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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
Solution :
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, $EF \parallel QR$.
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).
- 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
- 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
- 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 $\frac{AO}{BO} = \frac{CO}{DO}$.
- 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.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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