In $△ A B C$ , a line through A cuts the side BC at D such that BD : DC = $4 : 5$ . If the area of $△ A B D = 60 c m^{2}$ , then the area of $△ A D C$ is

62.50 c m^{2}

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58.25 c m^{2}

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93.75 c m^{2}

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75 c m^{2}

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Solution

The correct option is C $93.75 c m^{2}$
BD : DC = $4 : 5$
Area of $△ A B D = 60 c m^{2}$
$\frac{A r e a o f A B D}{A r e a o f A D C} = \frac{(B D)^{2}}{(C D)^{2}}$
$\frac{60}{A r e a o f A D C} = \frac{(4 x)^{2}}{(5 x)^{2}}$
$A r e a o f A D C = \frac{60 \times 25 x}{16 x} = 93.75 c m^{2}$

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