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
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).
@@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?
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.
@@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.
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!
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.
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)
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.
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.
@@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.
@@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.
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
By calculating the power of the intersection point of the proposed chords with respect to the circumference, we obtain the length of the upper section of the vertical chord: 2x6=3c ⇒ c=4. Taking this chord as height, we construct a rectangle inscribed in the resulting circumference with height H=3+4=7 and base B=6-2=4. The center of the constructed rectangle coincides with that of the circumference and its radius R is half the diagonal; its value is obtained by Pythagoras: R²=(B/2)²+(H/2)² = 2²+3.5² = 16.25 ⇒ R=√16.25 ⇒ R=4.0311
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.
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.
It could be done by properties of triangle[ABD] Ex-radius,R=abc/4Δ Here a,b,c are three side lengths and Δ is area of the triangle. So, a=6+2=8 b=√(6²+3²)=√45=3√5 c=√(3²+2²)=√13 Δ=(1/2)(8)(3)=12 Therefore, R =[8×(√13)×(3√5)]/[4×12] =[24√65]/[48] =[√65]/2 =4.031
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.
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
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 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.
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..
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
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.
I got to the diameter being sqrt(65) in my head by inspection of the thumbnail. I thought it worked because of the particular example given but have now worked out it holds for all similar problems showing two perpendicular chords. This was my approach: 1) the upper arm of chord CD is 4 (as cd=ab*, known property of perpendicular chords, in this case the arithmetic is trivial, see footnote). 2) (mentally) strike a mirror-image of CD giving a parallel chord, same length, and 2 units from point B. 3) The central section of chord AB, between chord CD and its mirror image, is 4 units (6-2). 4) The upper part of the mirror image chord is 4 units (symmetry to CD, calculated in 1)). 5) The distance between point C and the point where the top of the mirror-image chord touches the circle is also 4 (opposite side of square). 6) because the chords are perpendicular and parallel, the central top sector is a square (in this case a square but might be rectangle, method still holds). 7) because of 6) DC and the top horizontal chord form 90 degrees where they meet at point C. 8) The two chords in 7), being at 90deg, by definition form the legs of a right-triangle whose hypotenuse is the diameter. 9) Pythagoras' from 8) gives diameter^2 = 7^2 + 4^2 = 65. 10) diameter therefore sqrt(65), radius half that. * I used small case cdab to represent the respective distances between the upper case points CDAB and the chord intersection point. Derivation of formula from above steps: using w, x and y, x to label the chord parts (as defined at 1:10 in the video) the final triangle derived above has legs y+z and x-w. The hypotenuse of that triangle is the diameter so: d^2 = (y+z)^2 + (x-w)^2, which gives: d^2 = x^2 - 2wx + w^2 + y^2 + 2zy + z^2 The next step took me a while to spot but the perpendicular chord rule cd=ab (which translates to wx=yz here) means that -2wx and +2zy cancel each other leaving: d^2 = w^2 + x^2 + y^2 + z^2 Since the diameter is twice the radius, d^2 is (2r)^2, or 4r^2 so: 4r^2 = w^2 + x^2 + y^2 + z^2 (the formula given at 1:31 in the video) Therefore, the method described in steps 1-10 above should work for every instance of this problem.
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 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).
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
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.
Great! Many thanks! Another way solving the problem: w = 2 x = 6 x = 4 (via Chords Theorem) z = 3 N = point of intersection AB and CD = N(0; 0) y - z = 4 - 3 = 1 = 2(1/2) x - w = 6 - 2 = 4 → (1/2)4 = 2 → M = center of the circle = M(2;1/2) D = N(0;0) - (0;3) = D(0;-3) → r^2 = (2 - 0)^2 + (7/2)^2 = 65/4 → r = 2√65 🙂
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!
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.
I used the idea of circumcentre and it’s properties. Use the graph at 4:16 for example. Consider the the Triangle BCD. Let O be the center of the circle and also the triangle, R be the radius we want. We can easily know that CD=7, becuz O is the circumcentre of Triangle BCD, so the perpendicular of CD will pass through O. The distance from the point that bisects CD to the point which AB and CD bisects is 7/2-3=0.5. 0.5 is also the height of Triangle OAB, so now we simply use Pyth Theorem to get R=sqrt(4^2+0.5*2)=sqrt(65)/2
Call the intersection point P. Then triangles APD and BPC are similar and allow us to solve for CP=4. Then inscribe a rectangle in the circle with the bottom horizontal line as AB, and with a height of 1 (the height is chosen as CD - 2*DP = 7 - 2*3 = 1 since an inscribed rectangle must be centered vertically and horizontally in the circle). The diagonals of an inscribed rectangle are diameters of the circle. Since the rectangle is (8,1) in size, we know the diameter of the circle is (1^2 + 8^2)^0.5 by the PYTHAGOREAN THEOREM and thus the radius is sqrt(65)/2
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!
Applying the Intersecting Chords Theorem, the length of the remaining segment of the chord is (2 x 6) / 3 = 4. Join AC. Join BC. Let's say that the angle ABC is x. Then the angle subtended by the chord AC at the center of the circle is 2x. (Recall that the angle subtended by a chord at the center of a circle is twice the angle subtended by the chord at any point of the circle) Let the intersection point of the two chords be P, then from triangle CPB, we have sin x = 2 / sqrt(13). Mark the center of the triangle as O, then from triangle AOC we have 2r sin(2x / 2) = sqrt(2^2 + 4^2); or, 2r * (2 / sqrt(13)) = sqrt(20); or, r = sqrt(65) / 2. PS: Nice question, by the way. It feels good to have solved a geometry question using mainstream geometric theorems and not brute-forcing a system of equations to get the value of r.
I have another easier solution. Mark the intersection as origin, AB as x-axis CD as y-axis. Then we will get A(-2,0), B(6,0) and D(0,-3). According to some theorem in 9th/10th class the perpendicular bisector of cord passes through centre. Join BD. Find equations for perpendicular bisectors of cord BD and AB. Solve for x and y we will get the centre as (2,1/2) now find the distance between centre and B
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!
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
I ran across this problem on Twitter and did it this way, what do you think? Inscribed angles: Angle of a chord from a point on the circumference is half the angle of that chord from the centre So angle of AD from B is half the angle of AD from X (= centre of circle) Length AD (Pythagoras) = sqrt(3^2+2^2) Tan = Opposite / Adjacent Angle AD from B = atan(3/6) Angle AD from X = 2 * atan(3/6) Say Y is the centre of AD, length AY = sqrt(3^2+2^2) / 2 Make a right-angle triangle AX, AY, XY (As going at 90 degrees from the centre of a chord is a diameter, so goes through the centre) Angle AY to centre of circle = 2 * atan(3/6) / 2 = atan(3/6) AX = R = radius Sin = Opposite / Hypotenuse R = (sqrt(3^2+2^2)/2) / sin(atan(3/6)) = 4.0311
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.
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.
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.
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
I solved it with a system of equations. The circumference's equation is x² + y² + ax + by + c = 0, where a,b,c are the parameters that can define every circumference on the plane. The problem gives you the coordinates of three points on the circumference so I solved for the equation of the problem's circumference by solving for a, b, and c. The way to do it is to make a system of three equations of the form x² + y² + ax + by + c = 0, but substituting for x and y the values of every given point. The final equation was x² + y² - 4x + -y + 12 = 0, and the radius of any circle is given by the formula 0.5sqrt( a² + b² -4c ) which leads to 0.5 root 65.
Continue from time 3:50, using the other right triangle, one can get 4r^2 = (x + w)^2 + (y - z)^2 = (w^2 + x^2 + y^2 + z^2) + 2(wx - yz). So, solving the two equations, one can easily get 4r^2=w^2+x^2+y^2+z^2, & wx=yz.
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.
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.
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
1) Calculate remaining length as 4 2) Divide the horizontal line to two by drawing a perpendicular line towards it from the radius 3) Form a right angled triangle by merging the radius with point B. The equation is as follows: r^2 = 4^2 + height^2 4) To find the height, form a rectangle by drawing a perpendicular line towards vertical line from the radius. The vertical line is divided to two. Since its total length is 7, the distribution would be like 3, 0.5 (vertical side of the rectangle, hence the height), 3.5 when you look at the schematic carefully. 5) r^2 = 16 + 0.5^2 = 16.25 6) r = sqrt(16.25)
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 !
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
I can suggest another way that find AD =rt(13) and BD=rt(45) nd just apply R=(AD)(DB)(AB)/4 (areaADB) where area = 1/2 x 8 x 3 =12......(this formula comes from sin law..)....😀
Solved it with only the Pythagorean theorem. Draw center of Circle (C), Draw segment perpendicular intersection to horizontal chord. Call length of the segment "a". This bisects the chord, each half of the chord being 4. Draw a radius from C to intersection of chord with edge of circle. This gives a right triangle with sides, a and 4 and hypotenuse r. 4^2 + a^2 = r^2. Now draw perpendicular bisector from C to vertical chord. Notice that bisected chord length is 3 + a, and bisector is 2. From here you have another triangle with sides, 2 and 3+a, and hypotenuse r. 2^2 + (3+a)^2 = r^2. Solve for a, 1/2. Then plug in for r.
First see the missing length is 4 (because 2*6 = 3*x), then use the formula that the diameter is equal to the square root of all of the lengths squared and added, so √(3^2 + 4^2 + 2^2 + 6^2) = √(9 + 16 + 4 + 36) = √65. This means the radius is √65/2
@Willian Honório, - "Nice, how can I proof this?" Consider the figure at 4:19. Let a+b be the horizontal chord (AB), and c+d be the vertical chord (CD). Copy the vertical chord and shift it horizontally to the right, until it touches the circle circumference at F (at top) and E (at bottom). Then CDEF is a rectangle with diameter 2r, vertical side (c+d) and horizontal side (b-a). Using Pythagoras: (2r)² = (c+d)² + (b-a)² 4r² = c²+d² + 2cd + b² + a² - 2ab However, due to the intersecting chords theorem (which we also used to compute c = ab/d = 2*6/3 = 4), we know that ab = cd, so the terms 2cd and -2ab cancel eachother; and therefore we're left with 4r² = a² + b² + c² + d²
This is an easy task indeed. How do you construct the circle? Locus lines for the center of the circle are the perpendiculars of AD and DB. The coordinates of the 3 points can be read off directly, the straight lines AD and DB can be easily specified, the slopes of the vertical can be determined immediately. The midpoints of the routes remain to be determined (arithmetic mean of the respective coordinates). The point of intersection of the vertical lines results in the center of the circle. The radius is then e.g. the length of the line MB. In detail: AD: y = -1,5x-3 */* Midpoint (-1/-1,5) */* Vertical line y=(2/3)x-(5/6) */* DB: y=0,5x-3 */* Midpoint (3/-1,5) */* Vertical line y=-2x+4,5 */* point of intersection of the vertical lines M(2/0,5) */* radius r^2=MB^2=16,25 */* r=sqrt(16,25)=4,03 */*
The thing about math is that there are almost always multiple ways to solve a problem. I can typically understand the algebra involve in both proving and using a formula. My question is always the same: what is the motivation. That’s what math teachers need to be better at: explaining the why’s. Sometimes it’s good enough to say that you “guess” based on previous knowledge.
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.
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.
I solved it with the theorem of intersecting chords, then I drew a chord, parallel to CD, starting from point A, whitch by symmetry is equal to 1. Then I connect this new drawn 's vertex to B to get a right triangle. By the theorem that 2 chords, intersecting on a point of the perimeter of the circle , making a right angle, form a triangle with a hypotenuse that is equal to the diameter, I get that d=sqrt (1^1+8^2), whitch is d=sqrt (65). Then the radius is d/2=sqrt(65)/2, and that's my answer. Did YOU figure it out?! Thanks for seeing my solution, sub and ya...😂😀
I also solved this way. In fact, after finding y (= 4), you can flip the chords around either vertical or horizontal. The result is the same: r^2 = 8^2 + 1^2 = 7^2 + 4^2 = 65.
How I solved it: (keep in mind I didn't know the formula for finding the radius) Since the two chords are perpendicular, WX = YZ. Plug in the numbers: (2*6) = (Y*3) So Y has to equal 4. This means that the length of that chord will be 7 because 3+4=7. Now for a chord to be a diameter, it has to pass though the center point of the circle. If a chord splits another chord in half and it is perpendicular to that chord, it will pass though the center point. So I did half of 7 which is 3.5. Now, I let’s visualize the new picture. We have a circle with a chord that is split up into 2 segments, each of them equaling 3.5. We also have the chord that is bisecting this chord, with unknown lengths. Let’s label these unknown lengths A and B. This means that 3.5*3.5 = AB. This simplifies to 12.25 = AB. Now, we need to find what the values of A and B are. Since we started out with 6 and 2, and the the chord that we started out with and the chord that we have now are parallel, the 6 and the 2 would’ve had to have increased by the same number. Let’s call this number C. Therefore, A = 2+C and B = 6+C. I also rewrote the equation from the last paragraph: 12.25 = (c+2)(c+6). Simplify this equation and you get 12.25 = c^2 + 8C + 12. Now simplify it further (with quadratic formula or whatever) and you get -4 plus or minus the square root of 16.25. Since all chords are shorter or the same length as diameters, C will have to be positive. So you can reject -4 - the square root of 16.25 = C And only write -4 + the square root of 16.25 = C Now time to find the length of the chord segments. We wrote earlier that A = 2 + C. Plug in the numbers: A = 2 + -4 + the square root of 16.25 And you get A = -2 + the square root of 16.25. We also wrote earlier that B = 6 + C. Plug in the numbers again: B = 6 + -4 + the square root of 16.25 And you get B = 2 + the square root of 16.25. Now time to find the diameter. You do -2 + the square root of 16.25 + 2 + the square root of 16.25 And you get 2 times the square root of 16.25. But the problem asked for the radius. So you take half of 2 times the square root of 16.25 And you get the square root of 16.25 as your final answer. All this and I could’ve just solved this with one formula
A practical application for finding the center and edges of circles is CNC machining. There are radius probes that you can use, but I found they are clumsy and less accurate than just tapping off any 4 points on the edge and calculating the center and radius.
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!
Create a triangle using the length 8 chord and length 3 segment, then use the formula to find the circumcircle radius of a triangle abc/4[ABC]. took me way too long to figure out that I could do that though
Nice problem and nice methods ;-) But probably an even simpler solution approach is just applying the law of sines: a / sin(A) = b / sin(B) = c / sin(C) = 2*R = the circle's diameter! The drawn circle is simply the circumcenter of a triangle with base 2+6 and height 3. Hence, the sides of the triangle have lengths c = 8, b = sqrt(13) and a = sqrt(45), and the upper left angle reads sin(A) = sqrt(9/13) and the upper right angle reads sin(B) = sqrt(9/45). So in two different ways we arrive at 2*R = a / sin(A) = b / sin(B) = sqrt(13*45/9) = sqrt(65), so the radius R = 0.5*sqrt(65). No need to bother with coordinate systems anymore... :-)
I've just seen this problem. My solution: (a) compute the length of the upper part of CD is 4 by similar triangles (or the power law as mentioned). (b) Imagine the vertical diameter through the centre. The circle is symmetrical about this line. Draw the vertical line EF parallel to CD reflected in the vertical diameter. By symmetry it will intersect AB at a distance of 2 units from B, and thus will be 4 units along AB from CD. (c) Join CE, length 4, and consider the right-angled triangle CDE. DE will pass through the centre of the circle and thus be a diameter. Its length will be sqrt(CD^2+CE^2) = sqrt(65). Thus the radius is sqrt(65)/2.
@@ayeitsjoe8221 Are you sure? I can remember an article where a school teacher sent a student to detention because the student insisted the teacher was wrong. And why did the student insisted? Because the teacher came up with the idea/appraoch pi = 3 so calculation would be easier. So in the US everything is possible.
That would only be true if AB was a diameter though. But if AB was a diameter, then, due to the circle theorem of Thales, AB would be the hypotenuse of ADB and then AD^2+BD^2 would be equal to AB^2 the latter of which is 64. But that it isn't, because, also calculated via the theorem of Pythagoras: AD^2=3^2+2^2=9+4=13 and BD^2=3^2+6^2=9+36=45 and hence the sum of these two squares is 58 which clearly isn't 64. iow: the center of the circle lies not on the chord AB.
I have done similar to the 1st method. It goes by mirroring the image vertically and making a rectange and a triangle. I will try to explain it but make a sketch of the steps to see it more clearly. 1) Suppose the unknown section of the vertical line length is 3+x. If you flip vertically the image, the vertical sections will be swapped as well. It means the horizontal line has moved upwards by x. 2) Overlapping both images, you can see the normal image horizontal line and the mirrored image one. They are just parallel sections with a vertical distance of x. You can make a rectangle with the horizontal line as a base and the distance x as the height. By symmetry, the diagonal of the rectange is the diameter, so using Pitagoras theorem: (2+6)^2+x^2=(2 r)^2 3) Again, by symmetry, if you move now the horizontal line upwards by x/2 so both vertical segments will now have a length 3+x/2, it will be in the radial horizontal axis. We know the distance of the center of the horizontal line to the vertical line is just 2. now if we trace a segment from the center of the circle to the point D, it will be a radial one, so again, using Pitagoras: (3+x/2)^2+2^2=r^2. 4) Doing some algebra with both 2nd degree equations and taking the positive solution, we get r=sqrt(65)/2 and x=1 so the unknown segment length is 3+1=4.
I started by "mirroring" AB and CD, creating two rectangles. The diagonals create two right triangles ABB' (sides 8, x, 2r) and CDD' (sides x+6, 4, 2r). Using Pythagorean theorem I first solved the value of x and then the r.
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.. 😅
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
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.
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!
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.
(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.
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 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.
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
By calculating the power of the intersection point of the proposed chords with respect to the circumference, we obtain the length of the upper section of the vertical chord: 2x6=3c ⇒ c=4.
Taking this chord as height, we construct a rectangle inscribed in the resulting circumference with height H=3+4=7 and base B=6-2=4.
The center of the constructed rectangle coincides with that of the circumference and its radius R is half the diagonal; its value is obtained by Pythagoras:
R²=(B/2)²+(H/2)² = 2²+3.5² = 16.25 ⇒ R=√16.25 ⇒ R=4.0311
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.
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.
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
It could be done by properties of triangle[ABD]
Ex-radius,R=abc/4Δ
Here a,b,c are three side lengths and Δ is area of the triangle. So,
a=6+2=8
b=√(6²+3²)=√45=3√5
c=√(3²+2²)=√13
Δ=(1/2)(8)(3)=12
Therefore, R
=[8×(√13)×(3√5)]/[4×12]
=[24√65]/[48]
=[√65]/2
=4.031
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.
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
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
Nobody:
Problems in the 1970s: *handed out to unborn foetuses in Asia*
Peppermint Cookie 😂😂😂🤣🤣🤣👍👍✌🏼👍👍👍👍👍👍👍👍😂😂😂
foetuses lmao
Math should be esteemed.
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)
Very easy .I am 10 standard student and got right answer using simple chord theorems.
Challenger Approaching!
Presh Talwalker divides the competition!
Pomm popopoooomm pOpopOpoooOmm!!!
😂👌
You pick the longest and most obscure way to solve problems
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
Love the way you challenge the brain with these mind boggling problems. You are doing great work and I look forward to being challenged.
Thanks, now I know the radius of the Super Smash Bros logo
The co-ordinate geometry solution is beautiful
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 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. 😀😀
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.
I got to the diameter being sqrt(65) in my head by inspection of the thumbnail. I thought it worked because of the particular example given but have now worked out it holds for all similar problems showing two perpendicular chords. This was my approach:
1) the upper arm of chord CD is 4 (as cd=ab*, known property of perpendicular chords, in this case the arithmetic is trivial, see footnote).
2) (mentally) strike a mirror-image of CD giving a parallel chord, same length, and 2 units from point B.
3) The central section of chord AB, between chord CD and its mirror image, is 4 units (6-2).
4) The upper part of the mirror image chord is 4 units (symmetry to CD, calculated in 1)).
5) The distance between point C and the point where the top of the mirror-image chord touches the circle is also 4 (opposite side of square).
6) because the chords are perpendicular and parallel, the central top sector is a square (in this case a square but might be rectangle, method still holds).
7) because of 6) DC and the top horizontal chord form 90 degrees where they meet at point C.
8) The two chords in 7), being at 90deg, by definition form the legs of a right-triangle whose hypotenuse is the diameter.
9) Pythagoras' from 8) gives diameter^2 = 7^2 + 4^2 = 65.
10) diameter therefore sqrt(65), radius half that.
* I used small case cdab to represent the respective distances between the upper case points CDAB and the chord intersection point.
Derivation of formula from above steps:
using w, x and y, x to label the chord parts (as defined at 1:10 in the video) the final triangle derived above has legs y+z and x-w. The hypotenuse of that triangle is the diameter so:
d^2 = (y+z)^2 + (x-w)^2, which gives:
d^2 = x^2 - 2wx + w^2 + y^2 + 2zy + z^2
The next step took me a while to spot but the perpendicular chord rule cd=ab (which translates to wx=yz here) means that -2wx and +2zy cancel each other leaving:
d^2 = w^2 + x^2 + y^2 + z^2
Since the diameter is twice the radius, d^2 is (2r)^2, or 4r^2 so:
4r^2 = w^2 + x^2 + y^2 + z^2 (the formula given at 1:31 in the video)
Therefore, the method described in steps 1-10 above should work for every instance of this problem.
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.
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
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!
3:48 why is it w²+x² ??
Shouldn't there be a -2wx
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 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
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.
Use pythag. To find AD (=b) and BD (=a). Use a/sinA = b/sinB = 2R.
a/sinA = (3,606 x 6,708)/3 = 8,06 = 2R
R = 8,06/2 = 4,03
Great! Many thanks!
Another way solving the problem:
w = 2
x = 6
x = 4 (via Chords Theorem)
z = 3
N = point of intersection AB and CD = N(0; 0)
y - z = 4 - 3 = 1 = 2(1/2)
x - w = 6 - 2 = 4 → (1/2)4 = 2 → M = center of the circle = M(2;1/2)
D = N(0;0) - (0;3) = D(0;-3) → r^2 = (2 - 0)^2 + (7/2)^2 = 65/4 → r = 2√65 🙂
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!
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.
I did it with inverse trig functions and used the fact that arc ACB is equal to 2(arctan(2/3)+arctan(2))
I used the idea of circumcentre and it’s properties. Use the graph at 4:16 for example. Consider the the Triangle BCD. Let O be the center of the circle and also the triangle, R be the radius we want. We can easily know that CD=7, becuz O is the circumcentre of Triangle BCD, so the perpendicular of CD will pass through O. The distance from the point that bisects CD to the point which AB and CD bisects is 7/2-3=0.5. 0.5 is also the height of Triangle OAB, so now we simply use Pyth Theorem to get R=sqrt(4^2+0.5*2)=sqrt(65)/2
Call the intersection point P. Then triangles APD and BPC are similar and allow us to solve for CP=4. Then inscribe a rectangle in the circle with the bottom horizontal line as AB, and with a height of 1 (the height is chosen as CD - 2*DP = 7 - 2*3 = 1 since an inscribed rectangle must be centered vertically and horizontally in the circle). The diagonals of an inscribed rectangle are diameters of the circle. Since the rectangle is (8,1) in size, we know the diameter of the circle is (1^2 + 8^2)^0.5 by the PYTHAGOREAN THEOREM and thus the radius is sqrt(65)/2
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 click this on my recommendations.
Now i feel smart.
Yep
Nobody
Literally nobody:
Le me after watching this video: mothemoticon
Applying the Intersecting Chords Theorem, the length of the remaining segment of the chord is (2 x 6) / 3 = 4.
Join AC. Join BC. Let's say that the angle ABC is x. Then the angle subtended by the chord AC at the center of the circle is 2x. (Recall that the angle subtended by a chord at the center of a circle is twice the angle subtended by the chord at any point of the circle)
Let the intersection point of the two chords be P, then from triangle CPB, we have sin x = 2 / sqrt(13).
Mark the center of the triangle as O, then from triangle AOC we have 2r sin(2x / 2) = sqrt(2^2 + 4^2); or, 2r * (2 / sqrt(13)) = sqrt(20); or, r = sqrt(65) / 2.
PS: Nice question, by the way. It feels good to have solved a geometry question using mainstream geometric theorems and not brute-forcing a system of equations to get the value of r.
I have another easier solution.
Mark the intersection as origin, AB as x-axis CD as y-axis. Then we will get A(-2,0), B(6,0) and D(0,-3). According to some theorem in 9th/10th class the perpendicular bisector of cord passes through centre. Join BD. Find equations for perpendicular bisectors of cord BD and AB. Solve for x and y we will get the centre as (2,1/2) now find the distance between centre and B
4:21
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!
00:36
OA×OB=DO×OC
* O is the point of intersection
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
Okay TH-cam recommendations. I guess it’s big brain time.
This is one of the most liked channels by me.
I ran across this problem on Twitter and did it this way, what do you think?
Inscribed angles:
Angle of a chord from a point on the circumference is half the angle of that chord from the centre
So angle of AD from B is half the angle of AD from X (= centre of circle)
Length AD (Pythagoras) = sqrt(3^2+2^2)
Tan = Opposite / Adjacent
Angle AD from B = atan(3/6)
Angle AD from X = 2 * atan(3/6)
Say Y is the centre of AD, length AY = sqrt(3^2+2^2) / 2
Make a right-angle triangle AX, AY, XY
(As going at 90 degrees from the centre of a chord is a diameter, so goes through the centre)
Angle AY to centre of circle = 2 * atan(3/6) / 2 = atan(3/6)
AX = R = radius
Sin = Opposite / Hypotenuse
R = (sqrt(3^2+2^2)/2) / sin(atan(3/6)) = 4.0311
"These math videos avaliable for free on youtube, builds confidence for students"... wish I could say the same....
the feeling when you solved it with the easiest way possible, but you can't explain it due to language barrier
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.
تمرين جميل جيد . رسم واضح مرتب. شرح واصح جيد . شكرا جزيلا لكم والله يحفظكم ويحميكم ويجميعا . تحياتنا لكم من غزة غلسطين .
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.
Alternate title: "finding the radius of Australian super smash Bros logo!"
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.
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
I solved it with a system of equations. The circumference's equation is x² + y² + ax + by + c = 0, where a,b,c are the parameters that can define every circumference on the plane. The problem gives you the coordinates of three points on the circumference so I solved for the equation of the problem's circumference by solving for a, b, and c. The way to do it is to make a system of three equations of the form x² + y² + ax + by + c = 0, but substituting for x and y the values of every given point. The final equation was x² + y² - 4x + -y + 12 = 0, and the radius of any circle is given by the formula 0.5sqrt( a² + b² -4c ) which leads to 0.5 root 65.
Continue from time 3:50, using the other right triangle, one can get 4r^2 = (x + w)^2 + (y - z)^2 = (w^2 + x^2 + y^2 + z^2) + 2(wx - yz). So, solving the two equations, one can easily get 4r^2=w^2+x^2+y^2+z^2, & wx=yz.
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.
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
Normal people:math
Me, and intellectual: waluigi for smash
Intellectual mispelling *an.
Heh
1) Calculate remaining length as 4
2) Divide the horizontal line to two by drawing a perpendicular line towards it from the radius
3) Form a right angled triangle by merging the radius with point B. The equation is as follows: r^2 = 4^2 + height^2
4) To find the height, form a rectangle by drawing a perpendicular line towards vertical line from the radius. The vertical line is divided to two. Since its total length is 7, the distribution would be like 3, 0.5 (vertical side of the rectangle, hence the height), 3.5 when you look at the schematic carefully.
5) r^2 = 16 + 0.5^2 = 16.25
6) r = sqrt(16.25)
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 !
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
Now I can solve the radius of the Super Smash Bros logo
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
I can suggest another way that find AD =rt(13) and BD=rt(45) nd just apply R=(AD)(DB)(AB)/4 (areaADB) where area = 1/2 x 8 x 3 =12......(this formula comes from sin law..)....😀
Solved it with only the Pythagorean theorem.
Draw center of Circle (C), Draw segment perpendicular intersection to horizontal chord. Call length of the segment "a". This bisects the chord, each half of the chord being 4. Draw a radius from C to intersection of chord with edge of circle. This gives a right triangle with sides, a and 4 and hypotenuse r. 4^2 + a^2 = r^2.
Now draw perpendicular bisector from C to vertical chord. Notice that bisected chord length is 3 + a, and bisector is 2. From here you have another triangle with sides, 2 and 3+a, and hypotenuse r. 2^2 + (3+a)^2 = r^2. Solve for a, 1/2. Then plug in for r.
The new Smash DLC Fighter looks amazing!
First see the missing length is 4 (because 2*6 = 3*x), then use the formula that the diameter is equal to the square root of all of the lengths squared and added, so √(3^2 + 4^2 + 2^2 + 6^2) = √(9 + 16 + 4 + 36) = √65.
This means the radius is √65/2
slick. What's the name of that theorem?
pipertripp check the video
@@pipertripp idk, sorry. The place I learned it from never gave the name
@@Gandarf_ thx mate. Great little proof of it too.
I was able to do it, there is theorem that says
a^2+b^2+c^2+d^2=4R^2
Where a, b, c and d are chords
Yep, this is called the Presh Talwakar Theorem :)
Nice, how can I proof this?
@@w.jordan1736 I have a book, I wish I could send a picture
Switch two adjacent chords to prove instantly.
@Willian Honório,
- "Nice, how can I proof this?"
Consider the figure at 4:19. Let a+b be the horizontal chord (AB), and c+d be the vertical chord (CD). Copy the vertical chord and shift it horizontally to the right, until it touches the circle circumference at F (at top) and E (at bottom). Then CDEF is a rectangle with diameter 2r, vertical side (c+d) and horizontal side (b-a). Using Pythagoras:
(2r)² = (c+d)² + (b-a)²
4r² = c²+d² + 2cd + b² + a² - 2ab
However, due to the intersecting chords theorem (which we also used to compute c = ab/d = 2*6/3 = 4), we know that ab = cd, so the terms 2cd and -2ab cancel eachother; and therefore we're left with
4r² = a² + b² + c² + d²
This is an easy task indeed. How do you construct the circle? Locus lines for the center of the circle are the perpendiculars of AD and DB. The coordinates of the 3 points can be read off directly, the straight lines AD and DB can be easily specified, the slopes of the vertical can be determined immediately. The midpoints of the routes remain to be determined (arithmetic mean of the respective coordinates). The point of intersection of the vertical lines results in the center of the circle. The radius is then e.g. the length of the line MB. In detail:
AD: y = -1,5x-3 */* Midpoint (-1/-1,5) */* Vertical line y=(2/3)x-(5/6) */* DB: y=0,5x-3 */* Midpoint (3/-1,5) */* Vertical line y=-2x+4,5 */* point of intersection of the vertical lines M(2/0,5) */* radius r^2=MB^2=16,25 */* r=sqrt(16,25)=4,03 */*
The thing about math is that there are almost always multiple ways to solve a problem. I can typically understand the algebra involve in both proving and using a formula. My question is always the same: what is the motivation. That’s what math teachers need to be better at: explaining the why’s. Sometimes it’s good enough to say that you “guess” based on previous knowledge.
2R =BD/ ( sin(DAB)
Sin DAB = 3/( ROOT 13)
=> R = (root 65 )/2
excuse me, how
FratarGaspetra
Law of sines.
You can also use cosine rule and the fact that the angle at the circumference is half the angle at the center.
ngu thang I solved it this way as well
that feeling when I wrong on ≈0.031
6 + 2 /4?
Presh Talwalker smash reveal?
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.
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.
I used an extended version of The sine theorem a/sinA=abc/area=2R
I solved it with the theorem of intersecting chords, then I drew a chord, parallel to CD, starting from point A, whitch by symmetry is equal to 1. Then I connect this new drawn 's vertex to B to get a right triangle. By the theorem that 2 chords, intersecting on a point of the perimeter of the circle , making a right angle, form a triangle with a hypotenuse that is equal to the diameter, I get that d=sqrt (1^1+8^2), whitch is d=sqrt (65). Then the radius is d/2=sqrt(65)/2, and that's my answer. Did YOU figure it out?! Thanks for seeing my solution, sub and ya...😂😀
I also solved this way. In fact, after finding y (= 4), you can flip the chords around either vertical or horizontal. The result is the same: r^2 = 8^2 + 1^2 = 7^2 + 4^2 = 65.
How I solved it: (keep in mind I didn't know the formula for finding the radius)
Since the two chords are perpendicular, WX = YZ.
Plug in the numbers: (2*6) = (Y*3)
So Y has to equal 4.
This means that the length of that chord will be 7 because 3+4=7.
Now for a chord to be a diameter, it has to pass though the center point of the circle.
If a chord splits another chord in half and it is perpendicular to that chord, it will pass though the center point.
So I did half of 7 which is 3.5.
Now, I let’s visualize the new picture.
We have a circle with a chord that is split up into 2 segments, each of them equaling 3.5.
We also have the chord that is bisecting this chord, with unknown lengths.
Let’s label these unknown lengths A and B.
This means that 3.5*3.5 = AB.
This simplifies to 12.25 = AB.
Now, we need to find what the values of A and B are.
Since we started out with 6 and 2, and the the chord that we started out with and the chord that we have now are parallel, the 6 and the 2 would’ve had to have increased by the same number. Let’s call this number C.
Therefore, A = 2+C and B = 6+C.
I also rewrote the equation from the last paragraph: 12.25 = (c+2)(c+6).
Simplify this equation and you get 12.25 = c^2 + 8C + 12.
Now simplify it further (with quadratic formula or whatever) and you get -4 plus or minus the square root of 16.25.
Since all chords are shorter or the same length as diameters, C will have to be positive.
So you can reject -4 - the square root of 16.25 = C
And only write -4 + the square root of 16.25 = C
Now time to find the length of the chord segments.
We wrote earlier that A = 2 + C.
Plug in the numbers: A = 2 + -4 + the square root of 16.25
And you get A = -2 + the square root of 16.25.
We also wrote earlier that B = 6 + C.
Plug in the numbers again: B = 6 + -4 + the square root of 16.25
And you get B = 2 + the square root of 16.25.
Now time to find the diameter.
You do -2 + the square root of 16.25 + 2 + the square root of 16.25
And you get 2 times the square root of 16.25.
But the problem asked for the radius.
So you take half of 2 times the square root of 16.25
And you get the square root of 16.25 as your final answer.
All this and I could’ve just solved this with one formula
A practical application for finding the center and edges of circles is CNC machining. There are radius probes that you can use, but I found they are clumsy and less accurate than just tapping off any 4 points on the edge and calculating the center and radius.
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!
Create a triangle using the length 8 chord and length 3 segment, then use the formula to find the circumcircle radius of a triangle abc/4[ABC]. took me way too long to figure out that I could do that though
I clicked on this because the circle in the thumbnail looked like the Smash Ball.
I love your videos Press.
Would you mind to tell me what software you use to make your videos?
Thanks
Nice problem and nice methods ;-) But probably an even simpler solution approach is just applying the law of sines: a / sin(A) = b / sin(B) = c / sin(C) = 2*R = the circle's diameter! The drawn circle is simply the circumcenter of a triangle with base 2+6 and height 3. Hence, the sides of the triangle have lengths c = 8, b = sqrt(13) and a = sqrt(45), and the upper left angle reads sin(A) = sqrt(9/13) and the upper right angle reads sin(B) = sqrt(9/45). So in two different ways we arrive at 2*R = a / sin(A) = b / sin(B) = sqrt(13*45/9) = sqrt(65), so the radius R = 0.5*sqrt(65). No need to bother with coordinate systems anymore... :-)
I've just seen this problem. My solution: (a) compute the length of the upper part of CD is 4 by similar triangles (or the power law as mentioned). (b) Imagine the vertical diameter through the centre. The circle is symmetrical about this line. Draw the vertical line EF parallel to CD reflected in the vertical diameter. By symmetry it will intersect AB at a distance of 2 units from B, and thus will be 4 units along AB from CD. (c) Join CE, length 4, and consider the right-angled triangle CDE. DE will pass through the centre of the circle and thus be a diameter. Its length will be sqrt(CD^2+CE^2) = sqrt(65). Thus the radius is sqrt(65)/2.
3:55 (y + z) ^2 is not y^2 + z^2
Yeah it isn't,neither did he assume that,what's your point?
me: no, this can't be this easy
my brain: but, it is
me: then, how you solve that?
my brain: 6+2=8 8/2= 4, the radius is 4
But the line doesn’t cut directly through the center of the circle, so that won’t be accurate
@@ayeitsjoe8221 r/whoooooooshhhhh
@@ayeitsjoe8221 Are you sure? I can remember an article where a school teacher sent a student to detention because the student insisted the teacher was wrong. And why did the student insisted? Because the teacher came up with the idea/appraoch pi = 3 so calculation would be easier. So in the US everything is possible.
Lol I did the same. 4 was basically correct though
That would only be true if AB was a diameter though. But if AB was a diameter, then, due to the circle theorem of Thales, AB would be the hypotenuse of ADB and then AD^2+BD^2 would be equal to AB^2 the latter of which is 64.
But that it isn't, because, also calculated via the theorem of Pythagoras: AD^2=3^2+2^2=9+4=13 and BD^2=3^2+6^2=9+36=45 and hence the sum of these two squares is 58 which clearly isn't 64.
iow: the center of the circle lies not on the chord AB.
4.031 is rounded to 4 so I’m right and I done it all in my head, yayy!
I have done similar to the 1st method. It goes by mirroring the image vertically and making a rectange and a triangle. I will try to explain it but make a sketch of the steps to see it more clearly.
1) Suppose the unknown section of the vertical line length is 3+x. If you flip vertically the image, the vertical sections will be swapped as well. It means the horizontal line has moved upwards by x.
2) Overlapping both images, you can see the normal image horizontal line and the mirrored image one. They are just parallel sections with a vertical distance of x. You can make a rectangle with the horizontal line as a base and the distance x as the height. By symmetry, the diagonal of the rectange is the diameter, so using Pitagoras theorem: (2+6)^2+x^2=(2 r)^2
3) Again, by symmetry, if you move now the horizontal line upwards by x/2 so both vertical segments will now have a length 3+x/2, it will be in the radial horizontal axis. We know the distance of the center of the horizontal line to the vertical line is just 2. now if we trace a segment from the center of the circle to the point D, it will be a radial one, so again, using Pitagoras: (3+x/2)^2+2^2=r^2.
4) Doing some algebra with both 2nd degree equations and taking the positive solution, we get r=sqrt(65)/2 and x=1 so the unknown segment length is 3+1=4.
I started by "mirroring" AB and CD, creating two rectangles. The diagonals create two right triangles ABB' (sides 8, x, 2r) and CDD' (sides x+6, 4, 2r). Using Pythagorean theorem I first solved the value of x and then the r.