Question

Let z lies on the circle centred at the origin. If area of the triangle whose vertices are $z, \, \omega z$ and $z + \omega z$, where w is the cube root of unity $4\sqrt{3}$ sq. unit. Then radius of the circle is :

• A. 1 unit
• B. 2 units
• C. 4 units
• D. None of these

Answer 1

Answer: (C)

Let the point A represents z and B represents $\omega z$, then
$∠AOB = \frac{2\pi}{3}$. If $C$ represents $z + \omega z $ then $OACB$ is a parallelogram.
$\therefore $ Area of $\Delta ABC =Area of \Delta OAB$
$=\frac{1}{2}\left|z\right|\left|\omega z\right|sin\frac{2\pi}{3}$= $\frac{1}{2} \left|z\right| \left|z\right| \frac{\sqrt{3}}{2} \left[ \left|\omega z\right| = \left|z\right|\right] = \frac{\sqrt{3}}{4} \left|z\right|^{2}$
Now given that area of $\Delta ABC = 4\sqrt{3}.$
$\therefore \frac{\sqrt{3}}{4} \left|z\right|^{2} = 4\sqrt{3}$
$\Rightarrow \left|z\right| = 4$. That is z lies on a circle or radius $4$