Learn three different methods to find the radius of a circle if given 2 perpendicular lines. Utilize coordinate geometry, the pythagorean theorem, and the chords theorem. Step-by-step tutorial by Premath.com
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@242math3 жыл бұрын
The coordinate geometry method was very tedious. All of them tax your geometrical and algebraic skills. Your demonstration on solving this question using three methods is detailed and knowledgeable. Your steps are easy to follow and comprehend, excellent presentation.
@tnix803 жыл бұрын
Thank God for Pythagoras 3rd method is great though
@tnix803 жыл бұрын
@Shane Jericho why would you bother being with someone you don't trust? Better, why am I replying to spam?
@mathsinmarathibyanillimaye30833 жыл бұрын
Here value of h need not be calculated. Mere observations..seg OC IS PARALLEL to y axis. Dan is on
@larsjensen79583 жыл бұрын
The easy way is the distance ((((AB^2)*0.25)+(DC)^2))/(2DC)=r
@FRODOGOOFBALL2 жыл бұрын
I used a coordinate system with O as the origin. By using the cord theorem, it simplifies to a single variable problem, and can be solved much more quickly. Of course, this only worked because C is the center point of arc AB.
@rangaswamyks82872 жыл бұрын
Method seems to be very lengthy Just r-1^2+2^2=r^2 So r=2.5 We can calculate in mind sir But very interesting thank you sir.. God bless you sir
@channelsixtysix0662 жыл бұрын
Your videos are exciting and I've enjoyed every one I've watched. 👍
@timemerick7333 жыл бұрын
That was great. I enjoy seeing real application rather than just formula solving methods. Also I like seeing how different methods come up with the same answer.
@NASIR58able Жыл бұрын
Excellent Analysis Sir. By solving in 3 different ways.
@flavrt Жыл бұрын
Well done, Sir. Also useful as proofs for each theorem.
@user-oj3jc6py2n5 ай бұрын
I could understand the 3rd equation ok and I can apply it and use it. I'm using this to design the top of a camper. And it worked. Thanks a million Sir.
@jonathancapps11032 жыл бұрын
I never knew that third method. Thx!
@SuperBrainStorms Жыл бұрын
Thank you for the excellent video 😊
@shadrana12 жыл бұрын
You can generalise this problem by adopting Pythagoras Theorem. Extend CD through O to intersect the major sector of the circumference AB at F. Let AD=a,DB=b,CD=c and DF=d The radius of a circle always lies on the perpendicular bisector of a chord; DB=(a+b)/2 CO=(c+d)/2 DO=CO-CD=(c+d)/2-c=(c+d)/2-2c/2=(d-c)/2 OB = r say, Consider triangle DBO and apply the Theorem of Pythagoras to it; OB^2=OD^2+DB^2 r^2 =((d-c)/2)^2+((a+b)/2)^2 =(c^2+d^2-2dc)/4+(a^2+b^2+2ab)/4 4r^2 =c^2+d^2-2dc+a^2+b^2+2ab According to the Intersecting Chord Theorem ab=dc Therefore -2dc and 2ab vanish, Hence, 4r^2=a^2+b^2+c^2+d^2 This is a formula for the radius of a circle when two chords intersect at right angles to each other. I adapted this from a similar problem in 'Mind your Decisions' by Presh Talwalkar. This is a good place to stop and thanks for the problem and your solution.You are very clear in your solutions.
@ajaykumargupta43672 жыл бұрын
Well done professor.
@tintinfan0072 жыл бұрын
Superb! I just loved it.... Much better than gec
@murdock55372 жыл бұрын
Thank you very much - very interesting and well explained. Another method: Draw BC. tan alpha (DBC) = (1/2). CB = 5SR. Draw a line from origin O to the middle of BC (new point E, building two identical rectangle triangle OCE and OEB): CE = BE = (1/2)5SR. Because angle DCB = beta = 90 - alpha, angle COE = alpha (angle EOB is also alpha). tan alpha = (1/2) = CE/EO = ((1/2)5SR/a). Therefore a = 5SR. Do the math with Pythagorean theorem ((1/2)5SR) square + (5SR) square = 5 + (5/4) = (25/4) = r square. r = (5/2). Nice! Another way solving the problem (fast lane): 4r^2 = 2^2 + 2^2 + 1^2 + 4^2 = 25 → r = √(25/4) = 5/2 🙂
@MyTutoringBee2 жыл бұрын
Thank you! I loved seeing the 3 different methods. You explanation of each was very clear and easy to follow!
@robertberg1609 Жыл бұрын
Nice and clear solutions as always. I did like this: Look at triangle CDB. Pythagoras gives CB = sqrt(5). Now the triangle CBE is also right triangle due to Thales theorem. Those triangles can easily be proven to be congruent ( using sum of angles in a triangle). The long side is sqrt(5) times bigger then the short one. And CB is the long side in the small triangle and the short side in the big triangle thus; CE = sqrt(5)*CB = sqrt(5)*sqrt(5) = 5. This is the diameter so R = 5/2
@mozeenkhan80744 ай бұрын
Triangle can be proved similar by AA Similarity not congruent, just reply for the better understanding of others who read your solution, by the your approach is also good.
@williamwingo47402 жыл бұрын
Without peeking: Draw AO and OD to form right triangle AOD. AO is the radius r; OD is r - 1; and AD = 2. Then by the Pythagorean theorem: r^2 = 2^2 + (r - 1)^2 = 4 + r^2 - 2r + 1; subtract r^2 from both sides and collect terms to get 0 = 4 - 2r + 1 = 5 - 2r; add 2r to both sides to get 2r = 5; and finally, divide by 2 to get r = 5/2. Would have been quicker, but at first I spent a couple of minutes trying to use the difference-of-squares rule. Thank you, ladies and gentlemen; I'll be here all week. 😎
@maamjay89723 жыл бұрын
Thank you sir for sharing your knowledge..It refreshed my mind..Godbless
@prabirbhowmick87882 жыл бұрын
I am 55 years old. Enjoying your maths classes.
@satishbararia88603 жыл бұрын
Made it quite easy....👌
@shreyanshpatel97402 жыл бұрын
easiest method of all time!! join OC {since perpendicular to chord from radius bisec the chord} let OD=X OA=X+1 triangle ODA right angled (x+1)²=4+x² 2x=3 x=1.5 radius= 1.5+1 2.5
@theoyanto Жыл бұрын
Wow, there's a lot involved with the first method, however as always extremely interesting stuff.
@hansschotterradler37722 жыл бұрын
Very interesting. I'm a bridge engineer and a few years ago I have designed an arch bridge with a circular arch profile and a rise to span ratio of 1:4, similar to the arc segment ACB in this problem. For that 1:4 ratio, the radius ends up a nice even number as shown in the solution because triangles ADO and BDO turn out to be 3-4-5 triangles.
@q.e.d.91122 жыл бұрын
Sorry, Hans, but I think you’re wrong there. Both those triangles have shorter sides in a 1:2 ratio making the hypotenuse a factor of √5 no matter how you scale it. Definitely not a 3,4,5 triangle unless your rise/span ratio was 3/8.
@hansschotterradler37722 жыл бұрын
@@q.e.d.9112 OK, based on the solution OD = 1.5, AD = 2 and DO is 2.5. that makes it a 3-4-5 triangle.
@huwpickrell12092 жыл бұрын
What about the angles in a semicircle are 90 method too. Just copy the top cord and reflect it at the bottom. To create a rectangle. Sides of 4 and 2r-1. Joining the opposite corners would be the diameter since we have a right angle subsensed. Solve for r
@BubuMarimba2 жыл бұрын
In triangle ODB (R-1)²+2²=R² ; 2R=5 ; R=2.5
@AftabAli-op4sn Жыл бұрын
Simplest and quick method. 👍👍👍
@AnonimityAssured Жыл бұрын
My preferred method would definitely be the third of these, although the second method is also elegant and reasonably simple. I don't think I would even consider the first method, as it is too drawn out and elaborate, with numerous opportunities for possible slip-ups.
@Su4ji Жыл бұрын
Terimakasih soal matematikanya, bisa untuk latihan🙏
@patrickjacquiot90733 жыл бұрын
Very easy. First Pythagore √(2²+1²) twice. Then: Rcc=abc⁄√(2(a^2 b^2+b^2 c^2+c^2 a^2 )-(a^4+b^4+c^4)). Result : 2.5
@ryan3702 жыл бұрын
Cord theorom. 2*2=1*x. X=4. The diameter=1+4=5. R=2.5. This is the first one I was able to do instantly in my head
@huwpickrell12092 жыл бұрын
Me too
@govindashit65243 жыл бұрын
I love 2nd & 3rd method . 1st is too difficult. Thanks for the Video.
@Imran-tc6sn3 ай бұрын
Thankyou sir
@ronellmonieno43532 жыл бұрын
Imagine he is your math professor. So calm voice. Hope your students are not sleepy if they really like math.
@limfilms10892 жыл бұрын
Thank you. All methods are very interesting. I figured another one using the ratio of the sides of similar triangles. Hope this is correct :) 1. Draw OB=r 2. Draw CB 3. Triangle OCB is isosceles 4. CB is the Hypotenuse of right triangle CDB 5. CB^2 = 1^2+2^2=5, CB= ν5 (Pythagorean Theorem) 6. Draw OE altitude of the isosceles triangle OCB, it bisects CB at a right angle, thus CE=ν5/2 7. Right triangles CDB and EOC are similar (because each has one angle 90 and angle OCB is common in both triangles, therefore angle COE=CDB). 8. Take the ratio of the sides of the two triangles: OC/CE=CB/CD 9. Thus: r/ν5/2=ν5/1, 2r/ν5= ν5/1, 2r=5, r=2.5
@quattrocchialessandro47522 жыл бұрын
I solved it instantly First I considered the triangle ABC: it is a triangle inscribed in the circumference with radius r. there is a formula that links the inscribed triangle to the radius of the circumscribed circumference: r=abc/4A. The product of all sides, divided by 4 times the area of the triangle is equal to the radius of the circumscribed circumference. AC=BC=√5 (Pythagorean theorem) r=(√5×√5×4)/4×½×4×1= 20/8= 2,5
@tommar74233 жыл бұрын
Excellent El primer método fijando el centro en el origen Y tomando solo el punto B Es otra posibilidad La ecuación resulta semejante al segundo método
@pkumar-rd7py2 жыл бұрын
Can you provide us more questions like this. I want pdf of these questions
@boguslawszostak17842 жыл бұрын
Another way of solution 1 Let's look at the drawing and the designations of method one. Let us assume A (-2, 0), B (2, 0), C (0, 1). The center of the circle lies at the intersection of the Perpendicular bisectors of sides. The Perpendicular bisector of side AB is a line with the equation x = 0 The Perpendicular bisector of BC passes through the point P((2+0)/2 , (0+1)/2 so P (1, 0.5) and is perpendicular to BC, the vector BC has the coordinates [0-2, 1-0] = [-2, 1] The line perpendicular to the vector [-2, 1] passing through the point P (1, 0.5) has the equation (x-1) * (- 2) + (y-0.5) * 1 = 0. This line intersects the x axis at the point of which the y coordinate satisfies the equation (x-1) * (- 2) + (y-0.5) * 1 = 0; x = 0 2+ (y-0.5) = 0 => y = -1.5 so the center of the circle is O (0, -1.5) The circle radius is equal to the segment OC = 1 - (- 1.5) = 2.5
@tssaranlalbk73192 жыл бұрын
Third one is simple 😊😊
@trainingfoundation5.348 Жыл бұрын
The simply way on my side was to use the Patagonian theorem on time on right triangle ADO (R-1)^2 +2^2=R^2 After simplification -2R=-5 => R=5/2
@One-OH-12 жыл бұрын
Also, after finding out the value (h,k) = 2,-3/2, without doing third step it’s clear that radius is CD+ DO = 1+ 3/2 = 2.5 Therefore, R = 2.5
@pratapjadhao3883 жыл бұрын
Thanks Best way to demonstrate
@5p1n0za2 жыл бұрын
Even easier: ADE and ACD are similar triangles, so AD/CD = DE/AD; 2/1 = DE/2; DE = 4, etc.
@mansari6614 Жыл бұрын
Explain point E
@bfelten1 Жыл бұрын
As for the (albeit elegant) coordinate method; the equation for the circle is just a continuous use of Pythagoras. Put the origin at the center, and everything will be much easier and less tedious.
@tcratius17483 жыл бұрын
Regarding the Pythagorean method, roughly 13:50 - 14:00 minutes in, you expand the binomial (r - 1)^2, and I am wondering if the (a = r and b = 1), where the negative sign, "-" in the (r -1)^2 is captured by the negative sign, in "-" 2ab which is the righthand portion, a^2 -2ab +b^2, of (a - b)^2? Otherwise, if the negative sign is captured this way, (a = r, b = -1) then it would result in r^2 -2(r)(-1) +5 = r^2 leading to the answer being r = -5/2. I am on the right track?
@bournitolul88503 жыл бұрын
If u take b =-1 , (r-1)² is going to be of the forme (a+b)² which will lead to the same result r²+2*r*(-1)+1 . Eitherway r is a distance cant be negative
@kkyadav5326 Жыл бұрын
(R-1)^2+2^2=R^2 And solve because when we join the mid point of a chord from the centre of the circle it always perpendicular on the chord.
@jasobantarath69713 жыл бұрын
Nice presentation and steps to solve the given matter
@emadtammam27933 жыл бұрын
excellent
@nope_sup_yup2 жыл бұрын
It will be very simple if you take centre as origin in coordinate geometry.method
@ebi2ch3 жыл бұрын
今日も簡単だったぜ That was easy today, too.
@anasanasa6453 жыл бұрын
We ca use metric relation h^2= 1×(2r-1) think you
@farloverex30753 жыл бұрын
Thank you very much prof you refresh my brain
@gemalbenallie10073 жыл бұрын
i watched and liked the video
@laxmirajmoon43513 жыл бұрын
Good Morning 🌻❤️💕💓 Thank you for right explanation.
@mdchauhan54203 жыл бұрын
Great u r really great. U r way of explanation is superb.
@easymaths48973 жыл бұрын
Good work done, keep it up
@paulwomack58662 жыл бұрын
Sagitta calculations are very much "real world" in the building trade. What radius circle do I need to trace (with a trammel) to get a 3" high arch in a 38" wide doorway...
@muttleycrew2 жыл бұрын
Fun problem. You can of course also solve it with trig.
@rajendranarayandash87973 жыл бұрын
Excellent presentation and problem solving skill.
@NhanNguyen-wu3zp10 ай бұрын
Bạn đã gượng ép khi cho ODC thẳng hàng. Nếu AB và CD cùng nghiêng 1 góc thì ODC không thẳng hàng nữa. 😊
@SolveMathswithEase Жыл бұрын
In Right Angled Triangle ODB: r^2 = (r-1)^2 + 2^2 2r = 5 r = 2.5
@valdirsilva98423 жыл бұрын
I am brasilian, wonderful. Wonderful.
@PreMath3 жыл бұрын
Thanks Valdir for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from the USA!
@manjirikhanolkar58253 жыл бұрын
wonderful explaination. which program do you use for this online teaching? it's really good
@PreMath3 жыл бұрын
Thanks Manjiri You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA!
@andyandym753 жыл бұрын
I saw an school exam Q years ago; A circle with a chord of 10cm, find the radius.?
@portageglaprairie2 жыл бұрын
Cord therom method, solved in my head on about 15 seconds.
@huwpickrell12092 жыл бұрын
Me too
@st3althyone2 жыл бұрын
Definitely the easiest to solve, took the shortest time of all, although some people might not understand it and be more used to using the Pythagorean Theorem. Still, a wonderfully easy explanation using three possible ways.
@gehacktes3 жыл бұрын
you can calculate with tangens, this ist 4. Methode
@servenserov2 жыл бұрын
*From △ODB:* (r-1)²+2²=r²; r=2,5.
@DuaCreativeStudio Жыл бұрын
I solve this problem within 5 sec. By chord theorem of circle
@GetMeThere13 жыл бұрын
Wonderful!
@PreMath3 жыл бұрын
Thank you! Cheers! Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@danielnwn3 жыл бұрын
wow, amazing problem :D
@user-yu4xy8cw8w3 жыл бұрын
Продолжить CD до пересечения с окружностью. Произведения отрезков хорд равны. Задача решается устно.
@sigmamaleslogokijalegi66832 жыл бұрын
I did with a bit of geometry and a bit algebra
@user-ti2we9gc3b2 жыл бұрын
Какая длинная история.Не лучше ли продолжитьСД и использовать свойство перпендикуляра опущенного из точки окружности на диаметр.Коллега,вы слишком развезли!
@edilbertocortez8314 Жыл бұрын
But there is negative value for any unit/s, how come did you accept negative value for any linear measurement?
@India-jq7pi3 жыл бұрын
Thank you sir
@JarppaGuru Жыл бұрын
you have chord AB and height CD we only need those (AD*DB/CD)+CD=diameter 2*2/1+1=5 you know rest lol without mambo jumbo
@AmirgabYT21852 ай бұрын
r=2,5
@danielrousseau40703 жыл бұрын
utilisez les triangles semblables!!! CDBC = DBEB la règle de trois et hop.
@mumtazrasul82633 жыл бұрын
4th method.......suppose, X = 1/2 chord length i.e AD here as 2......P = riser i.e CD here as 1........Formula,,,,,,R = (X^2 + P^2) /2P.........check it please on some other examples.
@sivanaidoo5602 Жыл бұрын
Could have got h=2 from point C(2,1).
@boguslawszostak17842 жыл бұрын
Why complicate a simple task? Let's have a look at the picture and designations of the third method. According to Thales's theorem, the triangle of CBE is right-angled and its hypotenuse is CD . The result is that the triangles CBE, CDB and BDE are similar. b / c = d / b => d = b * b / c = 2 * 2/1 = 4 r = (d + c) / 2 = 5/2
@phonglam29603 жыл бұрын
Because B see CE a diameter so we have BE perdipencular BC. So BD ^2 = CD. DE, so DE = 4/1 = 4 => CE = 1 + 4 = 5 => r = 5/2
@sampathkumar16683 жыл бұрын
Your teaching is awesome.Sir can you explain Coordination method.. ?
@theophonchana50253 жыл бұрын
c^2 = 6.2(5)
@debajyotisaha55233 жыл бұрын
From given condition,how did you get cd is perpendicular to ab?
@nirmalajagdish89013 жыл бұрын
Awesome thanks
@dennisphilip75963 жыл бұрын
Please make a video on real life use of limits , mathematical induction,complex numbers As there is no use of just theory Please sir make a video on it
@devondevon43663 жыл бұрын
I first find the length of the cord using 1 and 2 and the Pythagorean theorem so 1^2 + 2^2= c^2' 5 = c^2 the square root of 5 = c since line cd=1, let line d to '0' the center of the circle = x hence the radius of the circle = 1 + x which implies that center '0' to x also = 1 + x, so the triangle formed is an isosceles which implies that the hypotenuse = 1 + x and the other two sides are 'x' and '2' therefore (1+x)^2 - x^2 =4 2x+1 =4 2x =3 x =3/2 since the radius is x+1, then 3/2 +1 = 5/2 Answer 10:44
@bolder992 жыл бұрын
Very good. Using cross-cross as a way to explain solving fractions is not the best way.
@gasparomagodostijolos46783 жыл бұрын
2,5.
@kennethkan32522 жыл бұрын
2.5
@user-wj9ku7px9r2 жыл бұрын
(r+r-1)*1=2*2 >>> r=2.5
@sorourhashemi3249Ай бұрын
2*2=1*(2r-1)===>r=5/2
@nirmalajagdish89013 жыл бұрын
Vv nice thanks
@ManojkantSamal6 күн бұрын
2.5 (may be )
@shashwatvats77862 жыл бұрын
Calculated in seconds r=5/2
@monmonxperia52413 жыл бұрын
❤️❤️❤️❤️
@NancyShabi-no2gm2 ай бұрын
Absolutely we can calculate the radius of the Earth planet too Is around 12,756 km ,,,Time to complete orbit of the Earth planet around the Sun 🌞 is 360 days or 1 full Year frome where 365 and 366 were invented. ?????? Each day moving ( Rotate/ shifting = One angle degree within circle orbits of 360 degree angles ) means we must have 360 days a Year Calendar exciting in 12 month of the Year each 30 Days if we dived 12000 of the earth Diameter over 12 results in 1000 each month we had leftover of 756 Meters of rotation on Land divide by 12 = 63 meter each month Dividing 63 meter/ 60 min = 1.5 monthly Differential over Time each Year calculating time set change hours time accurate to 360 days a year Obviously within one rotate circulation from ( 0 Degree to 360 degree) ,, From where came the Idea of 365 or 366 days in a Year Calendar All month must be set to 30 days .. depending on Eastern calendar of Shining Stars of the Sun 🌞 ( finding origin point and attach to Sun Shining) ...