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Q2(i):
$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (i) $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm
Solution :
Given: A triangle $PQR$ where $E$ is a point on side $PQ$ and $F$ is a point on side $PR$. The lengths are provided as follows:
- $PE = 3.9 \text{ cm}$
- $EQ = 3 \text{ cm}$
- $PF = 3.6 \text{ cm}$
- $FR = 2.4 \text{ cm}$
To Find: Determine whether $EF \parallel QR$.
Theoretical Basis: According to the 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, $EF \parallel QR$ if and only if:
$$\frac{PE}{EQ} = \frac{PF}{FR}$$
Step 1: Calculate the ratio of the segments on side $PQ$.
Given $PE = 3.9 \text{ cm}$ and $EQ = 3 \text{ cm}$.
$$\frac{PE}{EQ} = \frac{3.9}{3}$$
$$\frac{PE}{EQ} = 1.3$$
Step 2: Calculate the ratio of the segments on side $PR$.
Given $PF = 3.6 \text{ cm}$ and $FR = 2.4 \text{ cm}$.
$$\frac{PF}{FR} = \frac{3.6}{2.4}$$
To simplify, multiply both numerator and denominator by 10:
$$\frac{PF}{FR} = \frac{36}{24}$$
Divide both by their greatest common divisor, which is 12:
$$\frac{PF}{FR} = \frac{3}{2} = 1.5$$
Step 3: Compare the ratios.
From Step 1, $\frac{PE}{EQ} = 1.3$.
From Step 2, $\frac{PF}{FR} = 1.5$.
Since $1.3 \neq 1.5$, it follows that:
$$\frac{PE}{EQ} \neq \frac{PF}{FR}$$
Conclusion: Since the ratios of the segments are not equal, the condition for the Converse of the Basic Proportionality Theorem is not satisfied.
Final Answer: No, $EF$ is not parallel to $QR$.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.2
- Q1(i): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i).
- Q1(ii): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $AD$ in (ii).
- Q10: The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.
- 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
- 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
- Q3: In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
- Q4: In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.
- Q5: In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.
- Q6: In Fig. 6.21, $A$, $B$ and $C$ are points on $OP$, $OQ$ and $OR$ respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$.
- Q7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
- Q8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
- Q9: $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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