Check out the attached image.
The rectangle in the corner is 2cm * 4 cm. Find the radius of the circle.
http://www.vbforums.com/attachment.p...postid=1355681
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Check out the attached image.
The rectangle in the corner is 2cm * 4 cm. Find the radius of the circle.
http://www.vbforums.com/attachment.p...postid=1355681
I don't know for sure if it can be solved :p, but this is an inappropriate place for it. Post it in the Maths section :p
.
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? :p
Eleventy-six?
10 ?
SHOW YOUR WORK!!!
The radius is five :)
*** show my work...
how do you think i did it?
its a math question.. i used ratios :D
Seahag
It is a maths puzzle, that's why *pouts*Quote:
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? :p
.
32..yep
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?
Nope!!
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.
The only try so far. :)
An embarassing mistake...
I was calculating using 4cm and 8 cm :sheepish grin:
The answer is 10, but what is the method?
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?
Ladies and Gents, we have a wiener!Quote:
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?
Your prize is a special McBrain Burger with a hot dog.
i had 10 damit..
heres my work
5 squares fix across the top.. so at
4*5/2 = 10
seems less complicated
Wot no frogs legs?