Gobbling goat -- SUPER BONUS QUESTION I BET NO ONE CAN GET IT?

Question

a grassp feild is in the shape of a circle of a radius of 100m and is enclosed by a circular fence. a goat if attached by a rope at a fixed point on the fence. to stop the goat from gettin too fat, the farmer wants to make sure that it can only reach HALF of the grass on the feild. How long shoulf the rope be?

Answer 1

116m.

Answer 2

A good and non-trivial problem. To answer, make it as simple as possible: we have the problem of two circles of different radii intersecting. So to make it easy, assume that
1. the circle of radius r is at the origin
2. the cirlce of radius 100 m has its center on the positive x-axis and is tangent to the circle of radius r at the origin
3. consider the area portended to be in quadrant I so that we are looking at only one half the area
So, the area (A/2) that is portended is the area by the circle of radius 100 m is
integral from 0 to r of y1dx - integral from 0 to r y2dx where
y1 = sqr(r^2-x^2), the circle of radius r centered at the origin
y2 = sqr(100^2 - [x-r]^2)

Answer is the 1/2 the area is (1/2)(pi*100^2); let INT stand for integral, then

A/2 = INT[x:0->r]sqr(r^2-x^2)dx - INT[x:0->r]sqr(100^2-[x-r]^2)
= pi(100^2)/2
Solve integral from standard integration tables and solve for r.

Answer 3

Let length of rope = R.

Draw a picture, and draw in three lines from the centre of the field to the sheep and the two furthest points it can reach. Also join the two ends the sheep can reach.

We need the area of a big sector, minus the area of two small segments.

Call the angle opposite the R side in the small triangle x (in radians).

Then the area of the big sector is 0.5*R^2*(2x) = R^2 * x.

The other segments each have central angle pi - 2x, and radius 100. So each one has area 0.5*100^2*(pi-2x - sin(pi-2x)). Double that to get 100^2 (pi-2x-sin(pi-2x)).

Thus the total area is R^2 x + 100^2 (pi-2x-sin(pi-2x)).

Then, use the cosine rule on the triangle to get R^2 = 100^2 + 100^2 - 2*100*100*cos(pi-2x), or R^2 = 20000 - 20000cos(pi-2x).
Substitute that in:
20000x - 20000xcos(pi-2x) + 10000(pi-2x-sin(pi-2x)).
That has to equal 0.5*pi*100^2 = 5000pi.

Thats not solvable exactly, but numerically gives x = 0.952847.

Putting that in the formula for R gives R = 115.87m

Answer 4

If we define two circles such that area of circle 1 is half that of circle 2, we'll get our answer.

A(1) = pi * r(1)^2
A(2) = pi * r(2)^2

A(1) = 0.5 * A(2)
pi * r(1)^2 = 0.5 * pi * r(2)^2
r(1)^2 = 0.5 * r(2)^2

But we know that r(2) = 100 m, therefore

r(1)^2 = 0.5 * 100^2
r(1)^2 = 5000
r(1) = 70.7

Therefore, the length of the rope should be 70.7 m and should be placed in the middle of the circle (not more than 29.3 m away from the center).

Answer 5

This one is easy just think Logically I suggest Drawing it out

Answer 6

You're kidding right?

Answer 7

lol thats good.i wont give the answer tho they needa figure it out

Answer 8

50=pi X r^2
50/pi= r^2
sqrt(50/pi)=r
r= 3.989

approximately four meters.

Answer 9

I might be wrong, but I think it is 12.5 metres.

Answer 10

100m