# Class 10 NCRT Triangles Exercise – 6.2 pdf || UP Board

# __Triangles__

__Exercise – 6.2__

__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**** **Δ**PQR. 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

**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).**

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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