Need help with a math problem

[Joony]

click here

i got 25u^2 as my answer...i don't think i did it right, tell me how you got your answer step by step

 

[BornStar]

well, the width of the square should be 3+(3/sqrt(2)). We square that and get the answer. I'll post how I got to that in a second.

Alright, I hope I can explain this clearly. Draw a diagonal line through the centers of the circle. This line is obviously at 45 degrees. Now, go from the left side of the top circle to the center, that's one unit. From the center to where that circle touches the next circle is a 45 degree angle so if we use trig, it should be 1 cos 45 or 1/sqrt(2). We do this twice more as they both will be the same distance across and add them up. Thus we have 3+(3/sqrt(2)) as the length of a side. Let me know if you have a problem with something.

 

[Joony]

hmm...thinking about what you said...

 

[BornStar]

Whoops, hold on, let me go through it again.

 

[BornStar]

Sorry, that's not quite right. Probably shot my credibility to heck, but here we go. I was right for the top one and then repeat for the bottom one. Then for the middle one, we go from the intersection with the top to the intersection with the bottom. That distance is 2/sqrt(2) (1/sqrt(2) + 1/sqrt(2)). That should be 2+(4/sqrt(2)). Sorry about that.

 

[Joony]

uh...hmm....thinking

 

[silverpig]

<<2+(4/sqrt(2))>>


That's right, but it's only the length of a side. Square it to find the area... 4 + 16/sqrt(2) + 8 = 12 + 16/sqrt(2) = 12 + 8sqrt(2) units squared as your final answer.

 

[LordMaul]

I calculate it to be exactly 42.

 

[BornStar]

Here's my very cluttered work, might help you.

 

[Evadman]

Assuming I remember correctly, here is how you figure it out.

Ok, first you have the 3 circles with a Rad of 1. Then to that radius you must add the small sections of the square that the circle does not touch. Since the circle does not extend all the way into the corner, BornStar's answer will not work. You have to get that small section that is missed. If I remember form geometry, that squared off part of a circle is exactly 1/2 the radius, so you end up with 1/2(2) + 2(3) for a total of 7 for a bisect of the square from corner to corner. ( X shape )

now that you have that 7, you can make the square into 3 equalateral triangles. ( atleast they should be assuming that is a square. ) then you would end up with a side length of 3.5. 3.5 x 3.5 = 12.25

If you want to think it is not an equalateral, it will atleast be an iosilies, and you have 2 sides ( 3.5" and 3.5" ) and the degree of the angle between them (90) and take the cos (I think ) and you will get the length of the other side.

Did I do that right? It has been a while.

I get ya now Burnstar, I see what you are doing with that pic. I was typing this from right after yoru first post

 

[JW310]

edit: nm... didn't read the question right

JW

 

[BornStar]

<< Assuming I remember correctly, here is how you figure it out.

Ok, first you have the 3 circles with a Rad of 1. Then to that radius you must add the small sections of the square that the circle does not touch. Since the circle does not extend all the way into the corner, BornStar's answer will not work. You have to get that small section that is missed. If I remember form geometry, that squared off part of a circle is exactly 1/2 the radius, so you end up with 1/2(2) + 2(3) for a total of 7 for a bisect of the square from corner to corner. ( X shape )

now that you have that 7, you can make the square into 3 equalateral triangles. ( atleast they should be assuming that is a square. ) then you would end up with a side length of 3.5. 3.5 x 3.5 = 12.25

If you want to think it is not an equalateral, it will atleast be an iosilies, and you have 2 sides ( 3.5" and 3.5" ) and the degree of the angle between them (90) and take the cos (I think ) and you will get the length of the other side.

Did I do that right? It has been a while. >>



Uh uh, I wasn't trying for the length of the diagonal, I was going for the length of the side using multiple levels of the square. I just neglected to mention that my answer had to be squared to get the actual area of the square.

 

[Evadman]

<< How can the width of the square be >3 if there's some overlap in the way the circles are layed out? The width of the square is going to be slightly less than 3. >>


3 is the radius not the diameter.

 

[Joony]

me writing some stuff and scanning it in soon

 

[BornStar]

<< How can the width of the square be >3 if there's some overlap in the way the circles are layed out? The width of the square is going to be slightly less than 3.

JW >>



Quite easily, the measurement was a radius measurement. I amended my answer as it was incorrect.

edit: improper use of words

 

[JW310]

<<

<< How can the width of the square be >3 if there's some overlap in the way the circles are layed out? The width of the square is going to be slightly less than 3. >>


3 is the radius not the diameter. >>



Doh... realized that just after I posted... my mistake

JW

 

[silverpig]

Here it is spelled out a bit better.

Draw a line from the left edge of the square where it meets the top circle to the middle of the top circle. It should be horizontal like this: -

Next, draw a line from the centre of the top circle to it's edge where it meets the middle circle. It's on a 45 degree angle like this: \

Draw another line through the entire middle circle, from where it meets the top circle, to where it meets the bottom one. \

Draw a final line from the centre of the bottom circle to the edge where it touches the square on the right side. -

You should have something that looks like this (I hope the spacing works out)

_
..\
...\
....\
.....\_


You want to find the horizontal distance this goes, and you know the angle is 45 degrees, and the length of each segment is 1.

The horizontal lengths of the horizontal lines is easy; they're both 1. So you have 2 pieces, and 2(1) = 2.

Now, for the middle angled piece, you basically know the angle is 45 degrees and the length of the segment is 4(1) = 4. Use either sin or cos, it doesn't matter, but you'll have something like arccos 45 = x/4

arccos 45 is 1/sqrt2, multiply both sides by 4, and you get the horizontal distance of the middle angled line is 4/sqrt(2). Add that to the 2, and you get 2 + 4/sqrt(2). Square it like I just showed, and that's your area.

 

[Evadman]

<< edit: improper use of words >>



Why did that make me think of a bunch of people in OT?

Oh, and stupid queastion, but are we all coming up with the same answer of about 12.25 for a rad of 1?

 

[BornStar]

<<

<< edit: improper use of words >>



Why did that make me think of a bunch of people in OT? >>



Yeah, I try to use actual words, spell them and use them correctly and all of that. Sometimes my fingers just can't keep up with my brain though.

 

[Evadman]

First person I have ever seen who admitted to that!

 

[silverpig]

<<If I remember form geometry, that squared off part of a circle is exactly 1/2 the radius>>


You didn't remember correctly. That little squared off part is 1/sqrt(2) - 1/2 of the diameter if I did that right in my head...

 

[Joony]

I did it the easier way...

the wonders of special 45:45:90 triangles. Thanks for your help guys!

 

[Bulldozer]

I come up with A = 23.3 square units. I'll write it all out if somebody else doesn't come up with the same thing.

edit: I guess I was a little late.

 

[BornStar]

<< I did it the easier way...

the wonders of special 45:45:90 triangles. Thanks for your help guys! >>



Yeah, that way works to. I didn't feel like dealing with those little chunks in the two corners.

 

[silverpig]

That's right bulldozer, but the exact answer is 12 + 8sqrt(2) u^2