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:


(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

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.


(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

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.

 

Download Class 10 NCRT Triangles Exercise – 6.2 pdf