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 longest side of the barn measuring 60m by 30m. Over how much land can the cow graze.

the rope is on a hook located 10 meters away from the corner of the longest side.

Answer 1

I'm guessing you mean that the hook is 10 meters away from the corner of the barn, "even" with the 30m side. (If the hook was 10m from the wall 30m from either edge, the area would be smaller. The circumstance I'm first describing would be the maximum grazing area for the parameters you've given. I'll explore the minimum later.)

Using a coordinate plane to describe, I'll say the barn has corners
(0, 0), (-30, 0), (-30, -60), and (-60, 0).
The hook is at (10, 0).

The cow can graze an area ¾ of a circle of grass from
(-40, 0), through (10, 50) and (60, 0), and to (10, -50).
This circle has radius 50m and an area of π(50m)² = 2500πm².
¾ of this is 1875πm², or about about 5890.5m².

There are two additional sections of grass the cow can reach. One is a quarter-circle of radius 10m, arcing from (-40, 0) to (-30, -10).
The full circle has area π(10m)² = 100πm².
¼ of this is 25πm², or about 78.5m².

The other region is a sector of a 50m radius circle from
(10, -50) to (0, -20√6). [Thank you, Pythagoras.] This sector, with the help of the inverse sine of (10 / 50), measures approximately 11.537°.
The area would be (11.537 / 360) of the 50m radius circle, or about 251.7m².

The total grazing area would be approximately
5890.5 + 78.5 + 251.7 = 6220.7m².

Exactly, it would be 50π[38 + 5(Arcsin 0.2) / 36] square meters.

~~~~~ ~~~~~ ~~~~~ ~~~~~ ~~~~~

The minimum area for grazing would occur if the hook is at (10, -30). [The barn will have the same coordinates.] In this circumstance, the cow would be able to roam a 50m region just larger than a semicircle before the region is shortened by the corners of the barn.

Using the inverse tangent of (10 / 30), we can find the angles of the sector in addition to the semicircle. It's about 18.435° on either side. This makes the total 50m radius sector a total of [180 + 2(18.435)], or about 216.87°. The area of the full circle is 2500π, and the area of the sector is (216.87 / 360) of this, or about 4731.37m².

There are also two additional sectors, both the same area. Since the rope gets caught at (0, 0) from (10, -30), this distance is 10√10 meters. [Again, cheers to Pythagoras.] The length of the radius left after the rope catches on the barn is (50 - 10√10). Each of the two sectors would measure (90 - 18.435), or about 71.565°. These sectors would have an area of
π(50 - 10√10)²(71.565 / 360), or about 210.915m².

The total area would be 4731.37 + 2(210.915) = 5153.2m².

If the hook was anywhere between (10, -30) and (10, 0), the grazing area would be between 5153.2 and 6220.7 square meters.

Answer 2

I think it would help to know exactly where the hook is in relation to the longest side of the barn. For example, is the hook 10 m away from the midpoint of the long side of the barn, from the endpoint of the long side of the barn, or somewhere in between? Or can it be any of these?

If the hook is at the endpoint, note that the cow can graze behind the barn too.

Draw circles and other areas. Good problem!

Answer 3

10 meters away from where on the longest side? The midpoint of the longest side? The corner? Without this information, it is impossible to solve the problem.

Answer 4

It depends upon the placement of the tether point... is it 10m away from the middle? or an end? or where?... is the hook on the short side wall of the barn? it matters...

just figure out the circles involved... or partial circles.

Answer 5

It's a little vague, but you need to do area of circles. Start with the long side, then add like half or 1/4 of the little circle around the short side.

Answer 6

P---A--------------------
|.. /
|./
|\ C
|.\
|..\
|...\
Q--B--------------------

AC = 50, PC = 10 => Area(PAC) = 0.5*10*50 = 250
BC = 50, QC = 20 => Area(QBC) = 0.5*50*20 = 500
Angle(BCA) = pi- angle(PCA) - angle(QCB)

Area(ACB) = 50*angle(BCA).