Urgent help needed for weird math problem for 10 mks!!

[satish555]

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 help someone.

[Epsilon=One]

you need to find the radius of the circle.The radius is 5 cm.

You'll have to do the geometry to earn your credit.

[satish555]

please could ya tell me how cause i was slo told to write an explanation. please please. i dont understand it. could u help me pelase? could you please give an explanation and also if you could give a visual demonstaration like the one i have done, that would be very helpful. thnx.

[satish555]

i am not gonna ask u everything but i think i have a little clue.... i just found something. the diameter of cricle is 2x the radius. so say i cut the square into two using a diagonal, the rectangle is cut in 2 too. now i could find the length of the little peice of diagonal outside the circle of the rectangle. which is square root of 5 by pythagoras theoram. and u multiply this by 2 to get the two outside of the circle lengths of the diagonal so we know the length of the diagonal outside the circle but what next. can u please help?

[Epsilon=One]

i am not gonna ask u everything but i think i have a little clue.... i just found something. the diameter of cricle is 2x the radius. so say i cut the square into two using a diagonal, the rectangle is cut in 2 too. now i could find the length of the little peice of diagonal outside the circle of the rectangle. which is square root of 5 by pythagoras theoram. and u multiply this by 2 to get the two outside of the circle lengths of the diagonal so we know the length of the diagonal outside the circle but what next. can u please help?You are correct in the procedure. I believe you have noted that the diagonal of the square does not cut the rectangle into two even parts.

What you have is 3 similar triangles and you know:the side of the middle one is:
2–x;

and, the hypotenuse is:
√(2(x²–4x+4).

[satish555]

is it like this?
first cut the rectangle into two little squares so one has a dimension of 1x1. and the diagonal of that square will be sq.rt. of 2. and the diagonal of the big square will pass thru or connect to this one.
2. You have a triangle with sides - two equal sides= R & hypotenuse is R+(sqrt2)
3. Using pythogoras for this triangle gives -
R² +R²=(R +(sqrt2))²
R² +R²= R² +2xRx(sqrt2) +(sqrt2)²
2R² = R² +2xRx(sqrt2) + 2
2R² - R² = 2xRx(sqrt2) + 2
R² = 2xRx (sqrt2) +2
R² -2xRxsqrt2 - 2 =0
(sqrt2 = 1.414)
R² - 2.83R -2 =0
then solve this quadratic equation using quadratic formula.
R = 3.415cm?

[OfficeShredder]

How do you conclude the diagonal of the large square must be R+sqrt(2)?

You're close, but not quite there yet

[satish555]

no actually the diagonal of the large square is 2(R+Sqrt2) but the radius is r+sqrt2. u see i drew a triangle, from the center of circle to the little square measuring 1x1 which has a diagonal of sqrt 2. so the two sides of the triangle are equal but the third side is equal to one of the other sides + sqrt 2.

[OfficeShredder]

Oh, that makes more sense. For some reason I thought you were setting R as the diameter of the circle.

You've got it right in that case

[satish555]

but then epsilon one said the answer is 5.

[OfficeShredder]

Actually, no, you made the same mistake I did. You don't get 2R^2 = (R+sqrt(2))^2, because if you take the diagonal from the corner of the square, and extend it to the edge of the rectangle, it isn't at the circle yet.

[OfficeShredder]

Epsilon is right, the answer is five.

I'll give you a little hint.... extend the longer part of the rectangle across to the other side of the square

[Epsilon=One]

but then epsilon one said the answer is 5.You must solve for the ratios of the similar triangles with what you know: the sides of the rectangle are 1 cm and 2 cm.

[satish555]

well i got a bit of help from some other place but i cant understand it.here is what i got.

If X = radius, you can see that the lines (x-1) and (x-2) and x form a right angled triangle. This gives (x-1)^2 + (x-2)^2= x^2. You can either factorise it and solve it or use common knowledge of the Pythagorean triple 5, 4 ,3. Therefore, x = 5 !!

but i cant understand how u get the x-1 and the x-2. i think i really need a diagram or somehting to understand this.

[OfficeShredder]

Extend the line from the 2x1 rectangle that is horizontal across to the other side of the square. Draw a line from the center of the circle to the newly drawn horizontal line. Then draw a line from the center of the circle to the corner of the rectangle.

You just completed a triangle with hypotenuse r, and legs r-1 and sqrt(2r-1) (finding the third leg using pythagorean). Extending the horizontal leg, you can see sqrt(2r-1)+2=r. Hence, you solve that

[Epsilon=One]

Extend the line from the 2x1 rectangle that is horizontal across to the other side of the square. Draw a line from the center of the circle to the newly drawn horizontal line. Then draw a line from the center of the circle to the corner of the rectangle.

You just completed a triangle with hypotenuse r, and legs r-1 and sqrt(2r-1) (finding the third leg using pythagorean). Extending the horizontal leg, you can see sqrt(2r-1)+2=r. Hence, you solve thatBingo!!!

However, I don't believe satish555 has qualified to submit the procedure and gain any credit.

[satish555]

Could I Do It Like This:

http://maxupload.com/img/9091ADC7.jpg

Therefore using pythagoras theoram:

(x-1)^2 + (x-2)^2 =x^2
x^2 -2x +1 +x^2 -4x +4 =x^2
x^2 -6R +5 = 0

Solving using Quadratic formula:

here, a =1
b= -6
c=5

therefore X = [ -(-6) +_ sqrt((-6)^2 -4*1*5)]/ (2*1)
R= [6 +_ sqrt (36-20)]/2
R = [6 +_ 4]/2
R = [6+4]/2 or [6-4]/2
R =5 or 1

And The Radius Cant Be 1cm Because The Rectangle Already Has A Width Of 2cm, Therefore The Radius Is 5cm.

[Epsilon=One]

Could I Do It Like This:

http://maxupload.com/img/9091ADC7.jpgYes. Good solution when you can show how you arrived at x-1 and x-2 for the sides. Then, you qualify for the credit.