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Mathematics 15 Online
OpenStudy (anonymous):

A long time ago Dulani found an island shaped like a triangle with three straight shores of length 3km, 4km and 5km. He said nobody could come within 1km of his shore. What was the area of his exclusion zone?

jimthompson5910 (jim_thompson5910):

You basically have this triangle (which is the island itself)

jimthompson5910 (jim_thompson5910):

then you add on this buffer zone of 1 km around the entire island

jimthompson5910 (jim_thompson5910):

so you need to find the area of this red zone you can break it up into 3 rectangular pieces (shown in blue in this new attachment) and 3 circular sectors (shown in green)

jimthompson5910 (jim_thompson5910):

do you see how to get the answer?

OpenStudy (anonymous):

So like for the area of one of the rectangular pieces, do we just add 1km to each sides?

OpenStudy (anonymous):

oh wait kinda get it

jimthompson5910 (jim_thompson5910):

well the rectangular pieces are relatively easy to find 3*1 = 3 square km 4*1 = 4 square km 5*1 = 5 square km

jimthompson5910 (jim_thompson5910):

so we have 3 separate areas of 3, 4, and 5 square km

jimthompson5910 (jim_thompson5910):

this is all for the rectangular blue pieces

jimthompson5910 (jim_thompson5910):

the green circular pieces are going to be trickier to find, but it's doable

jimthompson5910 (jim_thompson5910):

you just need the central angles to find the area of each circular sector

jimthompson5910 (jim_thompson5910):

so how do you find those?

OpenStudy (anonymous):

is it you find the inside angle of the triangle and the you can find it? Using geometry..

jimthompson5910 (jim_thompson5910):

more like trigonometry (since you'll use sine, cosine or tangent)

jimthompson5910 (jim_thompson5910):

but yes, you find the inner angles of the triangle first

OpenStudy (anonymous):

Ok thanks, i just got mixed up a little! We kept changing math teachers in school and they never really finish each chapters..

jimthompson5910 (jim_thompson5910):

I gotcha, so how would you find the inner angles

OpenStudy (anonymous):

i got 120 for the inner angle of the sector

jimthompson5910 (jim_thompson5910):

the bottom left sector has an angle of 90 so that area is (90/360)*pi*1^2 = pi/4

jimthompson5910 (jim_thompson5910):

the top sector has an angle of 360 - 90 - 90 - arctan(3/4) = 143.1301 degrees so the top sector has an area of (143.1301/360)*pi*1^2 = 0.39758*pi

jimthompson5910 (jim_thompson5910):

finally, the bottom sector on the right has an angle of 360 - 90 - 90 - arctan(4/3) = 126.8699 degrees so the bottom right sector has an area of (126.8699/360)*pi*1^2 = 0.35242*pi

jimthompson5910 (jim_thompson5910):

so we have these 3 areas for the 3 rectangular pieces: 3, 4, 5 and these 3 areas for the 3 circular pieces: pi/4, 0.39758*pi, 0.35242*pi add them up and tell me what you get

OpenStudy (anonymous):

14.7

jimthompson5910 (jim_thompson5910):

adding up pi/4, 0.39758*pi, 0.35242*pi, you should get pi/4+ 0.39758*pi+ 0.35242*pi = 3.1415926535898 which is basically pi (more or less)

jimthompson5910 (jim_thompson5910):

this is because the circular sectors from each corner combine to make a circle of radius 1, which has an area of pi*r^2 = pi*1^2 = pi

jimthompson5910 (jim_thompson5910):

so your total area is 3+4+5+pi = 12 + pi

jimthompson5910 (jim_thompson5910):

12+pi is the total area of the shaded exclusion zone

OpenStudy (anonymous):

Yes yes, thank you so much!

jimthompson5910 (jim_thompson5910):

yw

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