Class 10 NCRT Triangles Exercise – 6.2 pdf || UP Board
Triangles
Exercise – 6.2
1. In Figure 6.17 (i) and (ii), DE ∥ BC.
Find EC in (i) and AD in (ii).
(i)
Let EC = x cm
Given that, In ΔABC, DE ∥ BC therefore,
From the Basic Proportionality Theorem,
AD/DB = AC/EC
1.5/3 = 1/x
x = (3 × 1)/1.5
x = 3/1.5 = 30/15 = 2
Here, EC = 2
(ii)
Let AD = x cm
Given that, In ΔABC, DE ∥ BC therefore,
From the Basic Proportionality Theorem,
AD/DB = AC/EC
x/7.2 = 1.8/5.4
x = (1.8 × 7.2)/5.4
x = 12.96/5.4 = 2.4
Here, AD = 2.4
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:
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm therefore
PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
Since, PE/EQ ≠ PF/FR Hence, EF and QR is not parallel.
Given that, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm therefore
PE/EQ = 4/4.5 = 40/45 = 8/9
PF/FR = 8/9
Since, PE/EQ = PF/FR Hence, by the converse of the Basic Proportionality Theorem,
EF and QR are parallel.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Given that, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm therefore
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
PE/EQ = 0.18/1.10 = 18/110 = 9/55
PF/FR = 0.36/2.20 = 36/220 = 9/55
Since, PE/EQ = PF/FR Hence, by the converse of the Basic
Proportionality Theorem (Theorem 6.2), EF and QR are parallel.
3. In
Figure 6.18, if LM ∥ CB and LN ∥ CD, prove that AM/AB = AN/AD
Given that, In ΔABC, LM ∥ CB therefore,
From the Basic Proportionality Theorem,
AM/AB = AN/AD …………….(1)
Similarly,
In ΔADC, LN ∥ CD therefore,
From the Basic Proportionality Theorem,
AN/AD = AL/AC ……………….(2)
From equation (1) and (2)
AM/AB = AN/AD
4. In
Figure 6.19, DE ∥ AC and DF ∥ AE Prove that BF/FE = BE/EC
Given that, In ΔABC, DE ∥ AC therefore,
From the Basic Proportionality Theorem,
BD/DA = BE/EC …………….(1)
Similarly
In ΔABC, DF ∥ AE therefore,
From the Basic Proportionality Theorem,
BD/DA = BF/FE ……………….(2)
From equation (1) and (2)
BF/FE = BE/EC
5. In
figure 6.20, DE ∥
OQ and DF ∥ OR. Show that EF ∥ QR
Given that, In ΔPOQ, DE ∥ OQ therefore,
From the Basic Proportionality Theorem,
PE/EQ = PD/DO …………….(1)
Similarly,
In ΔPOR, DF ∥ OR therefore,
From the Basic Proportionality Theorem,
PF/FR = PD/DO ……………….(2)
From equation (1) and (2)
PE/EQ = PF/FR
In ΔPOR,
PE/EQ = PF/FR
therefore,
From the converse of the Basic Proportionality Theorem,
EF ∥ QR
6. In
Figure 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.
Given that, In ΔPOQ, AB ∥ PQ therefore,
From the Basic Proportionality Theorem,
OA/AP = OB/BQ …………….(1)
Similarly,
In ΔPOR, AC ∥ PR therefore,
From the Basic Proportionality Theorem,
OA/AP = OC/CR ……………….(2)
From equation (1) and (2)
OB/BQ = OC/CR
In ΔOQR,
OB/BQ = OC/CR
therefore,
From the converse of the Basic Proportionality
Theorem,
BC ∥ QR
Let PQ be a line drawn through the mid-point P of side
AB of ΔABC, parallel to the other side BC. i.e. PQ ∥ BC therefore,
In ΔABC, From the Basic Proportionality Theorem
AP/PB = AQ/QC
AP/AP = AQ/QC [AP = PB]
AQ = QC
Hence, Q is the mid-point of side AC.
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).
Let PQ be the line passing through the mid-point P of
the side AB of ΔABC and the mid-point Q of the other side AC. therefore AP = PB and AQ = QC
AP/PB = 1,
AQ/QC = 1
Or,
AP/PB = AQ/QC = 1
Now In ΔABC,
AP/PB = AQ/QC
Hence From the converse of the Basic Proportionality Theorem,
PQ ∥ BC
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.
Through O and parallel to CD, draw a line EF so that
EF ∥
CD.
In ΔADC, EO ∥ CD
From the Basic Proportionality Theorem,
AE/ED = AO/OC ……….(1)
Similarly,
In ΔABD, EO ∥ AB
From the Basic Proportionality Theorem,
AE/ED = BO/OD ………….(2)
From equation (1) and (2),
AO/OC = BO/OD
AO/BO = CO/DO
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.
The diagonals of the quadrilateral ABCD intersect each
other at the point O in this way. A line EO is drawn through O and parallel to
AB. so that EO ∥
AB.
In ΔABD, EO ∥ AB
From the Basic Proportionality Theorem,
AE/ED = BO/OD ……………(1)
but given,
AO/BO = CO/DO
AO/CO = BO/DO
……………….(2)
From equation (1) and (2),
AE/ED = AO/OC
From the converse of the Basic Proportionality Theorem,
EO ∥ DC
AB ∥ OE
∥ DC
AB ∥ CD
Hence ABCD is a trapezium.
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