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Diagram 9 shows a piece of floor tiles centre at O

Diagram 9 shows a piece of floor tiles centre at O-example-1
User Abellina
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Notice that the perimeter in the figure consists of 2 sides equal to the radius of the circle and the arc that corresponds to a fraction of a whole circumference of a 14 cm radius.


\begin{gathered} P=14+14+\text{arc} \\ \text{and} \\ P=39 \\ \Rightarrow\text{arc}=39-14-14=11 \\ \Rightarrow\text{arc}=11 \end{gathered}

And,


\begin{gathered} \text{arc}=2\pi r((\theta)/(2\pi)) \\ \Rightarrow\text{arc}=r\theta \\ \Rightarrow\theta=(arc)/(r)=(11)/(14) \end{gathered}

Where theta is in radians.

The area of an arc is given by the equation:


A_{\text{arc}}=r^2\pi((\theta)/(2\pi))

Finally,


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User Sofo Gial
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