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Two goats are tied with a rope of length 40m outside of a rectangular shed of dimensions 50m X 30m. The goats are tied to different corners which lie on the opposite ends of a diagonal of the shed. What is the area in which the two goats can eat grass, if they choose not to eat in the common approachable area?

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  • $\begingroup$ Can the rope pass through the wall/fence? If not, I think that there is no common area. $\endgroup$ – ajotatxe Jul 20 '15 at 12:26
  • $\begingroup$ @ajotatxe no $\endgroup$ – Sumit Jha Jul 20 '15 at 12:30
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The area at reach looks like this:
enter image description here

Just sum all the quadrants.

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Looks like it should be just more than $1225\pi$ as the circle centered at C should intersect the circle centered at A. The point at which the circle intersect (say Q) upto that point the goat can eat the grass. Hence the goat in the right side will eat the grass of GHQE. Similarly we will do it for the goat on the left hand side.

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Consider one of the circle say $C_1$ to be centered at the origin $(0, 0)$ (end point of diagonal) & another say $C_2$ be centered at $(50\ m, 30\ m)$ (opposite end-point of diagonal) each having a radius of $40\ m$ Hence the equations of the circles can be given as $$C_1:\ x^2+y^2=40^2 \implies x=\pm \sqrt{1600-y^2}$$ $$C_2:\ (x-50)^2+(y-30)^2=40^2 \implies x=50\pm \sqrt{1600-(y-30)^2}$$ Now, using integration the common area $A_{c}$ between the two circles in the first quadrant can be calculated (by considering a horizontal rectangular slab) as follows $$A_c=\int(x_2-x_1)dy$$ $$\implies A_c=\int_{0}^{30}\left(\sqrt{1600-y^2}-(50-\sqrt{1600-(y-30)^2}\right)dy$$ $$\implies A_c=\int_{0}^{30}\left(\sqrt{40^2-y^2}+\sqrt{40^2-(y-30)^2}-50\right)dy$$ $$=\frac{1}{2}\left[y\sqrt{1600-y^2}+1600\sin^{-1}\left(\frac{y}{40}\right)\right]_{0}^{30}+\frac{1}{2}\left[(y-30)\sqrt{1600-(y-30)^2}+1600\sin^{-1}\left(\frac{y-30}{40}\right)\right]_{0}^{30}-[50y]_{0}^{30}$$ $$=\frac{1}{2}\left[30\sqrt{700}+1600\sin^{-1}\left(\frac{30}{40}\right)\right]+\frac{1}{2}\left[30\sqrt{700}+1600\sin^{-1}\left(\frac{30}{40}\right)\right]-[50\times 30]$$ $$\implies \color{blue}{A_c=1600\sin^{-1}\left(\frac{3}{4}\right)+300\sqrt{7}-1500}$$ Hence, the area of rectangular shed to be grazed by the goats $$=\text{total area of rectangular shed}-\text{area of common region within the rectangular shed}$$ $$=50\times 30-A_c$$ $$=1500-\left(1600\sin^{-1}\left(\frac{3}{4}\right)+300\sqrt{7}-1500\right)$$ $$=3000-300\sqrt{7}-1600\sin^{-1}\left(\frac{3}{4}\right)$$ $$\color{blue}{\approx 849.3752804\ m^2} $$ Note: above area is of rectangular shed only, the area outside the rectangular shed is not included here.

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