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This was a two stage question. Stage 1 was calculating the area of the below shape sans curve. Stage 2 is to calculate the area of the shaded region below.

We are told the curve has a radius of 3000

shape Note the diagram is not to scale.

So for stage 1, I calculated the area to be $22*15 (height) - 5*7 +1/2 (30+7-22)*15(height) = 407500 $ square units

So I believe if I were to subtract the white area from the diagram above I'd have the area for the shaded region, but I'm stuck in how to determine that area.

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  • $\begingroup$ Not sure how to best tag this, so open to re tagging $\endgroup$ – jamesmstone Oct 20 '17 at 23:56
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    $\begingroup$ How about you start with clearly stating the question. Then what work you have put in, and where you think you might be stuck? $\endgroup$ – Doug M Oct 20 '17 at 23:59
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    $\begingroup$ I think the answer is here (found by searching for area between tangents and circle). It tells you what to subtract from the answer to (1). brilliant.org/discussions/thread/… $\endgroup$ – Ethan Bolker Oct 21 '17 at 0:05
  • $\begingroup$ Thanks for suggestions, please see edits. $\endgroup$ – jamesmstone Oct 21 '17 at 0:06
  • $\begingroup$ I agree with Doug, you should mention which parts of the question you are stuck on, this will help us give a detailed answer for you :) $\endgroup$ – mdave16 Oct 21 '17 at 0:07
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enter image description here

\begin{align} [ABCDEVU] &= [ABCDEF] -[OVFU]+[OVU] ,\\ [OVFU]&= |OV|\cdot|VF| =R^2\cdot\cot(\tfrac12\angle VFU) ,\\ [OVU]&=\tfrac12\cdot R^2 \cdot\angle UOV =R^2\,(\tfrac\pi2-\tfrac12\angle VFU) . \end{align}

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Let $\Omega$ be the center of the circle $\gamma$ which gives you the "smooth angle". The value of actual sharp angle is $90°+45°=135°$ but we are more interested in its half, $\alpha:=\frac{135°}{2}$. Now, let $\beta=90°-\alpha$ the angle from $\Omega$ which sees half of the arc (the smooth angle) from one of the two tangent points to the intersection of the two tangents (the old edges). We get that the white area removed is obtained doubling the difference between a right triangle of area $\frac{R·(R\sin{\beta})}{2}$ - where $R$ is the given radius - and a circular sector of area $\frac{\beta}{360°}\pi R^2$. Now it suffices to get the nearest calculator :) Hope I've helped

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