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Q2(ii):
$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm

Solution :

Given: A triangle $\triangle PQR$ where $E$ is a point on $PQ$ and $F$ is a point on $PR$. The lengths are given as follows:

  • $PE = 4 \text{ cm}$
  • $QE = 4.5 \text{ cm}$
  • $PF = 8 \text{ cm}$
  • $RF = 9 \text{ cm}$

To Find: Determine whether $EF \parallel QR$.

P Q R E F

Theorem Used: Converse of Thales Theorem (Basic Proportionality Theorem). The theorem states that 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 ratio of the segments on side $PQ$.

The ratio is given by $\frac{PE}{QE}$.

Substituting the given values:

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

To simplify, multiply the numerator and denominator by $10$:

$\frac{PE}{QE} = \frac{40}{45}$

Divide both by their greatest common divisor, which is $5$:

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

Step 2: Calculate the ratio of the segments on side $PR$.

The ratio is given by $\frac{PF}{RF}$.

Substituting the given values:

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

Step 3: Compare the two ratios.

From Step 1, we have $\frac{PE}{QE} = \frac{8}{9}$.

From Step 2, we have $\frac{PF}{RF} = \frac{8}{9}$.

Since $\frac{PE}{QE} = \frac{PF}{RF}$, the condition for the Converse of the Basic Proportionality Theorem is satisfied.

Conclusion: Because the segments are divided in the same ratio, the line segment $EF$ must be parallel to the side $QR$.

Final Answer: Yes, $EF \parallel QR$ because $\frac{PE}{QE} = \frac{PF}{RF} = \frac{8}{9}$.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.2


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