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      <text>4.2.4.6 Area of sector (NOT IN Quiz, confusing)</text>
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      <text><![CDATA[<p class="HELM_exercise">Exercise<br></p>
<ol class="HELM_parts">
<li><p>What is the area \(A\) of a sector formed by a central angle \(\theta \) radians in a circle of radius \( r \)?</p>
<p> \(A =\) [[input:ans1]] [[validation:ans1]][[feedback:prt1]]</p><p>(Hint: By proportionality the ratio of the area of the sector to the total area of the circle equals the ratio of \( \theta \) to the total angle at the centre of the circle.)</p></li>
<p>
The shaded area \(S\) is the difference between a sector of radius \(R\) and a sector of radius \(r\), both of angle \(\theta\). What is the value of the shaded area \(S\) if \(\theta\) is measured (b) in radians, (c) in degrees?</p><p style="text-align: center;"><img src="@@PLUGINFILE@@/4.2%20fig%2016%20area%20of%20sector%20of%20annulus.svg" alt="Angular sector of angle theta" role="presentation" class="img-responsive atto_image_button_text-bottom" width="151" height="94"><br></p>
<li><p> \(S =\) [[input:ans2]] [[validation:ans2]][[feedback:prt2]] <br></p></li>
<li><p> \(S =\) [[input:ans3]] [[validation:ans3]][[feedback:prt3]]</p></li>
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      <text><![CDATA[<ol class="HELM_parts">
<li><p>Using the hint, <span>\(\dfrac{\theta}{2\pi}=\dfrac{A}{\pi r^2}\)</span>&nbsp;and from here we obtain \(A=\dfrac{\pi r^2\theta}{2\pi}=\dfrac{r^2\theta}{2}\)</p></li>
<li><p>With \(\theta\) in radians the shaded area is \(S=\dfrac{R^2\theta}{2}-\dfrac{r^2\theta}{2}=\dfrac{\theta}{2} (R^2-r^2)\)</p></li>
<li><p>If \(\theta\) is in degrees, then the proportion of the circle is \(\dfrac{\theta}{360^\circ}\) and we multiply this by the area of the the larger circle minus the smaller circle to give \(S=\dfrac{\pi\theta^\circ}{360^\circ}(R^2-r^2)\)</p></li>
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      <text>ta1: theta*r^2/2;
ta2: theta/2*(R^2 - r^2);
ta3: pi*theta/360*(R^2 - r^2)</text>
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