# Return to Answer

 4 corrected math edited Dec 10 '18 at 3:38 NigelMNZ 14066 bronze badges For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$$$A_{halfSlice} = \frac{xy}{2}$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$$$A_{halfSlice} = \frac{x^2tan(\theta_{slice})}{2}$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$$$\frac{\pi r^2}{n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$$$x = \frac{\sqrt{\pi}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ So for $$n=8$$ and $$r=15cm$$, we get $$x \approx 6.6467cm$$$$x \approx 9.399cm$$. Still unsure about $$n$$ below 8 though... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ So for $$n=8$$ and $$r=15cm$$, we get $$x \approx 6.6467cm$$. Still unsure about $$n$$ below 8 though... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = \frac{xy}{2}$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = \frac{x^2tan(\theta_{slice})}{2}$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\pi}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ So for $$n=8$$ and $$r=15cm$$, we get $$x \approx 9.399cm$$. Still unsure about $$n$$ below 8 though... 3 added 67 characters in body edited Dec 8 '18 at 23:39 NigelMNZ 14066 bronze badges For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ So for $$n=8$$ and $$r=15cm$$, we get $$x \approx 6.6467cm$$. Still unsure about $$n$$ below 8 though... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ Still unsure about $$n$$ below 8... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ So for $$n=8$$ and $$r=15cm$$, we get $$x \approx 6.6467cm$$. Still unsure about $$n$$ below 8 though... 2 added 9 characters in body edited Dec 8 '18 at 23:29 NigelMNZ 14066 bronze badges For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2} = x^2tan(\theta_{slice})$$$$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{tan(\theta_{slice})}}$$$$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ Still unsure about $$n$$ below 8... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{tan(\theta_{slice})}}$$ Still unsure about $$n$$ below 8... For $$n \geq 8:$$ Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $$x$$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice. $$A_{cake} = \pi r^2$$ $$A_{slice} = \frac{A_{cake}}{n}$$ $$\theta_{slice} = \frac{360^\circ}{n}$$ $$A_{halfSlice} = xy$$, where $$x$$ is the distance from the origin, and $$y$$ is the length of the cut. After finding $$y$$, we can then put $$A_{halfSlice}$$ in terms of $$x$$: $$y = xtan(\theta_{slice})$$ $$A_{halfSlice} = x^2tan(\theta_{slice})$$ Finally, we can solve for $$x$$, given that we know that $$A_{halfSlice} = \frac{A_{slice}}{2}$$: $$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$ $$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$ Still unsure about $$n$$ below 8... 1 answered Dec 8 '18 at 23:22 NigelMNZ 14066 bronze badges