Question
From adjoining figure, $\angle ABC =90^{\circ}, \angle DCB =90^{\circ}, AB =6, DC =8$, then $\frac{ A (\triangle ABC )}{ A (\triangle BCD )}=$ ?
✓
Answer
$\triangle A B C$ and $\triangle B C D$ have same base $B C$.
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{ AB }{ DC } \ldots[$ Triangles having equal base $]$
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{6}{8}$
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{3}{4}$
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{ AB }{ DC } \ldots[$ Triangles having equal base $]$
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{6}{8}$
$\therefore \frac{ A (\Delta ABC )}{ A (\Delta BCD )}=\frac{3}{4}$
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