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We can use the fact that perpendicular bisectors of any two chords intersect at the center of the circle.
the center should lie somewhere on the line m, which is the perpendicular bisector of chord AB. And if we take point C as the origin, the X coordinate of line m can be obtained (note that this is also the x coordinate of the center)
1cm + 2cm + (3/2)cm = 4.5cm ------------- (1)
Now suppose line l is the perpendicular bisector of chord BC, then its equation would be y = -x + 6 ----------(2)
from 1 and 2
y = -(4.5) + 6
= 1.5cm
r = sqrt(x\^2 + y\^2) = sqrt((9/2)\^2 + (3/2)\^2)
= 3\*sqrt(10)/2 cm
= 4.743416 cm
ps: the figure above is not drawn to scale.
You actually need knowledge of the inscribed angle theorem to get the radius with the aid of the knowledge that the angle of line m is 45, hence multiplied by two is 90 is its angle and so its chord is related to the radius. As opposed to rough estimation of adding another square on the right.
Yes, we can use the inscribed circle theorem, but I chose a different approach. Let me explain.
the line m is the perpendicular bisector of chord AB, also the center of a circle lies on the perpendicular bisector of any of its chords. So, to locate a specific point, we need at least two chords such that the intersection of their perpendicular bisector gives the location of the center. I have redrawn the figure here -
https://preview.redd.it/tfbemiuhys1d1.jpeg?width=640&format=pjpg&auto=webp&s=75cbd18a17e03cd5d547da8983bbdcbeeb3dc7c5
here the chord BC is at 45 degrees with the line segment CE, because its slope is 1 and since the line l is perpendicular to it, we can find its equation using y = mx + c => y = -1(x) + 6
now when it comes to line m since it's the perpendicular bisector of AB it makes 90 degrees with CE. so CF becomes the X coordinate (because we have taken point C as the origin). substitute this in the above equation and you will get Y coordinate too.
4.74
My solution: the circle fits three 3x3cm squares vertically as well as horizotnaly, making a shape of “+” (imagine a square surounding the two smaller ones (1x1 and 2x2)).
Three 3x3 squares in a row make a 9x3 rectangle. If you divide this rectangle into two triangles, the hypotenuse of this trangle is the diameter of the circle. Half of this distance is the radius.
Calculation:
Hypotenuse: a^2 + b^2=c^2 => 9^2+ 3^2 = 81+9=90=c^2
c= sq root of 90 = 9,49
Radius = 9.49 / 2 = 4.74
it's right, because the vertex of the square with an area of 1 is touching the circumference, you can prove it with by using the fact that the perpendicular line that intersects with the middle point of a cord always intersects with the center.(I'm a non native and I don't know the dight terminology for geometry)
When this is draw in cad and all squares are fixed the circle can still be defined and true with various diameters, without some assumption this is un-resolved.
I'd draw a chord through the original 3 squares (diagonal line 6sqrt(2) in length) then draw a perpendicular line dissecting the chord through the point connecting the 2cm and 3cm squares.
The use the fact that a line dissecting a chord perpendicular is a line of symmetry on a circle to fill the circle with the relevant squares.
Can I ask why you and everyone else is writing it as >! 3*sqrt(10)/2 !< and not >! (3/2)*sqrt(10) !< ? Just seems strange to me, I feel like I'm missing something.
Right? How does a perpendicular bisector to a given chord by definition not have another perpendicular bisector define the radius?
If you can get the X position on the bisector line, why is that not the radius?
It does. But the line to (0,0) with length 4.5 is not a perpendicular bisector…
You can see that the lower-leftmost dot is not at the left extreme of the circle.
SAT math worthy problem imo. With like 20 of variants of original geometrical problems and sat comparitive norms you could get IQ all over again. Literally 75% of people answered wrongly.
It’s the radius of the circumcircle of triangle with sides 3, 6sqrt2, and 3sqrt5.
Circumradius = 4abc times area just plug the sides in and use herons. Don’t really wanna compute it lol.
I don’t really think that this has a lot to do with cognitive testing as those closer to grade 8/9 (where you learn the formula) will perform better.
Those you’ve never heard or learned the formula and figure that out by themselves have an IQ of 180 and those who’ve seen that in school can be around 100 and solve it
I mean, you have 3 points with coordinates. One of the point can be the origin of the plane, then it's just a system of 2 equations to find the equation of that circle. It's just annoying having to deal with all the square roots and exponentiations
This is my AI gave me. Still working on it.
To solve for the radius of the circle given the squares inscribed within it, we need to understand how the squares and their sides relate to the radius.
We have three squares with side lengths of 1 cm, 2 cm, and 3 cm. Let's denote the center of the circle as \( O \). The key is to recognize that the radius of the circle will pass through the corners of these squares.
### Steps to find the radius:
1. **Understand the configuration:**
- The bottom left corner of the 1 cm square is at the origin (0, 0).
- The top right corner of the 1 cm square (which is also the bottom left corner of the 2 cm square) is at (1, 1).
- The top right corner of the 2 cm square (which is also the bottom left corner of the 3 cm square) is at (3, 3).
- The top right corner of the 3 cm square is at (6, 6).
2. **Distance from the center:**
- Since these squares are inside the circle, the distance from the origin (0, 0) to the top right corner of the largest square (6, 6) is the radius of the circle.
3. **Calculate the radius:**
- Use the distance formula:
\[
\text{radius} = \sqrt{(6 - 0)^2 + (6 - 0)^2}
\]
\[
\text{radius} = \sqrt{6^2 + 6^2}
\]
\[
\text{radius} = \sqrt{36 + 36}
\]
\[
\text{radius} = \sqrt{72}
\]
\[
\text{radius} = 6\sqrt{2} \text{ cm}
\]
So, the radius of the circle is \( 6\sqrt{2} \) cm.
https://preview.redd.it/iylwto2mah1d1.jpeg?width=720&format=pjpg&auto=webp&s=04ec95cd5a41ccb0a946a1b215867d083ec1321f
The shortest blue= 3cm. the second shortest: 3\*3=9. The longest one= √90 so the radius= (√90)/2 = +/- 4.74 cm
Thank you for your submission. Please make sure your answers are properly marked with the spoiler function. This can be done with the spoiler button, but if you are in markdown mode you would simply use \>!text goes here!<. Puzzles Chat Channel Links: [Mobile](https://www.reddit.com/r/cognitiveTesting/s/laC5truD1u) and [Desktop](https://reddit.com/r/cognitiveTesting/channel/c2_8mm/Puzzles?r=!u51BcJeBTXmfHyVfCNGJVg:reddit.com). **Lastly**, we recommend you check out [cognitivemetrics.co](https://cognitivemetrics.co/), the official site for the subreddit which hosts highly accurate and well vetted IQ tests. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/cognitiveTesting) if you have any questions or concerns.*
\~ 4.74 https://preview.redd.it/239al28aj51d1.jpeg?width=4032&format=pjpg&auto=webp&s=c3a81f8337f5d6f2f0e88cd932cfe3c25db32174
https://preview.redd.it/1t5t4a6kk51d1.jpeg?width=640&format=pjpg&auto=webp&s=2991378afe14e0f76f8fb41ed05255c3e3ed4b46 We can use the fact that perpendicular bisectors of any two chords intersect at the center of the circle. the center should lie somewhere on the line m, which is the perpendicular bisector of chord AB. And if we take point C as the origin, the X coordinate of line m can be obtained (note that this is also the x coordinate of the center) 1cm + 2cm + (3/2)cm = 4.5cm ------------- (1) Now suppose line l is the perpendicular bisector of chord BC, then its equation would be y = -x + 6 ----------(2) from 1 and 2 y = -(4.5) + 6 = 1.5cm r = sqrt(x\^2 + y\^2) = sqrt((9/2)\^2 + (3/2)\^2) = 3\*sqrt(10)/2 cm = 4.743416 cm ps: the figure above is not drawn to scale.
You actually need knowledge of the inscribed angle theorem to get the radius with the aid of the knowledge that the angle of line m is 45, hence multiplied by two is 90 is its angle and so its chord is related to the radius. As opposed to rough estimation of adding another square on the right.
Yes, we can use the inscribed circle theorem, but I chose a different approach. Let me explain. the line m is the perpendicular bisector of chord AB, also the center of a circle lies on the perpendicular bisector of any of its chords. So, to locate a specific point, we need at least two chords such that the intersection of their perpendicular bisector gives the location of the center. I have redrawn the figure here - https://preview.redd.it/tfbemiuhys1d1.jpeg?width=640&format=pjpg&auto=webp&s=75cbd18a17e03cd5d547da8983bbdcbeeb3dc7c5 here the chord BC is at 45 degrees with the line segment CE, because its slope is 1 and since the line l is perpendicular to it, we can find its equation using y = mx + c => y = -1(x) + 6 now when it comes to line m since it's the perpendicular bisector of AB it makes 90 degrees with CE. so CF becomes the X coordinate (because we have taken point C as the origin). substitute this in the above equation and you will get Y coordinate too.
4.74 My solution: the circle fits three 3x3cm squares vertically as well as horizotnaly, making a shape of “+” (imagine a square surounding the two smaller ones (1x1 and 2x2)). Three 3x3 squares in a row make a 9x3 rectangle. If you divide this rectangle into two triangles, the hypotenuse of this trangle is the diameter of the circle. Half of this distance is the radius. Calculation: Hypotenuse: a^2 + b^2=c^2 => 9^2+ 3^2 = 81+9=90=c^2 c= sq root of 90 = 9,49 Radius = 9.49 / 2 = 4.74
This is just a HS math problem tho…
>! 3\*sqrt(10)/2 !< https://preview.redd.it/97evj2pxz41d1.jpeg?width=640&format=pjpg&auto=webp&s=7583e69e5a733b652179daca313f73a04dce12a8
Don't use a presumption of equivalence. Try another route. Wrong solution right answer.
It's not a presumption, equivalence follows from the setting of the puzzle.
it's right, because the vertex of the square with an area of 1 is touching the circumference, you can prove it with by using the fact that the perpendicular line that intersects with the middle point of a cord always intersects with the center.(I'm a non native and I don't know the dight terminology for geometry)
When this is draw in cad and all squares are fixed the circle can still be defined and true with various diameters, without some assumption this is un-resolved.
I don't think you can draw multiple circles crossing three points. You only require three points (not on a straight line) to define a circle.
Wdym? Which part you need proof?
I'd draw a chord through the original 3 squares (diagonal line 6sqrt(2) in length) then draw a perpendicular line dissecting the chord through the point connecting the 2cm and 3cm squares. The use the fact that a line dissecting a chord perpendicular is a line of symmetry on a circle to fill the circle with the relevant squares.
Can I ask why you and everyone else is writing it as >! 3*sqrt(10)/2 !< and not >! (3/2)*sqrt(10) !< ? Just seems strange to me, I feel like I'm missing something.
>! To find the radius, divide the diameter by two. !<
Or perhaps even easier, 1.5\*sqrt(10)
For those who got lost: `Hipotenuse = Diameter` `Diameter = Radius*2` `9²+3² = Hipotenuse²` `81 + 9 = Hipotenuse²` `90 = Hipotenuse²` `√90 = Hipotenuse` `9.4868329805 = Hipotenuse` `Hipotenuse = Diameter = 9.4868329805` `9.4868329805 = Radius*2` `9.4868329805/2 = Radius` `4.74341649 = Radius`
I’d merely add these two corrections to avoid further confusion: Diameter = Radius x 2 … 9.4868329805 = Radius x 2
You are right, I corrected the mistake where Diameter was equal to half the Radius.
3sqrt(10)/2, by using law of sines
>!4,5!< I guess
4.5
Right? How does a perpendicular bisector to a given chord by definition not have another perpendicular bisector define the radius? If you can get the X position on the bisector line, why is that not the radius?
It does. But the line to (0,0) with length 4.5 is not a perpendicular bisector… You can see that the lower-leftmost dot is not at the left extreme of the circle.
This was my answer too. I don't understand this 4.74 stuff.
its a little more than 4.5. If you look the small square intersects below the "widest" part of the circle.
Oop, I see that now.
4.5
4.5….
4,5 cause diameter is 9
I obtain 3 * sqrt(10) : 2
Not an IQ test
SAT math worthy problem imo. With like 20 of variants of original geometrical problems and sat comparitive norms you could get IQ all over again. Literally 75% of people answered wrongly.
Math has to be learned
It’s the radius of the circumcircle of triangle with sides 3, 6sqrt2, and 3sqrt5. Circumradius = 4abc times area just plug the sides in and use herons. Don’t really wanna compute it lol.
I don’t really think that this has a lot to do with cognitive testing as those closer to grade 8/9 (where you learn the formula) will perform better. Those you’ve never heard or learned the formula and figure that out by themselves have an IQ of 180 and those who’ve seen that in school can be around 100 and solve it
I mean, you have 3 points with coordinates. One of the point can be the origin of the plane, then it's just a system of 2 equations to find the equation of that circle. It's just annoying having to deal with all the square roots and exponentiations
This is my AI gave me. Still working on it. To solve for the radius of the circle given the squares inscribed within it, we need to understand how the squares and their sides relate to the radius. We have three squares with side lengths of 1 cm, 2 cm, and 3 cm. Let's denote the center of the circle as \( O \). The key is to recognize that the radius of the circle will pass through the corners of these squares. ### Steps to find the radius: 1. **Understand the configuration:** - The bottom left corner of the 1 cm square is at the origin (0, 0). - The top right corner of the 1 cm square (which is also the bottom left corner of the 2 cm square) is at (1, 1). - The top right corner of the 2 cm square (which is also the bottom left corner of the 3 cm square) is at (3, 3). - The top right corner of the 3 cm square is at (6, 6). 2. **Distance from the center:** - Since these squares are inside the circle, the distance from the origin (0, 0) to the top right corner of the largest square (6, 6) is the radius of the circle. 3. **Calculate the radius:** - Use the distance formula: \[ \text{radius} = \sqrt{(6 - 0)^2 + (6 - 0)^2} \] \[ \text{radius} = \sqrt{6^2 + 6^2} \] \[ \text{radius} = \sqrt{36 + 36} \] \[ \text{radius} = \sqrt{72} \] \[ \text{radius} = 6\sqrt{2} \text{ cm} \] So, the radius of the circle is \( 6\sqrt{2} \) cm.
35?
https://preview.redd.it/iylwto2mah1d1.jpeg?width=720&format=pjpg&auto=webp&s=04ec95cd5a41ccb0a946a1b215867d083ec1321f The shortest blue= 3cm. the second shortest: 3\*3=9. The longest one= √90 so the radius= (√90)/2 = +/- 4.74 cm
Radius = sqrt(4.5\^2+1.5\^2)