Question
In figure 1.73, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then $\frac{ A (\triangle ABC )}{ A (\triangle DCB )}=?$
✓
Answer
$\text { We know that, Area of triangle }=\frac{1}{2} \times \text { base } \times \text { height }$
$ \Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DCB )}=\frac{\frac{1}{2} \times BC \times AB }{\frac{1}{2} \times BC \times DC }$.
$ =\frac{ AB }{ DC }$
$=\frac{6}{8}$
$=\frac{3}{4}$
$ \Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DCB )}=\frac{\frac{1}{2} \times BC \times AB }{\frac{1}{2} \times BC \times DC }$.
$ =\frac{ AB }{ DC }$
$=\frac{6}{8}$
$=\frac{3}{4}$
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