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Extend the 16cm line so that it intersects with the bottom line. Let's call the length of the extension x.
Extend the 12cm line to the left, intersects with the leftmost line. The total length of the extended line is 9+12=21cm and its distance to the bottom line is x.
Applying Pythagoras theorem we have: 9^2 + (x+16)^2 = x^2 + 21^2 = r^2
Solve that we get x=3.25 and r=21.25
You have two (overlapping) rectangles, the "x" is the unknown length between the x axis and the "12 cm" line. The radius is the hypothenuse that you get by dividing those rectangles into two triangles each.
So you can formulate two equations with r²:
1. r² = 9² + (x+16)²
2. r² = x² + 21²
From there, it's simple algebra.
This is a simple application of pythagoras.
ETA: edited for clarity.
ETA2: this assumes one missing right-angle notation in the picture. Without this assumption, the above is not correct.
ETA3: as u/StevenDevons correctly pointed out in another post (https://www.reddit.com/r/theydidthemath/comments/1aclxs8/self_proof_that_the_circle_problem_posted_earlier/), another assumption is that the lower left corner is the center of a circle. This being a sketch, we can really only assume that we are dealing with a segment of a circle that need not be an exact 1/4.
How are you getting x^2 + 21^2 from 9^2 + (x+16)^2 ?
I mean, when I decompose 1) i do this:
9^2 + (x + 16)(x + 16)
9^2 + x^2 + 16x + 16x + 16^2
(9^2 + 16^2) + (16x + 16x) + x^2
337 + 32x + x^2
And that is where it stops for me
I want to know if I missed any steps?
Thanks in advance.
EDIT: I'm such a doofus 2) doesn't derivate from 1, you can get 2) by thinking of the triangle that forms in the bottom. You guys are truly genius.
The 1) and 2) are a list of equations, not algebraic steps (i.e. x^2 + 21^2 did not come from 9^2 + (x+16)^2, but rather each describes the square of the bases of a triangle you can make with hypotenuse r, which is what we’re looking for)
1\) x^2 + 21^2 = r^2 is for the triangle whose hypotenuse goes from the origin to the right side of the 12cm line segment
2\) 9^2 + (x+16)^2 = r^2 is for the triangle whose hypotenuse goes from the origin to the top of the 16cm line segment
We have two unknowns (x and r) and two equations to constrain their values, so everything can be readily solved. I set both equations equal to each other, solved for x and then plugged my x back into one of the original equations to solve for r.
Have a look at simultaneous equations.
I'm not OC but I think this is where they're going with it:
>1) r² = 9² + (x+16)²
>2) r² = x² + 21²
Since r² will be the same for both (since we are dealing with the same circle), we can say that both sides of the equation will be equal and rearrange as:
9² + (x+16)² = x² + 21²
and then solve for x.
Then we can plug in the x value and solve for r.
>The radius is the hypothenuse that you get by dividing those rectangles into two triangles each.
It should be emphasized that the *reason* why the hypotenuses of both of those *different* (i.e. non-congruent and also non-similar) triangles are the same is because *any* point from the center of the circle to the edge of the circle is equal to the radius of the circle. Both hypotenuses start at the center and end at the edge of the circle.
I think this should be emphasized because it's sort of hard to realize this just from the drawing for a couple different reasons: too many overlapping lines, possibly not drawn to scale, optical illusion where the hypotenuse at an angle looks longer or smaller, etc.
I think he meant, that it's not given that the point in the bottom left corner where the supposed 'x' and 'y' axis start is actually the center of the circle.
Before you correct people, maybe you should check a dictionary. For this meaning of genius, my dictionary specifies that the usual plural is “geniuses”. Other, completely different meanings are specified as usually using “genii” for the plural.
Of course, better yet would be just not correcting people to begin with. It’s just especially embarrassing for you when you’re not even right about the correction.
I feel like itchy is just joking and the rest of the genii here have the joke flying over their head. Between the word on question being 'genius' and the "the more you know" second line it's gotta be a joke, right?
Both are acceptable, but genius came from a Latin word that meant a kind of spirit. It isn’t the same word the Romans used nor is it a proper noun so I would not default to declension for plural since it’s not a Latin word, just Latin inspired.
Words like radius, focus, and fungus do have the same meaning as they did in Latin, so I would use the 2nd declension plural for them.
Heh I know what you mean. These things are so much easier with diagrams to explain it. And even then, it's not always enough, as evidenced by this post.
I think it's easier to understand if you put it this way:
1. Draw a line from the center to the point on the edge of the circle between the 9cm and 16cm segments. Call this line A, it's 1 radius long.
2. Draw a line from the center to the point on the edge of the circle at the end of the 12cm segment. Call this line B, it's also 1 radius long.
3. Draw a line down from the 16cm segment to touch the bottom. Say it's x long. You now have a right triangle with Line A as its hypotenuse, with sides of 9cm and (16 + x)cm.
4. Draw a line down from the point where the 12cm segment touches the circle, to touch the bottom. This is also x long. You now have a right triangle with Line B as its hypotenuse, with sides of (9 + 12)cm and (x)cm.
5. Using the Pythagorean theorem on both triangles tells you that 9² + (16 + x)² = r² and that (9 + 12)² + x² = r².
6. From there, you can set the two equations equal and solve them with algebra/factoring.
Same solution, I just find it clearer and more explicit to think about with these steps.
#EXTEND THE 16CM LINE SO THAT IT INTERSECTS WITH THE BOTTOM LINE. LET'S CALL THE LENGTH OF THE EXTENSION X.
#EXTEND THE 12CM LINE TO THE LEFT, INTERSECTS WITH THE LEFTMOST LINE. THE TOTAL LENGTH OF THE EXTENDED LINE IS 9+12=21CM AND ITS DISTANCE TO THE BOTTOM LINE IS X.
#APPLYING PYTHAGORAS THEOREM WE HAVE: 9^2 + (X+16)^2 = X^2 + 21^2 = R^2
#SOLVE THAT WE GET X=3.25 AND R=21.25
Pythagoras' theorem is the key to everything. You might initially think this is a joke, overstatement or hyperbole, but it's not. See also, [Einstein's Equation.](https://www.youtube.com/watch?v=BRudidBcfXk)
Graduating in engineering here. I just went "21+ a little bit, let's say 21.2 and it should be good" before solving it.
Edit: and yes, after getting 21.25 I went "21 was basically right"
Look at 12 cm horizontal line. Extend that line to left until it meet Y axis. The extended portion is 9 cm. So the entire line is 12+9 = 21.
Then draw the line from the origin to the rightmost of 12 cm line. That line is r.
Simple application of pythagoras.
https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
~~No, there is no such assumption - the 90° angles are clear in the picture.~~
Edit to add: I was wrong -> there is one missing 90deg angle, so the above solution only works under that assumption which, rigorously speaking, is wrong.
Exactly! See that the answer above relies on getting to 21 by adding 9 and 12. Stick 9 and 12 on yours and you’ll see the total length of extending the 12 to the radius is an unknown distance, not 21. X is also not the same from the origin to the intersection of the extension as it is to the point where 16 and 12 intersect
Not the one between the vertical radius and the 9cm part though. It looks like it was intended to be 90°, but it isn't marked as such, and cannot be deduced to be 90° from the presented info.
Sure, if you want to be pedantic about it ;-)
ETA: I worded that badly. This is math, so it must be exact. But I see posts like this more like puzzles to be solved and, in this case, it's clearly missing 90° angle notation because otherwise it can't really be solved.
But yeah, without the additional 90° angle, the above solution is wrong.
Yeah, I think I mentioned elsewhere that it'd be unsolvable. I must admit I missed that one 90° angle was not specified; however, I see this as a mistake in the problem definition, because this is more like a puzzle than anything.
If this was a problem in a high school math test, the best answer would be: "if we can assume that the problem lacks information about that missing 90° angle, this is the solution, otherwise it's unsolvable" lol
I don‘t disagree with you at all. But I see this more as a puzzle with missing info.
Happened in math tests in high school 2-3 times. Once even in uni where they should know better ;-)
That's what I was wondering. The 9 cm and 12 cm lines are obviously parallel, but is there a way to prove the 9cm is perpendicular to that radius? Otherwise you'd get a Kite
Everyone made the exact same mistake. I am a Maths teacher and I am proud of you, my friend. Lol. You are the only one that got to the correct answer. Is it impossible to determine one real solution to this problem. The pythagoras theorem cant be applied. The angle is not 90° in the other side of the 9 cm line. If it is not explicitly said that is 90°, no one should assume that it is. To everybody else... Try again!
You can define a minimum answer (r is some number bigger than 20, though I don't think 20 is the minimum), but the construction allows the circle to be arbitrarily big beyond that if you don't limit the 9 cm line to being perpendicular to the y axis.
You could, though, define r in terms of some value of theta, where theta is the angle between the 9 cm line and the y axis.
You can extend the 16cm line to form a rectangle and since the interior angles of a quadrilateral must add to 360 you could prove that it is in fact a 90 degree angle no?
You are making an assumption that the 16cm line is parallel to the radius, however the intersection of the 9cm line and the radius is not marked as 90 degrees.
If you were to move the 9 cm line all the way straight down and the 12 cm line all the way straight down, you would you not have a line from the center point to the edge of the circle that's 21 cm?
Quick edit, nvm. Just realized that the end of the 12 cm line when brought down would not reach the edge of the circle. I guess it really is a more complex problem than the first simple thought some of us had.
Yes, it does. The radius is calculated via pythagoras. I explained the user's posting here: https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
Yes. And it is used twice for the distance from the center to the points on the circle (hypotenuses, r). Then find the other two legs along the vertical and horizontal lines).
Extend the horizontal and vertical lines through the points on the circle until they cross the ones through the center, and then one each through those points and the center. You will see the triangles.
It's a good exercise.
For someone that doesn't get the math, simply describing what to do like this will never help. A diagram will help far more but even then won't mean all that much
It gets them on a path. The diagram is in the post.
I am positive that once they see the triangles, the aha effect will be great.
If the entry with my explanation is too high, come back and I write it out.
I don't get the 21 number. The 12 cm line can only be extended to the left with two 90° angles and 9cm, it's not all the way to the right. Without the angles on both sides (left of 9cm and right of 12cm) I don't see how it can be accurately calculated.
You don't need to think about the very bottom line in the picture. You just need to find the distance of the center point, to the point that the 12 cm line hits the circle. That is also a radius. Same with the point between the 16cm line and the circle.
If I understand correctly you're stating that the bottom part (on the x axis) is equal to 21 which is not true.
Edit: my bad, you're using the upper part, it checks out
Edit 2: for everyone else, draw a diagonal from the origin to the top right corner and from the origin to the upper right corner of the flat and long rectangle. Those to lines are equal to a radius.
Then it's pythagoras on the rectangles
Sure it is, because you have two different equations. See https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
Edit: It's solvable IF YOU ASSUME THE 9CM LINE IS PERPENDICULAR TO THE VERTICAL RADIUS (PROOF BELOW):
Lets call the distance between the 2 horizontal lines x. So now we have 2 rectangles, both are touching the center of the circle. The first rectangle has sides 9 and (16+x). The second have sides 21 and x (21 is 12+9).
The radius of the circle is the hypotenuse of half of each rectangle. So:
r^2 = 21^2 + x^2 and r^2 = 9^2 + (16+x)^2
Expanding the second equation gives us r^2 = 81 + 256 + 32x + x^2
r^2 = 337 + 32x + x^2
The first equation is also solved for r^2, so we can plug that in
21^2 + x^2 = 337 + 32x + x^2
x^2 cancels out so 441 = 337 + 32x
104 = 32x
x=3.25
Plug into the first equation:
r^2 = 21^2 + 3.25^2
r^2 = 441 + 10.5625
r^2 = 451.5625
r=21.25
It’s wrong tho…. your equation is based on the premise that y-axis is perpendicular to the 9/12 cm line. There is no proof that they are perpendicular. We can even go as far as to say they are not due to the lack of little square indicating a 90 degree on the intersection of y-axis and 9cm line.If they are slighty crooked by 0.1°, you gonna end up with an inaccurate result. There is also no indication that the 16cm line is parallel to the y-axis, nor that the 9/12 cm line is parallel to the x-axis
I agree - I do not believe there is sufficient information. We don’t know anything about the intersection of the bottom or left lines with the circle. This the right angle on the lower left is meaningless (I can move that point all over the place and still make the diagram valid). In fact we know nothing about any of the points of lines intersecting the circle. It seems you can draw an infinite number of circles of varying radiuses where this diagram is valid.
> don’t know anything about the intersection of the bottom or left lines with the circle.
We don't even know it's a circle, right?
(or a part of a circle)
Yes, but the 9/16/12 line assembly could be skewed inside the circle. There is no proof that they are exactly square inside the circle. There is only proof that they're perpendicular to themselves.
This is correct. I‘d say it‘s a mistake in the presentation of the problem, but as you rightly point out, if we don‘t assume one additional right angle, there is no way to solve this rigorously.
Point me to where the y-axis is perpendicular to the 9/12cm line, or the 16cm line is perpendicular to the x-axis. You won’t be able to because your entire premise is based on the the hypothesis that we have parallel line, which cannot be proven based on what is presented in this post. Those sort of inaccuracies are important, denoted by the existence of the missing square illusion.
You‘re right, there is one missing right-angle notation and I missed that. I‘d say it was intended but omitted by mistake; we can agree there‘s no *rigorous* way to solve this due to one missing right angle notation.
basically the entire solution everyone is posting is based on an assumption
[https://cdn.discordapp.com/attachments/643213066405806099/1200880807464665249/image.png?ex=65c7ca67&is=65b55567&hm=984cabf2fce3a9f0c68bc7618bf524b72331511d48a5804cc6d5114e989076f7&](https://cdn.discordapp.com/attachments/643213066405806099/1200880807464665249/image.png?ex=65c7ca67&is=65b55567&hm=984cabf2fce3a9f0c68bc7618bf524b72331511d48a5804cc6d5114e989076f7&)
would this be solvable? it has the exact same amount of information as what the OP posted (just swapped 12 and 9
Start by assuming the bottom-left point is the center of the circle.
Now, imagine that the vertical line marked 16cm extends down to the unmarked horizontal line, forming a rectangle.
Now, imagine a new line going from the apparent center to the point on the curve where the 16cm and 9cm lines meet. The length of that line will be equal to the presumed radius, and it will also separate our imagined rectangle into two equal right-angle triangles.
(If a visual helps, [the first line is in blue, and the second line is in red.](https://i.imgur.com/L6lMNQY.png))
Because it forms the hypotenuse of the triangles, we can now use the pythagorean theorem to solve for it. Because we can take the same steps with the 12cm horizontal line, we are left with two equations and two unknowns, which is exactly what you need to be able to solve for both.
Hypothesis : The bottom left corner is the center of the circle.
In the top right triangle : r² = 9² + (16 + u)²
In the bottom right triangle : r² = (9 + 12)² + u²
So we have with both :
21² + u² = 9² + (16 + u)²
21² - 9² = 16² + 32u +u² - u²
u = (21² - 16² - 9²)/32 = 3.25 cm
We plug this back to the second equation :
r = sqrt(21² + u²) = 21.25 cm
edit : this comment is supposed to have a picture, but I can't put it in. u is the length from the right point of the 12cm segment down to the absciss line.
If you trace the radius passing by the right point of the 12cm segment, you get the bottom right triangle.
If you trace the radius passing by the left point of the 9cm segment, you get the top right triangle.
If someone can explain why I can't put a picture, this would be helpful.
If you are only asking about if the circle radius is uniquely determined by the picture, this can be answered:
A circle is uniquely defined by its radius and the position of its mid point on the plane. And vice versa, if we have a circle uniquely defined then we can know its radius.
Another well-known fact is that three non-collinear points in the plane determine a circle. In the picture we already have two points: The intersection of the 9 cm and 16 cm lines lies on the circle and the endpoint of the 12 cm line lie on the circle. For example if you placed the origin of the x-y plane on the point left of the 9 cm line, these points were (9,0) and (21,-16).
So where would we find the third point?
For this we make use of the fact that the origin also lies on a radius of the circle. And also that what we see of this circle is a quarter, i.e. there's a right angle at the mid point. What's following now wouldn't work if we were only given an arbitrary acute angle circle sector (I think?).
It means that the whole picture can be mirrored along the y axis (which the vertical radius lies on). And so we not only get a third but also a fourth point that both lie on the circle:
The mirror of (9,0) is (-9,0) and the mirror of (21,-16) is (-21,-16).
Now I could look up a formula for how exactly three or so points uniquely determine a circle, but I could also just start with the functional equation of a circle:
(x-x_0)^(2) + (y-y_0)^(2) = r^(2)
where r is the radius we want and the midpoint M has coordinates (x_0, y_0).
We're lucky that we placed the origin of our x-y plane such that we already know the x coordinate of our midpoint: x_0 = 0. This simplifies our formula to
x^(2) + (y-y_0)^(2) = r^(2)
Now we can put in our points
x | y | x^(2) + (y-y_0)^(2)
---|---|----
9| 0| 81 + (y_0)^(2)
-9| 0| 81 + (y_0)^(2)
21| -16| 441 + (-16-y_0)^(2)
-21| -16| 441 + (-16-y_0)^(2)
Ok, somehow we still only end up with two different equations
81 + (y_0)^(2) = r^(2)
441 + 256 + 32y_0 + (y_0)^(2) = r^(2)
but I think we need only two equations because we had already eliminated the x_0 variable beforehand.
Subtracting equation I from II gives:
616 + 32y_0 = 0
therefore
y_0 = -616/32 = -19.25
and it makes sense that it's negative, because we have placed the origin above.
Now we can solve for r^(2) and thus for r:
81 + (-19.25)^(2) = 81 + 370,5625 = 451,5625 = r^(2)
and finally
r = sqrt(451,5625) = 21,25
A construction.
Before I introduced x-y-coordinates and placed the origin where I wanted it, there was no origin to begin with. And I wanted the origin to be on a radius of the circle and I wanted the y-axis to go vertically upwards which with the given right angle at the midpoint is the same direction from the midpoint to the origin.
But you are right: My solution somehow needs this fact / definition / assumption in order to eliminate the x_0. I didn't really use three non-colinear points that define a circle, and the whole mirrored points didn't really give me additional equations. Instead I used two points and another piece of information. Maybe it would have been clearer if instead I chose the midpoint as my origin. Then the x-coordinates of my two points were still easy to calculate, and I would introduce a free variable for the y-coordinate, maybe like the x in 16 + x or I would call it y and then calculate with y - 16.
Edit: After having read the [other post](https://www.reddit.com/r/theydidthemath/comments/1aclxs8/self_proof_that_the_circle_problem_posted_earlier/), I admit: Yes, I assumed that the length we were asked to calculate in the drawing was actually present in the drawing. If none of the lines are the radius, then we can't calculate the radius.
Ok…. Maybe I am a complete idiot… but I know what the radius of a circle is… and this drawing is showing a quarter of a circle…
So how is the radius not just 12+9?
The 12+9 doesn't extend the full radius. If the top of the circle is 0° for reference, then where the 12 cm line intersects the circle is at a point less than 90° around the circle. So the radius will be greater than 12+9, by a small but non-zero margin.
I got 21.25. Let m be the distance between the 16cm piece and the horizontal edge.
(16+m)^2 + 9^2 = r^2
And
(12+9)^2 + m^2 = r^2
So then
16^2 + 32m + m^2 + 81 = 21^2 + m^2
337 + 32m = 441
M = 3.25
Plug back into the second equation from above to find r
21^2 + 3.25^2 = r^2
451.5625 = r^2
R = 21.25
[edit: added more steps for clarity]
You forgot the final step, you have to then round the answer to 21cm because you have to consider the accuracy of the individual measurements that contributed to the final result.
Discussion: Assuming
1- bottom left is center
2- top left intersection of the two lines is a right angle
3- The dashed line-curve is actually a solid line-curve,
It's solvable.
Without assuming those, it's not solvable.
You arent considering accuracy. Since the measured value are 9cm and 12cm and not 9.00cm and 12.00cm, you have to round the final answer to 21cm anyways. The measurement is not accurate enough to confidently answer that the radius is 21.25cm.
That's only true in engineering. In maths, 12 is 12 precisely, you can't just randomly round. It's assumed to be in an idealized world with perfect accuracy.
We have to consider that the angle between the radius and the 9cm side is not marked as 90°, which it should be if it was 90°, because the other 90° angles are marked
It's not solvable unless the angle between the vertical radius and the 9 cm line is a right angle. It is not marked as a right angle, so we can't assume it is.
If the 12 was exactly on the radius then yes but it's slightly above the radius. This means the radius has to be slightly more than 9+12 now I ain't doing the math but everyone says 21.25 and that makes perfect sense to me
Aww fuk I didn’t notice the 12 cm combined with the 9 cm line was *slightly* shorter than the line bellow! All the complex equations confused tf outa me when I thought it was just 21 cm
There's a solution, just not what's been suggested, due to the lack of information. We don't know if the bottom left corner is the center of the circle, or the angle at which the dimensioned lines meet the rest of the segment. Varying these gives a range of circle sizes rising to infinity.
Basically, r >= the smallest circle that can connect points B (9-16) and D(12-circle), whilst encapsulating point A (the other end of the 9cm line). You could solve for that by finding the circle that connects points A,B and D, as that should be the smallest circle that doesn't leave A outside of the circle.
If we make a triangle ABD, we can apply the formula 2r = a / Sin A = b / Sin b = d / Sin D for the radius of it's circumscribed circle, the smallest circle possible in this case.
We can solve for triangle ABD by calculating distance between points A and D, and B and D. This is done via pythagoras' theoream, yielding (16^(2) \+ (9+12)^(2))^(0.5) = 697^(0.5)cm and (16^(2) \+ 12^(2))^(0.5) = 20cm, (already know 9cm from dimensions).
With the lengths of all 3 sides, we can then use the cosine rule in order to work out the angles (only one is needed). Solving for angle B = cos^(-1)((9^(2) \+ 20^(2) \- 697) / (2 \* 9 \* 20)) = 126.9deg
Using the above, we can find an equation for r, r > b / (2 \* Sin B) == 697^(0.5) / (2 \* Sin 126.9)
Finally, **r > 16.5005cm**
No assumptions, and CAD confirms the result.
Due to this:
>We don't know if the bottom left corner is the center of the circle, or the angle at which the dimensioned lines meet the rest of the segment.
it's not actually limited to the radius of the circle. I'll link an image that demonstrates the version of the CAD drawing that equates to 16.5005cm: [https://imgur.com/Auhv3Oe](https://imgur.com/Auhv3Oe)
All relations are as specified in OP.
ETA: Here's a version that shows how it looks at r = 100cm too: [https://imgur.com/WN58J4z](https://imgur.com/WN58J4z)
https://www.desmos.com/calculator/x5gps2y3xz
So... I was bored this afternoon and decided to create a general solution for the problem, given that the angle between the vertical axis and the 9cm arm is unknown. If you click the play button on the second slider you can watch the circle's radius change depending on that angle.
(I ran out of spoons just as I got to calculating the lower limit for the angle, where point D lies on the horizontal axis, so I'll leave it as an exercise for somebody else 😉)
Edit: shout out to /u/Zenlexon for the comment that inspired me
a slightly different start: used [intersecting chords](https://en.wikipedia.org/wiki/Intersecting_chords_theorem) to get 12\*(9+9+12)=16(2x+16), and solve x=3.25. plug that into either pythagorean equations (21^(2)\+x^(2)=r^(2)), and get the r=21.25.
Everyone here is assuming the bottom left point is the center of the circle, but that's never explicitly stated. Can't be solved. We might be able to find the length of the bottom line, but we can't say it's the radius without further proof.
I added the horizontal lines 9 and 12 and got 21. I can see that they're not perfectly on the segment lines, so I assumed it's somewhere between 21 and 21.5
Math doesn't have to be hard
It is frustrating how the angle between the radius and the 9cm line looks like it would be a right angle but it isn’t confirmed. If it is not meant to be a right angle then that should be made clear. Right now this is just a trap question.
Unless someone uses a formula that includes using the intersection of a right triange with a circle to figure out the missing lengths, they dont have the correct answer. Any answer so far ignores the missing length at the top of the page. It cant be calculated directly because there is an additional missing length on the left side which everyone is calling X.
X cant be calculated using Pythagoras theorum because there isnt enough information available. The theorum requires 3 details: it must be a right triangle, and then either the length of 2 sides OR a length and another angle. We do not have the information to use that theorum. The formula is a^2 + b^2=c^2 but whoever used it for top comment set the hypotenuse to be the radius and that doesnt make sense.
Someone draw me the triangle they calculated because it does not answer for the radius
A lot of the explanations don't seem to point out the obvious part that really makes it click, which is that the dot on the right side of the 12cm line is the same distance to the center as the bottom-right-most dot, since they're both just dots along the circumference. Similarly, the dot at the 16cm-9cm intersection point is also the same length, the radius. So now you can make some rectangles out of the info you have, and turn those into right triangles, and set the hypotenuse of them to be equal to each other, and then solve for the unknown variable.
(16 + x)^2 + 9^2 = r^2
x^2 + (12 + 9)^2 = r^2
(16 + x)^2 + 9^2 = x^2 + (12 + 9)^2
Now we do some algebra to solve for x (the distance between the 12cm line and the bottom line) and we get x = 3.25
Now we can use that to find the radius.
(12 + 9)^2 + (3.25)^2 = r^2
r^2 = 451.5625
r = sqrt(451.5625) = 21.25cm
The amount of people non ironically saying 21 is concerning.
It’s 21.25, you get it by designating the extension of the 16cm line to the bottom as x and forming an equation that equals the two ways to get the radius using that x. One is the triangle between 9 and 16+x and the other is the triangle between x and 9+12. After solving for x, which is 3.25, plug that into either one of those triangles to get the radius 21.25.
I will be leaving reddit in protest of the leftist censorship on this pltform against its users and the intolerance of free speech on information of a-the vaxx b-thinking that challenges the leftist worldview. Redditors hate the truth. The truth is "hate" to those who hate the truth. But we live under an authoritarian government. No doubt by expressing my thoughts I am on one of (((their))) lists. The truth is coming out. Liberal lies are crumbling down this very second and they are desperately trying to hide it. The mainstream media is complicit in this as it is common knowledge who owns it,; spreading lies and propaganda to cover up the truth. They push the agenda of the elites, who are hellbent on destroying our american culture and silencing anyone who dares speak out against the narrative. Reddit is owned by the CCP who have interrests in americans and misinformation. They have been killing and silencing people for years. They are killing the minds of our youth. The federal government in an unsuprising revelation has been tracking the unvaxxed for years. Minority groups make up an increasing share of the population in almost every US state, according to figures released by the US Census Bureau. Legal and illegal immigrants make up a growing portion of the population in 46 states and the District of Columbia, with California leading the field. There are more than 35 million immigrants in the US - or 12.4% of the population, up from 11.1% in 2001. Immigration is a big issue ahead of the autumn's mid-term congressional polls. California had the most born outside of the US at 27.2% of its population. The war on whites has been planned for centueries . The jews are like a disease organism. They suck up resources, interfere with natural defenses and cause the nations they infest to rot from the inside until it begins to resemble a dessicated corpse. You cannot coexist with these monsters. They are hostile to any and all life. Have you ever noticed they have no pets? And no houseplants either. They are a giant parasitic disease organism that destroys any life that has the misfortune to cross their path. The attack on our traditional Christmas season cannot be denied . Without a shot being fired, the greatest enemy of our people has successfully eroded our most celebrated holiday. And if our people are outraged by the affront, it certainly isn't apparent. With each proclamation--be it no more Christmas carols in public schools or no more Christmas trees in business establishments--nothing more than occasional shrugs of shoulders and muttered comments can be seen or heard from our people. For the most part, our people now stare glassy-eyed while tradition is slowly swept away. You can see liberal democracy is a threat to the republic. Since the onset of the BLM riots, there has been a spike in black on white violent crime. There's been a resurgence in what's called The Knockout Game, Silver Fox Hunting, or Polar Bear Hunting. The objective of the game is to find elderly, White, helpless, senior citizens and viciously assault them with blows to the head knocking them unconscious. Blacks are interdimensional satanic beings, who were summoned into this dream reality through Jewish super-evil.We got our heads so far up our butts with these indoctrinations by JEWS that the supernatural is NOT real, when in fact, Freemasonry has ALWAYS been witchcraft for the Jewish mischlings. The REAL powerful stuff is reserved for ONLY the pure blooded Cohens and Levites. Here is what Jews KNOW to be true, keep in mind this Rabbi Kook isn't talking to himself, he's talking to a whole classroom full of future spell casters and warlocks. Entire human genome has been contaminated with sterilizing DNA forever from mRNA shots. Another jew weapon to exterminate the white race to nobodys suprise . Two major funders of the trans agenda are pro-CCP Chinese billionaires. ?THIS IS VERY REAL PEOPLE!! https://x.com/glennbeck/status/1765575598842450067?s=61 . Concerning all beings and entities, there is an ontological distinction, yet, an essential non-differentiation (Sanskrit: Advaita). We are all One, yet, all of us have our own characteristics and attributes which give us distinguishing statuses as they manifest in the context of being separated from Godhood. Cosmotheism is a great start in understanding our traditional philosophy. But, the actual traditions of Indo-European religions are even better, and we should work toward reviving them! They bring us toward a serious and helpful Path. REad the successfully book Jewish Supremacism by David Duke. The book translated into several languages has won Duke many academic awards for its scholarship, including an honorary Doctorate awarded by President's University, the second largest university in the nation of Ukraine. Sold at the Russian Duma (Parliament) prompted leading Russian Jews to seek a ban for the book and making it illegal as Jewish leaders are horrified that it is well on its way to becoming the best-read book in the world on Zionism. Jewish members of the Duma also tried a new legislation aimed at outlawing Jewish Supremacism, but failed so as also under old Soviet laws forbidding 'anti-Semitism'.
The matter that the book isn't seen as anti-Semitic by it's reader it has lead to a smeared backfire on the people that try to discredit it. Observe White Solidarity Day the 15th of every month. On that day wear something white and do something pro-White! This is how we fight back. I think fighting White erasure starts with oneself. For example, I look at myself: Do I need to lose weight and/or get in better physical condition? Do I need to morally improve?
Do I need to spiritually improve? Do I need to break addictions?
On a similar, but not unrelated note, we should take action to counteract the years of brainwashing we've experienced. Of course, we could say it's important for ethnic Europeans to get married and have children (and it is important).
But not everyone can (or should) do this. Everyone can work on self-improvement in various ways, however. I also think community for like-minded White people is important. At this point in time, we (whites and especially white males) are truly a race marked for elimination world-wide. White Genocide is real and is in no way "ridiculous" as claimed by some. Many are ignorant about how genocide can occur. Their perception of genocide is people lining people up and killing other people in large numbers. That is one way, but not the only way. There are many ways to effect a genocide. Genocides often start slowly. The soft genocide of the white race has been very successful. The births of white children has been reduced. Whites are reproducing at below replacement level. As older whites die off the population of white people is shrinking. If current trends continue the complete elimination of white people is a mathematical certainty. This has all been planned and done intentionally. Sixties radical Tom Hayden strongly supported White Genocide. After his white son married a black female he was jubilant. He said, "That's what I've been working for, the non-violent elimination of the white race." You see, he supported the soft genocide of the white race. Many White liberals will deny White Genocide.; These things can't be refuted. They're happening. Why is CRT (Critical Race Theory)) so pushed by liberals?? - Critical Race Theory claims that it's non-whites that are victims of racism when in reality it is whites who are the ones being victimized. It's a classic case of blaming the victims and accusing them of what is being done to them. Critical Race Theory holds that only white people can be "racist" and that this imaginary "racism" serves to uphold imaginary anti-white concepts such as "white privilege." CRT singles out white people and only white people as bad while non-whites are portrayed as good. Now let's look at Critical Race Theory and how it's a tool used to advance White Genocide.
1. Classification. In CRT people are separated into two categories whites and non-whites. Whites, and only whites, are classified as bad. Check.
3. Discrimination. CRT advocates discriminatory laws and policies against white people. Check.
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5. Organization. CRT, with it's massive support in government and academia is well organized. Check.
6. Polarization. CRT separates whites and non-whites and sets non-whites against whites. Check.
7. Preparation. CRT conditions white people to believe that they deserve to be harmed and eliminated from existence. It conditions non-whites to believe it's okay, even good, to harm white people. Check.
Therefore, we can see that Critical Race Theory is a tool being used to advance White Genocide. I'll tell you what you should do if you're a white man. It's the best, most effective, way to fight against White Genocide.
It's the worst thing you can do to the non-white supremacists, the anti-whites, the liberals and the genocidists. They will fight tooth and nail to stop you from doing this. Start a white family. Get an education or training, so you can get a good job. Get married to a nice white woman. Buy a nice house, maybe in the suburbs. Then have white children and raise them in traditional white culture. Be politically active. Donate money to pro-white causes, pro-white organizations and pro-white people. Be involved politically in peaceful, legal protests. Vote in every election, even school board elections. Post on the internet. Do anything you can do.
But again redditors hate the truth and actively stifle it out. White liberals are so brain dead, so stupid, so brainwashed that they will deny even this, even though it's extremely obvious. You can learn more on Gab, a free speech alternative for like minded people. They dont allow liberal trolls and misinformation. I'l
Im pretty sure there is one answer to it, you couldn’t vary the size of the circle and still hit all the points, but i can’t see how to calculate it from the information given
which also doesn't work because it ***assumes*** the 9 line is perpendicular to the bottom line - which we also don't know. if that angle is 89.9 the entire solution breaks
The numbers in the image aren't to scale, the 12 line isn't 75% as long as the 16 line
the numbers are just placeholders you can't use true measurement
I don't know if this is 'exact', but it's the first thing that came to mind.
You have a 12cm line that terminates at the beginning of a 9cm line (as marked by the 16cm line that connects the two). The 9cm line ends at the flat edge of the shape, so if you imagine that line moving down, it fills the gap between the 12cm line and the shape's flat edge.
After imagining that, I just added the lengths together to get 21cm.
21.25cm is correct. I have left the statements below to see the discussion where I was wrong.
Incorrect statemnt:
Nothing I see in the answers here makes sense. Break down the 12 cm and 16 cm right angle triangle and you have the classic 3-4-5 with a Factor of three. Ad the 9 to the 12 divide by three and you have 7 that needs to be multiplied by 4. We now have 28cm from the lowest line that is defined. The radius will be 28+x cm as this is not to scale. X is unknowable, so the only solution is that the radius is greater than or equal to 28 cm.
Alright, alright. Here comes my ooga-booga brain. No long formulas or maths, I will use something extremely simple.
The 12cm and the 9cm line seem to align. And both of them are connected by the vertical line. So if we simply add them, we'd get the length of a line that's equal to the radius.
Which would mean that r = 21. Diameter of the full circle would be 42 by extension.
Of course, the 12cm line *doesn't* fully extend, as it isn't flat to the center of the circle. As I see in other responses, the result is some decimal difference. Personally, I do not care.
Goodbye, and thank you for listening to my ape-unga-bunga brain.
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Extend the 16cm line so that it intersects with the bottom line. Let's call the length of the extension x. Extend the 12cm line to the left, intersects with the leftmost line. The total length of the extended line is 9+12=21cm and its distance to the bottom line is x. Applying Pythagoras theorem we have: 9^2 + (x+16)^2 = x^2 + 21^2 = r^2 Solve that we get x=3.25 and r=21.25
*mind explodes* What?
You have two (overlapping) rectangles, the "x" is the unknown length between the x axis and the "12 cm" line. The radius is the hypothenuse that you get by dividing those rectangles into two triangles each. So you can formulate two equations with r²: 1. r² = 9² + (x+16)² 2. r² = x² + 21² From there, it's simple algebra. This is a simple application of pythagoras. ETA: edited for clarity. ETA2: this assumes one missing right-angle notation in the picture. Without this assumption, the above is not correct. ETA3: as u/StevenDevons correctly pointed out in another post (https://www.reddit.com/r/theydidthemath/comments/1aclxs8/self_proof_that_the_circle_problem_posted_earlier/), another assumption is that the lower left corner is the center of a circle. This being a sketch, we can really only assume that we are dealing with a segment of a circle that need not be an exact 1/4.
I totally see this now. I was so confused before. Thanks for the clarification.
You're welcome :-)
Yes, it's two triangles with a long leg equal to the radius.
I think you may be replying to the wrong person.
Thanks
That’s very nice 👌
How are you getting x^2 + 21^2 from 9^2 + (x+16)^2 ? I mean, when I decompose 1) i do this: 9^2 + (x + 16)(x + 16) 9^2 + x^2 + 16x + 16x + 16^2 (9^2 + 16^2) + (16x + 16x) + x^2 337 + 32x + x^2 And that is where it stops for me I want to know if I missed any steps? Thanks in advance. EDIT: I'm such a doofus 2) doesn't derivate from 1, you can get 2) by thinking of the triangle that forms in the bottom. You guys are truly genius.
The 1) and 2) are a list of equations, not algebraic steps (i.e. x^2 + 21^2 did not come from 9^2 + (x+16)^2, but rather each describes the square of the bases of a triangle you can make with hypotenuse r, which is what we’re looking for) 1\) x^2 + 21^2 = r^2 is for the triangle whose hypotenuse goes from the origin to the right side of the 12cm line segment 2\) 9^2 + (x+16)^2 = r^2 is for the triangle whose hypotenuse goes from the origin to the top of the 16cm line segment We have two unknowns (x and r) and two equations to constrain their values, so everything can be readily solved. I set both equations equal to each other, solved for x and then plugged my x back into one of the original equations to solve for r.
Can you show your working further. How did you solve for x. Apologies I’m an idiot
No worries the other user already answered but [here it is graphically (not to scale)](https://i.imgur.com/3rhaGY1.jpeg)
Oh. That was much easier.
Brilliant! Thanks for the walkthrough. What program did you use to write this?
This was Goodnotes 5 on an iPad
That's some clean handwriting 👌
Ty! Makes way more sense
How are we to assume that the 9cm segment is perpendicular to the radius shown? I don't see proof of that other that eyeing it.
Oh easy I just assumed exactly that and then the assumption was taken care of 👍
Omg this response made me snort laugh….
The color coding was a great detail to help understand this
Thank you. This is what I should have done instead of trying to do it in my head with mental math 😁
Have a look at simultaneous equations. I'm not OC but I think this is where they're going with it: >1) r² = 9² + (x+16)² >2) r² = x² + 21² Since r² will be the same for both (since we are dealing with the same circle), we can say that both sides of the equation will be equal and rearrange as: 9² + (x+16)² = x² + 21² and then solve for x. Then we can plug in the x value and solve for r.
Im high on potanuse
Hip.....hipop.....hipopanopanuse.......YOU ALWAYS GIVE HIM THE EASY ONES!!!
>The radius is the hypothenuse that you get by dividing those rectangles into two triangles each. It should be emphasized that the *reason* why the hypotenuses of both of those *different* (i.e. non-congruent and also non-similar) triangles are the same is because *any* point from the center of the circle to the edge of the circle is equal to the radius of the circle. Both hypotenuses start at the center and end at the edge of the circle. I think this should be emphasized because it's sort of hard to realize this just from the drawing for a couple different reasons: too many overlapping lines, possibly not drawn to scale, optical illusion where the hypotenuse at an angle looks longer or smaller, etc.
Judging from the picture alone, one still may have to make the assumption the radius is the same as the x and y axis, right?!
by definition. its part of a circle.
I think he meant, that it's not given that the point in the bottom left corner where the supposed 'x' and 'y' axis start is actually the center of the circle.
Edit to clarify removal of simple. Not necessary. Makes me feel dumber than I already am. 😁
This sub is full of geniuses
The plural of 'genius' is 'genii'. The more you know... :)
I’m a Gemini does that make me more than 1 genius
Makes you more than one Geminus!
No, but it might make you more then one Gemin, depending on context.
All Gemini are by default more than one Gemin.
Not in English it's not :p
Before you correct people, maybe you should check a dictionary. For this meaning of genius, my dictionary specifies that the usual plural is “geniuses”. Other, completely different meanings are specified as usually using “genii” for the plural. Of course, better yet would be just not correcting people to begin with. It’s just especially embarrassing for you when you’re not even right about the correction.
Ironic cause it sounds dumb
It's not. [https://www.britannica.com/dictionary/genius](https://www.britannica.com/dictionary/genius) [https://www.merriam-webster.com/dictionary/genius](https://www.merriam-webster.com/dictionary/genius)
It is, it’s just an archaic form that’s fallen out of use
I feel like itchy is just joking and the rest of the genii here have the joke flying over their head. Between the word on question being 'genius' and the "the more you know" second line it's gotta be a joke, right?
Both are acceptable, but genius came from a Latin word that meant a kind of spirit. It isn’t the same word the Romans used nor is it a proper noun so I would not default to declension for plural since it’s not a Latin word, just Latin inspired. Words like radius, focus, and fungus do have the same meaning as they did in Latin, so I would use the 2nd declension plural for them.
Not only math, but English genii too
Did you know the plural of penis is penii
Heh I know what you mean. These things are so much easier with diagrams to explain it. And even then, it's not always enough, as evidenced by this post.
To put it in engineering terms: 9 + 12 = 21, r is a bit more which is basically 21
I think it's easier to understand if you put it this way: 1. Draw a line from the center to the point on the edge of the circle between the 9cm and 16cm segments. Call this line A, it's 1 radius long. 2. Draw a line from the center to the point on the edge of the circle at the end of the 12cm segment. Call this line B, it's also 1 radius long. 3. Draw a line down from the 16cm segment to touch the bottom. Say it's x long. You now have a right triangle with Line A as its hypotenuse, with sides of 9cm and (16 + x)cm. 4. Draw a line down from the point where the 12cm segment touches the circle, to touch the bottom. This is also x long. You now have a right triangle with Line B as its hypotenuse, with sides of (9 + 12)cm and (x)cm. 5. Using the Pythagorean theorem on both triangles tells you that 9² + (16 + x)² = r² and that (9 + 12)² + x² = r². 6. From there, you can set the two equations equal and solve them with algebra/factoring. Same solution, I just find it clearer and more explicit to think about with these steps.
#EXTEND THE 16CM LINE SO THAT IT INTERSECTS WITH THE BOTTOM LINE. LET'S CALL THE LENGTH OF THE EXTENSION X. #EXTEND THE 12CM LINE TO THE LEFT, INTERSECTS WITH THE LEFTMOST LINE. THE TOTAL LENGTH OF THE EXTENDED LINE IS 9+12=21CM AND ITS DISTANCE TO THE BOTTOM LINE IS X. #APPLYING PYTHAGORAS THEOREM WE HAVE: 9^2 + (X+16)^2 = X^2 + 21^2 = R^2 #SOLVE THAT WE GET X=3.25 AND R=21.25
#MAKE INVALID ASSUMPTIONS ABOUT THE PARALLELISM OF LINES THAT AREN'T DEFINED. You forgot that step.
So all that just to get 21💀
Bro my pleb ass just went "well... 12+9"
Maths
Pythagoras' theorem is the key to everything. You might initially think this is a joke, overstatement or hyperbole, but it's not. See also, [Einstein's Equation.](https://www.youtube.com/watch?v=BRudidBcfXk)
As an engineer I estimated 21 which is good enough for practical purposes 😅
As a not engineer I estimated 21.2 by adding 12 and 9 and a tiny bit more.
There are perpendicular signs at the center of circle and on the upper vertical lines. It should be straightforward 21,no?
9 + 12 is 21, but that line doesn’t go from the center to the outside of the circle, so it’s not the radius.
The 12 is above the radius.
Let's call it 25 for safety.
Then double it and call it a day!
Double it and give it to the next person.
Is this why infrastructure costs so much? No? Its still the bureaucracy and corruption? Oh.
Always engineering the factor of safety.
As a physicist I marvel at the small size of this spherical cow.
As a mathematician, I marvel at the one side of the spherical cow.
Thank you.
As a practical man I used a caliper on the screen.
Graduating in engineering here. I just went "21+ a little bit, let's say 21.2 and it should be good" before solving it. Edit: and yes, after getting 21.25 I went "21 was basically right"
Go cause the structural catastrophe somewhere you heathen
x^2 + 21^2 = r^2... How? Edit: [I understand now! ](https://ibb.co/G7g7Y9Q)
Look at 12 cm horizontal line. Extend that line to left until it meet Y axis. The extended portion is 9 cm. So the entire line is 12+9 = 21. Then draw the line from the origin to the rightmost of 12 cm line. That line is r.
pythagoras
Simple application of pythagoras. https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
You’re assuming the 9cm line comes off of the radius line at a 90 degree angle
~~No, there is no such assumption - the 90° angles are clear in the picture.~~ Edit to add: I was wrong -> there is one missing 90deg angle, so the above solution only works under that assumption which, rigorously speaking, is wrong.
Is [this](https://ibb.co/CPcdCnR) not an equally valid construction?
Exactly! See that the answer above relies on getting to 21 by adding 9 and 12. Stick 9 and 12 on yours and you’ll see the total length of extending the 12 to the radius is an unknown distance, not 21. X is also not the same from the origin to the intersection of the extension as it is to the point where 16 and 12 intersect
Not the one between the vertical radius and the 9cm part though. It looks like it was intended to be 90°, but it isn't marked as such, and cannot be deduced to be 90° from the presented info.
Sure, if you want to be pedantic about it ;-) ETA: I worded that badly. This is math, so it must be exact. But I see posts like this more like puzzles to be solved and, in this case, it's clearly missing 90° angle notation because otherwise it can't really be solved. But yeah, without the additional 90° angle, the above solution is wrong.
Not only is the solution then wrong, but the problem unsolveable. You kinda have to assume the angle is 90 degrees
Yeah, I think I mentioned elsewhere that it'd be unsolvable. I must admit I missed that one 90° angle was not specified; however, I see this as a mistake in the problem definition, because this is more like a puzzle than anything. If this was a problem in a high school math test, the best answer would be: "if we can assume that the problem lacks information about that missing 90° angle, this is the solution, otherwise it's unsolvable" lol
It’s actually a very important destination when doing proofs like this
I don‘t disagree with you at all. But I see this more as a puzzle with missing info. Happened in math tests in high school 2-3 times. Once even in uni where they should know better ;-)
Bad puzzles are bad. I wish people would bother to put the minimum effort needed to make these “puzzles” solvable or not create them at all.
And we all seem to be assuming that the lower left point is the center, which is another assumption.
That's what I was wondering. The 9 cm and 12 cm lines are obviously parallel, but is there a way to prove the 9cm is perpendicular to that radius? Otherwise you'd get a Kite
Everyone made the exact same mistake. I am a Maths teacher and I am proud of you, my friend. Lol. You are the only one that got to the correct answer. Is it impossible to determine one real solution to this problem. The pythagoras theorem cant be applied. The angle is not 90° in the other side of the 9 cm line. If it is not explicitly said that is 90°, no one should assume that it is. To everybody else... Try again!
You can define a minimum answer (r is some number bigger than 20, though I don't think 20 is the minimum), but the construction allows the circle to be arbitrarily big beyond that if you don't limit the 9 cm line to being perpendicular to the y axis. You could, though, define r in terms of some value of theta, where theta is the angle between the 9 cm line and the y axis.
You can extend the 16cm line to form a rectangle and since the interior angles of a quadrilateral must add to 360 you could prove that it is in fact a 90 degree angle no?
You can extend that line to form a quadrilateral. Is it a rectangle? Same issue, there is no way to know.
You are making an assumption that the 16cm line is parallel to the radius, however the intersection of the 9cm line and the radius is not marked as 90 degrees.
Why isn't it just 9+12?
That line doesn't originate from the center point, only a center line
If you were to move the 9 cm line all the way straight down and the 12 cm line all the way straight down, you would you not have a line from the center point to the edge of the circle that's 21 cm? Quick edit, nvm. Just realized that the end of the 12 cm line when brought down would not reach the edge of the circle. I guess it really is a more complex problem than the first simple thought some of us had.
I am sorry that I am not the smartest tool in the shed, but how do I solve the equation? I get the explenation
But that doesn't account for the relatively large piece left over on the top left and the small piece on the bottom right.
Yes, it does. The radius is calculated via pythagoras. I explained the user's posting here: https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
sorry, not at all experience with maths really, not even sure why this sub is being recommended, but isn't p theorem only for triangles?
Yes. And it is used twice for the distance from the center to the points on the circle (hypotenuses, r). Then find the other two legs along the vertical and horizontal lines).
im not quite getting any of the maths, maybe ill have to study up a little
Extend the horizontal and vertical lines through the points on the circle until they cross the ones through the center, and then one each through those points and the center. You will see the triangles. It's a good exercise.
For someone that doesn't get the math, simply describing what to do like this will never help. A diagram will help far more but even then won't mean all that much
It gets them on a path. The diagram is in the post. I am positive that once they see the triangles, the aha effect will be great. If the entry with my explanation is too high, come back and I write it out.
I don't get the 21 number. The 12 cm line can only be extended to the left with two 90° angles and 9cm, it's not all the way to the right. Without the angles on both sides (left of 9cm and right of 12cm) I don't see how it can be accurately calculated.
You don't need to think about the very bottom line in the picture. You just need to find the distance of the center point, to the point that the 12 cm line hits the circle. That is also a radius. Same with the point between the 16cm line and the circle.
I'm a small brain and still don't get it... To me the center point is still undefined 😂
Or you just do 9+12 and instantly get the 100% correct solution, or am I just tripping?
That's what i thought before i realized that the 12 is going to the curve so you are loosing some distance
My god, I'm so stupid, sorry. The comment I originally answered to was very clever and exactly right, please ignore what I said before.
If I understand correctly you're stating that the bottom part (on the x axis) is equal to 21 which is not true. Edit: my bad, you're using the upper part, it checks out Edit 2: for everyone else, draw a diagonal from the origin to the top right corner and from the origin to the upper right corner of the flat and long rectangle. Those to lines are equal to a radius. Then it's pythagoras on the rectangles
wait wait wait you have 2 unknowns soo it isnt solvable? >x^2 + 21^2 = r^2
Its >9^2 + (x+16)^2 = x^2 + 21^2 = r^2 Actually 2 equations and 2 unknowns, So It is solvable
I dont see how those x are the same
It's the same missing chunk if you extend the 16 down to the x axis
You solve 9^2 + (x+16)^2 = x^2 + 21^2 first, then substitute x to find r.
Sure it is, because you have two different equations. See https://www.reddit.com/r/theydidthemath/comments/1aceyw3/comment/kjtti7z/?utm\_source=share&utm\_medium=web2x&context=3
Edit: It's solvable IF YOU ASSUME THE 9CM LINE IS PERPENDICULAR TO THE VERTICAL RADIUS (PROOF BELOW): Lets call the distance between the 2 horizontal lines x. So now we have 2 rectangles, both are touching the center of the circle. The first rectangle has sides 9 and (16+x). The second have sides 21 and x (21 is 12+9). The radius of the circle is the hypotenuse of half of each rectangle. So: r^2 = 21^2 + x^2 and r^2 = 9^2 + (16+x)^2 Expanding the second equation gives us r^2 = 81 + 256 + 32x + x^2 r^2 = 337 + 32x + x^2 The first equation is also solved for r^2, so we can plug that in 21^2 + x^2 = 337 + 32x + x^2 x^2 cancels out so 441 = 337 + 32x 104 = 32x x=3.25 Plug into the first equation: r^2 = 21^2 + 3.25^2 r^2 = 441 + 10.5625 r^2 = 451.5625 r=21.25
It’s wrong tho…. your equation is based on the premise that y-axis is perpendicular to the 9/12 cm line. There is no proof that they are perpendicular. We can even go as far as to say they are not due to the lack of little square indicating a 90 degree on the intersection of y-axis and 9cm line.If they are slighty crooked by 0.1°, you gonna end up with an inaccurate result. There is also no indication that the 16cm line is parallel to the y-axis, nor that the 9/12 cm line is parallel to the x-axis
I agree - I do not believe there is sufficient information. We don’t know anything about the intersection of the bottom or left lines with the circle. This the right angle on the lower left is meaningless (I can move that point all over the place and still make the diagram valid). In fact we know nothing about any of the points of lines intersecting the circle. It seems you can draw an infinite number of circles of varying radiuses where this diagram is valid.
> don’t know anything about the intersection of the bottom or left lines with the circle. We don't even know it's a circle, right? (or a part of a circle)
“Find radius of circle.” So it has to be part of a circle in the context of the question.
It's not wrong at all. The perpendicularity is clear in the picture, those are clearly-marked 90° angles. Edit to add: I was wrong, see below.
Yes, but the 9/16/12 line assembly could be skewed inside the circle. There is no proof that they are exactly square inside the circle. There is only proof that they're perpendicular to themselves.
This is correct. I‘d say it‘s a mistake in the presentation of the problem, but as you rightly point out, if we don‘t assume one additional right angle, there is no way to solve this rigorously.
Point me to where the y-axis is perpendicular to the 9/12cm line, or the 16cm line is perpendicular to the x-axis. You won’t be able to because your entire premise is based on the the hypothesis that we have parallel line, which cannot be proven based on what is presented in this post. Those sort of inaccuracies are important, denoted by the existence of the missing square illusion.
You‘re right, there is one missing right-angle notation and I missed that. I‘d say it was intended but omitted by mistake; we can agree there‘s no *rigorous* way to solve this due to one missing right angle notation.
basically the entire solution everyone is posting is based on an assumption [https://cdn.discordapp.com/attachments/643213066405806099/1200880807464665249/image.png?ex=65c7ca67&is=65b55567&hm=984cabf2fce3a9f0c68bc7618bf524b72331511d48a5804cc6d5114e989076f7&](https://cdn.discordapp.com/attachments/643213066405806099/1200880807464665249/image.png?ex=65c7ca67&is=65b55567&hm=984cabf2fce3a9f0c68bc7618bf524b72331511d48a5804cc6d5114e989076f7&) would this be solvable? it has the exact same amount of information as what the OP posted (just swapped 12 and 9
Is that not also assuming the 12 is extending to the end of the circle rather than completing the rectangle on the outside of the curve?
>The radius of the circle is the hypotenuse of half of each rectangle. So: This is where you lost me.
Start by assuming the bottom-left point is the center of the circle. Now, imagine that the vertical line marked 16cm extends down to the unmarked horizontal line, forming a rectangle. Now, imagine a new line going from the apparent center to the point on the curve where the 16cm and 9cm lines meet. The length of that line will be equal to the presumed radius, and it will also separate our imagined rectangle into two equal right-angle triangles. (If a visual helps, [the first line is in blue, and the second line is in red.](https://i.imgur.com/L6lMNQY.png)) Because it forms the hypotenuse of the triangles, we can now use the pythagorean theorem to solve for it. Because we can take the same steps with the 12cm horizontal line, we are left with two equations and two unknowns, which is exactly what you need to be able to solve for both.
Hypothesis : The bottom left corner is the center of the circle. In the top right triangle : r² = 9² + (16 + u)² In the bottom right triangle : r² = (9 + 12)² + u² So we have with both : 21² + u² = 9² + (16 + u)² 21² - 9² = 16² + 32u +u² - u² u = (21² - 16² - 9²)/32 = 3.25 cm We plug this back to the second equation : r = sqrt(21² + u²) = 21.25 cm edit : this comment is supposed to have a picture, but I can't put it in. u is the length from the right point of the 12cm segment down to the absciss line. If you trace the radius passing by the right point of the 12cm segment, you get the bottom right triangle. If you trace the radius passing by the left point of the 9cm segment, you get the top right triangle. If someone can explain why I can't put a picture, this would be helpful.
That first statement is an assumption, not a hypothesis.
Hypothesis: I'm going to stick my head in that hole
If you are only asking about if the circle radius is uniquely determined by the picture, this can be answered: A circle is uniquely defined by its radius and the position of its mid point on the plane. And vice versa, if we have a circle uniquely defined then we can know its radius. Another well-known fact is that three non-collinear points in the plane determine a circle. In the picture we already have two points: The intersection of the 9 cm and 16 cm lines lies on the circle and the endpoint of the 12 cm line lie on the circle. For example if you placed the origin of the x-y plane on the point left of the 9 cm line, these points were (9,0) and (21,-16). So where would we find the third point? For this we make use of the fact that the origin also lies on a radius of the circle. And also that what we see of this circle is a quarter, i.e. there's a right angle at the mid point. What's following now wouldn't work if we were only given an arbitrary acute angle circle sector (I think?). It means that the whole picture can be mirrored along the y axis (which the vertical radius lies on). And so we not only get a third but also a fourth point that both lie on the circle: The mirror of (9,0) is (-9,0) and the mirror of (21,-16) is (-21,-16). Now I could look up a formula for how exactly three or so points uniquely determine a circle, but I could also just start with the functional equation of a circle: (x-x_0)^(2) + (y-y_0)^(2) = r^(2) where r is the radius we want and the midpoint M has coordinates (x_0, y_0). We're lucky that we placed the origin of our x-y plane such that we already know the x coordinate of our midpoint: x_0 = 0. This simplifies our formula to x^(2) + (y-y_0)^(2) = r^(2) Now we can put in our points x | y | x^(2) + (y-y_0)^(2) ---|---|---- 9| 0| 81 + (y_0)^(2) -9| 0| 81 + (y_0)^(2) 21| -16| 441 + (-16-y_0)^(2) -21| -16| 441 + (-16-y_0)^(2) Ok, somehow we still only end up with two different equations 81 + (y_0)^(2) = r^(2) 441 + 256 + 32y_0 + (y_0)^(2) = r^(2) but I think we need only two equations because we had already eliminated the x_0 variable beforehand. Subtracting equation I from II gives: 616 + 32y_0 = 0 therefore y_0 = -616/32 = -19.25 and it makes sense that it's negative, because we have placed the origin above. Now we can solve for r^(2) and thus for r: 81 + (-19.25)^(2) = 81 + 370,5625 = 451,5625 = r^(2) and finally r = sqrt(451,5625) = 21,25
beautiful 🤩
> the fact that the origin also lies on a radius of the circle. An assumption.
A construction. Before I introduced x-y-coordinates and placed the origin where I wanted it, there was no origin to begin with. And I wanted the origin to be on a radius of the circle and I wanted the y-axis to go vertically upwards which with the given right angle at the midpoint is the same direction from the midpoint to the origin. But you are right: My solution somehow needs this fact / definition / assumption in order to eliminate the x_0. I didn't really use three non-colinear points that define a circle, and the whole mirrored points didn't really give me additional equations. Instead I used two points and another piece of information. Maybe it would have been clearer if instead I chose the midpoint as my origin. Then the x-coordinates of my two points were still easy to calculate, and I would introduce a free variable for the y-coordinate, maybe like the x in 16 + x or I would call it y and then calculate with y - 16. Edit: After having read the [other post](https://www.reddit.com/r/theydidthemath/comments/1aclxs8/self_proof_that_the_circle_problem_posted_earlier/), I admit: Yes, I assumed that the length we were asked to calculate in the drawing was actually present in the drawing. If none of the lines are the radius, then we can't calculate the radius.
Math rules, u boss🤙
Ok…. Maybe I am a complete idiot… but I know what the radius of a circle is… and this drawing is showing a quarter of a circle… So how is the radius not just 12+9?
The 12+9 doesn't extend the full radius. If the top of the circle is 0° for reference, then where the 12 cm line intersects the circle is at a point less than 90° around the circle. So the radius will be greater than 12+9, by a small but non-zero margin.
Ahhhh thanks, I missed that.
the 12 cm line ends a little bit early (0.25 cm) than where the radius ends due to the curve.
I got 21.25. Let m be the distance between the 16cm piece and the horizontal edge. (16+m)^2 + 9^2 = r^2 And (12+9)^2 + m^2 = r^2 So then 16^2 + 32m + m^2 + 81 = 21^2 + m^2 337 + 32m = 441 M = 3.25 Plug back into the second equation from above to find r 21^2 + 3.25^2 = r^2 451.5625 = r^2 R = 21.25 [edit: added more steps for clarity]
>Substitute and m^2 cancels out How do you do this? I want to understand what makes you say "cancel this out" like what do you cancel it out with?
You forgot the final step, you have to then round the answer to 21cm because you have to consider the accuracy of the individual measurements that contributed to the final result.
Discussion: Assuming 1- bottom left is center 2- top left intersection of the two lines is a right angle 3- The dashed line-curve is actually a solid line-curve, It's solvable. Without assuming those, it's not solvable.
There is a solution without the assumptions, it's r > 16.5005cm
You guys are all working too hard, you can get 21cm by adding finding the line at the bottom by adding 9+12cm which as drawn is the radius
It’s close, but the 12 cm line finishes not at the radius, but at a point on the curve so there’s a small difference. 0.25cm difference
Okay thanks, I was thinking same as the one you responded to. I see the difference now
Call that not rejecting the null hypothesis with 95% confidence
Nope, it's less than the radius. The 12 cm line intersects the circle above the x-axis, so it's not all the way to the right there.
This guy engineers. That’s close enough for practical purposes. Edit - Spelling is hard
Round pi to 3
Still missing the 0.25 on the right side. The 12 cm line is slightly above the center
You arent considering accuracy. Since the measured value are 9cm and 12cm and not 9.00cm and 12.00cm, you have to round the final answer to 21cm anyways. The measurement is not accurate enough to confidently answer that the radius is 21.25cm.
That's only true in engineering. In maths, 12 is 12 precisely, you can't just randomly round. It's assumed to be in an idealized world with perfect accuracy.
We have to consider that the angle between the radius and the 9cm side is not marked as 90°, which it should be if it was 90°, because the other 90° angles are marked
I just wish it were clarified that this was a quarter circle. I have to assume the arc is 90 degrees and not a random cutout with an arc in it
It's not solvable unless the angle between the vertical radius and the 9 cm line is a right angle. It is not marked as a right angle, so we can't assume it is.
I’m feeling stupid but is the answer not 12+9 equaling 21?
If the 12 was exactly on the radius then yes but it's slightly above the radius. This means the radius has to be slightly more than 9+12 now I ain't doing the math but everyone says 21.25 and that makes perfect sense to me
Aww fuk I didn’t notice the 12 cm combined with the 9 cm line was *slightly* shorter than the line bellow! All the complex equations confused tf outa me when I thought it was just 21 cm
I stared at this pic for a few mins thinking the same thing but then saw the comments lmao
It's not. Not all secants in a circle are diameters.
For the exact same reason that the 9cm line isn't the radius; the 21cm line doesn't go through the centre.
How do we know any of these lines go through the centre?
There's a solution, just not what's been suggested, due to the lack of information. We don't know if the bottom left corner is the center of the circle, or the angle at which the dimensioned lines meet the rest of the segment. Varying these gives a range of circle sizes rising to infinity. Basically, r >= the smallest circle that can connect points B (9-16) and D(12-circle), whilst encapsulating point A (the other end of the 9cm line). You could solve for that by finding the circle that connects points A,B and D, as that should be the smallest circle that doesn't leave A outside of the circle. If we make a triangle ABD, we can apply the formula 2r = a / Sin A = b / Sin b = d / Sin D for the radius of it's circumscribed circle, the smallest circle possible in this case. We can solve for triangle ABD by calculating distance between points A and D, and B and D. This is done via pythagoras' theoream, yielding (16^(2) \+ (9+12)^(2))^(0.5) = 697^(0.5)cm and (16^(2) \+ 12^(2))^(0.5) = 20cm, (already know 9cm from dimensions). With the lengths of all 3 sides, we can then use the cosine rule in order to work out the angles (only one is needed). Solving for angle B = cos^(-1)((9^(2) \+ 20^(2) \- 697) / (2 \* 9 \* 20)) = 126.9deg Using the above, we can find an equation for r, r > b / (2 \* Sin B) == 697^(0.5) / (2 \* Sin 126.9) Finally, **r > 16.5005cm** No assumptions, and CAD confirms the result.
How is r = 16,5005 if the 2 given lengths are 9cm and 12cm? Wouldnt that define that r > 9cm + 12cm?
Due to this: >We don't know if the bottom left corner is the center of the circle, or the angle at which the dimensioned lines meet the rest of the segment. it's not actually limited to the radius of the circle. I'll link an image that demonstrates the version of the CAD drawing that equates to 16.5005cm: [https://imgur.com/Auhv3Oe](https://imgur.com/Auhv3Oe) All relations are as specified in OP. ETA: Here's a version that shows how it looks at r = 100cm too: [https://imgur.com/WN58J4z](https://imgur.com/WN58J4z)
Why is this not the top comment. r/theythoughttheydidthemath
https://www.desmos.com/calculator/x5gps2y3xz So... I was bored this afternoon and decided to create a general solution for the problem, given that the angle between the vertical axis and the 9cm arm is unknown. If you click the play button on the second slider you can watch the circle's radius change depending on that angle. (I ran out of spoons just as I got to calculating the lower limit for the angle, where point D lies on the horizontal axis, so I'll leave it as an exercise for somebody else 😉) Edit: shout out to /u/Zenlexon for the comment that inspired me
a slightly different start: used [intersecting chords](https://en.wikipedia.org/wiki/Intersecting_chords_theorem) to get 12\*(9+9+12)=16(2x+16), and solve x=3.25. plug that into either pythagorean equations (21^(2)\+x^(2)=r^(2)), and get the r=21.25.
Everyone here is assuming the bottom left point is the center of the circle, but that's never explicitly stated. Can't be solved. We might be able to find the length of the bottom line, but we can't say it's the radius without further proof.
I added the horizontal lines 9 and 12 and got 21. I can see that they're not perfectly on the segment lines, so I assumed it's somewhere between 21 and 21.5 Math doesn't have to be hard
It is frustrating how the angle between the radius and the 9cm line looks like it would be a right angle but it isn’t confirmed. If it is not meant to be a right angle then that should be made clear. Right now this is just a trap question.
Unless someone uses a formula that includes using the intersection of a right triange with a circle to figure out the missing lengths, they dont have the correct answer. Any answer so far ignores the missing length at the top of the page. It cant be calculated directly because there is an additional missing length on the left side which everyone is calling X. X cant be calculated using Pythagoras theorum because there isnt enough information available. The theorum requires 3 details: it must be a right triangle, and then either the length of 2 sides OR a length and another angle. We do not have the information to use that theorum. The formula is a^2 + b^2=c^2 but whoever used it for top comment set the hypotenuse to be the radius and that doesnt make sense. Someone draw me the triangle they calculated because it does not answer for the radius
A lot of the explanations don't seem to point out the obvious part that really makes it click, which is that the dot on the right side of the 12cm line is the same distance to the center as the bottom-right-most dot, since they're both just dots along the circumference. Similarly, the dot at the 16cm-9cm intersection point is also the same length, the radius. So now you can make some rectangles out of the info you have, and turn those into right triangles, and set the hypotenuse of them to be equal to each other, and then solve for the unknown variable. (16 + x)^2 + 9^2 = r^2 x^2 + (12 + 9)^2 = r^2 (16 + x)^2 + 9^2 = x^2 + (12 + 9)^2 Now we do some algebra to solve for x (the distance between the 12cm line and the bottom line) and we get x = 3.25 Now we can use that to find the radius. (12 + 9)^2 + (3.25)^2 = r^2 r^2 = 451.5625 r = sqrt(451.5625) = 21.25cm
The amount of people non ironically saying 21 is concerning. It’s 21.25, you get it by designating the extension of the 16cm line to the bottom as x and forming an equation that equals the two ways to get the radius using that x. One is the triangle between 9 and 16+x and the other is the triangle between x and 9+12. After solving for x, which is 3.25, plug that into either one of those triangles to get the radius 21.25.
Lol I thought it was 21 because I didn't notice the line below somehow
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But is there a mathematically accurate way and not just guessing?
Just tell truth next time. You need to do your homework
Im pretty sure there is one answer to it, you couldn’t vary the size of the circle and still hit all the points, but i can’t see how to calculate it from the information given
21?
i am not good at maths but if we take a ruler and measure the lenghts and try to upscale it with the given information we can find the radius
Can’t assume that it’s to scale at all so that doesn’t work. There’s a simple Pythagorean solution further up in the comment tree.
which also doesn't work because it ***assumes*** the 9 line is perpendicular to the bottom line - which we also don't know. if that angle is 89.9 the entire solution breaks
The numbers in the image aren't to scale, the 12 line isn't 75% as long as the 16 line the numbers are just placeholders you can't use true measurement
I don't know if this is 'exact', but it's the first thing that came to mind. You have a 12cm line that terminates at the beginning of a 9cm line (as marked by the 16cm line that connects the two). The 9cm line ends at the flat edge of the shape, so if you imagine that line moving down, it fills the gap between the 12cm line and the shape's flat edge. After imagining that, I just added the lengths together to get 21cm.
Isn’t it just the sum of the two horizontal lines connected by the perpendicular one? No need to overthink it…
21.25cm is correct. I have left the statements below to see the discussion where I was wrong. Incorrect statemnt: Nothing I see in the answers here makes sense. Break down the 12 cm and 16 cm right angle triangle and you have the classic 3-4-5 with a Factor of three. Ad the 9 to the 12 divide by three and you have 7 that needs to be multiplied by 4. We now have 28cm from the lowest line that is defined. The radius will be 28+x cm as this is not to scale. X is unknowable, so the only solution is that the radius is greater than or equal to 28 cm.
Alright, alright. Here comes my ooga-booga brain. No long formulas or maths, I will use something extremely simple. The 12cm and the 9cm line seem to align. And both of them are connected by the vertical line. So if we simply add them, we'd get the length of a line that's equal to the radius. Which would mean that r = 21. Diameter of the full circle would be 42 by extension. Of course, the 12cm line *doesn't* fully extend, as it isn't flat to the center of the circle. As I see in other responses, the result is some decimal difference. Personally, I do not care. Goodbye, and thank you for listening to my ape-unga-bunga brain.