Can you find the Radius of the circle? | (Triangle inscribed in a circle) |

Рет қаралды 283,957

6 ай бұрын

Learn how to find the radius of the circle. Blue triangle inscribed in a circle. Side lengths of the triangle are 13, 14, and 15. Important Geometry skills are also explained: Pythagorean Theorem; similar triangles; Thales' theorem. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find the Radiu...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you find the Radius of the circle? | (Triangle inscribed in a circle) | #math #maths
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindRadius #ThalesTheorem #Radius #GeometryMath #PythagoreanTheorem
#MathOlympiad #SimilarTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #ExteriorAngleTheorem #PythagoreanTheorem #IsoscelesTriangles #AreaOfTriangleFormula #AuxiliaryLines
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #CompetitiveExams #CompetitiveExam
#MathematicalOlympiad #OlympiadMathematics #LinearFunction #TrigonometricRatios
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Olympiad Question
Find Area of the Shaded Triangle
Geometry
Geometry math
Geometry skills
Right triangles
Exterior Angle Theorem
pythagorean theorem
Isosceles Triangles
Area of the Triangle Formula
Competitive exams
Competitive exam
Find Area of the Triangle without Trigonometry
Find Area of the Triangle
Auxiliary lines method
Pythagorean Theorem
Square
Diagonal
Congruent Triangles
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

Пікірлер: 180

@bigm383 6 ай бұрын

Great work, Professor!❤😀

@PreMath 6 ай бұрын

Thank you so much! 😃❤️

@wackojacko3962 6 ай бұрын

Area of triangle can be found using Heron's Formula. Then (13 × 15 × 14) ÷ (4 × Area) = R. 🙂

@PreMath 6 ай бұрын

True! Thanks ❤️🌹

@sr2291 5 ай бұрын

Why does that work?

@venkateshwarlujanga9653 4 ай бұрын

Properties of Triangles Area of triangle (Δ)= abc/4R where R is circumradius @@sr2291

@JanPBtest 4 ай бұрын

@@sr2291 The radius of the circumcircle equals abc/(4*area). For proof, see Wikipedia for "Law of sines".

@sr2291 4 ай бұрын

@@JanPBtest That's really cool. Thanks.

@michaelpantano3324 4 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

@harrymatabal8448 Күн бұрын

Excellent work. Thanks

@prashant245100 7 күн бұрын

Hello sir..I m from India..and watch your videos...I really like it and encourage others to watch

@samsheerparambil 6 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)

@theoadieye 5 ай бұрын

nice explanation thank you

@aaziz11 2 ай бұрын

nicely explained

@Abby-hi4sf 6 ай бұрын

Excelent explanation

@PreMath 6 ай бұрын

Glad you liked it❤️ Thanks ❤️🌹

@riccardobaiocco9766 3 ай бұрын

Very clear. You are a very good teacher

@madaxeybuufis3085 2 ай бұрын

Good job

@Ibrahimfamilyvlog2097l 6 ай бұрын

Great work sar❤❤❤❤

@PreMath 6 ай бұрын

Thanks a lot ❤️

@parthtomar6987 6 ай бұрын

Nice solution sir

@PreMath 6 ай бұрын

Thanks and welcome❤️🌹

@rangaswamyks8287 29 күн бұрын

GREAT WORK PROFESSOR.. YOU ARE MY FAVOURITE MATHS GURUJEE AFTER A LONG TIME IAM WATCHING YOUR VIDEOS.. BLESS ME SIR.

@mathpro926 Ай бұрын

good explanation thank you teacher

@suyogbakliwal11 5 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

@dickroadnight 6 ай бұрын

Think outside the triangle! The sides of a circumscribed circle are chords of that circle and the perpendicular chord bisecting lines pass through the centre O of the circle. The lines from the corners to the centre of the circle are radii. Make D the midpoint of AB, and E the midpoint of AC. Extend EO to F on AB extended. By the cos rule A = 67.18 degrees. The radius is AO = r. AEF is a right angle triangle. AE = AC/2 = 6.5 AD = AB/2 = 7 AF = AE/cosA = 16.9 DF = AF-AD = 9.9 AFE = A-90 degrees OD = DF tan(90-A) = 4.125 r^2 = AD^2 + OD^2 By Pythagoras’ therom Radius = 8.125

@PreMath 6 ай бұрын

Thanks ❤️🌹

@ybodoN 6 ай бұрын

Easy solution: Heron's formula gives the area of the triangle.Then the radius of the circle is abc / 4A.

@Abby-hi4sf 6 ай бұрын

Will you elaborate it please

@ybodoN 6 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.

@PreMath 6 ай бұрын

Great! Thanks ❤️🌹

@chintamanisatyamurthy5342 4 ай бұрын

Very well explained. Though different methods r also there, the way u explained in ur own method is superb.

@quabledistocficklepo3597 2 ай бұрын

So what? I knew that,. but how would that

@tombufford136 3 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

@Fatjack-jy8gs 4 ай бұрын

I did enjoy that. I have not done any Maths for far too long. I hardly know any of it any more.

@shirish11 5 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.

@marcelowanderleycorreia8876 6 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-nj6qq 26 күн бұрын

Please elaborate

@PrithwirajSen-nj6qq 26 күн бұрын

Please elaborate

@michaelstahl1515 5 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.

@marcuscicero9587 4 ай бұрын

very nice

@miriamvianaesilva1118 3 ай бұрын

Too remember. Thank you

@CalvinLXVII 26 күн бұрын

Un ejercicio muy interesante! Saludos!

@giuseppemalaguti435 6 ай бұрын

R=ABC/4S...S=(Erone)=4*3*7=84...R=65/8

@PreMath 6 ай бұрын

Wow! Thanks ❤️🌹

@Marco-AurelioHernandez 4 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

@mathematicsbyraunaksir9778 6 ай бұрын

We can solve it by using formula a*b*c/4*area of triangle 13*14*15/4*84

@PreMath 6 ай бұрын

Sure! Thanks ❤️🌹

@user-sw1zo5cy9x 2 ай бұрын

Cosine rule to find one angle then doing sine rule =2R and you find R

@Imran-tc6sn Ай бұрын

Thankyou sir

@weipenglim7840 Ай бұрын

You can use cosine rule to solve for angle alpha

@aryabhattagamharia5586 6 ай бұрын

Excellent presentation sir

@PreMath 6 ай бұрын

Thanks dear ❤️🌹

@uwelinzbauer3973 5 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!

@geometer6121 5 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_Zahur Ай бұрын

How did you derive the measure 84?

@prossvay8744 6 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 . 🙏❤❤

@Copernicusfreud 6 ай бұрын

That is how I did it.

@PreMath 6 ай бұрын

Super!!! Thanks ❤️🌹

@sudhangshubhattacharya4991 2 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

@sudhangshubhattacharya4991 2 ай бұрын

The one of the intelligent problems found in u tube videos

@jkevincolligan8317 2 ай бұрын

⁠@@sudhangshubhattacharya4991 Great geometric problem !!😅😅😅

@edwardwestenberger1890 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?

@prime423 2 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.

@hailelove1 10 сағат бұрын

You are amazing.want to say hi from Ethiopia

@user-ts5ne9qd4o Ай бұрын

Nice explanation but we could use herons formula and area of triangle is equal to abc /4R to find the circumradius

@STEAMerBear 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

@hemantdikshit3008 3 ай бұрын

Beyond words

@jacobcombs1106 5 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

@MindPowerAsYourChildsCompass Ай бұрын

If you calculate this, you get 8.0829. See my more long winded explanation above. Thanks

@alexcwagner 2 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.

@ahmadesteitieh 10 күн 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

@kkulkulkan5472 4 ай бұрын

Used trig to solve it (cosine and then sine law) which in hindsight wasn’t necessary.

@user-ii5vr3um9r 6 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

@user-vi9zj6nh5c 6 ай бұрын

2R = 65/4 so R = 65/8

@BarrieHughes 2 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😉

@harikatragadda 6 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

@PreMath 6 ай бұрын

Great! Thanks ❤️🌹

@VIKING-dl7wz 3 ай бұрын

CIRCUMRADIUS = multiplication of sides/ 4 times the area of triangle. 65/8

@anildeshkar906 4 ай бұрын

Will this solution hold good if 'O' is outside of triangle ABC?

@jarikosonen4079 4 ай бұрын

That seems interesting question...

@ccdsah 2 ай бұрын

S=p*R. S= area using Heron's formula. p= semiperimeter

@dhrubajyotidaityari9240 6 ай бұрын

∆ =84 R=abc/4∆=13.14.15/4.84 =65/8 sq units

@PreMath 6 ай бұрын

Thanks ❤️🌹

@phidesx6099 4 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

@hariprasadchakrala7900 12 күн 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 🤔

@abdoomar554 4 ай бұрын

Use cose rule to get the angel. Next a'÷Sina =2r

@jimwinchester339 2 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.

@sesa2910 2 ай бұрын

You just found the diameter. To get the radius you must devide the diameter by 2. Thanks

@murdock5537 6 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

@PreMath 6 ай бұрын

Great job! Thanks ❤️🌹

@user-sk9oi9jl2g 6 ай бұрын

Можно найти площадь треугольника по формуле Герона. R=a*b*c/(4*s)=8.125.

@PreMath 6 ай бұрын

Great! Many thanks ❤️🌹

@STEAMerBear 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!

@user-sk9oi9jl2g Ай бұрын

@@STEAMerBear хочется решить не только правильно, но и быстро. Спасибо.

@PrithwirajSen-nj6qq 26 күн 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.

@murvetaykac7041 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.

@Aditya_196 6 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

@pralhadraochavan5179 6 ай бұрын

Good morning sir

@shivachaturvedhi8840 3 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

@Salman_Zahur Ай бұрын

Hello Dr Tahir

@thetenniszone123 6 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

@michaelgarrow3239 6 ай бұрын

Just keep watching. Pause at the beginning and try to think of an answer. 👍

@dickroadnight 6 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.

@dongxuli9682 4 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.

@TheEulerID 5 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.

@User-jr7vf 3 ай бұрын

There is a far quicker way of solving this, and I'm surprised nobody in the comments mentioned it.

@MaheshKumar-lx1ku 3 ай бұрын

Please share 🙏

@User-jr7vf 3 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-lx1ku 3 ай бұрын

@@User-jr7vf thanks

@justarandomnerd3360 2 ай бұрын

@@User-jr7vfno trigonometry

@kashifjamal5455 4 ай бұрын

R (radius of circumcircle)=abc/4∆ where ∆ is the area of triangle

@johnplong3644 2 ай бұрын

I forgot this well is has been close to over 45 years I knew I was in trouble well at least I can tutor Algebra 1 and 2 and some Trigonometry for get pre-calculus and calculus I have forgotten 90% of that

@nalinweerasinghe8475 4 ай бұрын

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.

@susennath6035 2 ай бұрын

After long time. Can u remember me?

@AmirgabYT2185 4 ай бұрын

R=8,125

@aR3DFgH8kLj2pOqIuVx5 Ай бұрын

Someone explain why at 5:59 the triangle 12, 14, 15 is right is 144+196 is not 225.

@misterenter-iz7rz 6 ай бұрын

area^2=21 6 7 8=84^2, area=84=1/2 13 15 (7/r), r=7 13 15 /(2 7 12)=(13 5)/(2 4)=65/8.😊

@PreMath 6 ай бұрын

Thanks ❤️🌹

@CMTattoos 17 күн бұрын

why the radius is 65 divided by 8 ?it was divided by 4 ..on previous step ?

@kmgshop1570 4 күн бұрын

It would be more interesting if 15 is the radius so the problem gives just 2 sides and the question is find the radius of the circle

@mintusaren895 20 күн бұрын

R not mention by pai

@beaumatthews6411 2 ай бұрын

Very good video, but the constant uhs do really distract

@radedraskovic425 2 ай бұрын

8.13

@axelbehr43 Ай бұрын

in the very beginning of the video, the assumption that D is 90 degrees is guess work !

@JSSTyger 6 ай бұрын

I believe r = 65/8. I tried this in my head so I could be way off.

@frommytravels Ай бұрын

that was a lot of work! hahaha

@lezgi53 15 күн бұрын

R=abc/4S R=8.125

@elye3701 Ай бұрын

I was expecting to find a general answer.

@borutcigale778 2 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-dq3nx 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!

@borutcigale778 Ай бұрын

@@rey-dq3nx You don't need to calculate sine. When you get cosine you just use formula sinx=sqrt(1-(cosx)^2).

@danishkhan5836 Ай бұрын

8.125

@GillesF31 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!

@hansnagle5764 2 ай бұрын

r = a/(2 x sin alpha) and alpha can be found by using cosine rule and lies opposite of a. Note: Scrolling down I just discovered Michael Pantano used the same method. A pity for me😢😢.

@deepskytrekkin9600 Ай бұрын

16.25

@Anandkumar-tq2go 2 ай бұрын

There is a mistake in calculation, according to your calculation radius should be half of 65/4 not 65/8 that u Wright's on next step

@itamar4586 2 ай бұрын

You're over complicating it. You need to use the cos formula and then the sin formula, and its ratio is going to be equal to 2R, divide by two and you got it.

@thebig12conference73 26 күн бұрын

Engineer here I can solve without looking at the video. Take the base of the triangle 14 which is the diameter and divide by 2 answer is 7. She did not say find the area she said find the radius. Arewanis pie r squares the formula you know but you don’t use it