Question

E and F are points on the side PQ and PR respectively of $△\mathrm{PQR}$.For the following case , state whether EF $\parallel$QR
PE = 3.9 cm, EQ = 1.3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer 1

Here,
PE = 3.9 cm, EQ = 1.3 cm, PF = 3.6 cm and FR = 2.4 cm



$$\begin{gathered} \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{1.3}=\frac{3}{1} \\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=\frac{3}{2} \\ \frac{\mathrm{PE}}{\mathrm{EQ}}\ne \frac{\mathrm{PF}}{\mathrm{FR}} \end{gathered}$$

According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Here,

$\frac{\mathrm{PE}}{\mathrm{EQ}}\ne \frac{\mathrm{PF}}{\mathrm{FR}}$

Therefore, EF is not parallel to QR.