Learn 3 Different Methods to Find the Radius of a Circle | In-Depth Explanation
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- čas přidán 27. 03. 2021
- 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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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.
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
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.
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
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.
Excellent Analysis Sir. By solving in 3 different ways.
In triangle ODB (R-1)²+2²=R² ; 2R=5 ; R=2.5
Simplest and quick method. 👍👍👍
Thank you! I loved seeing the 3 different methods. You explanation of each was very clear and easy to follow!
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
I never knew that third method. Thx!
Thank you for the excellent video 😊
Well done, Sir. Also useful as proofs for each theorem.
Well done professor.
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.
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.
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
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 🙂
Thank you sir for sharing your knowledge..It refreshed my mind..Godbless
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
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.
Superb! I just loved it.... Much better than gec
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.
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. 😎
I am 55 years old. Enjoying your maths classes.
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
Made it quite easy....👌
Wow, there's a lot involved with the first method, however as always extremely interesting stuff.
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.
Terimakasih soal matematikanya, bisa untuk latihan🙏
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
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.
Even easier: ADE and ACD are similar triangles, so AD/CD = DE/AD; 2/1 = DE/2; DE = 4, etc.
Explain point E
In Right Angled Triangle ODB:
r^2 = (r-1)^2 + 2^2
2r = 5
r = 2.5
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
Thankyou sir
I love 2nd & 3rd method . 1st is too difficult.
Thanks for the Video.
Thanks
Best way to demonstrate
Third one is simple 😊😊
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
It will be very simple if you take centre as origin in coordinate geometry.method
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
Nice presentation and steps to solve the given matter
Imagine he is your math professor. So calm voice. Hope your students are not sleepy if they really like math.
*From △ODB:*
(r-1)²+2²=r²;
r=2,5.
Thank you very much prof you refresh my brain
Какая длинная история.Не лучше ли продолжитьСД и использовать свойство перпендикуляра опущенного из точки окружности на диаметр.Коллега,вы слишком развезли!
Good work done, keep it up
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
Great u r really great. U r way of explanation is superb.
Can you provide us more questions like this.
I want pdf of these questions
Good Morning 🌻❤️💕💓
Thank you for right explanation.
今日も簡単だったぜ That was easy today, too.
excellent
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...
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
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!
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!
Fun problem. You can of course also solve it with trig.
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
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.
i watched and liked the video
Продолжить CD до пересечения с окружностью. Произведения отрезков хорд равны. Задача решается устно.
Bạn đã gượng ép khi cho ODC thẳng hàng. Nếu AB và CD cùng nghiêng 1 góc thì ODC không thẳng hàng nữa. 😊
We ca use metric relation h^2= 1×(2r-1) think you
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) ...
I solve this problem within 5 sec. By chord theorem of circle
I did with a bit of geometry and a bit algebra
Excellent presentation and problem solving skill.
you can calculate with tangens, this ist 4. Methode
Wonderful!
Thank you! Cheers!
Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
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
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
Thank you sir
But there is negative value for any unit/s, how come did you accept negative value for any linear measurement?
r=2,5
utilisez les triangles semblables!!! CDBC = DBEB la règle de trois et hop.
2*2=1*(2r-1)===>r=5/2
(r+r-1)*1=2*2 >>> r=2.5
wow, amazing problem :D
Could have got h=2 from point C(2,1).
I saw an school exam Q years ago; A circle with a chord of 10cm, find the radius.?
From given condition,how did you get cd is perpendicular to ab?
2.5
Calculated in seconds r=5/2
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.
c^2 = 6.2(5)
#Pythagoras #PythagoreanTheorem
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
Very good. Using cross-cross as a way to explain solving fractions is not the best way.
2,5.
AD*DB = CD*(2*(CD+DO) = 4 so Radius = 2.5
R = 2.50 units
Radius 9 becausE 4 is onE unit Less than midPoinT radiuS --- buT when 5 5 to PoinT 0 becomes less ThaTs 4 --- radius 9
2×2=(2r-1)×1
❤️❤️❤️❤️
#Radius
Your teaching is awesome.Sir can you explain Coordination method.. ?
Pythagorean Theorem