Q2(iii):

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm

Given:

In $\triangle PQR$, $E$ is a point on $PQ$ and $F$ is a point on $PR$. The given measurements are:

$PQ = 1.28 \text{ cm}$

$PR = 2.56 \text{ cm}$

$PE = 0.18 \text{ cm}$

$PF = 0.36 \text{ cm}$

To Find:

Determine whether $EF \parallel QR$.

Theorem Used:

Converse of Thales Theorem (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. That is, if $\frac{PE}{EQ} = \frac{PF}{FR}$, then $EF \parallel QR$.

Step 1: Calculate the lengths of segments $EQ$ and $FR$.

Since $E$ lies on $PQ$, $EQ = PQ - PE$.

$EQ = 1.28 \text{ cm} - 0.18 \text{ cm} = 1.10 \text{ cm}$

Since $F$ lies on $PR$, $FR = PR - PF$.

$FR = 2.56 \text{ cm} - 0.36 \text{ cm} = 2.20 \text{ cm}$

Step 2: Calculate the ratios $\frac{PE}{EQ}$ and $\frac{PF}{FR}$.

Ratio 1: $\frac{PE}{EQ} = \frac{0.18}{1.10}$

Multiply numerator and denominator by 100 to simplify: $\frac{18}{110} = \frac{9}{55}$

Ratio 2: $\frac{PF}{FR} = \frac{0.36}{2.20}$

Multiply numerator and denominator by 100 to simplify: $\frac{36}{220}$

Divide both by 4: $\frac{36 \div 4}{220 \div 4} = \frac{9}{55}$

Step 3: Compare the ratios.

Since $\frac{PE}{EQ} = \frac{9}{55}$ and $\frac{PF}{FR} = \frac{9}{55}$, it follows that:

$\frac{PE}{EQ} = \frac{PF}{FR}$

Step 4: Conclusion based on the Converse of the Basic Proportionality Theorem.

Because the segments are divided in the same ratio, the line $EF$ must be parallel to the side $QR$ by the Converse of the Basic Proportionality Theorem.

Final Answer: Yes, $EF \parallel QR$.