Question

Chord AB subtends an angle of 120º at the centre O of the circle with radius \(\frac{21}{2}\) cm. Find the perimeter of shaded segment ACB. (Use \(\sqrt{3} = 1.7\))

Hint:

Always use the exact value of \(\sqrt{3}\) specified in the question.
Using the standard \(1.73\) instead of the given \(1.7\) will result in a slightly different decimal answer, which can lead to a loss of marks!

Answer 1

Step 1: Find the arc length using the radian formula.
Converting $120^\circ$ to radians: $\theta=120^\circ\times\frac{\pi}{180^\circ}=\frac{2\pi}{3}$. Arc length $l=r\theta=\frac{21}{2}\times\frac{2\pi}{3}=\frac{21}{2}\times\frac{2}{3}\times\frac{22}{7}=22$ cm.
Step 2: Find the chord length using the law of cosines.
In triangle $OAB$, $AB^2=OA^2+OB^2-2(OA)(OB)\cos120^\circ=2r^2\left(1+\frac12\right)=3r^2$, so $AB=r\sqrt3$.
Step 3: Substitute the radius and given value of $\sqrt3$.
$AB=\frac{21}{2}\times1.7=10.5\times1.7=17.85$ cm.
Step 4: Add arc and chord for the perimeter.
Perimeter $=22+17.85=39.85$ cm.
\[ \boxed{39.85\text{ cm}} \]