chetan2u wrote:

What is the ratio of area of triangle BOC to area of triangle AOC , if the coordinates of vertices is as given in the attached figure?

(1) a = 2b

(2) X = 4 and a = 2

Question: find the ratio of \(\frac{Area.BOC}{Area.AOC}\)

Area of a triangle: \(\frac{base.height}{2}\)

Distance between a point and the origin: \(\sqrt{x^2+y^2}\)

Triangle BOC: \(Base=\sqrt{(0^2+b^2)}=b\), \(Height=\sqrt{(X^2+0^2)}=x\) and \(Area=\frac{b*X}{2}\)

Triangle AOC: \(Base=\sqrt{(0^2+a^2)}=a\), \(Height=\sqrt{(X^2+0^2)}=x\) and \(Area=\frac{a*X}{2}\)

Ratio of triangle areas: \(\frac{b*X}{2}*\frac{2}{a*X}=\frac{b}{a}\), so we need to find the values of \(a\) and \(b\).

(1) a = 2b. Replace \(a\) and ratio is: \(\frac{b}{a}=\frac{b}{(2b)}=\frac{1}{2}\), sufficient.

(2) X = 4 and a = 2. This only provides the value of \(a\) and nothing of \(b\), so insufficient.

(A) is the answer.