Yet another puzzle

[mendhak]

Check out the attached image.

The rectangle in the corner is 2cm * 4 cm. Find the radius of the circle.

[honeybee]

I don't know for sure if it can be solved , but this is an inappropriate place for it. Post it in the Maths section

.

[mendhak]

WELL

There have been puzzles here before.

you're not going to go there and tell me to post it in chit chat, are you?

[crptcblade]

Eleventy-six?

[SeaHag]

10 ?

[mendhak]

SHOW YOUR WORK!!!

[plenderj]

The radius is five

[SeaHag]

*** show my work...


how do you think i did it?

its a math question.. i used ratios


Seahag

[honeybee]

Originally posted by mendhak
WELL

There have been puzzles here before.

you're not going to go there and tell me to post it in chit chat, are you?

It is a maths puzzle, that's why *pouts*

.

[Arc]

32..yep

[Hu Flung Dung]

4*2^(1/2)

4 times the square root of 2

How?

The rectangle is 8 cm^2.

Each side of a square with the same area would have a length of the square root of 8, or 2*2^(1/2).

The lower-right corner of the small square with these dimensions would be directly on the circle (not inside or outside), and half-way between the top edge and the left edge of the circle.

Therefore, each side of the small square would be 1/4 the the length of each side of the big square, or 1/2 the radius of the circle.

2*2^(1/2) = 4*2^(1/2) = 4 times the square root of 2!

Is that right?

[mendhak]

Nope!!

[Hu Flung Dung]

Oh, yeh! Hmm. Your right, the lower-right edge of such a small square wouldn't be on the circle. It would be inside it.

Well, at least its the best try so far.

[mendhak]

The only try so far.

[mendhak]

An embarassing mistake...
I was calculating using 4cm and 8 cm :sheepish grin:

The answer is 10, but what is the method?

[Simon Caiger]

Difficult to explain in words but :

1. Draw two radii (length R), one vertical and one horizontal to form a quadrant at the top left of the circle.
2. Drop a vertical line from the bottom right hand corner of the box to the horizontal radius.
3. Draw a horizontal line from the bottom right hand corner of the box to the vertical radius.

Let's call the height of the box X and the width of the box Y

Using Pythagoras

(R - X)^2 + (R - Y)^2 = R^2

Substituting the values for X and Y (2cm and 4cm respectively)

(R - 2)^2 + (R - 4)^2 = R^2

Expanding
R^2 - 2R + 4 - 2R + R^2 - 4R + 16 - 4R = R^2
Simplify
R^2 - 12R + 20 = 0

using x = -b +/- sqr(b^2 - 4ac)/2a

gives R= 10cm

Any prizes this time?

[mendhak]

Originally posted by Simon Caiger
Difficult to explain in words but :

1. Draw two radii (length R), one vertical and one horizontal to form a quadrant at the top left of the circle.
2. Drop a vertical line from the bottom right hand corner of the box to the horizontal radius.
3. Draw a horizontal line from the bottom right hand corner of the box to the vertical radius.

Let's call the height of the box X and the width of the box Y

Using Pythagoras

(R - X)2 + (R - Y)2 = R^2

Substituting the values for X and Y (2cm and 4cm respectively)

(R - 2)2 + (R - 4)2 = R^2

Expanding
R^2 - 2(2)R + 22 + 2R - 2(4)R + 42 = R^2
Simplify
R^2 - 12R + 20 = 0

using x = -b +/- sqr(b^2 - 4ac)/2a

gives R= 10cm

Any prizes this time?

Ladies and Gents, we have a wiener!

Your prize is a special McBrain Burger with a hot dog.

[SeaHag]

i had 10 damit..

heres my work

5 squares fix across the top.. so at
4*5/2 = 10

seems less complicated

[Simon Caiger]

Wot no frogs legs?