Question 23
If in $$\triangle ABC$$, D and E are the points on AB and BC respectively such that DE $$\parallel$$ AC, and AD : AB = 3: 8, then (area of $$\triangle BDE$$) : ( area of quadrilateral DECA) = ?
Solution
$$\triangle$$ BDE and $$\triangle$$ ABC are similar.
BD = AB - AD = 8 - 3 = 5
$$\frac{area of BDE}{area of ABC} = (\frac{BD}{AB})^2$$
$$\frac{area of ADE}{area of ABC} = (\frac{5}{8})^2 = \frac{25}{64}$$
Area of quadrilateral DECA = area of ABC - area of BDE = 64 - 25 = 39
(area of $$\triangle BDE$$) : ( area of quadrilateral DECA) = 25 : 3
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