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.
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.
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.
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
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.
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. 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.
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. 😎
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.
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
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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
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 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
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.
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) ...
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.
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
