No it's its hard work To do smart work use properties of circle and chord Answer expected to come within 6-7 calculations
@BenFranklin177610 ай бұрын
Pick any angle in the trianle to be theta. Lets say oppose of 15. Law of cosine: Cos(theta) = (13^2 + 14^2 - 15^2) / (2*13*14) = 5/13 Sin^2 + cos^2 = 1, so sin(theta) = 12/13 (You can note here we are getting the 5,12,13 triangle you had) Extended law of sines says 2R = a/sin, so 2R = 15/sin(theta) R = 1/2 * 15/(12/13) = (13*15) / (2*12) = 65/8
@wackojacko396211 ай бұрын
Area of triangle can be found using Heron's Formula. Then (13 × 15 × 14) ÷ (4 × Area) = R. 🙂
@PreMath11 ай бұрын
True! Thanks ❤️🌹
@sr229110 ай бұрын
Why does that work?
@venkateshwarlujanga965310 ай бұрын
Properties of Triangles Area of triangle (Δ)= abc/4R where R is circumradius @@sr2291
@JanPBtest10 ай бұрын
@@sr2291 The radius of the circumcircle equals abc/(4*area). For proof, see Wikipedia for "Law of sines".
@sr229110 ай бұрын
@@JanPBtest That's really cool. Thanks.
@ybodoN11 ай бұрын
Easy solution: Heron's formula gives the area of the triangle.Then the radius of the circle is abc / 4A.
@Abby-hi4sf11 ай бұрын
Will you elaborate it please
@ybodoN11 ай бұрын
@@Abby-hi4sf The second formula follows from the law of sines: a / sin α = b / sin β = c / sin ɣ = abc / 2A = 2R. So, A = √(21 (21 − 13)(21 − 14)(21 − 15)) = √7056 = 84. Then R = (13)(14)(15) / ((4)(84)) = 2730 / 336 = 65 / 8.
@PreMath11 ай бұрын
Great! Thanks ❤️🌹
@evbdevy3522 ай бұрын
That's the short way.I think so❤
@harrymatabal84485 ай бұрын
Excellent work. Thanks
@ΟμηροςΝεοφυτιδης8 ай бұрын
Cosine rule to find one angle then doing sine rule =2R and you find R
@suryanarayanabadithamani768625 күн бұрын
Well explained!👌👌👌
@samueltso129115 күн бұрын
I used a much quicker method. By cosine rule on triangle CAB, angle CAB = 67.38 degrees. Based on angle properties of circle, angle COB = 2 x angle CAB = 134.76 degrees. Using cosine rule on triangle COB, OC or OB (radius of the circle) = 8.125. I worked out the answer within a minute.
@Fatjack-jy8gs10 ай бұрын
I did enjoy that. I have not done any Maths for far too long. I hardly know any of it any more.
@prossvay874411 ай бұрын
area of the triangle: A=√s(s-a)(s-b)(s-c) s=a+b+c/2=13+14+15/2 s=42/2=21 A=√21(21-13)(21-14)(21-15)=84 84=1/2(13)(15)sin(x) x=59.5 2x=2(59.5)=119 Cos(119)=r^2+r^2-14^2/2r^2 r=8.12 . 🙏❤❤
@Copernicusfreud11 ай бұрын
That is how I did it.
@PreMath11 ай бұрын
Super!!! Thanks ❤️🌹
@sudhangshubhattacharya49918 ай бұрын
In brief the area of the square in circle to be found out then diagonal of that square will be the diameter of the circle and half of it will find the radius
@sudhangshubhattacharya49918 ай бұрын
The one of the intelligent problems found in u tube videos
@jkevincolligan83177 ай бұрын
@@sudhangshubhattacharya4991 Great geometric problem !!😅😅😅
@prashant2451005 ай бұрын
Hello sir..I m from India..and watch your videos...I really like it and encourage others to watch
@parthtomar698711 ай бұрын
Nice solution sir
@PreMath11 ай бұрын
Thanks and welcome❤️🌹
@Abby-hi4sf11 ай бұрын
Excelent explanation
@PreMath11 ай бұрын
Glad you liked it❤️ Thanks ❤️🌹
@giuseppemalaguti43511 ай бұрын
R=ABC/4S...S=(Erone)=4*3*7=84...R=65/8
@PreMath11 ай бұрын
Wow! Thanks ❤️🌹
@tombufford1369 ай бұрын
At a quick glance: The centroid of the triangle and the three medians are coincident. where h is the height. X1 and Y1 are the x and y coordinates from A of the Centroid. X1 = 14/2 = 7. Y1 = h/3. AD =x . r^2=7^2+h^2/9 , x^2+h^2=169 and h^2=225-(14-x)^2. R^2=49+(169-x^2)/9. h^2=29-28x-x^2. h^2=169-x^2 then 140-28x=0 and x=5. Then h = 12 and the radius = SQRT(49+ 144/9)=8.1
@ajeethpandey5 ай бұрын
R=a.b.c/4.∆ which gives R=65/8
@riccardobaiocco97668 ай бұрын
Very clear. You are a very good teacher
@suyogbakliwal1111 ай бұрын
We know that 🔼 = abc / 4R, Where 🔼 is the Area of Triangle ABC, a,b,c are the sides of the triangle, and R is the circumradius of the circle. The area of a triangle can be found using Heron's formula. and we can find R consequently
@chintamanisatyamurthy534210 ай бұрын
Very well explained. Though different methods r also there, the way u explained in ur own method is superb.
@quabledistocficklepo35978 ай бұрын
So what? I knew that,. but how would that
@mathematicsbyraunaksir977811 ай бұрын
We can solve it by using formula a*b*c/4*area of triangle 13*14*15/4*84
@PreMath11 ай бұрын
Sure! Thanks ❤️🌹
@mathematicsbyraunaksir97784 ай бұрын
Always helpful videos thanks for doing it @@PreMath
@Marco-AurelioHernandez9 ай бұрын
I used the Rule of Cosines to find the angles. Let a = 13, b = 14, c = 15. So angle A = 53.1, B = 59.5, C = 67.4. Then for the circumcircle radius R, I used the Rule of Sines with any side: a/sin A = 2R. Thus, 13/2(0.799)) = 8.13
@shirish1110 ай бұрын
Join OA and OB. OA= OB = R. Angle AOB = twice of angle ACB. Using cosine law for triangle ABC, we can find cos C. From here, find cos 2C , that is , cos AOB. Now again use cosine law for triangle AOB to find R.
@rangaswamyks82876 ай бұрын
GREAT WORK PROFESSOR.. YOU ARE MY FAVOURITE MATHS GURUJEE AFTER A LONG TIME IAM WATCHING YOUR VIDEOS.. BLESS ME SIR.
@UNFORGIVEN18215 ай бұрын
A simpler solution is to proportionally find the arc to the length of each side of the triangle that is proportional. Then the correspondence in degrees of each unit of length taking into account the 360 degrees of the circle. Finally you find the length that corresponds to 180 degrees, this is the diameter of the circle and dividing by two you find the radius.
@mathewpv6815 ай бұрын
We can find the area of the triangle by herons theorem. Since area is half x base x altitude, we can find an altitude. From altitude and a side we can find the sine of the included(base and side) angle. Then we can use the formula, 2R=a/sine(A) to find the radius. With a lttle effort, since the values of the sides given are simple, we can do the calculation mentally.
@alokranjan41493 ай бұрын
Beautiful solution , without using the heron's formula & R = abc/ area
@geometer612111 ай бұрын
Great work for instructional purposes, but it would be much easier to use the formula of the circumradious: R = ABC / 4[area of triangle]. In this case: R =(13)(14)(15) / 4[84] = 2,730 / 336 = 8.125
@Salman_Zahur7 ай бұрын
How did you derive the measure 84?
@falomasbu4 ай бұрын
How about using Euclidean concepts, instead....constr OB=OA, then first angle ACB by cosine rule. Then AOB will be twice ACB. Repeat step above and find AOB, the radius
@PrithwirajSen-nj6qq6 ай бұрын
We may join OA OB OC The large triangle will be splitted into three small isosceles triangles Each of these three isosceles triangles area may be computed with the help of radius and each of the sides. Then their sum will be with one unknown that is radius. This sum is equal to the area of the big triangle. The area of the big triangle may be computed by Heron's theorem of 🔺. Then we may form an equation from which we may get the RADIUS. Thanks for ur solution.
@murvetaykac70416 ай бұрын
We can use the tringles area formula.The first formula is S=a.b.c/4R and second is S=(u.(u-a).(u-b).(u-c)^1/2 .We find the result useing thees formulas.R=65/8 thank you very much.
@phidesx609910 ай бұрын
Extended law of sines: (1) a / sin α = b / sin β = c / sin γ = abc / 2A = 2R, where A = area of the triangle, R = radius of the circumscribed circle Heron's formula: (2) Area of the triangle: A = square root of p(p - a)(p - b)(p - c), where p = semiperimeter of the triangle p = (13 + 14 + 15) / 2 = 21 A = square root of 21(21 - 13)(21 - 14)(21 - 15) = 84 From (1) : abc / 2A = 2R -> R = abc / 4A = 13 x 14 x 15 / 4 / 84 = 8.125
@KipIngramАй бұрын
Sure - no problem. The half-perimeter of the triangle is 21. So the area of the triangle is (by Heron's formula) area = sqrt(21*(21-13)*(21-14)*(21-15)) = 84 This has to be equal to 0.5*base*height = 7*height, so height = 12 Now we can figure out the angle C; it's C = arccos(12/13) + arccos(12/15) = 59.4898 degrees Now by the law of sines the diameter is diameter = 14 / sin(59.4898) = 16.25 So the radius is 8.125.
@alexcwagner8 ай бұрын
I'm not claiming my way is the best way, but I had to work with what I could remember. I know that angle AOB is double the angle of ACB, and that the perpendicular bisector of AB passes through O. So, I considered the right triangle A-mid(A,B)-O, where A-mid(A,B) is 7, AO is r, and the angle at O is gamma (where gamma is the angle at C). So, from that triangle, we get r = 7/sin(gamma), and from the big triangle, we can use the law of cosines to get 14^2 = 13^2 + 15^2 - 2(13)(15)cos(gamma), which yields cos(gamma) = 33/65. From cos(gamma) we can derive sin(gamma) = 56/65, so r = 7/(56/65) = 65/8 or 8.125.
@Ibrahimfamilyvlog2097l11 ай бұрын
Great work sar❤❤❤❤
@PreMath11 ай бұрын
Thanks a lot ❤️
@BalaSribhaskarImmidi7 ай бұрын
Nice explanation but we could use herons formula and area of triangle is equal to abc /4R to find the circumradius
@dhrubajyotidaityari924011 ай бұрын
∆ =84 R=abc/4∆=13.14.15/4.84 =65/8 sq units
@PreMath11 ай бұрын
Thanks ❤️🌹
@STEAMerBear7 ай бұрын
So we need to find the circumcenter. That’s the concurrent point where the perpendicular bisectors meet. All the vertices fall on the inscribing circle from that point, so they are equidistant from it as well as radii of the circle. So the distance from the circumcenter to a vertex is r. r =(abc)/sqrt((a+b+c)(b+c−a)(c+a−b)(a+b−c)) =8.125
@dawit-y7s2 ай бұрын
Thanks mr.
@edwardwestenberger18907 ай бұрын
I have been working on a (carpentry) problem (on and off) for about a month. My solutions have been: 1. An iterative solution using newton's method 2. A forth order equation using Sin(theta) as the variable: it creates 3 extraneous roots in addition to the desired solution. 3. A forth order equation that uses Cos(theta) as the variable... same problem as above. 4. A forth order equation using Tan(theta) as the variable... Same problem The basic equation is: A = B / Cos(theta) + C * Sin(theta) where A, B & C are to be treated as constants and theta is the variable to be solved for. I hope this is the right forum, can you help me?
@extremovolador5 ай бұрын
Sorry, I had a mistake, the answer is A=B/{sqrt(1+C^2))[(1/sqrt(1+C^2))*cos(theta)+(C/sqrt(1+C^2))Sin(theta))}->A=B/[Sqrt(1+C^2)*Cos(phi+theta)], Cos(phi)=1/Sqrt(1+C^2), Sin(phi)=C/Sqrt(1+C^2), or phi=Arctan(C), therefore Cos(phi+theta)=B/[A*Sqrt(1+C^2)]->theta=ArcCos(B/[A*Sqrt(1+C^2)])-phi
@murdock553711 ай бұрын
Nice! ∆ ABC → AB = 14 = AR + BR = 7 + 7; AO = BO = r → sin(ARO) = 1; BC = 15; AC = 13; BCA = δ = ROA 198 = 394 - 2(13)(15)cos(δ) → cos(δ) = 33/65 → sin(δ) = √(1 - cos^2(δ)) = 56/65 = AR/r = 7/r → r = 65/8
@PreMath11 ай бұрын
Great job! Thanks ❤️🌹
@shivachaturvedhi88408 ай бұрын
answer short cut formulka: fatorize ->term1: (6.5+7.5)^2-(7^2) = 7^2 * 3 -> term23: (7^2)-(7.5-6.5)^2 = 48 = 4^2*3 sqrt of product of above two terms: sqrt ( 7^2 * 3^2 * 4^2) = 3*4*7 next calculate : 2* 13*14*15 / (3*4*7) = 65 so answer is 65/8 = 8.125 Explanation: term1 = (b+c)^2-a^2 Term2 = a^2-(b-c)^2 factors = sqrt(term1*term2) R = 1/8 * (2ABC/factors) FYIP, sides A,B,C, half sides a,b,c
@aljawadАй бұрын
Whenever I see a problem that involves a circle, I proceed with the application of the equation of the circle ((diff in X)^2 + (diff in Y)^ = Radius^2). In this case I assumed point D was the original (0,0), and after finding the values of x and h, I proceeded to give points A, B and C horizontal and vertical coordinates, and applied the equation of the enclosed circle to arrive to the same value for the radius, plus a bonus was the original of the circle at point O. But in this case, there is a famous ratio: product of the sides/(4*area)!
@anildeshkar90610 ай бұрын
Will this solution hold good if 'O' is outside of triangle ABC?
@jarikosonen40799 ай бұрын
That seems interesting question...
@幕天席地-w9c11 ай бұрын
Cosine rule:15^2=13^2+14^2-2*13*14*cos(A), so cos(A)=5/13, sin(A)=12/13; Sine rule: a/sin(A)=2R, R is radius of circumscribed circle, so R=65/4
@강운이-l1p11 ай бұрын
2R = 65/4 so R = 65/8
@MKPoqdPDeamimzpygxDjg7sJY597 ай бұрын
Someone explain why at 5:59 the triangle 12, 14, 15 is right is 144+196 is not 225.
@jeyaramkrishanth29882 ай бұрын
In triangle BCD , BD isn't equal to 14. BD is 14-X (which is 5) so BD is 9. Now 144+81=225
@hariprasadchakrala79006 ай бұрын
🙂👌🏻🙏He followed the geometrical approach. Trigonometrical approach is different which is derived from geometrical approach for variable values 🎉here geometrical approach is appropriate I think to test the basic knowledge 🤔
@weipenglim78407 ай бұрын
You can use cosine rule to solve for angle alpha
@ahmadesteitieh6 ай бұрын
Find the area of the triangle using half perimeter then deduce the length of the heght you drew and find the radius as you proceeded
@Aditya_19611 ай бұрын
4:23 here u could have just set x² + h² = 13 get x and find out the answer i personally would prefer ur method cuz i really hate that herons formula
@prime4238 ай бұрын
13 and 15 are a giveaway for a9-12-15 and 5-12-13.Note 12 is the altitude. There is a well known formula for the circle given the area and sides of the triangle. A good mathlete could solve this in a minute or two. Like anything, being a good Mathlete takes time and lots of practice.Fortunately,there are many resources for mathletes.I could give the formula but lets see if anyone can discover it.Its not so easy.
@jimwinchester3398 ай бұрын
Didn't know Thale's formula. But the inscribed angles theorem doesn't seem right: as E approaches B, the angle CEB approaches a right angle.
@harikatragadda11 ай бұрын
By Cosine rule in ∆ABC , Cos∠ABC = 3/5 Sin∠ABC = 4/5 Draw a line AD passing through the center O. ∆ACD is a Right triangle, with ∠ADC = ∠ABC Hence Sin∠ADC = 13/2R = 4/5 R = 65/8
@PreMath11 ай бұрын
Great! Thanks ❤️🌹
@aaziz118 ай бұрын
nicely explained
@ccdsah8 ай бұрын
S=p*R. S= area using Heron's formula. p= semiperimeter
@jacobcombs110610 ай бұрын
I solved it differently. I took each of the triangle sides over the sum of the sides multiplied by 360 to get their angles. 14 was 120° which is perfect. I then drew a triangle with angles 120/30/30 where one side was 14 and the other two sides were r. I then bisected it creating 2 equal 30/60/90 triangles with a side of 7 and a hypotenuse of r. I then used the relationship of 30/60/90 triangles where the side opposite the 60 is x√3 and the hypotenuse is 2x to solve. X=7/√3 therefore r=2×7/(√3)=14/(√3)=8.1
@MindPowerAsYourChildsCompass6 ай бұрын
If you calculate this, you get 8.0829. See my more long winded explanation above. Thanks
@michaelstahl151511 ай бұрын
I enjoyed your video. My suggestion is using the cosinus theorem for calculating angel alpha at point A . I got alpha nearly 67, 3 . After that I used a theorem for calculating the radius of the circle around the triangle . r = a / 2 * sin (alpha) = 15 / 2 * 0,92 = 8,125.
@thangnguyeninh63173 күн бұрын
use heron to cal the area, we have S=abc/4R, haizzz 60s
@BarrieHughes8 ай бұрын
Use cos rule to work out angle A. BOC is double that. Use cos rule again to get OC=R in triangle BOC😉
@VIKING-dl7wz9 ай бұрын
CIRCUMRADIUS = multiplication of sides/ 4 times the area of triangle. 65/8
@orliestutorials5813 ай бұрын
R=a*b*c/sqrt((a+b+c)*(a+b-c)*(a+c-b)*(b+c-a))
@madaxeybuufis30858 ай бұрын
Good job
@getahundesta65014 ай бұрын
I want to buy the book .where can I get it? and what is its name?
@marcelowanderleycorreia887611 ай бұрын
Many diferent forms to do. I used the heron formula in order to find the area of the triangle ABC and the proporcionaly of the chords in a circle, in order to find the radus.
@PrithwirajSen-nj6qq6 ай бұрын
Please elaborate
@PrithwirajSen-nj6qq6 ай бұрын
Please elaborate
@mathpro9267 ай бұрын
good explanation thank you teacher
@theoadieye10 ай бұрын
nice explanation thank you
@mathbynisharsir558611 ай бұрын
Excellent presentation sir
@PreMath11 ай бұрын
Thanks dear ❤️🌹
@uwelinzbauer397311 ай бұрын
My idea was: the perpendicular lines through the mids of the sides of a triangle intersect in the center point of the outer circle. To find the radius of the circle, I used the cosine and the sine rules. Luckily I found that same correct solution in the end 😅 Nice challenge, a little hard to work out, but challenges make us 💪 😀 Greetings!
@Salman_Zahur7 ай бұрын
Hello Dr Tahir
@samsheerparambil11 ай бұрын
By combining sine law and area of triangle we can say R=abc/4*Area hope this helps. Area= square root of s(s-a)(s-b)(s-c)
@_dr.mayhemАй бұрын
This video shouldn't be longer than 1 minute, use Heron's formula for area, then substitute values of side and area into area formula for triangle using circumradius, Which is, Area= (axbxc)/4r.
@kkulkulkan547210 ай бұрын
Used trig to solve it (cosine and then sine law) which in hindsight wasn’t necessary.
@sesa29108 ай бұрын
You just found the diameter. To get the radius you must devide the diameter by 2. Thanks
@wasimahmad-t6c4 ай бұрын
Full area 207.3942 & tenagale area 15×11.2=168÷2 =84
@TheEulerID10 ай бұрын
My method was a bit different. I set the co-ordinates of A as 0,0 and B as 14,0 (note, that I can do that as the whole figure can be rotated if necessary to make the line A-B to be horizontal, even if the drawing isn't to scale). I then worked out the perpendicular height of the triangle from the AB baseline to C to be 12 using Pythagoras on the two right angles with one shared side. That also gives the x co-ordinate of the vertical line to be 5. Thus the co-ordinate of C is 5,12. If we take the generalised formula for a circle, x^2 + y^2 + 2gx + 2fy + c = 0, then we can see as the circle passes through point A (0,0) then c=0. Thus we have x^2 + y^2 + 2gx + 2fy = 0. Now plug in the co-ordinates of point B (14,0) and we get 196 + 14x = 0, therefore x = -7. That means the equation for our circle is x^2 + y^2 - 14x + fy = 0. Plug in the co-ordinates for point C (5,12) and we get 25 + 144 - 70 + 24f. Re-arrange and you get 24f = 99, thus f = -33/8. We now have a circle centred at 7,33/8 which passes through 0,0. The radius is sqrt(g^2 + f^2 - c), which is sqrt(49 + (33/8)^2), which works out at 65/8 or 8.125.
@kashifjamal545510 ай бұрын
R (radius of circumcircle)=abc/4∆ where ∆ is the area of triangle
@abdoomar55410 ай бұрын
Use cose rule to get the angel. Next a'÷Sina =2r
@thetenniszone12311 ай бұрын
how do you approach such tough sums? when i see the solutions i understand the problem but have no clue how to start such problems. pls help
@michaelgarrow323911 ай бұрын
Just keep watching. Pause at the beginning and try to think of an answer. 👍
@dickroadnight11 ай бұрын
Just think what rules or theorems will get you from the dimensions you have to the answer you want. You can Google e.g “circle formulae”. I “has been” a draftsman - so I think how I could use the given info to draw it. In this case, you could use a compass to bisect the chords and find the centre.
@dongxuli968210 ай бұрын
A triangle is fully defined by 3 side length; angles are fully defined by sides (law of cosines); length to sine of the opposite angle ratio is the diameter of its circumcircle. No trick is needed.
@hailelove15 ай бұрын
You are amazing.want to say hi from Ethiopia
@User-jr7vf9 ай бұрын
There is a far quicker way of solving this, and I'm surprised nobody in the comments mentioned it.
@MaheshKumar-lx1ku9 ай бұрын
Please share 🙏
@User-jr7vf9 ай бұрын
@@MaheshKumar-lx1ku ok. First use the law of cosines to find the angle ACB. Now draw lines from O to A and from O to B, and note that these lines are the radii of the circle. There's a theorem (I forgot the name) which states that the angle AOB is twice the angle ACB. Then by using the law of cosines once more, you can find the radius.
@MaheshKumar-lx1ku8 ай бұрын
@@User-jr7vf thanks
@justarandomnerd33607 ай бұрын
@@User-jr7vfno trigonometry
@marcuscicero95879 ай бұрын
very nice
@ОльгаСоломашенко-ь6ы11 ай бұрын
Можно найти площадь треугольника по формуле Герона. R=a*b*c/(4*s)=8.125.
@PreMath11 ай бұрын
Great! Many thanks ❤️🌹
@STEAMerBear7 ай бұрын
Спасибо! Это то, что я тоже сделал. Я удивлен, увидев так много людей, выполняющих слишком много работы. Хотя на вершину может быть много путей, ни одна полезная тропа не приведет нас к подножию горы! == translated from == Thank you! This is what I did too. I'm surprised to see so many people doing too much work. Although there may be many paths to the top, no useful path will lead us to the bottom of the mountain!
@ОльгаСоломашенко-ь6ы7 ай бұрын
@@STEAMerBear хочется решить не только правильно, но и быстро. Спасибо.
@CMTattoos6 ай бұрын
why the radius is 65 divided by 8 ?it was divided by 4 ..on previous step ?
@CalvinLXVII6 ай бұрын
Un ejercicio muy interesante! Saludos!
@miriamvianaesilva11189 ай бұрын
Too remember. Thank you
@Imran-tc6sn7 ай бұрын
Thankyou sir
@hemantdikshit30088 ай бұрын
Beyond words
@beaumatthews64118 ай бұрын
Very good video, but the constant uhs do really distract
Brahmagupta's Formula (also known as Heron's equation) area of triangle ABC = S = sqrt (s(s-a)(s-b)(s-c)) where s = half-perimeter = (13+14+15)/2 = 21, a= 13, b=14 and c=15. so S = 84 We can also right S = abc/4R where R is the radius of a circle that inscribed triangle ABC. So R = (13 x 14 x 15) / (4 x 84) = 8.125 unit.
@MubarekHussien-xe4mm4 ай бұрын
Than you I like you
@mintusaren8956 ай бұрын
R not mention by pai
@wasimahmad-t6c4 ай бұрын
16.25÷2=8.125×8.125=66.015625×3.14159268=207.3942 full area sarcol
@GillesF317 ай бұрын
I think we can get the radius value in 2 steps only (universal method). See below: ----- step #1 ----- ABC triangle area (HERON method): • half-perimeter: (13 + 14 + 15)/2 = 42/2 = 21 • area = √[21·(21 - 13)·(21 - 14)·(21 - 15)] = √(21·8·7·6) = √7056 = 84 ----- step #2 ----- Radius computation (circumscribed circle of a triangle): • formula: radius = (triangle_side_product)/(4·triangle_area) • R = (13·14·15)/(4·84) • R = 2730/336 -------------------- | R = 8.125 | -------------------- 🙂 Note: There is not enough place here to develop the formula >. Sorry!
@borutcigale7787 ай бұрын
You totally overcomplicated it. You can just find one angle cosa=(b2+c2-a2)/2bc. Then find sine value of alpha and then devide a by sina. And then by tvo and you got radius. Very simple.
@rey-dq3nx7 ай бұрын
So where’s your answer using your over simplistic approach? Ah, you know all the sines and cosines of all the angles by heart, I see!
@borutcigale7787 ай бұрын
@@rey-dq3nx You don't need to calculate sine. When you get cosine you just use formula sinx=sqrt(1-(cosx)^2).
@arvindarkumar94732 ай бұрын
.. very good solution And intresting 😊😊 ARVINDAR YADAV New Delhi.. India
@JSSTyger11 ай бұрын
I believe r = 65/8. I tried this in my head so I could be way off.
@NikitaYadav-he1kc5 ай бұрын
It's not Pythagoras theorem it is baudhayan theorem