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Geometry 18 Online
OpenStudy (anonymous):

Pappy's Pond is located close to an elementary school. The city plans to fence it in to keep small children from accessing the water. What is the perimeter of a triangular-shaped fence the city can place around the pond, using integer coordinates (no decimals) on the grid? You must show all work to receive credit.

OpenStudy (anonymous):

OpenStudy (anonymous):

you gotta enclose the whole thing right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

oh i thought it was a rectangle, but it has to be a triangle damn

OpenStudy (anonymous):

darn

OpenStudy (anonymous):

looks like \((0,0)\) to \((2,8)\) to \((10,0)\) would do it doesn't it?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

k then we can do that one i think find the perimeter i mean

OpenStudy (anonymous):

idk how

OpenStudy (anonymous):

from \((0,0)\) to \((10,0)\) you just count and get \(10\)

OpenStudy (anonymous):

the other two require the distance formula do you know it ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

it is pythagoras basically from \((0,0)\) to \((2,8)\) you have a triangle with one side \(2\) and the other side \(8\) so the distance (hypotenuse) is \[\sqrt{2^2+8^2}=\sqrt{4+64}=\sqrt{68}\]

OpenStudy (anonymous):

not sure how you can do this problem without the distance formula or some such thing does that ring a bell at all?

OpenStudy (anonymous):

then from \((2,8)\) to \((10,0)\) the distance is \[\sqrt{8^2+8^2}=\sqrt{64+64}=\sqrt{128}\] or if you prefer \[8\sqrt2\]

OpenStudy (anonymous):

is that the answer

OpenStudy (anonymous):

does this look at all familiar to you?

OpenStudy (anonymous):

to get an answer you would have to add them up \[10+\sqrt{68}+\sqrt{128}\]

OpenStudy (anonymous):

i think you would get a different answer if you used different points so there would be no one right answer

OpenStudy (anonymous):

8.246+10+11.313?

OpenStudy (anonymous):

29.559?

OpenStudy (anonymous):

i guess if you used a calculator

OpenStudy (anonymous):

but thats the answer

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

?

OpenStudy (anonymous):

looks good to me

OpenStudy (anonymous):

thanks could you help w another

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