In the given figure, $DE || BC$. If $DE = 3\ cm$, $BC = 6\ cm$ and $\text{ar}(\triangle\text{ADE})=15\text{cm}^2,$ find the area of $\triangle\text{ABC}.$
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It is given that $DE || BC$
$\therefore\angle\text{ADE}=\angle\text{ABC}$ (Corresponding angles)
$\angle\text{AED}=\angle\text{ACB}$ (Corresponding angles)
By AA similarity, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
$\therefore\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$\Rightarrow\frac{15}{\text{ar}(\triangle\text{ABC})}=\frac{3^2}{6^2}$
$\Rightarrow\text{ar}(\triangle\text{ABC})=\frac{15\times36}{9}$
$=60\text{cm}^2$
Hence, area of triangle ABC is $60cm^2$
$\therefore\angle\text{ADE}=\angle\text{ABC}$ (Corresponding angles)
$\angle\text{AED}=\angle\text{ACB}$ (Corresponding angles)
By AA similarity, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
$\therefore\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$\Rightarrow\frac{15}{\text{ar}(\triangle\text{ABC})}=\frac{3^2}{6^2}$
$\Rightarrow\text{ar}(\triangle\text{ABC})=\frac{15\times36}{9}$
$=60\text{cm}^2$
Hence, area of triangle ABC is $60cm^2$
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