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23 May 2007, 15:59
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When we are solving for the radius in these two circumstances:
1)Square is inscribed in a circle. The diagonal creates a 90 degree angle and we have sides with a ration of 1:1: sq root of 2. If the diagonal is equal to 6 then we have sides of 6:6: 6 * sq root of 2. and a radius of 3* sq. root of 2.
2) Circle is inscribed inside of a square. Let's say the Diagonal is equal to 4
I know to get the radius we use the triangle formula:
y^2 + y^2 = r^2
2y^2 = (4)^2
y^2 = 16
y = 2* sq root of 2 and the radius is equal to sq root of 2.
Here is my question. When I'm approaching these two problems. For the second problem I understand how to calculate it because I've memorized it, but I'm trying to visually make sense of the difference between 1 and 2. For 2 if I didn't know any better I would assume that the diagonal 4 would create a 1:1: square root of 2 ration similar to the first problem and we would have 4:4: 4 * sq. root of 2 and a radius of 2 *sq. root of 2. Can someone explain the difference?
1)Square is inscribed in a circle. The diagonal creates a 90 degree angle and we have sides with a ration of 1:1: sq root of 2. If the diagonal is equal to 6 then we have sides of 6:6: 6 * sq root of 2. and a radius of 3* sq. root of 2.
2) Circle is inscribed inside of a square. Let's say the Diagonal is equal to 4
I know to get the radius we use the triangle formula:
y^2 + y^2 = r^2
2y^2 = (4)^2
y^2 = 16
y = 2* sq root of 2 and the radius is equal to sq root of 2.
Here is my question. When I'm approaching these two problems. For the second problem I understand how to calculate it because I've memorized it, but I'm trying to visually make sense of the difference between 1 and 2. For 2 if I didn't know any better I would assume that the diagonal 4 would create a 1:1: square root of 2 ration similar to the first problem and we would have 4:4: 4 * sq. root of 2 and a radius of 2 *sq. root of 2. Can someone explain the difference?
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24 May 2007, 10:58
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My logic is this (I need your help to continue):
Square is inscribed in a circle
If the diagonal of the square is 6 , then the circle radius will be 3 ! the pink line in the attachment. I don't understand your logic. please explain !
Circle is inscribed inside of a square
If the diagonal of the square is 6, then the we get as you rightly said a isosceles triangle with the proportions of 1:1:sqrt(2) where the proportions are r:r:3 and r*sqrt(2) = 3
solving for r will give us:
r=3/sqrt(2)
r = the green line = the circle radius !!
please comment ! I feel like i'm missing something and not helping !
Square is inscribed in a circle
If the diagonal of the square is 6 , then the circle radius will be 3 ! the pink line in the attachment. I don't understand your logic. please explain !
Circle is inscribed inside of a square
If the diagonal of the square is 6, then the we get as you rightly said a isosceles triangle with the proportions of 1:1:sqrt(2) where the proportions are r:r:3 and r*sqrt(2) = 3
solving for r will give us:
r=3/sqrt(2)
r = the green line = the circle radius !!
please comment ! I feel like i'm missing something and not helping !
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inscribed.JPG [ 7.86 KiB | Viewed 2082 times ]
25 May 2007, 07:59
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Your pic is actually very helpful. I was trying to apply the diagonal to the whole square for the circle inscribed inside of a square, so breaking down helped me, but here is the specific question that I'm struggling with.
For the circle insribed inside a square: This is how I'm solving and getting the problem wrong.
Diagonal is 6 for the whole square. Cutting that in half gives us a hypotenuse of 3-pink line and we have a 45:45:90 triangle for 1:1:sq root of two
so taking the hypotenuse of 3-pink line times the ratio ....we have 3*sq root of two for the hypotenuse-pink line
so solving for the green line or one of the sides: I used two y's since both sides are equal
y^2 + y^2 = (3* sq root of 2)^2
2y^2 = 18
y^2 = 9
y= 3 and therefore the radius is equal to 3-green line.
You are saying: Diagonal is 6. Half of it is 3. 1:1:sq root of 2 ratio
so r*sq root of 2 = 3...solving for r gives us sq root of 2 divided by 3. This is where I'm a little confused as you can see. I'm taking the number given for the hypotenuse and always multiplying by the sq. root of 2 and solving for the remaining sides. Is this not right?
Thanks for your patience. I feel like I'm pretty close to figuring this out.
For the circle insribed inside a square: This is how I'm solving and getting the problem wrong.
Diagonal is 6 for the whole square. Cutting that in half gives us a hypotenuse of 3-pink line and we have a 45:45:90 triangle for 1:1:sq root of two
so taking the hypotenuse of 3-pink line times the ratio ....we have 3*sq root of two for the hypotenuse-pink line
so solving for the green line or one of the sides: I used two y's since both sides are equal
y^2 + y^2 = (3* sq root of 2)^2
2y^2 = 18
y^2 = 9
y= 3 and therefore the radius is equal to 3-green line.
You are saying: Diagonal is 6. Half of it is 3. 1:1:sq root of 2 ratio
so r*sq root of 2 = 3...solving for r gives us sq root of 2 divided by 3. This is where I'm a little confused as you can see. I'm taking the number given for the hypotenuse and always multiplying by the sq. root of 2 and solving for the remaining sides. Is this not right?
Thanks for your patience. I feel like I'm pretty close to figuring this out.
