



 
Hi James, I'm going to write an expression for the area of the "cap" $ABED$ as the area of the sector $CABE$ minus the area of the triangle $CAE.$ The area of a sector of a circle of radius $r$ is $\large \frac12 \normalsize r^2 \theta$ where $\theta$ is the central angle measured in radians. In this case the central angle is the angle $ECA$ which is twice the angle $DCA.$ Since triangle $ADC$ is a right triangle \[\cos\left(\frac{\theta}{2}\right) = \frac{t}{r} \] and hence \[\frac{\theta}{2} = \cos^{1}\left(\frac{t}{r}\right) \] and thus the area of the sector $CABE$ is \[r^2 \cos^{1}\left(\frac{t}{r}\right).\] Since triangle $ADC$ is a right triangle Pythagoras Theorem tell us that the length of $DA$ is $\sqrt{r^2  t^2}$ and hence the area of the triangle $CAE$ is \[t \sqrt{r^2  t^2}\] and thus the area of the cap $ABED$ is \[r^2 \cos^{1}\left(\frac{t}{r}\right)  t \sqrt{r^2  t^2}\] and hence the area of the light brown slice you want to remove from the cake is \[\pi \; r^2  2 r^2 \cos^{1}\left(\frac{t}{r}\right) + 2 t \sqrt{r^2  t^2}.\] In the situation you describe with a 12 inch diameter cake from which you want to remove a centre slice with surface area one eighth of the cake you have \[\frac18 \pi \; \times 36 = \pi \; \times 36  2 \times 36 \; \cos^{1}\left(\frac{t}{6}\right) + 2 t \sqrt{36  t^2}\] and you want to solve for $t.$ This expression can be simplified somewhat to give the equation \[ \frac{63}{4} \pi  36 \; \cos^{1}\left(\frac{t}{6}\right) + t \sqrt{36  t^2}= 0\] but unfortunately this equation can't be solved for $t$ to give the "simple" solution you desire. Harley  


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