What is the area of the square? They did the math but did you? Reddit geometry r/theydidthemath
Vložit
- čas přidán 28. 12. 2023
- I saw this geometry question on Reddit r/theydidthemath. We have a square that's placed with two corners touching a circle (as shown in the diagram). Can we find the area of the square based on just the given information?
See the original post here: / lsotnmnddg
Shop my math t-shirts & hoodies on Amazon: 👉 amzn.to/3qBeuw6
-----------------------------
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 / blackpenredpen . Every bit of support means the world to me and motivates me to keep bringing you the best math lessons out there! Thank you!
-----------------------------
#math #algebra #mathbasics
Keeping the people maths ready (GCSE geometry problem, Reddit r/GCSE) czcams.com/video/5y5ugd-ZmDI/video.html
Here is a quicker path to the solution: Apply the intersecting chords theorem, which states that if two chords of a circle intersect within the circle, the product of the segment lengths of one chord equals the product of the segment lengths of the other chord. Let one chord be the vertical diameter of the circle, and let the other chord be the top side of the square. Then (1)(a) = (a/2)(a/2) --> a = a^2/4 --> 4a = a^2. Hence a=4 and the area of the square is 16.
Since a can't equal zero, you can factor out the a early which you normally can't do in an algebra equation like that which is nice. Of course, some teachers might require you to show why a can't equal zero, too.
Lol yeah I was like this video is an extremeeeeely long way around this problem.
it is not quicker if you don't know the theorem.
Oh that was probably the intended solution
How do you know the rates radius splits the top side into 2 equal parts?
At a^2=4a, a faster way to isolate a would be to simply divide both sides by a (this is allowed, as we know that a ≠ 0), giving a=4 right away.
I did similar, it becomes a*a = 4a and obvious that a must be equal to 4.
Exactly, what's up with all that quadratic BS?
While it works in this context, sometimes dividing both sides by the variable will remove an answer. You should try not to get in this habit!
@@andersonjoshua18if you end up at a^2, you know there are two possible answers. In this case, one of the answers wouldn't make sense, but as the other person mentioned it's a bad habit because thats not always the case (more often than not)
@@arthurl4300
I think a better habit would be "any time you divide by [] you write out 'assuming [] ≠ 0'"
This way you get both solutions without either or
Here is my solution:
First, using Thales’s theorem, you can draw a right triangle whereby the right angle is at the upper right corner of the square and the hypotenuse is the vertical diameter of the circle.
The right triangle altitude theorem now gives you the equation 1*a=(a/2)^2, which you can easily solve.
Everyone: *tries to solve*
Me: a²
😂😂
The decision is wrong. What if the red line we extended does not go through the center of the circle? How do we know from the problem condition that it will pass through the center and divide side a in half?
my thoughts exactly
Physics teacher:- 16 what? Bananas? Apples?
Apples 🍎
Cause Apple iPhone, right? Or just because?
@@bprpmathbasics Lol.
16 units, as the problem did not specify which unit it is appropriate to answer unit less. If you typically measure in bananas, that will do fine.
@richardhole8429 Wouldn't it be units squared, since we are dealing with the area?
Great stuff!😊🎉
Bisect the circle along the 1 unit segment. Form right triangle from (a) bottom of circle to (b) intersection of bisecting line and top of square to (c) intersection of either top corner of square and circle. Find ratios of triangle sides using trig. Apply ratios to new (similarly proportioned) triangle formed from (b), (c), and intersection at top of circle. Since the line segment = 1, then we can find the area of the square, which is 16.
I solved it in a bit different way:
- I created a coordinate system in the center of the circle.
- I drew the radius of the circle, that's connected to the top left corner of the square.
- I defined the angle between the x axis and that new radius. Let's call it fi, although it wouldn't really be used.
- I defined a = d - 1 (d is the diameter of the circle, while a is the side of the square)
- Therefore cos(fi) = (a/2)/r, and sin(fi) = (r - 1)/r
- Then I used the property of sin and cos: sin^2(fi) + cos^2(fi) = 1 And I got a quadratic equation after replacing the relevant parts.
Please do more geometry!! I hated it in high school, and although I've graduated I can't seem fully understand it 😭
The question doesn't give enough information. For example, we had to assume the point of tangency, center of circle and point on the circle that intersects the segment of length 1 are colinear.
Right
More accurately, we have to assume that the line of length 1 is a perpendicular bisector of the square's side, and the implied circle, square/rectangle and points of intersection are as they seem.
Those are the only assumptions necessary.
When you get 4a = a² you could divide by a as a is not zero. Nice that it gave so nice anger, not any weird pi stuff.
Power of a point can instantly solve
Amazing!
There is well known fact that when a chord is divuded by diameter into two halves with length c1, and, in the same time, the diameter is divided by that particular chord into two lines d1 and d2, then these are in relation such as: c1^2 = d1*d2.
In our case c1 = a/2, d1 = 1, d2 = a, all by construction, then (a/2)^2 = 1*a -> a =4 -> A = a^2 = 16.
In this sort of problems I always try to find the ratio of two areas, in this case, between the circle and the square. I'm doing that simply because the ratio is always dimentionless and gives just a bit more perspective on the problem.
So, here ratio = C/A = pi*(r^2) / (a^2) => ratio = pi*(5/2)^2 / 16 => ratio = 25*pi / 64 = 1.22718....
Also, it is preferable to solve the problem in the general case,, not right away for some particular values of the variables. Once it is solved, it's solved forever, you know.
Here, instead of d1=1 we must have d1= b, then solve the problem, then substitute b = 1 in the answer, if we have to.
So, (a/2)^2 = b*a => a =4*b => A = 16*b^2, and with the ratio we have: ratio = (pi*(a+b)^2/4 )/(16*b^2) = (pi/64)*(1+a/b)^2 = 25*pi / 64 = 1.22718....
As we can see, the ratio remains the same regardless of the values of a and b. Now we learned something new.
Well done! I got stuck on the answer simply not being solvable calling it a^2.
I did it through multiplication of lengths of segments of intersecting chords.
(a/2)^2 = a*1
=> a/4 = 1
=> a = 4
=> S = a^2 = 16
How to proof that segment of lenght 1 is situated on the diameter?
Good point. Changing the original problem to specify that the top a is ½a from corner to that middle segment would fix that yes? But unless there's further specification in the question that the square to circle intersections are tangents it's an assumption.
@@cr1197Well, that and some right angle markers to make it clear that all the depicted angles are 90 degrees.
@@ZipplyZane The question does say it's a square so that part is covered.
@@cr1197 That covers the four angles of the square, yes. But not the two angles formed by the line segment connecting the square to the circle.
@@ZipplyZane That'd be redundant with the other qualifications necessary though no? It being 90° isn't by itself helpful, and the information necessary to prove it as a radial line & the square being corner to corner within the circumference would also by necessity prove it as 90°. Unless it being 90° works by itself as proof I'm unaware of?
Of course not saying it couldn't be labeled as such. Just that it's extraneous info in much the same way as the square angles.
EDIT: Realized that a right angle marker along with marking ½a length would be the simplest way to show the line as radial. So yes you're right I was thinking stating it in words rather than showing.
yo this is like actual math magic
I drew the same triangle, but I expressed everything in r and had to use the whole quadratic formula then calculate a.
Well one shorter step. When we see 4a = a^2, we can see that 4a=a*a => a=4. Of course using the general method to solve a quadratic works, but this case just 'jumped out' at me as an obvious thing.
You can't divide by zero, meaning it would be wrong for every problem where a can equal zero, so not so obvious.
@@m-h1217Rewriting a^2 as a*a, we have 4*a = a*a. We don't necessarily have to 'divide by a' to compare terms on each side. The first term on left is '4', the first term on right is 'a'. All other terms are identical on both sides, so it's clear that the first terms must also be equal. The only way in this instance that 'a' could be zero is if we had 0*a = a^2.
@@mikefochtman7164
No.
If a = 0, the equation 4a = a^2 is fulfilled, i.e 0 = 0.
What you just explained is a mental division by a. Which is just as bad as showing division by a for two reasons.
1. You're potentially dividing by zero
2. It is a jumped conclusion that there aren't any other solutions by multiplication.
Yeah, dividing by a works out in this particular case, since clearly the square has a non-0 side length. However, 4*a = a*a is also solved with 4*0 = 0*0, that is 0 = 0.
@@m-h1217 But it's obvious a doesn't equal zero from the geometry
Thanks, I'm ashamed to admit that as an adult, I'm very bad with math and hated the subject in school. These puzzles rekindle my desire to learn it, and you explain well.
I didn't hate math persay, but had a sucky math teacher who traumatized me right after getting a handle on algebra so, geometry, trig and calculus were triggering....
Hope you had a nice life Cecilia....
This video is great but I didn't understand how at 1:26 the perpendicular from centre of Circle to the side of square is a/2. How do we prove that the diameter bisected the chord which was a side of the square (wouldn't that only be true if the diameter is perpendicular to the given chord)
I intuited it's 16 from the image... The 1 looks like the small side of a rectangular triangle with the other side of 2... so the side of the square us 2x2, so the area is 4x4.
Use circle intersecting chord Theorem: 1*a=(a/2)(a/2) or a=4
1:27 Y A Y
...and as I was always told when answering a question about area, the answer should be expressed in square units so the correct answer for full marks should be 16 sq units.
Here's a quicker solution I found. first, construct a diameter like in the video. then, using thales's theorem, you can construct a right triangle between one of the upper corners of the square (point A) and each end of the diameter (the top will be point B and the bottom will be C). The intersection between the diameter and the upper edge of the square will be called point D.
△ABD ~ △ACD, by angle-angle similarity. because their corresponding parts are congruent, the ratio between the legs is also equal. as such, a/(a/2) = (a/2)/1. This simplifies to a=4, and so the area of the square is 16 square units. QED
These similar triangles is how I solved it too!
I am still trying to understand how that line at 1:27 is a/2?!
here's another way, you take the inscribed rectangle of edge lengths a and a-1, by definition his diameter equals the circle's diameter. sqrt(a^2 + (a-1)^2) = a + 1 => a^2 - 2a = 2a => a(a-4) = 0 => a=4
a = 0 makes sense to me as well if instead of 1 we had x, we'd have a(a-4x) = 0 => a=0 or a=4x so if x = 0 => a = 0 meaning the circle and square (and everything else) collapse to a point. Geometry breaks but the solution is still valid I think!
Indeed, when you got that right triangle of (r-1, r-1/2, r), you can notice that the edges form an arithmetic sequence. Thus, the triangle must be a smaller version of (3, 4, 5).
Odd question but what was the app you used in this video?
Maybe it could be seen in a bit of simpler way if you write the side length of the square in two different ways in terms of R (radius) and then equate. From the radius and chord length 1, you can write the side length to be 2R - 1. And from the triangle and Pythagorus you can rewrite the horizontal length of the square to be 2√(2R-1). Equate these two expressions, and you will get R = 5/2 and the rest is easy.
Who is Pythagorus?
@@azzteke🤓🤓
The heck did you do here?
Draw a segment from the "1" to the other side. Now we can use chord properties. Using symmetry, top chord is bisected. Either side = x. Remaining part of the other chord = 2x since it's a square. Then from intersecting chords formula: (1)(2x) = (x)(x), 2x = x^2, x = 2. Side of square = 4 and area = 16.
Please, what tablet and program application do you use to write on the screen?
iPad + Apple Pencil + GoodNote + iPad screen record.
you have two triangles.
tan (alpha) = a/2a = 1/2 for one triangle
tan (alpha) = 1/(a/2) for the other triangle
so 1/(a/2)=1/2 so a=4
Find the function by finding values of constants using given max/min points or inflection points. czcams.com/video/RW55-BN8LA0/video.html
How do we know for sure this is the answer, when we have to make assumptions about the problem?
draw a line corner to bottom, corner to top, top to bottom, this gives a right angle triangle so
((a/2)^2 + a^2 ) + (a/2^2 +1)=(a+1)^2
6a^2 = 4a^2 + 8a
a^2 = 4a
a=4 since a not zero, the zero case is when diameter=1 so it's a boring square
a=4, A=16
3:50 = You didn't need to expand it to a QUADRATIC - AX^2+BX+C..There is NO C - so it was SIMPLY AX^2+BX.... 4A=A^2 : Divide thru by A and you get A=4.. No need for 4 other steps....
how is that line a/2?
The 4a = a² solutions seems overly complicated. You could just do:
4a = a * a
Which you can divide by a to get
4 = a or a = 4
Or maybe you cannot divide by a because it might be 0
@@blaaeekeI mean you can do it anyway, stating that the answer only works if a ≠ 0
the information does not specify that the 1 unit line (if extended ) is a diameter, does it? Nor that the 1 unit line intersects square at 90 degrees.
how do we know the centre of the circle lies along that bisecting line?
The "bisecting line" is literally a diameter. I imagine it'd be quite difficult to find a circle whose center doesn't lie on there.
It is established the bottom of the square is tangent to the circle. That in mind, the radius connecting the center to the tangent point is along the bisection of the square.
@@dashyz3293 There is no indication of:
- The line segment at the top bisecting the top edge of the square
- The line segment at the top extending into a radial line
- The line segment at the top being perpendicular to the top edge of the square
Are these reasonable assumptions to make given that this problem is supposed to have a numerical solution (and not something expressed in terms of at least two angles)? Yes. But they're still assumptions and should be pointed out as such.
All of this is based on the assumption that the 1 unit long segment is collinear with the radius. There is nothing suggesting that this would be the case. If we knew that it bisects the chord at 90 degrees then we could deduce that.
Can someone explain to me something about this question? If we say that a is equal to 2r minus 1 and we change the values a little, we get that r is equal to a/2 plus 1, but if we say that a is 4, we contradict these statements because if we change a=4 in r=a/2 plus 1 we obtain that the radius is 3 but if we put that r is equal to 3 in a=2r minus 1 we obtain that 4=5, but if we do it the other way around starting with a=2r minus 1 we obtain that r=3 and if we put that r is equal to 3 in r=a/2 plus 1 we obtain that a=5 so could someone help me a little with this
How are you allowed to assume that the line labeled 1 is going to be intersecting the square perpendicularly and that the line labeled 1 will create a line that goes through the center of the circle giving a diameter?
I used the wrong method to get the right answer anyway lol. I saw a right angled triangle with short side length of 1, assumed the other sides to be 2 and rt3 as a natural triangle. That gives me 1/2*a = 2.
many people think that you should've used some theorem but by my wording you know that i have no clue what they mean. so thank you for showing a simple way to this problem. i probably would've turned off before the end otherwise
I just eyeballed it
Your method of math is nice, I was taught wrong in teh start of algebra, the lady would always state you move the number ot the other side and change the sign, when in reality, nothing moves in the equation, we jsut do one thing and then the other and it would solve itself, and allows a mini work area to see ones steps and process... So the foundaiton of math is important... So obviosuly I had left overs adn moving errors, but if I do this method of what you do to one side you do the otehr, and you keep it separate, it makes way more sense and becomes way cleaner... This sin't a revelation here, jsut struggling with mat hwhen your foudnation is taught wrong at certain steps impedes your process to articualte and udnerstand waht one is meaning. Seeing the work makes more sense... the teacher even used arrows to demonstrate movemetn and this was a few of them in teh public system. So college these same kinds of women were over paid and screaming I am a evil white male so.. I am assuming that these humans are part of some destructive cult or club of nonsense...
It's much easier using diameter
x² + (x - 1)² = (x + 1)²
Where:
x is the side of the square
x - 1 is the side of the square bounded by the circumference
x + 1 is the hypotenuse formed, which is the diameter!
here it is:
x² + x² - 2x + 1 = x² + 2x + 1
Reducing:
x² - 4x = 0
x = 4
This is what I did, after proving that the side of the square was in fact a - 1
Just use intersecting chord theorem
a × 1 = (a/2) × (a/2)
a = 4 or 0 but of course not zero
Area of square = a^2
= 4^2 = 16
Easiest and fastest solution: Apply Height's Theorem, which states that in a semicircle, the square of the height is equal to the product of the two sides, thus 1^2 = (a/2)(a/2). Hence, a = 2 and the area equals 4
wrong
@@rk-ds4vl Why?
@@danivalles1856 Height = 1 and sides = a/2 is not a semicircle.
@@Robbatog1 Yup, that was my bad, I got the theorem wrong, but the reasoning still stands. The true Height's Theorem (that's the name in Spanish, maybe I didn't translate it correctly yo English) states that in a given right triangle, the square of the height measured over the hypotenuse equals the product of the protections of the two sides over the hypotenuse.
So, as in that area between the circle and the square, que can imagine a right triangle, and measure it's height right in the middle, then the formula 1^2 = (a/2)(a/2) is correct.
Sorry for the mistakes in the previous comments. When I search it in english, the Height's Theorem doesn't come up, in Spanish it is "Teorema de la Altura" in case you're interested.
@@danivalles1856 I appreciate your response, but it's still wrong because the triangle you're describing between the circle and square is not a *right* triangle. The "top" angle is not 90 degrees.
Looking at the question, my immediate answer was a^2. Prove me wrong.
Now if they'd asked for the value of a, I'd have done a bit more work.
Not me thinking about using photoshop to measure the pixels of the known info and use the ratios instead of doing all the math I haven’t seen yet
Let the side of the square be 2x. When chords intersect, the product of the two sections of each chord is the same for each chord. So 1 * 2x = x * x. Solution is x=0 or x=2. Discard the first solution, and you now know the area of the square is 4x^2 = 16. Mental arithmetic and doesn't take 5 minutes.
This is the most elegant solution I've seen! Well done!
edit: Whenever the solution(s) is trivial, you should ask yourself "are there other solutions hiding?" so in this case you can add x has utmost 2 solutions since we're looking for roots of a 2nd degree polynomial, so 0 and 2 are the only solutions!😊
0:30 how do we know at ðe radius and ðe 1 unit long segment are colinear?
The same way we know the quadrilateral figure is a square and the round one is a circle. By assuming right angles and regularity everywhere.
The "top" side of the square is a chord of the circle.
The length 1 segment meeting both the circle and the side at right angles is the corresponding sagitta.
@@christianbarnay2499 þank you
TIL: a perpendicular segment from a chord to its circunference is called a sagitta
@@tomasbeltran04050 Not any perpendicular to the chord. The one in the middle, which is the longest one.
@@christianbarnay2499 sorry, I should have known
can someone explain what the solutions for x are here? x=√ (4^2)
Hands up if your first reaction to seeing the question was same as mine: "A = a^2" Took me a few moments to twig they were after a numerical value.
Boss said: I pay for Solid Works, so U better use it!!1
I think u can just do easier before the end,
like this:
4a = a²
4 = a (divide by a on both side)
Then we will get a must equal to 4
and not possible for 0
if a = 0, you can't divide by a, so dividing by a presupposes a not equal to 0.
a = 0 *is* a solution.
Circle diameter 1 and zero sized square.
@@dougaltolan3017 wrong
@@Hanible can you show me the mathematics that states a zero size square isn't a square.
I fully agree that divide by zero can only be manipulated with extreme care, as in calculus.
@@dougaltolan3017 never said that part was wrong!
Up to 3:50 everything was good. But why all that stuff you do next? Just divide both parts by a (since a is not 0 by geometry) and you get that a = 4.
Some teachers do want you to explain that away though. If you make an algebra move that is normally incorrect to do, you sometimes need to show on either your homework or test why you did what you did.
@@MrSeezero why isn't it a correct algebra move? Even if it wasn't geometry related, it would still be correct if we knew that a is not zero.
Well, you never know with some teachers. I have had both kinds of teachers. Some teachers want you to show all your work. Some are cool with your showing at least some of your work.
@@F1r1at By incorrect, I mean not considering all the possible answers. For example, if you have the equation x^2 = 4, you can't just go x^2 = 4; x = 2. You have to remember that each number that is not zero has two square roots. So, instead you would go x^2 = 4; x = 2, -2. In this video, you would show that a can equal zero, and then briefly explain why the possibility a = 0 can be eliminated.
@@F1r1at I learned a hard lesson about not explaining everything when I worked on a computer program as a project for a college computer science class. I had finished a program that ran perfectly well with no errors, but I only got a 80 percent, because my instructor felt that I did not put enough comments to explain the meaning of all the variables and procedures that I was using for my program. If you ever take a computer programming class, be sure to include comments to explain what you did in the program and what some of the variables are being used for if the their names don't explain what is going on on their own. For example, "triangle_count" would be a better variable name than "rty23".
I would have just used a ruler.
3:55 just 4a=a^2 -- just divide both sides by a, a=4. no quadratic about it
Yes instead, when you have an equation like that: ax²+bx=0 this is a quadratic equation. Then you factorize the polynomial and you get x(ax+b)=0 when try to find the solutions one of these is always 0 and the other solution is -b/a, but because this is a geometry problem you have to cancel the first answer because a length can't be 0
Cant divide by zero
I dont get your explanation of 1:40 where does r-1 come from? I dont get how you know its r-1 based on known information
It is the radius minus the line that's equal to 1 on top of the square.
@@m-h1217oh I'm dumb lol, I forgot that the square line is parallel to the diameter line drawn, so you can get r-1 from the side in red.
Thanks 🙏
I miss addition
16 what? If I zoom in the screen ( for example) the area increase with the same proportions between the square and the circle! I don’t get it. Someone’s help Me making sense of that please?
Lost on why the first part is 2r and the second is r^2. 🤦🏻♀️
A = a^2
3:54 cant you just divide by a on both sides?
You could for this particular problem since a can not equal 0, but in general you can not.
At 4, just divide by a on both sides. a=4. a^2=16
Yeah, but technically wrong because a could be zero, and you cannot divide by zero. In this case it's no big deal because we discard the a=0 solution anyway, but generally you should avoid dividing by a variable
@@msolec2000it's not technically wrong,we already know it can't be zero
@msolec2000 a could not be zero given you know this is a geometry problem. You know this all through this problem, so it doesn't block you from dividing by a.
That is Davinci man.
Hello! I need basic instructions with (r-1)sqrd and the result on the right hand side
I'm using ^2 for squaring and * for multiplication.
(r-1)^2
= (r-1)*(r-1)
= r*(r-1) + -1*(r-1)
= r^2-r + -r+1
= r^2 -2r + 1
The rest of this is probably unnecessary, but here's a couple sample problems to show what's going on more clearly if that was hard to follow. ('a' and 'b' are just placeholder variables here. This 'a' is not the 'a' from the circle/square problem he was solving.)
Sample problem 1:
(a+3)*b
= a*b + 3*b
This is called factoring. You can try putting in different values for 'a' and 'b' and check that these two things are equal.
Sample problem 2:
(Subtraction is interchangeable with the addition of a negative.)
(r-1)*a
= (r + (-1))*a
= r*a + (-1)*a
= r*a + -a
= ra - a
Sample problem 3:
(a+3)*(b+1)
= a*(b+1) + 3*(b+1)
= (b+1)*a + (b+1)*3 (optional rewrite for consistency with sample problem 1)
= b*a + 1*a + b*3 + 1*3
= ab + a + 3b + 3
(This last step is rewriting it with implicit multiplication. Its just leaving out the multiplication symbol before multiplying by a variable.)
@@Elrog3 Thank you so much! I appreciate the lessons and samples.
This is super easy, the area is a^2.
8th or 9th grade math standard?
1:26 - how that length is a/2? The diagram does not look symmetrical
Basic geometry. It's a square.
The bottom edge of the square is tangent to the circle, so a perpendicular at the point of contact will pass through the center of the circle (the radius is always perpendicular to a tangent line). Continue this radial line through the center to the other edge, it will intersect the top edge of the square at a 90' angle as well. The top edge of the square is a chord of the circle, the radius perpendicular to a chord will always bisect the chord.
In other words, a rectangle and a circle with these points of contact will always be symmetrical.
The only "assumption" that is really made about the diagram is that the length of 1 is perpendicular to the square and touches the peak of the arc. But without this assumption there's no way to solve the problem as the 1 is the only information we really have about the circle and so must be useful in some way.
@@Ashley.D a perpendicular line that touches the peak of the arc would necessarily be part of the radius. So yes, I did cover that.
@@phiefer3Why must a perpendicular line go through the center?
@@phiefer3 Yes, but there is no guarantee from the diagram the the perpendicular is touching the peak of the arc. Plus, technically, there are no right angle indicators to show it is perfectly perpendicular.
The best form would have at least one little square on top to indicate a right angle, and somehow label that each line segment on either side is equal--either with variables, or that little tick mark.
I could see a version of that has each line segment as a, and labels the vertical side as 2a.
Halfway through the video i already solved the remaining part in my head... But then he goes for another 2 minutes very detailed... And i was taught how to teach people so I know why he does that, but damn the length teacher need to go to make sure to instill good habits into students...
For me these are a refresher course and I don't need quite the detail that new students require. So I will scroll ahead when I remember what he teaches.
Intersecting chords:
1*a = a/2 * a/2
a = a^2 /4
a^2 - 4a = 0
a > 0 , so a = 4 and A = 16
4a=a^2....4=(a^2/a)....4=a....a=4.....what am I missing here? Why did he solve it as a quadratic?
root(i)*root(-i)
=root(i(-i))
=root(-i^2)
=root(-(-1))
=root(1)
=1
Or root(i)*root(-i)
=root(i)root(i^3)
= root(i)i*root(i)
=i*(root(i))^2
=i*i
=-1
where is the mistake please help
Maybe before solving the question, could you analyze the diagram and state any assumptions one has to make beforr solving? E.g. the line of length 1 is cutting the side of the square exactly into 2.
Yes. State any assumptions. This one can best be stated "assuming that the line segment of length one is on a diameter and that it intersects the top of the square at the centerpoint.
Amazing, now look for r
2.5
The length of the circle is a + 1
So the radius is a/2 + 1/2
a = 4 so
4/2 + 1/2 = 5/2 = 2.5
Bout tree fiddy
You could also sketch the picture in F360 and measure side of square, result is also 4.
No, just no
Is this like theoretical math? All the solutions I see assume something at some point
a = 0 is a valid solution. It is both trivial and degenerate BUT I don't understand your exclusion!
I dont think squares can have 0 length because they wouldnt be squares anymore
@@killianobrien2007 :), that is what degenerate is... your logic is 'backwards' and erroneous here.
You are adding a choice here. Formally
a = 0 is discarded because it is zero
OR
predefine range of a as a>0
Three different outcomes
1) incomplete
2)inconsistent
or
3) nontrivial
Your logic leads to inconsistences. I fall into the same trap, which is why I asked my question.
I confirm 0 is a perfectly valid solution. The argument that because it's a geometry question it can't have zero as an answer is far fetched. The only restriction we have in a geometry question is that we can't have a length that is negative or non real. But any positive real number is valid.
And actually one way of proving that 2 points A and B are the same point is to prove that the segment AB has length zero.
How do we get a-4=0 from 0=a(a-4)?
null factor law, if 2 or more terms that are multiplied equal 0 then that means at atleast one of the terms must be equal to zero, here we have 0=a(a-4) that means that either 'a' is 0 or 'a-4' is 0, therefore we get 2 solutions a=0 and a-4=0 but we already know that 'a' cant be 0 since 'a' is the sidelength of a cube and a cube cannot have sidelengths of 0.
Dividing through by a...
0/a = 0
a(a-4)/a = a-4
If you look at it you'll see that "0=a(a-4)" is just an equation. There are two terms: the first term "a" and the second term "a-4." When multiplied together they equal 0. Ask yourself how do you get a solution of 0? Any number multiplied by 0 is equal to 0. So essentially you are finding how to make each term equal to 0. And thus "a=0" and "a-4=0"
@@GamezGuru1cant divide by 0
Wel how do you get 0? That's only the case if one of the terms is zero. So you check each case by making it equal to zero
The answer is a^2
From 4a = a^2, can't you just divide both sides by a to get a = 4 straight away?
That's what I did - low and behold it's correct, and saves half the maths down in this video...
i think the issue is that it wouldn’t be rigorous enough? after all it doesn’t show why a=0 is not a valid solution
In a general polynomials equation this would be a bad call since it eliminates one of the solutions. But if you do it while acknowledging that you don't care about the 0 solution then it's fine.
@@randomminecraftplayer6857 yea since you already know a can't be 0, you can simply divide by a
@@CharlieSprague-br4ijyour teacher will object as it is a bad habit to discard a solution by Division. When you do it anyway, do document a=0 as an invalid solution. Cover your tracks!
Here s a quicker solution: 4.
CZcams recommendations reminding me that I am a failed Asian
Funny how one mathematician can reject 0 as an answer because "this is a geometry problem". All the while another math teacher on another video just explained that a degenerate triangle can have an angle of 0 degree making both segments of the triangle the same length of the hypothenus.
Technically a=0 can be rejected because we know that the square is not a degenerate square (as otherwise the circle couldn't exist) and thus has a side length greater than zero.
@@ZipplyZane The square CAN be a degenerate square, and the circle would be absolutely fine. Imagine a circle with diameter 1 and a degenerate square on one point of the circle.
can't you just divide by _a_ at the end instead of doing the quadratic formula?
Bad habit, you could either divide by zero (which luckily a can't be for this particular problem), or you could miss another answer for a.
The area is a^2 🧠
How do we know the side of the right triangle is r-1? 1:43
Look at the rectangle that contains that triangle, the left and right edges of that rectangle must be the same length, the left edge goes from the center of the circle to the edge of the square, let's call that e. The total distance from the center to the edge of the circle is r, and the distance from the edge of the square to the edge of the circle is 1. Therefore: e+1=r -> e=r-1
2r=a+1 is an incorrect assumption
how
If you agree to the assumptions that the round figure is a circle, lines that seem straight are actually straight, lines that seem perpendicular are actually perpendicular, 2 corners of the square are on the circle and the opposite side is tangent to the circle then 2r=a+1 is an obvious proven result coming from known relations between diameter, chord and tangent.
Bprp now be doing geometry
Thanks to my iPad. 😆
hear me out its easier i you just move the lines
asnwer=50 isit