Question

Let a line passing through point $A$ divides the square $ABCD$ into two parts so that area of one portion is double the other, then the length of portion of line inside the square is

• A. $\frac{\sqrt{10}}{3}$
• B. $\frac{\sqrt{13}}{3}$
• C. $\frac{\sqrt{11}}{3}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{\sqrt{13}}{3}$

Consider the given figure consisting of a square of unit area.
Therefore by ratio and proportions the are of the triangle (smaller area) will be half of the larger area as given in the question, while the total area is 1 square unit.
Considering the area of the right angled triangle to be x we get
$x+2x=1$
$x=\frac{1}{3}$
Hence the area of the triangle is $\frac{1}{3}$
Since $area=\frac{bh}{2}$
Therefore
$\frac{\left(1\right).\left(b\right)}{2}=\frac{1}{3}$
$\Rightarrow b=\frac{2}{3}$
Since the triangle is a right angled triangle right angled at D, we get the Length of the hypotenuse as
$\sqrt{1^2+{\frac{2}{3}}^2}$
$=\frac{\sqrt{13}}{3}$