• The rainbow shown in the photo is an arc of a circle. One of the application is to determine the length of the arc of the rainbow.
• The cross-section of a train tunnel is usually in the form of a major arc of a circle. The arc length and the area of this cross-section tunnel can be calculated using the formulas for circular measure.

The diagram below shows the sectors \(OAB\) and \(OCD\) with a common center \(O\).

Find

(a) the value of \(k\) if the area of the shaded region is \(13.09\) cm\(^2\),
(b) the perimeter of the shaded region.

(a)

Change the angle to radian,

\(75^\circ\times \dfrac{\pi}{180^\circ}=1.3092\text{ rad}.\)

Use the formula of area of sector,

\(\text{Area}=\dfrac{1}{2}r^2\theta\)

Area of shaded region:

\(=\text{Area of sector }OAB-\text{Area of sector }OCD\)

\(\begin{aligned} 13.09&=\dfrac{1}{2}(3k)^2(1.3092)-\dfrac{1}{2}(2k)^2(1.3092) \\\\ 13.09&=\dfrac{1}{2}(9)(1.3092)k^2-\dfrac{1}{2}(4)(1.3092)k^2 \\\\ 13.09&=5.8914k^2-2.6184k^2 \\\\ 13.09&=3.273k^2 \\\\ k&=\sqrt{\dfrac{13.09}{3.273}}\\\\ k&=1.9998\approx2\text{ cm}. \end{aligned}\)

(b)

Use the formula of arc length:

\(s=r\theta\)

Perimeter of the shaded region:

\(=k+\text{Arc }CD+k+\text{Arc }AB\)

\(\begin{aligned} &=2+2+2(2)(1.3092)+3(2)(1.3092) \\ &=4+5.2368+7.8552 \\ &=17.092\text{ cm}. \end{aligned}\)

• The rainbow shown in the photo is an arc of a circle. One of the application is to determine the length of the arc of the rainbow.
• The cross-section of a train tunnel is usually in the form of a major arc of a circle. The arc length and the area of this cross-section tunnel can be calculated using the formulas for circular measure.

The diagram below shows the sectors \(OAB\) and \(OCD\) with a common center \(O\).

Find

(a) the value of \(k\) if the area of the shaded region is \(13.09\) cm\(^2\),
(b) the perimeter of the shaded region.

(a)

Change the angle to radian,

\(75^\circ\times \dfrac{\pi}{180^\circ}=1.3092\text{ rad}.\)

Use the formula of area of sector,

\(\text{Area}=\dfrac{1}{2}r^2\theta\)

Area of shaded region:

\(=\text{Area of sector }OAB-\text{Area of sector }OCD\)

\(\begin{aligned} 13.09&=\dfrac{1}{2}(3k)^2(1.3092)-\dfrac{1}{2}(2k)^2(1.3092) \\\\ 13.09&=\dfrac{1}{2}(9)(1.3092)k^2-\dfrac{1}{2}(4)(1.3092)k^2 \\\\ 13.09&=5.8914k^2-2.6184k^2 \\\\ 13.09&=3.273k^2 \\\\ k&=\sqrt{\dfrac{13.09}{3.273}}\\\\ k&=1.9998\approx2\text{ cm}. \end{aligned}\)

(b)

Use the formula of arc length:

\(s=r\theta\)

Perimeter of the shaded region:

\(=k+\text{Arc }CD+k+\text{Arc }AB\)

\(\begin{aligned} &=2+2+2(2)(1.3092)+3(2)(1.3092) \\ &=4+5.2368+7.8552 \\ &=17.092\text{ cm}. \end{aligned}\)