Question : In $\triangle \mathrm{ABC}$, $AB=20$ cm, $BC=7$ cm and $CA=15$ cm. Side $BC$ is produced to $D$ such that $\triangle \mathrm{DAB} \sim \triangle \mathrm{DCA}$. $DC$ is equal to:
Option 1: 9 cm
Option 2: 8 cm
Option 3: 10 cm
Option 4: 7 cm
Correct Answer: 9 cm
Solution :
Given:
AB = 20 cm
BC = 7 cm
CA = 15 cm and
$\triangle DAB\sim\triangle DCA$
Let $CD\ =\ x$ and $AD\ =\ y$
Since $\triangle DAB\sim\triangle DCA$
⇒ $\frac{AB}{AC}\ =\ \frac{DB}{AD}\ =\ \frac{AD}{DC}$
Take the first two parts of the equation, we get
$\Rightarrow \frac{20}{15}\ =\ \frac{7\ +\ x}{y}$
$\Rightarrow \frac{4}{3}\ =\ \frac{7\ +\ x}{y}$
$\Rightarrow y\ =\ \frac{21\ +\ 3x}{4}$..................(1)
Take the first and third part
$\Rightarrow \frac{y}{x}\ =\ \frac{4}{3}$
$\Rightarrow y\ =\ \frac{4x}{3}$........... (2)
from equation (1) and (2)
$\Rightarrow \frac{21\ +\ 3x}{4}\ =\ \frac{4x}{3}$
$\Rightarrow 16x\ =\ 63\ + 9x$
$\Rightarrow 7x\ =\ 63$
$\Rightarrow x\ =\ 9$
So, the length of DC is 9 cm.
Hence, the correct answer is 9 cm.
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