# Thread: Geometry - The area the goat can graze?

1. ## Geometry - The area the goat can graze?

I need some help with this one.

The area the goat can graze - Google Search=

3. Let $\displaystyle A_1=\pi{(15.7)}^2-\frac{\frac{\pi}{4}}{2\pi}\pi{(15.7)}^2=\pi{15.7}^ 2-\frac{\pi(15.7)^2}{8}=\frac{7\pi(15.7)^2}{8}$ be the area the goat can graze in a circle around the pen from when the rope (fully extended) is paralel to BC until it is parallel to AB.

Let $\displaystyle A_2=\frac{1}{4}\pi{(15.7-\sqrt{162})^2}$ be the remaining quarter circile the goat can graze after its tether makes contact with point A.

Then the area the goat can graze is given by:

$\displaystyle A_1+A_2=...$

4. Originally Posted by VonNemo19
Let $\displaystyle A_1=\pi{(15.7)}^2-\frac{\frac{\pi}{4}}{2\pi}\pi{(15.7)}^2=\pi{15.7}^ 2-\frac{\pi(15.7)^2}{8}=\frac{7\pi(15.7)^2}{8}$ be the area the goat can graze in a circle around the pen from when the rope (fully extended) is paralel to BC until it is parallel to AB.

Let $\displaystyle A_2=\frac{1}{4}\pi{(15.7-\sqrt{162})^2}$ be the remaining quarter circile <<< (?) the goat can graze after its tether makes contact with point A.

Then the area the goat can graze is given by:

$\displaystyle A_1+A_2=...$
I don't want to pick at you but I've got a slightly different result:

The remaining sector with radius $\displaystyle r \approx 3$ has the area:

$\displaystyle A_2=\dfrac{\bold{\color{red}3}}{\bold{\color{red}8 }}\pi{(15.7-\sqrt{162})^2}$

But probably I have misunderstood the question - as usual ...

5. Agree with Earboth (maybe we both misunderstood!).

p = pi ; r = 15.7 ; h = sqrt(162)

A = (p/8)(10r^2 - 6hr + 3h^2) = ~687.9812

6. Yeah, earboth is right. I gave it the old college try though.