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In the adjoining figure, $DE \parallel BC$ and
$\frac{DE}{BC} = \frac{1}{3}$.
If $AD = 1.5$ cm, then find the length of BD.
$\frac{DE}{BC} = \frac{1}{3}$.
If $AD = 1.5$ cm, then find the length of BD.
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Ans. Let $BD = x$ cm
$\triangle ADE \sim \triangle ABC$ (by AA Similarity) (1/2 Mark)
$\frac{AD}{AB} = \frac{DE}{BC}$ (1/2 Mark)
$\frac{1.5}{1.5+x} = \frac{1}{3}$ (1 Mark)
$x = 3$, i.e., $BD = 3$ cm (1/2 Mark)
$\triangle ADE \sim \triangle ABC$ (by AA Similarity) (1/2 Mark)
$\frac{AD}{AB} = \frac{DE}{BC}$ (1/2 Mark)
$\frac{1.5}{1.5+x} = \frac{1}{3}$ (1 Mark)
$x = 3$, i.e., $BD = 3$ cm (1/2 Mark)