Learn 3 Different Methods to Find the Radius of a Circle | In-Depth Explanation
2021 ж. 27 Нау.
210 237 Рет қаралды
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
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.
Thank God for Pythagoras 3rd method is great though
@Shane Jericho why would you bother being with someone you don't trust? Better, why am I replying to spam?
Here value of h need not be calculated. Mere observations..seg OC IS PARALLEL to y axis. Dan is on
The easy way is the distance ((((AB^2)*0.25)+(DC)^2))/(2DC)=r
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.
Your videos are exciting and I've enjoyed every one I've watched. 👍
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.
Excellent Analysis Sir. By solving in 3 different ways.
Well done, Sir. Also useful as proofs for each theorem.
Thank you for the excellent video 😊
Thank you! I loved seeing the 3 different methods. You explanation of each was very clear and easy to follow!
I never knew that third method. Thx!
Well done professor.
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
Superb! I just loved it.... Much better than gec
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.
Made it quite easy....👌
Thank you sir for sharing your knowledge..It refreshed my mind..Godbless
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.
Terimakasih soal matematikanya, bisa untuk latihan🙏
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 🙂
Wow, there's a lot involved with the first method, however as always extremely interesting stuff.
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
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.
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.
@@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.
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
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.
Thankyou sir
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. 😎
Thanks Best way to demonstrate
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
I love 2nd & 3rd method . 1st is too difficult. Thanks for the Video.
Nice presentation and steps to solve the given matter
Can you provide us more questions like this. I want pdf of these questions
I am 55 years old. Enjoying your maths classes.
Good work done, keep it up
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.
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
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
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...
Third one is simple 😊😊
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
Excellent presentation and problem solving skill.
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.
It will be very simple if you take centre as origin in coordinate geometry.method
Great u r really great. U r way of explanation is superb.
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
Me too
(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.
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
Even easier: ADE and ACD are similar triangles, so AD/CD = DE/AD; 2/1 = DE/2; DE = 4, etc.
Explain point E
Good Morning 🌻❤️💕💓 Thank you for right explanation.
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?
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
In triangle ODB (R-1)²+2²=R² ; 2R=5 ; R=2.5
Simplest and quick method. 👍👍👍
Imagine he is your math professor. So calm voice. Hope your students are not sleepy if they really like math.
Thank you very much prof you refresh my brain
excellent
i watched and liked the video
wonderful explaination. which program do you use for this online teaching? it's really good
Thanks Manjiri You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA!
Cord therom method, solved in my head on about 15 seconds.
Me too
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.
Bạn đã gượng ép khi cho ODC thẳng hàng. Nếu AB và CD cùng nghiêng 1 góc thì ODC không thẳng hàng nữa. 😊
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
We ca use metric relation h^2= 1×(2r-1) think you
But there is negative value for any unit/s, how come did you accept negative value for any linear measurement?
I did with a bit of geometry and a bit algebra
Fun problem. You can of course also solve it with trig.
In Right Angled Triangle ODB: r^2 = (r-1)^2 + 2^2 2r = 5 r = 2.5
I am brasilian, wonderful. Wonderful.
Thanks Valdir for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from the USA!
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
I saw an school exam Q years ago; A circle with a chord of 10cm, find the radius.?
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
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
you can calculate with tangens, this ist 4. Methode
Продолжить CD до пересечения с окружностью. Произведения отрезков хорд равны. Задача решается устно.
*From △ODB:* (r-1)²+2²=r²; r=2,5.
今日も簡単だったぜ That was easy today, too.
From given condition,how did you get cd is perpendicular to ab?
I solve this problem within 5 sec. By chord theorem of circle
Какая длинная история.Не лучше ли продолжитьСД и использовать свойство перпендикуляра опущенного из точки окружности на диаметр.Коллега,вы слишком развезли!
Could have got h=2 from point C(2,1).
Thank you sir
Wonderful!
Thank you! Cheers! Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
wow, amazing problem :D
r=2,5
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.
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
Awesome thanks
utilisez les triangles semblables!!! CDBC = DBEB la règle de trois et hop.
Very good. Using cross-cross as a way to explain solving fractions is not the best way.
It can also be solved by intersecting chord theorm we can produce CD passing through o to the circumference of circle at m also da = db = 2 therefore by this theorem ad ×db=cd×dm therefore dm =4 now 4r² =(2)²+(2)²+(1)²+(4)² From this r²=25/4 r=5/2 which is nothing but 2.5
Your teaching is awesome.Sir can you explain Coordination method.. ?
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
Calculated in seconds r=5/2
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
c^2 = 6.2(5)
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
#Pythagoras #PythagoreanTheorem
Vv nice thanks
(r+r-1)*1=2*2 >>> r=2.5
R = 2.50 units
2,5.
2.5
The answer is 5/2.
#Radius
c = 2.5
Radius 9 becausE 4 is onE unit Less than midPoinT radiuS --- buT when 5 5 to PoinT 0 becomes less ThaTs 4 --- radius 9
AD*DB = CD*(2*(CD+DO) = 4 so Radius = 2.5