The base BC of a triangle ABC is divided at D such that BD = half of DC. The area of triangle ABC is ___________ times the area of triangle ADC.

Correct option is B

Given:

BD = $$\frac{1}{2}$$ DC, meaning BD:DC = 1:2.

We need to find the ratio of the area of △ABC to △ADC.

Formula Used:

​$$\text{Area ratio} = \frac{\text{Base of } \triangle ABC}{\text{Base of } \triangle ADC}$$​​

Since △ABC and △ADC share the same height from A, their area ratio is directly proportional to their bases:

​$$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ADC} = \frac{BC}{DC}$$​​

Solution:

The area of a triangle is proportional to its base if they share the same height.

Since BD: DC = 1:2, the entire base BC is divided into three equal parts, meaning:

BC = BD + DC = 1+2 = 3 parts

Hence, DC is $$\frac{2}{3}$$​of BC, and BD is$$\frac{1}{3}$$​​ of BC.

​$$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ADC} = \frac{3}{2}$$​​

Thus, the area of △ABC is $$\frac{3}{2}$$​ times the area of △ADC.

Option (B) is right.