cow math problem?

Question

a cow is tethered by a rope 50 meters long. The rope is fastened to a hook located 10 meters away from the corner of the longest side of the barn measuring 60m by 30m. Over how much land can the cow graze?

Answer 1

Thomas F, do your own work next time, instead of stealing mine!
:-P

foodme12, my work is here: http://answers.yahoo.com/question/index;_ylt=Al9HKcZ41rtBjeqAWKt29Jzzy6IX?qid=20060724081211AAJTp4h

For Andrew G below, I hadn't thought of that possibility when solving this problem before: "A cow is tethered by a rope 50 meters long. The rope is fastened to a hook located 10 meters away FROM THE LONGEST SIDE OF THE BARN measuring 60m by 30m. Over how much land can the cow graze?"

With this restating of the problem as "10 meters away FROM THE CORNER," you're right on the money. It's:
(½ · π · 50²) + (¼ · π · 40²) + (¼ · π · 10²)
= 1250 π + 400 π + 25 π
= 1675 π square meters, or approx. 5262.1677m².

Answer 2

I have to agree with Andrew G... I think that the problem is meant to read "... 10 meters away from the corner *on* the longest side..." otherwise mentioning this side would be irrelevant.

Draw the situation, you will see that the cow can reach half a circle keeping the rope straight. By folding the rope around the corner, she has another quarter circle (R = 40) at the side of the barn, and by taking another corner a quarter circle at the opposite long side (R = 10).

Total area is therefore

pi/4 * (50^2 + 40^2 + 10^2) = 1050 pi = 3299 m2.

Answer 3

First, start by drawing the rectangle that will be your barn. Next, draw the point on the barn that is described in the problem. Now, figure out the half circles created by the problem (for a half circle, Area is equal to one-half of pi times the radius squared or 1/2*pi*r^2. Add up the area of the half circles, et voila, you have your answer.

Answer 4

A*********************T******B
______________________
|******************* **********|
|******************* **********|
|***************** ************|
|**************** *************|
|************** ***************|
|______________ _______ |
C********N************M******D

I do not think any of the answers above is correct. To really tackle the problem, we need the above drawing. To make the drawing display correctly, I have to use * to filll up the space. When you read it, assume where * appears is just empty.

I assume the cow ia teathered at point T. AB=CD=60, AC=BD=30. TB=10. N is
the point that saparates where the cow can touch on CD from where it can not touch, i.e., cow can not eat between CN, but can eact grass between ND. Note TN=50.

Note that AT=50 so cow can reach A. The area the cow can mow is then TNDB + part of the circle that covers the triangle ATN.

Notice that NM=40 because TN=50, TM=30, i.e., 40^2 + 30^2=50^2.


Area of TNDB = (TB + ND) * BD / 2 = (10 + 50) * 30 / 2 = 900m^2.

Now assume the angle NTM is s. Then sin(s) = 0.8. s = 53.131degrees. angle ATN is then 90-s=36.869 degrees.

The area of part of the circle is PI * 50^2 * 36.869 / 360 = 804.36m^2.

So the total area is about 1704.36m^2.

Answer 5

Great answer!

If, however, the hook is located on the wall of the barn, 10m down from the corner, then you'd half of the area of the circle with radius 50m, plus the area of one quarter of a circle with radius 40m, plus the area of one quarter of a circle with radius 10m.