NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.2

Question 2

Concept used: Converse of Basic Proportionality Theorem (Thales’ theorem).

If a line through the sides \(PQ\) and \(PR\) of \(\triangle PQR\) cuts them at points E and F such that

\[\frac{PE}{PQ} = \frac{PF}{PR},\]

then \(EF \parallel QR\). If the ratios are not equal, then \(EF\) is not parallel to \(QR\).

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

First find full side lengths:

\[PQ = PE + EQ = 3.9 + 3 = 6.9\,\text{cm}\]

\[PR = PF + FR = 3.6 + 2.4 = 6.0\,\text{cm}\]

Compute the ratios:

\[\frac{PE}{PQ} = \frac{3.9}{6.9} = \frac{39}{69} = \frac{13}{23}\]

\[\frac{PF}{PR} = \frac{3.6}{6.0} = \frac{36}{60} = \frac{3}{5}\]

Clearly,

\[\frac{13}{23} \neq \frac{3}{5}\]

Conclusion: The ratios are not equal, so \(EF\) is not parallel to \(QR\). Answer: No.

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

Find full side lengths:

\[PQ = PE + QE = 4 + 4.5 = 8.5\,\text{cm}\]

\[PR = PF + RF = 8 + 9 = 17\,\text{cm}\]

Compute ratios:

\[\frac{PE}{PQ} = \frac{4}{8.5} = \frac{4}{\frac{17}{2}} = \frac{8}{17}\]

\[\frac{PF}{PR} = \frac{8}{17}\]

So,

\[\frac{PE}{PQ} = \frac{PF}{PR} = \frac{8}{17}\]

Conclusion: Ratios are equal, so by converse of BPT, \(EF \parallel QR\). Answer: Yes.

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

Here entire side lengths are already given: \(PQ = 1.28\,\text{cm}\), \(PR = 2.56\,\text{cm}\).

Compute the ratios:

\[\frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}\]

\[\frac{PF}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}\]

So,

\[\frac{PE}{PQ} = \frac{PF}{PR} = \frac{9}{64}\]

Conclusion: Ratios are equal, so \(EF \parallel QR\). Answer: Yes.