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Inside a very large field there is a shed in the shape of a regular pentagon of side $12$ m. A goat is tied at one vertex of the pentagon by a rope of length $16$ m. The goat cannot access the area inside the shed. Find the area that can be accessed by the goat inside the field.

No figure was provided in the original question. Hence note that the figure attached is my own interpretation. Please correct the figure if it seems wrong.
Figure:
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As I have interpreted we need to find the area of the shaded region of the circle with radius $16$ m.
The problem I am facing is that I am not allowed to use the values of any trigonometric ratios besides the standard $45^{\circ},30^{\circ},60^{\circ},90^{\circ},0^{\circ}$. I could have solved it had a calulator been allowed but no value except these has to be used.

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  • $\begingroup$ There is some discussion of a similar problem at sosmath.com/CBB/viewtopic.php?t=28503 $\endgroup$ Oct 12, 2013 at 11:23
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    $\begingroup$ Your main mistake is that you assume that the outer boundary is a piece of circle. If you think a bit about what happens when the goat goes around the corner, you'll realize that the problem requires no trigonometry at all, just the general observation that the area of a circular sector is proportional to its aperture. $\endgroup$
    – fedja
    Oct 12, 2013 at 11:49
  • $\begingroup$ @fedja i can't follow you Could you please be a little more descriptive. $\endgroup$
    – Suy
    Oct 12, 2013 at 11:52
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    $\begingroup$ Take a box and a piece of rope (or thread) longer than the side of the box. Attach one end of the rope to one corner of the box (barn) and attach the other end to your finger (goat). Start walking the goat around the box and watch what happens with the rope. It is better to see it once than to hear about it 100 times. :-) $\endgroup$
    – fedja
    Oct 12, 2013 at 12:11
  • $\begingroup$ @fedja thanks I got it $\endgroup$
    – Suy
    Oct 12, 2013 at 14:16

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As fedja said you assumption that the area that the goat can reach is a piece of circle is wrong, because the rope can't go through the walls of the shed, that means that when the rope will touch the wall and we want go into wall's direction, then we would have another anchor point at the another vertex. Because the side is of length $12$ and the rope of length $16$. We have rope of length $4$ left, i.e. we can make a circle of radius $4$ from the new anchor point. I hope it's clearer to you with picture:

enter image description here

You can calculate the area of the sector $GDF$ using the formula:

$$P = \frac{r\pi\theta}{180} $$

We know that the interior angle of pentagon is $108^{\circ}$ so the angle $\theta = 360^{\circ} - 108^{\circ} = 252^{\circ}$

And for the sectors $FEH$ and $GCI$ use the same formula, but this time use the exterior angle of pentagon, which we know that is $72^{\circ}$. And to get the final result just add those 3 areas together.

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  • $\begingroup$ yeah I had solved it from fedja's hint but thanks. $\endgroup$
    – Suy
    Oct 13, 2013 at 6:23

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