Learn 3 Different Methods to Find the Radius of a Circle | In-Depth Explanation

preview_player
Показать описание
Рекомендации по теме
Комментарии
Автор

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.

math
Автор

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

rangaswamyks
Автор

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.

hansschotterradler
Автор

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.

timemerick
Автор

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

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

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.

shadrana
Автор

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

quattrocchialessandro
Автор

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

shreyanshpatel
Автор

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

devondevon
Автор

In triangle ODB (R-1)²+2²=R² ; 2R=5 ; R=2.5

BubuMarimba
Автор

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

limfilms
Автор

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.

AnonimityAssured
Автор

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.

ojjcpyn
Автор

Your videos are exciting and I've enjoyed every one I've watched. 👍

channelsixtysix
Автор

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

ryan
Автор

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

huwpickrell
Автор

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 🙂

murdock
Автор

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. 😎

williamwingo
Автор

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

boguslawszostak
Автор

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

robertberg
Автор

Excellent Analysis Sir. By solving in 3 different ways.

NASIRable