In triangle ABC, a line is drawn from the vertex A to a point D on BC. If BC = 9 cm and DC = 3 cm, then what is the ratio of the areas of triangle ABD and triangle ADC respectively?
Given : BC = 9 cm and DC = 3 cm and AE = $$h$$ cm
=> BD = 9-3 = 6 cm
Now, area of $$\triangle$$ ABD = $$\frac{1}{2}\times(BD)\times(AE)$$
= $$\frac{1}{2}\times6\times h=3h$$ $$cm^2$$
Similarly, area of $$\triangle$$ ADC = = $$\frac{1}{2}\times3\times h=1.5h$$ $$cm^2$$
$$\therefore$$ Required ratio = $$\frac{3h}{1.5h}=\frac{2}{1}$$
=> Ans - (B)
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