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(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

Answer :

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.

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