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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,174

**A cow is tied to one corner ( vertex ) of a 20 m per side square barn. The cow has a 50 m length of chain to keep it from escaping while it is grazing. How big is the cow's grazing area?**

This problem or similar ones appear a couple of times in this forum as well as on the internet. It does have an exact answer but that is for another thread, here we will see what geogebra can do.

1) Draw points A(0,50) and B(50,0).

2) Use the circle with radius tool to draw a circle with A as the center and radius 50.

3) Draw a line through point s A and B.

4) Draw a perpendicular line to AB through point A.

5) Draw a circle with center A and radius 20.

6) Get the points of intersection with this smaller circle and the line AB and the line perpendicular to AB. Your drawing should look like Fig 1.

7) Hide points E and D and the y axis.

8) Draw a line parallel to AFB through C and a line through F parallel to AC.

9) Get the point of intersection of the two new lines that go through C and F. It will be labeled G. Check Fig 2.

10) Hide the small circle and the 4 lines and use the polygon tool and click on A, F, G, and C and a poly1 will created with sides 20 and area 400 square feet.

11) Use the insert text tool to label the diamond. Call it "Barn".

12) Use the circle with radius tool to draw a circle with radius 30 and center C. Do the same with center F.

13) Get the point of intersection with the small leftmost circle and the larger one. Point H will be created there.

14) Get the point of intersection with the small rightmost circle and the larger one. Point I will be created there.

15) Get the point intersection of the two small circles, points K and J will be created. Hide K. For clarity color H, J and I red and make them a little bigger. See Fig 3.

16) Rename A, to "Cow".

17) We need the the blackened area at the bottom. See Fig 4. We will use the integralbetween command.

18) Enter in the input bar,

IntegralBetween[-sqrt(-x² - 20sqrt(2) x + 700) - 10sqrt(2) + 50, 50 - sqrt(2500 - x²), x(H), x(J)]

This is nothing but the area of the leftmost little circle minus the area of the big circle from H to J.

19) Immediately in the algebra pane you will see i = 120.33845614907142. This corresponds to the red area in Fig 5.

20) To get the entire area that the cow can graze enter in the input bar,

2500π - 400 - 2i. The answer will pop up in the algebra pane:

j = 7213.30472167634

which correct to all displayed digits. We are done!

For completeness the exact answer is

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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