A hexagon is inscribed in a circle of radius r. Two of its sides have length 1, two have length 2 and last two sides have length 3. Then radius r satisfies the equation

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 r^{3} + 7 r - 3 = 0$

$2 r^{3} - 7 r + 3 = 0$

$2 r^{3} + 7 r + 3 = 0$

$2 r^{3} - 7 r - 3 = 0$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

If side of length i subtends angle $α_{i}$ , at the
centre, then $2 α_{1} + 2 α_{2} + 2 α_{3} = 360^{\circ}$

Also $\sin (\frac{α_{1}}{2}) = \frac{1}{2 r}, \sin (\frac{α_{2}}{2}) = \frac{1}{r}, \sin (\frac{α_{3}}{2}) = \frac{3}{2 r}$
$\begin{array}{l} \cos \frac{α_{1}}{2} = \frac{\sqrt{4 r^{2} - 1}}{2 r}, \cos (\frac{α_{3}}{2}) = \frac{\sqrt{r^{2} - 1}}{r} \\ \frac{α_{1}}{2} + \frac{α_{2}}{2} = 90^{\circ} - \frac{α_{3}}{2} \\ \Rightarrow \cos \frac{α_{1}}{2} \cos \frac{α_{2}}{2} - \sin \frac{α_{1}}{2} \sin \frac{α_{2}}{2} = \sin \frac{α_{3}}{2} \\ \Rightarrow \frac{\sqrt{4 r^{2} - 1} \sqrt{r^{2} - 1}}{2 r^{2}} - \frac{1}{2 r^{2}} = \frac{3}{2 r} \\ \Rightarrow \sqrt{(4 r^{2} - 1) (r^{2} - 1)} = 3 r + 1 \\ \Rightarrow 4 r^{4} - 5 r^{2} + 1 = 9 r^{2} + 6 r + 1 \\ \Rightarrow 2 r^{3} - 7 r - 3 = 0 \end{array}$