Question

Let An $A_0{A}_1{A}_2{A}_3{A}_4{A}_5$ As be a regular hexagon inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments $A_0{A}_1,A_0{A}_2$ and $A_0{A}_4$ is

• A. $\frac{3}{4}$
• B. $3\sqrt{3}$
• C. $3$
• D. $\frac{3\sqrt{3}}{2}$

Answer 1

The correct option is C $3$

Let $O$ be the centre of the circle of unit radius and the coordinates of $A_0$ be $(1,0)$.

Since each side of the regular hexagon makes an angle of ${60}^{\circ }$ at the centre $O$.
Coordinates of $A_1$ are $(\cos {60}^{\circ },\sin {60}^{\circ })=(\frac{1}{2},\frac{\sqrt{3}}{2})$

$A_2$ are $(\cos {120}^{\circ },\sin {120}^{\circ })=(-\frac{1}{2},\frac{\sqrt{3}}{2})$
$A_3$ are $(-1,0)$ $A_4$ are $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$and $A_5$ are $(\frac{1}{2},-\frac{\sqrt{3}}{2})$

.Now $A_0{A}_1=\sqrt{{(1-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
$A_0{A}_1=\sqrt{{(1+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{3}=A_0{A}_4$

So that $\left(A_0{A}_1\right)\left(A_0{A}_2\right)(A_0{A}_4.)=3$