Question 2

Converse of Basic Proportionality Theorem (BPT): If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. So if $\frac{PE}{EQ} = \frac{PF}{FR}$, then EF ∥ QR.

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

Step 1 — Find ratio on side PQ

$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$

Step 2 — Find ratio on side PR

$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$

Step 3 — Compare

Since $1.3 \neq 1.5$, the ratios are not equal.

EF is not parallel to QR.

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

Step 1 — Find ratio on side PQ

$\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$

Step 2 — Find ratio on side PR

$\frac{PF}{RF} = \frac{8}{9}$

Step 3 — Compare

Since $\frac{8}{9} = \frac{8}{9}$, the ratios are equal. By the Converse of BPT:

EF ∥ QR.

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

Here PQ and PR are full side lengths, so we compare PE/PQ and PF/PR.

Step 1 — Find ratio on side PQ

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

Step 2 — Find ratio on side PR

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

Step 3 — Compare

Since $\frac{9}{64} = \frac{9}{64}$, the ratios are equal. By the Converse of BPT:

EF ∥ QR.

Answer

(i) EF is not parallel to QR. (ii) EF ∥ QR. (iii) EF ∥ QR.