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Let $\mathrm{ABC}$ be a triangle with $\angle \mathrm{B}=90^{\circ}$. Let $\mathrm{AD}$ be the bisector of $\angle \mathrm{A}$ with $\mathrm{D}$ on $\mathrm{BC}$. Suppose $\mathrm{AC}=6 \mathrm{~cm}$ and the area of the triangle $\mathrm{ADC}$ is $10 \mathrm{~cm}^{2}$. Then the length of $\mathrm{BD}$ in $\mathrm{cm}$ is equal to
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Verified Answer
The correct answer is:
$\frac{10}{3}$
From angle bisector theorem
$\begin{array}{l}
\frac{r}{6}=\frac{p}{q} \\
q r=6 p ...(1)
\end{array}$
Area of $\triangle \mathrm{ADC}=10 \mathrm{~cm}^{2}$
$\begin{array}{l}
\frac{1}{2}(\mathrm{DC})(\mathrm{AB})=10 \\
\frac{1}{2}(\mathrm{q})(\mathrm{r})=10 \\
\mathrm{q} \mathrm{r}=20
\end{array}$
From (1)
$\begin{array}{l}
\Rightarrow 20=6 \mathrm{p} \\
\mathrm{p}=\frac{20}{6}=\frac{10}{3}
\end{array}$
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