Help Needed For Geometry Based Math Problem.[Urgent]?

Question

can anyone help me with this one? it looks a bit though.

Here's a rough sketch of the problem which i made in photoshop:

http://maxupload.com/img/E1818F63.jpg

ill still type the details though. first there is a big square. then a circle inside with touches the edges of the square. then there is a little rectangle on the top right which has an edge (right bottom) touching the circle.and the dimensions of that little rectangle are 1x2. you need to find the radius of the circle. note the diagram is not drawn to scale.

Please Explain in detail and also if u can give a visual demonstration like i did. i need to submit this in 2 days and its worth good marks so please help.

can anyone help me with this one? it looks a bit though.

Here's a rough sketch of the problem which i made in photoshop:

http://maxupload.com/img/e1818f63.jpg...

ill still type the details though. first there is a big square. then a circle inside with touches the edges of the square. then there is a little rectangle on the top right which has an edge (right bottom) touching the circle.and the dimensions of that little rectangle are 1x2. you need to find the radius of the circle. note the diagram is not drawn to scale.

Please Explain in detail and also if u can give a visual demonstration like i did. i need to submit this in 2 days and its worth good marks so please help.

EDIT: What Does This Sign (^) mean? And Also everyone has given different answers who shud i believe? and could u also give a visual demonstration too please?

EDIT: What Does This Sign (^) mean? And Also everyone has given different answers who shud i believe? and could u also give a visual demonstration too please?

EDIT: What Does This Sign (^) mean? And Also everyone has given different answers who shud i believe? and could u also give a visual demonstration too please?

please explain in real detail i am just 15 and dont know much. mean maths teacher just gives tough homework.

i understand until this part:

(a+2d)^2=a^2+a^2
(a+2d)^2=2a^2
a^2+4d^2+4ad=2a^2
4d^2+4ad=a^2
a^2-4ad-4d^2=0
d=sqrt(5)
a^2-4a*sqrt(5)-20=0

why do u have to square all these things? but i am begining to understand much better thoough. will hit the paper tomorow morning and try to put all those equations to work.

i think that your best friends answer is the closest except that the diagonal of the square will not pass through the diagonal of the rectange. so i shud cut the rectangle to a square to have dimensions 1x1 and then apply the same method from there. the answer shud be around 3.42.

well i have also posted this on other sitres and majority say 5 is the answer and majority have working like what zax has. but the used (x-1)^2 + (x-2)^2= x^2 . but i just dont understand how u get points x-1 and x-2. i am not that gud at geometry so could someone do a quick diagram in paint to show the x-1 and x-2 or explain how u get it. thnx.

Answer 1

This is turning out to be an interesting problem. I don't really know how to solve this, but messing around I came up with a radius of 5, since I wasn't sure I arrived at that correctly, I drew the figure in a CAD program using 5 for the radius and it fit. So My answer agrees with Catbarf. I don't quite understand "Your best friend's" "proof" that 5 is wrong, but I disagree with it.

I'd recommend, studying Catbarf's answer and trying to understand it (that is what I am going to do, since as i said, I don't know how i came up with 5, I just know that it does work when the figure is constructed)

BTW, the "^" means Exponentiation, e.g. 2^2 is 2 squared, if you happen to use the Windows calculator, the "button: for square is {x^2} on many pocket calculators it will be [x²]

EDIT: well I figured out what "Your best friend" did wrong he is adding apples and oranges, or in this case rectangles and squares, his method would work if the small rectangle was a square, but it isn't.

2nd Edit:
Well, I still don't understand the algebra in involved in rearanging the equation, so I can't add anything about _how_ to find the answer. However, the votes are 3 to 2 for the answer being 5.
If you still prefer "Your best friends" answer, you could do a rough graphical check. While Photoshop is not a CAD program and is definitely not the program I'd recommend, since you say did use Photoshop for your illustration, I'll point out that since Photoshop does have guides, grids and rulers, you could (approximately) test your answer by drawing it (if you have a ruler and a compass you could also do it on paper) . If you do, you'll find that with a radius of 5.4 the rectangle does not touch the circle "quo errat demonstrator" However if the radius is 5 the rectangle appears to touch,. Due to the lack of precision using Photoshop for this we can't say for sure that the answer IS 5, but it does show 5 is closer than 5.4, "quod erat demonstrandum". Using a CAD program, the rectangle comes within the precision of the programs math library to touching, the measured mismatch is, 0.0000009 which is pretty close and is just math error..

To sum it up, the answer is 5, "quod ego dico"

3rd edit;
You ask; "but i just don't understand how u get points x-1 and x-2. i am not that good at geometry so could someone do a quick diagram in paint to show the x-1 and x-2"

The 1 and 2 are the corner of the rectangle that is touching the circle, remember it is 1cm by 2cm. I don't have anywhere to upload a picture, and I don't really want to register at maxupload. But we can work with your picture. Imagine that the upper left hand corner (where the small rectangle meets the square) is the origin, with the coordinates 0,0. For convenience we will use positive numbers for our coordinates, usualy going down from the origin the coordinates would be negative numbers, but that would just confuse things here. Since one corner of the rectangle is at 0,0, and it is 2 units wide by 1 high, the coordinates of the corner where it touches the circle are 2,1. Since the perimeter of the circle touches the square, the coordinates of the center of the circle are equal to the radius (which is being indicated by x, presumably because it is unknown, I think x is a poor choice because of possible confusion with x,y coordinates, but what ever) So the center of the circle is at x,x. Therefore, the coordinates of the point we know (the corner of the rectangle) = x -2, x-1.

From that we get (x-1)^2 + (x-2)^2= x^2, how we get from there to 5, I still don't really understand, sorry.

Answer 2

^ means to the power of. It is common syntax in programming, and it is often used on this forum.

2 to the second power or 2 squared is 2^2.

I am not sure what you mean by visual demonstration.

The relationship to note are that you have the pythagorean theorem which states that for a right triangle, the sum of the squares of the sides equals the square of the diagonal. There are two diagonals in this problem, the big square and the small rectangle. If you notice, the diagonal of the big square is the sum of two small rectangle diagonals and the diameter of the circle.

You know what the diagonal of the small rectangle is from the pythagorean theorem and the dimensions of 1 cm and 2 cm. It is 1*1+2*2=c*c; c=sqrt(5) or 2.236 (this can be shown as 1^2 + 2^2 = c^2). And, you know that the sides of the big square are equal to the diameter of the circle. Now, you just have to put the things you know into relationships and choose appropriate variables.

I doubt that anyone will go to the effort of making a diagram like you did in photoshop.

The first answer is not wrong. But, I will use the same variables to make it clear from a different perspective.

The diagonal is a+2d not a+d

So...

(a+2d)^2=a^2+a^2
(a+2d)^2=2a^2
a^2+4d^2+4ad=2a^2
4d^2+4ad=a^2
a^2-4ad-4d^2=0
d=sqrt(5)
a^2-4a*sqrt(5)-20=0

Use the quadratic equation to solve.

a=1; b=-4*sqrt(5); c=-20

I get a=10.7967 cm = diameter

so, the radius of the circle is about 5.3984 cm

When I checked my first two answers, I was off. I know I made a mistake somewhere. But, I think you get the idea of the analysis. And, always check your answer. The above answer is correct. I expect that 5.4 cm is close enough for your purposes.

The answer that states that the radius is 5 is also wrong. If it was 5, then the diameter would be 10 and a side of the square would be 10. Plus we know the diagonal of the rectangles is sqrt(5)=2.23607 or so.

We get 10^2+10^2=sqrt(200)=14.14
And, we get 2*sqrt(5)=4.47+10=14.47

These numbers are not equal. So, the answer cannot be 5.

Answer 3

Let's put an axis thru the center of the circle such that the circle's diameter is 2m, and the circle has the equation x^2+y^2= m^2. The square then is defined by the lines y= +/- m and x= +/- m. The right corner of the square is at (m,m), and the point of the rectangle that touches the circle is (m-2, m-1). Since that point is on the circle, we can substitute to get (m-2)^2+(m-1)^2 = m^2.
Then m^2-4m+4+m^2-2m+1 = m^2
and m^2 -6m + 5 = 0
which factors as (m-5)(m-1).
m=5 is the real answer, m=1is sort of a trivial answer.

Answer 4

From the point where the rectangle touches the circle
define the distance to the horizontal axis as a -1
and define the distance to the vertical axis as a - 2 .
This will form a congruent triangle where the radius can
be calculated from ( a - 1 ) ² + ( a - 2 ) ² = a ² and a = 5 cm.

Answer 5

Let the side of the square be 2a

Then the radius of the circle is a
Let the diagonal of the ractangle be d
Now using pythagoras theorem :
(a+d)^2 = a^2 + a^2
(a+d)^2 = 2a^2
but d = sqrt(5)
a^2 + d^2 + 2a sqrt 5 = 2a^2
5 + 2a sqrt 5 = a^2
or
a^2 - 2 sqrt 5 a - 5 = 0

Solving this equation and rejecting the negative root :

a = (1+sqrt(2)) sqrt (5) = 5.3983 = 5.4

Answer 6

r=5.

Only look at the uper-left 1/4 of your plot, that is the square of size r by r. Then denote the upper-left vertex of as point A, center of the circle as point O, the touching point of rectangle and circle as point B, and the upper-right vertex of the rectangle as point C.

Then angle CAB: tan(CAB)=1/2 ==> CAB=26.565 deg
Angle OAB= 45-26.565==18.435

Start from B draw a perpendicular line to OA, meet OA at point D. Then
BD= AB*sin18.435=0.707 cm
AD=AB*cos18.435=2.1213 cm

DO=AO-AD=1.414r-2.1213

DO^2+BD^2=BO^2
(1.414r-2.1213)^2+0.707^2=r^2
==> r=5, (throw away r=1)