How To Solve For The Radius. Challenging 1970s Math Contest!
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- Опубліковано 28 чер 2024
- There doesn't seem to be enough information to solve this, but the answer works out magically. Thanks to all patrons! Special thanks this month to: Richard Ohnemus, Michael Anvari, Shrihari Puranik, Kyle.
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Maths By Amiya : Perpendicular Chords and Radius
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I just noticed the video length is 6:28 on mobile--what an accidentally perfect time. Thanks to all patrons! Special thanks this month to: Richard Ohnemus, Michael Anvari, Shrihari Puranik, Kyle. I also credit patrons Pradeep Sekar and Nestor Abad for finding a typo in my original video--thanks! (You can get early access to videos by supporting on Patreon--such support makes a huge difference.) www.patreon.com/mindyourdecisions
Lmao, good job though
I think there's a typo at 4:56, it's supposed to be (-2+p)²+q²=r² not (-2-p)²+q²=r². That or I didn't understand what you did.
5:40 how can you get the value of q with r's value unknown?
@@chunfengmugu he didn't show but he said something about eliminating the r^2 term by subtracting
I have best method for finding the radius... it's by using ptolemy's theorem.. 😅
the secret trick to maths problems, pull some obscure formula from out of nowhere
The formula isn't obscure nor pulled out of nowhere.
It isn't that hard to find either. But you have to work a bit to find it and prove it.
If you can't prove the formula you use, you shouldn't use the formula at all, since you don't know what you are doing.
RodelIturalde shut up nerd
@@Marcel-vz7vp lol.
Sorry for knowing stuff and actually trying to learn something.
I should probably go ignorant like you and call everyone who knows more then basic addition and multiplication nerds.
RodelIturalde *than. Damn can’t even use then and than properly.
@@RodelIturalde Well said.
My method to solve it was: 1.) Worked out the line going up to C is a length of 4 (similar triangles in circle geometry)
2.) Taking the bisecting 2 lines as (0,0), I then used coordinate geometry to find circle center as (2,1/2)
3.) Then perpendicular bisected the y axis to form a triangle and used Pythagorean theorem to find the radius = root(3.5^2 + 2^2) = root(65) / 2
Maths is awesome! So many ways to solve!
Second question: How to solve the value of the vertical segment, which is not shown in the figure?
Greetings from Paris.
professeur essef, in mathematics (active for over a year, on YT & Wiki, mainly in astronomy and astrophysics).
@@lecinquiemeroimage angles ADC and CBA are equal.
@@Jerryfan271 Not at all!
Neither for the angles ADC and ACB (at least, you make two confusions) ...
Answer (to justify): let us write I the intersection of the two lines (AB) and (CD); then IC = 4 (different from ID = 3)
P.S: angles ADC and ACB would be equal [to 90°] if [A;B] were a diameter, and then IC = ID.
@@lecinquiemeroimage um, you conclude IC=4 but that in itself implies SAS similarity of ADI and CBI meaning ADC and ABC must be the same angle.
I used same-arc subtending to get that result. Since arc AC subtends both ADC and ABC, and D and B are on the major arc, the angles are equal. So IC=4 iff ADC=ABC.
if I am somehow wrong and ADC=ABC is false then why not give a counterexample?
wait whaaaat u doin maths now a days ? niceeee
You pick the longest and most obscure way to solve problems
Very easy .I am 10 standard student and got right answer using simple chord theorems.
I immediately thought 4. But it’s never that easy
6+2=8 (diameter) then 8÷2=4 (radius)
That's what I thought
@@janus9148 he said chord AB, so 6+2 is not the diameter. It just happens in this case that they're similar, which means this chord is very close to the centre of the circle.
@@dennisdegouveia5439yeah, i thought the chord AB was in the center
Janus no not unless the center point is on the line which would make it the diameter
It is never that easy. No matter how sure you are.
Super Math Bros Ultimate
Kyledude252 I wanted to make that joke
Bruh
Don't you make fun of people with lisps. /s
Numbers weave into a tower of math
I visited this video just to make sure someone commented that
Looking at 4r² = w² + x² + y² + z² I realize that if you draw a cross somewhere into a circle, making each of the cross's quadrants a square, that's the same area as a square drawn around the circle. Geometrical beauty!
that's actually 4 times the area of the squares drawn from the quadrants but nice way to see it :)
@@ultimatedeatrix9149 jbglaw stated it correctly.
I think this is wrong??
It is correct.
a² is the size of the surface of a square with sides a, b² same for b, and r² for a square with sides r.
A square around the circle would have sides of the length of the diameter of the circle, that's 2r.
Its surface is (2r)² = 4r².
thankyou so much for that,that is much easier to visualize!
Nobody:
Problems in the 1970s: *handed out to unborn foetuses in Asia*
Peppermint Cookie 😂😂😂🤣🤣🤣👍👍✌🏼👍👍👍👍👍👍👍👍😂😂😂
foetuses lmao
Math should be esteemed.
You can also use the circumradius theorem of the triangle ABD. The formula is simply R = abc/4L where a,b,c are the side lengths and L is the area. The area is 8*3/2=12. a, b and c is the length of BD, AB and AD in any order. By phytagoras, you get AD = √13 and BD = 3*√5. Thus, plugging in to the formula, we get R = 8*√13*3*√5/(4*12)= √65/2
Don't get it wrong though, your solution is also amazing, I'm just showing my approach when I first saw this problem.
By the way, you have a great channel, thank you for your amazing content and presentation.
Edit: I forgot to say what L stands for.
This is also what I did. I was very happy to have the same result
It is also a secret and hidden formula.by the way can u pls derive it for me❤❤❤❤
@@devanshusharma9768 If you draw the line (new chord) AD and bisect it, the extended bisection will go through the circle's centre. Likewise the bisection of AB will go through the centre, parallel to CD but 2 units distant. You end up with lots of CONGRUENT triangles and with PYTHAGORAS it's easy to solve without any formulas, just the most rudimentary algebra.
@Ayush Ojhau did very intresting approach to the problem
I did the same
I now know the radius of Super Smash Bros.
Thank you
That’s what I thought also lol
Lmao
Ben Talks i clicked this video because of that
at least I'm not the only one who noticed
I actually solved this in a third way! I used the Pythagorean theorem to solve for BD and AD. Then, I used the cosine theorem with the triangle ABD to find the value of the cosine of the angle ADB, and consequently its sine. At this point, all I needed to do was remember that given a circle and a chord, the chord in equal to 2r sin(Alpha). We obtain AB = 8 = 2r sin(ADB), r = AB/2sin(ADB)
Hmm good idea. Clean and simple
Just solved in the same way. Never knew "the power of a point", etc. Using phone calculator and twice proving -- 30 min. If using sin() = sqrt(1-cos()^2) instead of sin(acos()) the same formula is obtained as that of the presenter.
This was my approach too
I only understood half of this. Please someone explain 🥺
Or you could find the sine of angle ADB by O/H i.e. 3/BD as you have just calculated BD.
Love the way you challenge the brain with these mind boggling problems. You are doing great work and I look forward to being challenged.
(x-x0)^2+(y-y0)^2=r^2 : three variables with three conditions (-2,0),(6,0),(0,-3)
Yeah, too easy
@@ilPescetto Yeah I used that method too. I was just worried about mistakes during solving but it turned out alright for me ^^"
I solved it by using this method too.
@@ilPescetto o
Just learned that couple weeks ago in school.
The co-ordinate geometry solution is beautiful
This is pretty simple if you know the properties of a circle.
I solved this problem in two steps:
1) Joined the points A and D and used the property "angles of same segment in a circle are equal" to got the relation between the angles ACO and DBO (named O as the intersection point of the chords) and OAC and ODB as ACO=DBO and OAC=ODB. Then applied the concept of similarity to derive the relation: OC/OB = AO/OD after proving triangles ACO and DBO similar by 'AA' similarity criterion. Hence by substituting the known parameters in the above relation got the value of OC=4.
2) applied the property of the the perpendicular bisection of chords by the radius of the circle to get MC=MD=3.5 (named M as the bisection point of chord DC) and AN=NB=4 (named N as the bisection point of chord AB). Then subtracted MC from OC to get OM=0.5 and finally joined XB (named X as the centre of the circle) and applied Pythagoras Theorem : XN^2 + NB^2 = XB^2 (Radius^2). Now from the diagram XN=ON. Therefore substituted it's value in the equation and got the answer as sqrt.(16.25)
I enjoyed the video quite much because when I was in college and doing research on writing unimportant bits of a computational software, we needed an algorithm that takes in the coordinates of three points on a plane and spit out the coordinate of the center of the circle that passes through all the points, and the radius. I remember pulling a neat linear algebra trick where I used the property that the vector that describes the direction of a chord must be orthogonal to the vector of its perpendicular bysector, which is expressed in terms of the unknown circle center coordinate. The two orthogonality conditions directly translate into a linear system and you never even touch r to get the center coordinate, just linear algebra. I pulled out the method, dusted it off and used it to solve this problem. 10 minutes well spent. Thanks for posting these problems; they make quite neat brain exercises.
Challenger Approaching!
Presh Talwalker divides the competition!
Pomm popopoooomm pOpopOpoooOmm!!!
😂👌
I used formulas for triangle areas and Pythagoras theorem, I combined S=ah/2 with S=abc/4R and found R.
I used formula DO*OC=AO*OB to find CO (O is intersection of AB and CD), drew triangle CBD, used Pythagoras theorem to find sides, calculated area of triangle and calculated R.
Thank you for your nice explanation.
I have another approach by using basic geometry.
1/Finding the point O, the center of the circle.
Label H as the meeting point of AB and CD, and M and N as the midpoints of AB and CD respectively. Then draw the two bisectors lines from two midpoints which meet at point O. O is the center of the circle. Now we have a rectangle HNOM.
2/Calculating the hypotenuse OH and the radius of the circle:
Using chord theorem: CHxHD=AHxHB---> CH=2x6/3=4---> HN=(7/2)-3=1/2 and HM=AM-2=2
Using Pythegorean theorem: sq OH=sqHN+sq HM= sq (1/2) +sq 2= (1/4)+4=17/4----> OH= (sqrt of17)/2.
EXtend OH to the other sides we have the diameter of the circle (radius=R)
Using chord theorem: (R-OH)x(R+OH)= AHxHB=2x6=12-----> sqR - sqOH= 12-----> sq R - (17/4 )=12 ---> sq R= 12+(17/4)= (48+17)/4=65/4.
Thus the answer is R=(sqrt of 65)/2
I had never seen that theorem before, but my mind immediately went to perpendicular bisectors of chords intersecting in the center. I suppose I would have rediscovered it that way!
You can also do it by:
1) calculate the lengths AD and DB using Pythagoras
2) calculate the angle ADB using trigonometry + the three lengths
3) Lets call the centre point O, the circumflex angle at the centre subtended by the arc AB is twice ADB. So the interior angle AOB is 360 - the angle we just calculated.
4) Because the triangle ADB is isoceles, the angle OAB is (180 - AOB) / 2
5) Use trigonometry to calculate length AO, which is the radius.
Not being familiar with the power of a point, I had a different approach to this.
I duplicated both lines, rotated 180°.
The line A(A`) is a diameter of the circle, and also the hypotenuse of the triangle AB(A`).
AB has length 8, and B(A`) is unknown. Representing B(A`) as x, and using Pythagoras, we get a diameter of √(8² + x²).
CD(C`) has lengths 4 and (6+x), so C(C`) gives the diameter a value of √(4² + (x + 6)²).
Then we can solve (x + 6)² + 16 = x² + 64
Which gives x = 1.
Substituting this into the aforementioned formula for the diameter A(A`), we get √(8² + 1²) = √65
So r = (√65)/2
Glad I found this channel. Love these type of problems.
Hi Presh, Thanks again for the great videos. i think you can solve this problem by using simple triangle properties without using coordinate systems or having to know other formulae. This is what I did:
- I labelled the 4 vertices A, B, C and D clockwise starting from top. P is the point of intersection of the chords.
- GIVEN: DP=2, CP=3, BP=6. SOLVE radius R= ?
- Mark the center of the circle as O. Connect O to A,B,C & D
- OA=OB=OC=OD = R.
Drop a perpendicular line from O to chords AC and BD. Label the lengths of these as x & y.
- Now, AP*3 = 2*6 , AP= 4.
- Triangles AOC and BOD are isosceles. so it follows:
4-y/R = 3+y/R ; y = 0.5
2+x/R = 6-x/R; x = 2.
Now applying Pythagoras theorem , we get
R^2 = (4-y)^2 + x^2
R^2 = 3.5^2 + 2^2 = 16.25
==> R = 4.031.
This is what I did. Easy.
Thank you for this. I also used the co-ordinate method! I was surprised and pleased to learn of the perpendicular chords theorem - that method gives the fastest solution.
and here i am solving math problems by gut feeling
Then congrats.....you are a human
My gut feeling immediately said "4" and I was only off by .031 so that worked out well enough.
I noticed that the horizontal line is not going exactly through the center. So, my solution was not 4, but „4 and a little bit“. So, I was exactly on spot. „A little bit“ might not exactly be a mathematical term but in real life it works pretty well in solving real life problems (as opposed to the brain masturbation presented here).
@@MothaLuva good enough for the girls I go with
So excellent. Enjoyed it thoroughly! Maths is so much fun.
Just subscribed to your channel! Proposing another solution: Connecting CB and AD (or AC and BD), we can also see that we get two similar triangles through inscribed angles based on the same pair of points. This allows us to get the top length to be 4. From there, we can use Pythagorean theorem on the 3.5 and 2 to get the final answer as well.
I solved it without knowing that special formular and that wx=yz
draw the center (approximately) of the circle and mark it with O
AB=8 so the bisector is 4
OB=r
distance from chord AB to O labeled with x
this is our first triangle: r²=x²+4²
secound triangle:
OD=r
distance chord CD to O labled with y = AB/2 - 2 = 2
third side is x+3
so: r²=(x+3)² + 2²
equaling both formulars
x²+4²=(x+3)² + 2²
x²+4²=x²+6x+3²+2²
4²-3²-2²=6x
x=(16-9-4)/6
x=1/2
r²=x²+4²
r²=(1/2)² + 4²
r²=1/4 + 16
r²=(1+16*4)/4
r=root(65)/2
Same thing. Takes about a minute cause it is rather easy and no special knowledge needed.
Yep, that's the one I went with too.
Thank you! I think this way is more intuitive and easy to follows without knowing any special formula, etc.
Same 😁
Exactly , I too did it on my first attempt. 😀😀
I did it another way; note that you have a triangle with base 2+6=8 and height 3 and thus area 12 at the bottom of the diagram, with sides 2+6=8, sqrt(13), and 3sqrt(5). then the circumradius of that triangle is 1/2 abc/A = 1/2 * 24sqrt(65)/12 = sqrt(65) as desired.
I like what you did, but you ended up with the diameter.
Interesting relations discovered along the way... Thank you!
That first formula is amazing. So useful!
I worked out the angle ABD using trigonometry. I worked out length AD using Pythagoras' theorem. The angle AOD (O being the centre of the circle) is 2 times angle ABD. I then used the cosine rule to work out the radius (2 of the side lengths of triangle AOD). Gave me the same answer.
We use the same way to solve the problem🙌🏻
(6+2):2
First time I finally solved a challenge from talwalker. Now i can finally go to sleep.
I did it with circumcircle and circumradius. The answer matched!
That radius formula's proof is impressive. Nice!
I m a Indian and too much interested in mathematics and your videos today 5 may I sat for one of the thoughest exam of india to pursue BSC in MATHEMATICS and this question appeared in exam thank you Mind your decision .....U guys r doing a fantastic job..
Which exam was it?
I did it with inverse trig functions and used the fact that arc ACB is equal to 2(arctan(2/3)+arctan(2))
Nice. I used the coordinate method but set my origin at the midpoint of the horizontal chord, so I knew the centre was (0,c) and it had to pass through (4,0) and (-2,-3). From there it fell out. Very pleasing!
So elegant and so soo beautiful yet so simple !!!!
i used Trigonometry, arc tan twice to find angle ADB, then angle at center = twice angle at circumference. then split angle intto 2 (isosceles triangle) & cosine to find the Radius
for those puzzled at Intersecting Chords Theorem wx=yz , u can prove it using Similar Triangles ACP & DBP , where P is the intersection point
Ditto. Inscribed Angle Theorem + Law of Cosines FTW
I never knew id get to knw a new formula !!
i wanted it to be solved by geometry n was aghast when the formula was mentioned thinking the geometrical deduction wld be skipped ... Thanks for actually showing how the formula is derived !
Great problem. I solved it using geometry, knowing that wx=yz, but not knowing the formula for r^2. Then I did the general case, and derived the formula that you used and proved. Very neat & satisfying!
We can use the sine rule in the ABD triangle:
2R=AD/sinB...etc
ye thats what i did. got the answer pretty quick that way.
That's trigonometry method but we need a pure geometric sol
@@jitendramohan7500 1) trigonometry IS geometry. 2) Who says we _need_ a purely geometric solution?
I love your videos Press.
Would you mind to tell me what software you use to make your videos?
Thanks
Amazing.I respect your ability of thinking.
Ur Voice is so Curious Than Math problem.
I really loved it !!
I really enjoy your videos, thank you for your efforts and keep going!
With trigonometry you can find another way to solve:
1) calculate the angle CDB as the arctan (6/3)
2) arc CB = 2 CDB
3) the angle ADC = arctan (2/3)
4) arc AC = 2 ADC
5) as the whole circle has an arc of 2*PI rad, then arc ADB = 2*PI - AC - CB
6) if O is the center of the circle, then angle AOB = arc ADB
7) in the triangle AOB we have AO = OB = r and angle AOB = 165.78
8) solving the triangle AOB you can find that r = 4.03
Do you have a video on "Power of a Point"? I've never encountered that before! Great math exercise. I love circle geometry. Thanks for presenting it, Presh.
He'll surely have a Powerpoint presentation about it :)
And he just takes it as a given. You might not know the other equation, but surely you know this! Um, no?
I think the w*x=y*z formula comes from the fact that the triangle with sides w and y is a similar triangle to the triangle with sides z and x. Two sides of 2 similar triangles are proportional to each other which gives the formula y/w=x/z. From which you can get the w*x=y*z formula.
co ordinate geometry method impressed me sir,thank you very much.
Thank You, doing the same process every time but not thought of a theorem🙏.
The new Smash DLC Fighter looks amazing!
I think there is an easier way that uses both geometry and coordinate geometry.
Let's label the origin point O. Since AB and DC are along the axes (this would also work if they were parallel to the axes), The center of the circle will be the middle x coordinate of AB and the middle y coordinate of DC (because a normal from the center to AB or DC would cross their middle).
By intersecting chords, we know that CO = 4, so point C is (0, 4). Therefore, the coordinates of the center will be ( (-2 + 6) / 2, (4 - 3) / 2 ) which gives us (2, 0.5).
Then, the radius is simply the distance between the center point and any of the points A, B, C, or D.
Example with A: r = sqrt( (2 - (-2))^2 + (0 - 0.5)^2 ) = sqrt(65 / 4) = sqrt(65) / 2.
I'm so proud of myself. I got this one right away!
In my solution I also went for the coordinate approach. However I used the center of the circle as the coordinate origin which gave me much simpler formulas. You will of course always get to the same solution regardless of what reference point is used. However it can pay off if you give some thought to finding the appropriate reference system for a given problem because that can simplify the calculation considerably. (especially when you are sitting in an exam and the clock is ticking)
Okay UA-cam recommendations. I guess it’s big brain time.
I would've used trigonometry, but your method is way cooler and more elegant.
Anch'io ho usato la trigonometria! Chiamando R l'incrocio delle corde, trovi facilmente RC, poi BC, quindi usi il teorema della corda col triangolo BCD, in cui l'angolo alfa = atan(2); da lì trovi r=sqrt(13)/sin(atan2) Su questo profilo c sono una bella serie di problemi, molto più tosti di questo...saluti dall'Italia!
@@61rmd1 grazie per le informazioni! Sono uno studente di ingegneria matematica, questi video sono per me il pane 👍👍
I liked how you solved this one, even though other solving tricks are cool as well.
Thanks for the video!!!
I used an extended version of The sine theorem a/sinA=abc/area=2R
A more “human” approach is:
1. Let r be the radius
2. Add a chord EF parallel and symmetric to CD, and another chord GH parallel and symmetric to AB;
3. CF and BG should pass through the centre of the circle, i.e. they are diameters, so equals to 2r
4. |> CDF and |> ABG should be right angle triangles
5. For |>ABG, let AG = a; then
for |>CDF, CD = 3 + a + 3 = a + 6
6. For |>ABG, AG^2 + AB^2 = BG^2
for |>CDF, CD^2 + DF^2 = CF^2
7. For |>ABG, a^2 + 8^2 = (2r)^2
for |>CDF, (a+6)^2 + 4^2 = (2r)^2
8. a^2 + 8^2 = (2r)^2. ..........i
(a+6)^2 + 4^2 = (2r)^2 ......ii
9. ii - i, 12a + 36 + 16 - 64 = 0, => a = 1
10. Using i, 1+ 64 = 4r^2,
Answer: r = sqrt(65) / 2
Yes this is much simpler - no advanced theorems needed, just symmetry and Pythagoras! I expected Presh to solve it this way and was surprised he didn't. I can even imagine him describing it: "Let's make a second copy of the chords, rotated by 180 degrees, now draw a red triangle here, and a blue triangle here". Much more elegant.
Loved it!
I pulled out the first equation from my consciousness, knowing the sum of the hypotenuse would be the diameter, but the power of the points was new and fascinating to me.
Presh Talwalker smash reveal?
"These math videos avaliable for free on youtube, builds confidence for students"... wish I could say the same....
I've solved it using circumcircle of triangle ADB. Took me a while though as I was first trying to solve it using a set of equations using chord's angle and a radius.
Cool video I initially started with Euclidean geometry and I wanted to establish whether the horizontal line could be seen as the diameter of the circle. I went with the assumption but the those theorems simply does not agree with the assumption. I then used co-ordinate geometry and got the same answer
You can just find some angles (using tan (a)) and use the law of sines on one of the triangles that lays on the circle... Easy.
I did this way too. i didn't even understand why was it hard
I don't understand
@@btdpro752 read about the law of sines, and I'm sure you'll see it
Alternate title: "finding the radius of Australian super smash Bros logo!"
Your method is simple and creative :)
It can be directly solved in one step using radius of circle around triangle formula, and Pythagorean theorem of course. More work to simplify the answer but very straight forward.
I click this on my recommendations.
Now i feel smart.
Yep
Nobody
Literally nobody:
Le me after watching this video: mothemoticon
I use geometry+congruent triangles+Pythagoras and got the answee
We have learnt this furmola when we were 12. Such a useful one
I thought of the method of finding the lengths of AD and BD through Pythagoras, finding the sin of ABD. Then the triangle is inside the circle - you can use the sin theorem with 2R
the feeling when you solved it with the easiest way possible, but you can't explain it due to language barrier
00:36
OA×OB=DO×OC
* O is the point of intersection
I found OD by the first formula mentioned in the video and then created a rectangular where I found the cosine angle ADB, I turned the cosine into sine by the fundamental theorem of trigonometry and then plunged everything in the triangle ABD into the sin theorem.
Solved using a mix of trigonometry and circle geometry. Used circle geometry (angle at center = 2x angle at radius), then used cosine rule to calculate the radius.
Can also be done using concept of "power of point".
How? Please explain
I solved it like this:
Area of the triangle ABD is
|AB|•|BD|•|AD| ÷ (4R)
where R is radius of the circle
|AB| = 8
|BD| = 3 * sqrt(5)
|AD| = sqrt(13)
And area is of course 8*3/2 = 12
Solving for R is easy
So did I!💕
Same
@@mevnesldau8408 I thought you were dead due to pneumonia , welcome back from realm of dead , btw big fan here
Where the 4R come from?
IMHO-
It's much more easier and more obvious then the example in video
This page is amazing👍👍👍
I used coordinate geometry too, but used the point slope equation for lines to find the perpendicular bisectors of AD (point (-1, -1.5) with a slope of 2/3) and DB (point (3,-1.5) with a slope of -2). Solved those to find the center (2, .5), then calculated the distance between the center point and B (sqrt((6-2)^2 + (.5-0)^2)).
Clearly, the other part of the vertical cord = 4. (3×4 = 2×6). Now if we take that point of intersection as our Origin of the cartesian plane with vertical chord as Y-axis and horizontal one as X-axis, this problem becomes way to easy.
The center lies on (0.5, 2) [centre perpendicularly bisects every chord] and one of the circumference points is (0,4). We can also use (6,0) or another two of those 4 points.
EDIT: The geometry method is similarly derived too. Both the methods use the same concept
Shadow Ninja idc nerd
For this Just Construct perpendicular on both the chords and you will get .5 a and 4 unit of the side of triangle in which the hypotenuse is the radius.
Exactly: I did it in my head not knowing the formula; .5 sq + 4 sq = 16.25. Take the square root = 4.031. Why make things so complicated. You can easily see the half way point of the line 7 is 3.5, which is .5 from the centre and the base is 4.
Hello. I'm Sofien from Tunisia. Thank you for these very interesting mathematical problems. I think that we can solve this problem with trigonometry. We can apply the sinus theorem in a triangle (with base 6+2 = 8 and height 3), in fact sin a / A = sin b / B = sin c /C = 1 / 2R. we have tan a = 3/6 = 1/2 and and A²= 2²+3² = 13. sin a = tan a / sqrt( 1 +tan² a),
then sin a = 1/2 /sqrt(5/4) = 1/sqrt(5) and A = sqrt(13). We find that 2 R = A/sin(a) and thus R = sqrt(5)*sqrt(13)/2 = sqrt(65)/2.
Just solved it and noticed, that the video was published on my birthday :D
Normal people:math
Me, and intellectual: waluigi for smash
Intellectual mispelling *an.
Heh
The intersection of the two lines I called point E. Then I shove the line CD to the middle so that AE= 4 & BE=4. There thene is ane equal distance (x) berween point C and the top of the circle and point D and the bottem of the circle. So (3 + x)•(4+x)=16 and (3+x+4+x)/2=r. Then I solved for x
Initially, I was looking for a method like this (shifting the chords and reapplying intersecting chords theorem), but I couldn't figure it out (I didn't think of labeling x). Cool method! I'm glad to see that it can indeed be done that way.
I did look over a hundred or so of the 2.3 thousand comments. My solution was similar to yours. I posted it already. So I thought I would pass it along to you personally:
There have been over two thousand comments to this problem over five years. So, the approach I took has probably already been submitted. Regardless, this is my solution to the problem:
The two parts of the vertical line segment CD have lengths three and four. So the midpoint of the vertical cord is 3.5 units; a horizontal line drawn through that midpoint is the diameter of the circle. If the horizontal cord AB is "pushed up"so that it coincides with the above diameter line, then its new length will be (2+x) + (6+x).
So, by the Intersecting Chords Theorem: (2+x)(6+x) = (3.5)(3.5)
The solutions to the resulting quadratic equation, x² + 8x -.25 = 0, are .03 and -8.03.
x = .03 is the useable root.
The diameter length is (2+x) + (6+x) = 8 + 2x = 8 + .06 = 8.06
Therefore the circle’s radius = 4.03
You could alao use properties of intersecting chords and get the actual answe =4. AB=CD
Thank you for solving the radius of the smash ball
I solved for The lengths of ABD triangle sides. And then applied The formula for The radius of The circumscried triangle R=abc/(4S)
What is S?
@@PatrickPitso Aria of the triangle
that feeling when I wrong on ≈0.031
6 + 2 /4?
Fantastic explanation to arrive at the Answer 👍
You can also set them in relation to each other:
x/w = y/z
When you draw the perpendicular bisector of two non parallel chords their intersection point always be the centre of circle, it's mentioned in class 10 NCERT
Yeah bhai... Mein abhi 12 mein hu aur mujhe sharam aa rhi h ki ye mere se hua nahi kese
12 mein aate aate sirf calculus calculus hota h
Siempre me creí una eminencia en la matemática y ahora que veo todos tus videos me doy cuenta de que soy bien bruto 🤦🏻♂️
Solo sé que no sé nada 😂😭
Wow, great video!!
I used power of points first to get the other part of the vertical, then found the total length of each chord. I set the intersection of the chords as my origin, then bisected each to find the coordinates of the center of the circle. Using that, I calculated the distance to the point (-2,0), which is the radius 4.031.