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This question belongs to Arithmetic Ability Triangles
Triangles
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In ΔABC, D and E are points on AB and AC respectively such that DE || BC and DE divides the ΔABC into two parts of equal areas. Then ratio of AD and BD is

Answer & Solution
Correct Answer: Option B
Triangles mcq solution image
ar ADE = ar DEBC
So, ar ΔADE = 1 unit2 and ar ABC = 2 unit2
$$\eqalign{ & \frac{{{\text{ar}}\,\Delta ADE}}{{{\text{ar}}\,\Delta ABC}} = \frac{{A{D^2}}}{{A{B^2}}} \cr & \frac{1}{2} = {\left( {\frac{{AD}}{{AB}}} \right)^2} \cr & \frac{1}{{\sqrt 2 }} = \frac{{AD}}{{AB}} \cr & \therefore \frac{{AD}}{{DB}} = \frac{1}{{\sqrt 2 - 1}} \cr & \left( {\therefore DB = AB - AD = \sqrt 2 - 1} \right) \cr & {\text{So,}}\,AD:BD = 1:\sqrt 2 - 1 \cr} $$
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