Question

In the figure above, point $O$ is the center of the circle, line segments $LM$ and $MN$ are tangent to the circle at points $L$ and $N$, respectively, and the segments intersect at point $M$ as shown. If the circumference of the circle is $96$, what is the length of minor arc $\stackrel{}{LN}$ ?

• A. $30.12$
• B. $31.98$
• C. $32$
• D. $32.30$

Answer 1

The correct option is B $31.98$

The circumference of the circle is $96$ units.

Therefore, $2\pi \times r=96$

$\Rightarrow r=\frac{96\times 7}{2\times 22}=15.27$ units

Now, in the quadrilateral $OLMN,\angle OLM+\angle LMN+\angle MNO+\angle NOL={360}^o$

Given that LM and MN are tangents to the circle, $\angle OLM=\angle ONM={90}^o$

$\therefore \angle NOL=360-90-90-60={120}^o$

The length of arc is given by $r\theta$ where $\theta$ is the angle made by the arc at the centre in radian.

Thus, arc length $LN=15.27\times \frac{120\pi }{180}=31.98$ units.