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Q1(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:

A B C D E AD = ? BD = 7.2 AE = 1.8 EC = 5.4

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


CBSE Solutions for Class 10 Mathematics Triangles


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