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# E and F are points on the sides PQ and PR respectively of a $$\triangle PQR$$. 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 = 0.63 cm

Given, in $$\triangle PQR$$, E and F are two points on side PQ and PR respectively.

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,
$${{PE} \over {EQ}} = {{3.9} \over {3}} = {{39} \over {30}} = {{13} \over {10}} = 1.3$$
And
$${{PF} \over {FR}} = {{3.6} \over {2.4}} = {{36} \over {24}} = {{3} \over {2}} = 1.5$$
$$\therefore$$ $${{PE} \over {EQ}} \ne {{PF} \over {FR}}$$

Hence, EF is not parallel to QR.

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

Therefore, by using Basic proportionality theorem, we get,

$${{PE} \over {QE}} = {{4} \over {4.5}} = {{40} \over {45}} = {{8} \over {9}}$$
And,
$${{PF} \over {RF}} = {{8} \over {9}}$$

So, we get here,

$${{PE} \over {QE}} = {{PF} \over {RF}}$$

Hence, EF is parallel to QR.

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

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And,
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
$$\therefore$$ $${{PE} \over {EQ}} = {{0.18} \over {1.10}} = {{18} \over {110}} = {{9} \over {55}}$$………….(i)
And,
$${{PE} \over {FR}} = {{0.36} \over {2.20}} = {{36} \over {220}} = {{9} \over {55}}$$…………(ii)

So, we get here,
$${{PE} \over {EQ}} = {{PF} \over {FR}}$$

Hence, EF is parallel to QR.