**Question 1 :**

In the figure given below the sides DE and BC are parallel. Find the length of the side EC ?

**Solution :**

Let EC = x

Given that :

The sides DE and BC are parallel.

By using basic proportionality theorem, we have

(AD/DB) = (AE/EC) ----(1)

AD = 1.5 cm, DB = 3 cm, AE = 1 cm

By applying the known values in (1), we get

(1.5/3) = (1/x)

1.5 x = 3

x = 3/1.5

x = 2 cm

Hence, the length of side EC is 2 cm.

**Question 2 :**

In the figure given below the sides DE and BC are parallel. Find the length of the side AD ?

**Solution :**

Let AD = x

Given that :

The sides DE and BC are parallel.

By using basic proportionality theorem, we have

(AD/DB) = (AE/EC) ----(1)

AD = x DB = 7.2 cm AE = 1.8 cm EC = 5.4 cm

(x/7.2) = (1.8/5.4)

5.4 x = 1.8 x 7.2

x = (1.8 x 7.2)/5.4

x = 2.4 cm

**Question 3 :**

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, EQ = 4.5 cm, PF = 8 cm and FR = 9 cm

(iii) PQ = 1.28 cm PR = 2.56 cm PE = 0.18 cm and PF = 0.36 cm

**Solution :**

Given that :

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

(PE/EQ) = (PF/FR) ---(1)

(3.9/3) = (3.6/2.4)

1.3 ≠ 1.5

So, the sides EF and QR are not parallel.

(ii) PE = 4 cm , EQ = 4.5 cm, PF = 8 cm and FR = 9 cm

**Solution :**

According to basic proportionality theorem, we have

(PE/EQ) = (PF/FR)

(4/4.5) = (8/9)

0.88 = 0.88

So, the sides EF and QR are parallel.

(iii) PQ = 1.28 cm PR = 2.56 cm PE = 0.18 cm and PF = 0.36 cm

**Solution :**

PQ = PE + EQ

1.28 = 0.18 + EQ

EQ = 1.28 - 0.18 = 1.1

PR = PF + FR

2.56 = 0.36 + FR

2.56 - 0.36 = FR

FR = 2.2

According to basic proportionality theorem, we have

(PE/EQ) = (PF/FR)

(0.18/1.1) = (0.36/2.2)

0.1636 = 0.1636

So, EF is parallel to QR.

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