Simple math problem - need help or explanation

[sivamit]

Simple math problem



_Sharon

 

[Chinaski]

Re: Simple math problem

16.25 cm

 

[subharpe]

Re: Simple math problem

Join the corners connected to the blue lines. it is a diagonal passing through the center of the circle, as the square is exactly at the middle of the circle O. Then let
OD = xcm

Then radius = OC = (10+x)cm =OA
Also OB = x cm
AB = 15cm. Then,
Also because the two blue lines are both 15cm the lines must be perpendicular to BO. Then from Pythagoras,
AB²=AO² - OB²
or 15² = (10+x)² - x².

This can be solved to be
x = 4.25.
Therefore radius = 16.25 cm.

See the figure for reference.

 

[sivamit]

Re: Simple math problem

What if the angle is not 90 "(the lines are parallel but the angle with the square is not 90) You cannot use the theorem..right?
I think, that information is missing..(?!)

_Sharon:

 

[subharpe]

Re: Simple math problem

The angles will be 90 degrees because of symmetry. The two blue lines are parallel and of same length. The red line is passing through the centre and therefore is along the radius.

 

[sivamit]

Re: Simple math problem

The angles will be 90 degrees because of symmetry. The two blue lines are parallel and of same length. The red line is passing through the centre and therefore is along the radius.

But we can also draw 3 lines parallel to each other,without passing thru the box's centre point..:?:

 

[subharpe]

Re: Simple math problem

You have to consider the symmetry of the structure. As the two blue lines are of same length they must be at same distance from the central line (red). This comes from the symmetry of the structure

 

[sivamit]

Re: Simple math problem

You have to consider the symmetry of the structure. As the two blue lines are of same length they must be at same distance from the central line (red). This comes from the symmetry of the structure

yes u r right if you consider the particular case..yes..

 

[subharpe]

Re: Simple math problem

I still persist to say, it is not a special case. The red line has no option but to pass through the centre of the circle as the two ble lines are parallel and of the same length. They can be parallel bit not of the same length if they had not been perpendicular to the diagonal.

 

[sivamit]

Re: Simple math problem

I still persist to say, it is not a special case. The red line has no option but to pass through the centre of the circle as the two ble lines are parallel and of the same length. They can be parallel bit not of the same length if they had not been perpendicular to the diagonal.

Yes you are right..
The blue lines are of equal length because they are touching the circle.If it was nt a circle..?? for instance, a square not aligned with the inner square...in that case the lines may be parallel but not of equal length.Or equal length but may not be parallel...right..?

_Sharon

 

[Jone]

Re: Simple math problem

Here is another way. First rotate the figure so that all lines become vertical. If the width of the square is a then its diagonal is a/√2. Thus the radius is r = a/√2+10. The intersection point of the blue lines and the circle is (x, y) = (±a/√2, 15). Substituting this for the equation of the circle, x^2+y^2 = r^2 we get r = 16.25.