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Q1(ii):
In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $AD$ in (ii).
In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $AD$ in (ii).
Solution :
Given:
In $\triangle ABC$, we are given that $DE \parallel BC$.
The lengths provided for figure (ii) are:
- $AE = 1.8 \text{ cm}$
- $BD = 7.2 \text{ cm}$
- $EC = 5.4 \text{ cm}$
To Find:
The length of $AD$.
Visual Representation:
Step 1: Stating the Relevant Theorem
According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Therefore, in $\triangle ABC$, since $DE \parallel BC$, we have:
$\frac{AD}{DB} = \frac{AE}{EC}$
Step 2: Substituting the Given Values
Let $AD = x$. Substituting the known values into the equation:
$\frac{x}{7.2} = \frac{1.8}{5.4}$
Step 3: Solving for $x$
First, simplify the fraction on the right side:
$\frac{1.8}{5.4} = \frac{18}{54} = \frac{1}{3}$
[Since $18 \times 3 = 54$]
Now, the equation becomes:
$\frac{x}{7.2} = \frac{1}{3}$
To isolate $x$, multiply both sides by $7.2$:
$x = \frac{7.2}{3}$
$x = 2.4$
Step 4: Conclusion
Since $x$ represents the length of $AD$, we conclude that $AD = 2.4 \text{ cm}$.
Final Answer: $AD = 2.4 \text{ cm}$
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).
- 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
- 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 $\frac{AO}{BO} = \frac{CO}{DO}$.
- 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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