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Mathematics 20 Online
OpenStudy (scarlettred):

Can someone help me with these two problem

OpenStudy (scarlettred):

OpenStudy (scarlettred):

OpenStudy (scarlettred):

@mathmate

OpenStudy (mathmate):

If we join the three warehouse locations together, we get a triangle. The location which is equidistant from the three supermarkets is the centre of the circumcircle.

OpenStudy (mathmate):

|dw:1405978815668:dw| The circumcentre equidistant to A, B and C can be found at the intersection of the perpendicular bisectors of the sides AB and BC. Since these two sides are parallel to the axes, their equations should be quite easy to figure out. Hope that helps, and feel free to post your answer for verification.

OpenStudy (scarlettred):

Thank you @mathmate but can you also help me with the second one

OpenStudy (mathmate):

For the first problem about the park, the idea is to find the total length of the two bike paths that cross each other. ok so far?

OpenStudy (scarlettred):

yep thank you but how do you find that

OpenStudy (mathmate):

We do not get to know the individual lengths, but just need the total length.

OpenStudy (mathmate):

Have you done cosine rule before?

OpenStudy (mathmate):

@scarlettred

OpenStudy (scarlettred):

not that i remember

OpenStudy (mathmate):

Well, here's a link that gives the details: http://en.wikipedia.org/wiki/Law_of_cosines Back to our problem: |dw:1405979665830:dw| Please study the figure and tell me if you agree with it. Our objective is to find the sum of the lengths AC & BD

OpenStudy (scarlettred):

It looks right to me and for the other question is the answer (0,-1)

OpenStudy (mathmate):

For the other question, the x-coordinate does not look right. can you check? For the bike path problem, we will need the cosine rule, Consider triangle ABC, \(AC^2=AB^2+BC^2-2~AB~BC~cos(x)\) and consider triangle BCD, \(BD^2=BC^2+CD^2-2~BC~CD~cos(x)\) But we know BC=12 AB=CD=16 and cos(x)=-cos(180-x) So we can add up the two equations to get: \(AC^2+BD^2=12^2+16^2-2*12*16*cos(x)+12^2+16^2+2*12*16*cos(x)\) \(AC^2+BD^2=800\) That's all we get to know without knowing x. Sorry, I get stuck here. It doesn't look like the problem is solvable without knowing the angle x. It is obvious if the park is a rectangle, then the total length would be 2*20=40 miles, which is the maximum total length. :(

OpenStudy (mathmate):

Oh, I just realized, the road is on the perimeter, nothing to do with the bike paths. So use the equation for the perimeter: Perimeter=2(L1+L2) and you will get the answer in a jiffy. No more cosine rules (for the bike paths).

OpenStudy (scarlettred):

ohh thank you i thought we had to include the bike path thank you for clarifying that @mathmate

OpenStudy (mathmate):

You're welcome. Sorry for having the messy details about the bike paths. :(

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