**Download NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2**

**1. In given figure, (i) and (ii), DE||BC. Find EC in (i) and AD in (ii)**** **

**Solution: **

**(i)** Let EC = x cm

Since DE||BC

By Basic proportionality theorem, we get

x =

x = 2

Therefore, EC = 2 cm

**(ii)** Let AD = x cm.

Since DE||BC in the figure (ii)

By Basic proportionality theorem, we get

x =

x = 2.4

Therefore, AD = 2.4 cm

**2. 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:**

**(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**

**Solution:**

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

= 1.3

= 1.5

Since

Hence in this case EF is not parallel QR.

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

Since

Hence in this case EF is parallel to QR.

**(iii)** Given: PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm, and PF=0.36 cm

Since

Hence in this case EF is parallel to QR.

**3. In this figure if LM||CB and LN||CD, prove that**** **

**Solution:** Given that LM || CB then using the basic proportionality theorem we have

……(1)

Also, LN || CD

……(2)

From equation (1) and (2) we get

Hence Proved

**4. In the figure DE||AC and DF||AE. Prove that**** **

**Solution:** In ∆ABC we have DE||AC

Then by basic proportionality theorem, we have

….(1)

In ∆ABE we have DF||AE

Then by basic proportionality theorem, we have

……(2)

From equation (1) and (2) we get

**5. In the figure, DE||OQ and DF||OR. Show that EF||QR.**

**Solution:** In ∆ POQ, DE||OQ

By using basic proportionality theorem, we have

……(1)

In ∆ POR, DF||OR

By using basic proportionality theorem, we have

……(2)

From equation (1) and (2) we have

By using the converse of basic property theorem, we have EF||QR.

**6. In the figure A, B and C are the points on OP, OQ and OR respectively such that AB||PQ and AC||PR. Show that BC||QR.**

**Solution:** In ∆POQ, we have AB||PQ

By using basic proportionality theorem, we have

……..(1)

In ∆POR, we have AC||PR

By using basic proportionality theorem, we have

……(2)

From equation (1) and (2), we get

By using the converse of basic proportionality theorem, we have

BC||QR

**7. Using the basic proportionality theorem prove that the 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 in class IX).**

**Solution: **

Let ∆ABC be any triangle and DE be the line drawn through the mid-point of the side AB and any point E on AC. Here DE||BC

By using basic proportionality theorem, we have

[since D is the mid-point of AB]

AE = EC

Hence E is also the mid-point of AC.

**8. Using the converse of basic proportionality theorem, prove that the line joining the mid-points of any sides of a triangle is parallel to the third side.**

** Solution:**

Let ∆ABC be any triangle and DE be the line segment joining mid-points D and E of the sides AB and AC respectively. Hence AD=DB and AE=EC

Therefore, and

Hence,

By converse of basic proportionality theorem, we have DE||BC

**9. ABCD is a trapezium in which AB||DC and its diagonal intersect each other at the point O. Show that **

**Solution: **

Draw a line EF passes through point O, such that EF||CD

In ∆ACD, EO||CD

By basic proportionality theorem, we have

…….(1)

In ∆ABD, EO||AB

By basic proportionality theorem, we have

…….(2)

From equation (1) and (2), we have

**10. The diagonals of the quadrilateral ABCD intersect each other at the point O such that**

**. Show that ABCD is a trapezium. **

**Solution: **

Draw a line EF parallel to AB.

In ∆ABD, EO||AB

By basic proportionality theorem, we have

……(1)

[Given]

…….(2)

From equation (1) and (2), we have

By converse of basic proportionality theorem, we have EO||DC

Hence, AB || OE || DC

⇒ AB || DC.

Therefore, ABCD is a trapezium.

**Download NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2**

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