD = AL/AC ………………2
From equation 1 and 2, we get
AM/AB = AN/AD
Question : 4:In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Answer :
Given, in triangle ABC, DE || AC, therefore
BD/DA = BE/EC ………………1
Similarly, in triangle ABC, DE || AE, therefore
According to Thales theorem, we have
BD/DA = BF/FE ………………2
From equation 1 and 2, we get
BF/FE = BE/EC
Question : 5:In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Answer :
Given, in triangle POQ, DE || OQ, therefore
PE/EQ = PD/DO ………………1
Similarly, in triangle POR, DF || OR, therefore
According to Thales theorem, we have
PF/FR = PD/DO ………………2
From equation 1 and 2, we get
PE/EQ = PF/FR
In triangle PQR,
PE/EQ = PF/FR [Proved]
Hence, according to converse of Thales theorem, EF || QR.
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.
Answer :
Given, in triangle POQ, AB || PQ, therefore
Similarly, in triangle POR, AC || PR, therefore
According to Thales theorem, we have
OA/AP = OC/CR ………………2
From equation 1 and 2, we get OB/AP = OC/CR
In triangle OQR,
OB/AP = OC/CR [Proved]
Hence, according to converse of Thales theorem, BC || QR.
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 :
Let PQ is a line through the mid-point of AB which is parallel to BC intersects Ac at Q
i.e. PQ || BC,
According to converse of Thales theorem, we have
AP/PB = AQ/QC
⇒ 1 = AQ/QC [Since AP = PB]
⇒ AQ = QC
Hence, Q is the mid-point 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 :
Let PQ is a line which passes through the mid-point of AB and AC.
⇒ AP/PB = 1 and AQ/QC = 1
⇒ AP/PB = AQ/QC = 1
Now, in triangle ABC,
AP/PB = AQ/QC
Hence, according to converse of Thales theorem, we have PQ || BC.
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 :
A line is drawn through the point O parallel to CD such that EO || AB.
According to Thales theorem, we have
AE/ED = AO/OC ……….1
Similarly, in triangle ABD, EO || AB
According to Thales theorem, we have
AE/ED = BO/OD ……….2
From equation 1 and 2, we get
AO/OC = BO/OD
⇒ AO/OB = OC/OD
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 :
A line is drawn through the point O parallel to AB such that EO || AB.
In triangle ABD, EO || AB
AE/ED = BO/OD ……….1
But given that
AO/OB = CO/OD
⇒ AO/CO = OB/OD ………2
From equation 1 and 2, we get
AE/ED = AO/OC
According to converse of Thales theorem, we have
⇒ ED || DC
⇒ AB || OE || DC
⇒ AB || CD
Hence, ABCD is a trapezium.
