i couldn't follow your instructions... it's nearly 4 am here.
I constructed your scenario (square with a 16 diagonally, blue lines 24) and I got a less than 16 red line. I draw in a 1:4 size (4 cm in the LC = 1 cm on paper). I didn't get a straight 16 with the other two given scenarios either (mine, where the diagonally is 24 and the other one given where the red and blue can form a 90° triangle with the last side being the length of the square. Those last two were closer to 16 than yours, so I'll guess it's either 27 or 28.
Honestly... it's getting too late (early) for me and I think I don't really care anymore. Maybe I'll just accept a (possibly) wrong answer and go with 28. Hopefully next week will be easier again. I don't see the mathematical way without making at least ONE assumption that can't be proven.
Ok then. I will not critisize your decision. I may not have been clear enough.
hints
I don't recommend trying to work from a scale drawing, because
we don't know the scale of the circle nor the size of the square
relative to the given lines that emanate from the corners of the
square.
The answer can be calculated by using analytic geometry.
Two hints:
1. It helps (a lot) if you use the right coordinate system.
2. Two equations in two unknowns.
I got 28 as the radius. Print it out, draw a second set of blue lines one of which crosses one of the original lines. Not you have a right triangle where you have 2 given sides (12 cm) and on unknown. After using Pythagorean theorem, you get the hypotenuse which is 16.79. Use that as you new given, use pythagrean theorem again to get the diagonal of the square which is 23.99. Divide that by 2 and add 16 (the red line) to it, and you get 27.995. Round that up and you get 28. Not sure how good my math is because I really didn't put to much effort in it, I already have the avvie anyways. So I wouldn't count on my answer, but whatev. Have fun solving it yourself if you think my answer is wrong, which it very well might be.
devineatheist wrote:I got 28 as the radius. Print it out, draw a second set of blue lines one of which crosses one of the original lines. Not you have a right triangle where you have 2 given sides (12 cm) and on unknown....
im not sure about my geometry, therefore not sure about a few things.
are you sure that then you draw the second blue line perpendicular to the first one, that is bisects that line into 2 segments of 12?
the answer is 26,
if one of the blue lines is 24 and the inner square is centered then it would be 24 the reverse way giving us a chord length of 48. the segment hieght would then be the 16.
can anyone take it beyond that?
Roly_Poly_Sandwiches wrote:the answer is 26,
if one of the blue lines is 24 and the inner square is centered then it would be 24 the reverse way giving us a chord length of 48. the segment hieght would then be the 16.
can anyone take it beyond that?
Well the thing is... the chord length is not 48 BECAUSE THERE IS THE SQUARE IN BETWEEN!
devineatheist wrote:I got 28 as the radius. Print it out, draw a second set of blue lines one of which crosses one of the original lines. Not you have a right triangle where you have 2 given sides (12 cm) and on unknown.
The only problem with that is, I already measured the picture with rulers and IT IS NOT TO SCALE! therefore you cannot definitively SAY that it is 12 cm. That was my first method of trying to solve it, but after seeing that it didn't work, I used the method posted 2 pages ago.
yo devineatheist we got the same answer. someone can check this if they want. anyways, that's what i got! ciao.
jonathanasdf wrote:Well the thing is... the chord length is not 48 BECAUSE THERE IS THE SQUARE IN BETWEEN!
The square isnt in the way. the chord is 48. if you extend the blue line all the way to the end, the CORNER of the square would mark the middle of the chord. 24*2 would be 48
This is correct, but how can you use this information to solve the problem?
smokeybear - your logic works but where do you get the 24 you used when you solved for a?
i used the law of sines, with the 24 from one of the given blue lines.
but doesn't that mean that for your calculation of a you are assuming c = 24?
Sorry tooter. I didn't look at the image and forgot how it was.
klo you are right. For sine law to work there can't be 2 unknowns in the equation, so he substituted 24 for c, so he produced a circular argument.
Sin90/c=Sin45/a
c=sqrt(2a^2)
1/(sqrt(2)xa) x a = sin45
1/sqrt(2)=sin45
sin45=sin45
LOL!
I'm not quite sure what you're asking, c was not used in the formula.. i hope this clears that step up! but maybe i am wrong altogether, it's been a while since i've done geometry
smokeybear wrote:i used the law of sines, with the 24 from one of the given blue lines.
Your formula for the Radius is incorrect.
It should be +32 since you need the other red line
jonathanasdf wrote:Sorry tooter. I didn't look at the image and forgot how it was.
klo you are right. For sine law to work there can't be 2 unknowns in the equation, so he substituted 24 for c, so he produced a circular argument.
Sin90/c=Sin45/a
c=sqrt(2a^2)
1/(sqrt(2)xa) x a = sin45
1/sqrt(2)=sin45
sin45=sin45
LOL!
I never used c for 24. 24 is given in the equation. i have 1 unknown in the formula.
