Question

Chord AB of a circle with centre O and radius 21 mm subtends an angle of \(120^\circ\) at the centre. Find the perimeters of the shaded region. (Use \(\sqrt{3} = 1.73\))

Hint:

You can also find the chord length in an isosceles triangle \(\Delta OAB\) with angle \(120^\circ\) by dropping a perpendicular from the center \(O\) to the chord \(AB\).
This perpendicular bisects the angle into two \(60^\circ\) angles and bisects the chord into two equal segments.
Using simple right-triangle trigonometry, each half of the chord is \(r \sin 60^\circ\), which gives the total chord length as \(2r \sin 60^\circ = r\sqrt{3}\) directly!

Answer 1

Step 1: Find the chord length using a dropped perpendicular.
Drop a perpendicular from centre O to chord AB, this bisects the \(120^\circ\) angle at O into two \(60^\circ\) angles and bisects AB into two equal halves. Each half of the chord is \(r \sin 60^\circ = 21 \times \frac{\sqrt{3}}{2}\), so the full chord is: \[ AB = 2 \times 21 \times \frac{\sqrt{3}}{2} = 21\sqrt{3} = 21 \times 1.73 = 36.33 \text{ mm} \]
Step 2: Find the arc length using the standard fraction formula.
\[ \text{Arc } AB = \frac{120^\circ}{360^\circ} \times 2\pi r = \frac{1}{3} \times 2 \times \frac{22}{7} \times 21 = 44 \text{ mm} \]
Step 3: Add the straight and curved parts.
\[ \text{Perimeter} = \text{Arc } AB + \text{Chord } AB = 44 + 36.33 \]
Step 4: Compute the final value.
\[ \text{Perimeter} = 80.33 \text{ mm} \]
\[ \boxed{\text{Perimeter} = 80.33 \text{ mm}} \]