Question

Let $A_0,A_1,A_2,A_3,A_4,A_5$ be a regular hexagon inscribed in a circle of unit radius. 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 $3$

Consider the problem

Here $O{A}_0=1$

Then,

$O{A}_1=O{A}_2=O{A}_3=O{A}_4=O{A}_5=1$

Since, it is regular hexagon

therefore,

All sides are equal

And,

Each side of hexagon makes an angle ${60}^{\circ }$ at the centre $O$ of the circle coordinates of $A_1,A_2,A_4,A_5$ are $(\cos {60}^{\circ },\sin {60}^{\circ }),(\cos {120}^{\circ },\sin {120}^{\circ }),(-\cos {60}^{\circ },\sin {60}^{\circ }),(-\cos {120}^{\circ },-\sin {120}^{\circ })$ respectively.

$A_1=(\frac{1}{2},\frac{\sqrt{3}}{2})$

$A_2=(-\frac{1}{2},\frac{\sqrt{3}}{2})$

$A_4=(-\frac{1}{2},-\frac{\sqrt{3}}{2})$

$A_5=(\frac{1}{2},-\frac{\sqrt{3}}{2})$

And

$A_3=(-1,0)$ and $A_0=(1,0)$ (given circle is of radius.)

Now,

By distance formula

$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$(A_0{A}_1{)}^2=(\frac{1}{2}-1{)}^2+(\frac{\sqrt{3}}{2}-0{)}^2$

So,

$A_0{A}_1=1$

Now,

$(A_0{A}_2{)}^2=(-\frac{1}{2}-1{)}^2+(\frac{\sqrt{3}}{2}-0{)}^2$

$=3$

then,

$A_0{A}_2=\sqrt{3}$

And,

$(A_0{A}_4{)}^2=(-\frac{1}{2}-1{)}^2+(\frac{\sqrt{v}3}{2}-0{)}^2$

$=3$

$A_0{A}_4=\sqrt{3}$

And,

the product of lengths of the line segments $A_0{A}_1,A_0{A}_2$ and $A_0{A}_4$ is

$=1\times \sqrt{3}\times \sqrt{3}$

$=3$

Hence the option $C$ is the correct answer.