Great work, Professor!❤😀

@PreMath

5 ай бұрын

Thank you so much! 😃❤️

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

@PreMath

5 ай бұрын

True! Thanks ❤️🌹

@sr2291

4 ай бұрын

Why does that work?

@venkateshwarlujanga9653

3 ай бұрын

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

@JanPBtest

3 ай бұрын

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

@sr2291

3 ай бұрын

@@JanPBtest That's really cool. Thanks.

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

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.

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

5 ай бұрын

That is how I did it.

@PreMath

5 ай бұрын

Super!!! Thanks ❤️🌹

@sudhangshubhattacharya4991

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

Ай бұрын

The one of the intelligent problems found in u tube videos

@jkevincolligan8317

26 күн бұрын

⁠@@sudhangshubhattacharya4991 Great geometric problem !!😅😅😅

nice explanation thank you

good explanation thank you teacher

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)

nicely explained

Nice solution sir

@PreMath

5 ай бұрын

Thanks and welcome❤️🌹

Excelent explanation

@PreMath

5 ай бұрын

Glad you liked it❤️ Thanks ❤️🌹

Very clear. You are a very good teacher

Good job

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

Great work sar❤❤❤❤

@PreMath

5 ай бұрын

Thanks a lot ❤️

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

@quabledistocficklepo3597

Ай бұрын

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

Thankyou sir

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

@Abby-hi4sf

5 ай бұрын

Will you elaborate it please

@ybodoN

5 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

5 ай бұрын

Great! Thanks ❤️🌹

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.

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

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

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

Too remember. Thank you

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

5 ай бұрын

Thanks ❤️🌹

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

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!

very nice

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

@PreMath

5 ай бұрын

Wow! Thanks ❤️🌹

You can use cosine rule to solve for angle alpha

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.

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

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.

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?

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

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

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.

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

19 күн бұрын

How did you derive the measure 84?

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

Beyond words

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

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

@PreMath

5 ай бұрын

Sure! Thanks ❤️🌹

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

@PreMath

5 ай бұрын

Thanks ❤️🌹

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

Use cos rule to work out angle A. BOC is double that. Use cos rule again to get OC=R in triangle BOC😉

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

5 ай бұрын

Great job! Thanks ❤️🌹

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

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

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

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.

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!

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

5 ай бұрын

Great! Thanks ❤️🌹

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

@MaheshKumar-lx1ku

2 ай бұрын

Please share 🙏

@User-jr7vf

2 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

Ай бұрын

@@User-jr7vf thanks

@justarandomnerd3360

Ай бұрын

​@@User-jr7vfno trigonometry

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

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

4 ай бұрын

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

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

@jarikosonen4079

3 ай бұрын

That seems interesting question...

Excellent presentation sir

@PreMath

5 ай бұрын

Thanks dear ❤️🌹

R = abc/(4T) ;T is the area Using Heron's formula, T = sqrt[p(p-a)(p-b)(p-c)] where p = (a+b+c)/2 is half the side length ; gives R= 2730/336 ; and abbreviated by 42 R=65/8 ; alternative and faster solution if exact result is not required: use the cosine relation to find angle A, A=E, use sine for angle E thus obtaining: R= 0.5(15/sin (67.38)) = 8,13

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

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

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

Good morning sir

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

@PreMath

5 ай бұрын

Great! Many thanks ❤️🌹

@STEAMerBear

21 күн 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

21 күн бұрын

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

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

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

5 ай бұрын

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

@dickroadnight

5 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

3 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.

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

Hello Dr Tahir

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

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.

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😢😢.

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

5 ай бұрын

Thanks ❤️🌹

After long time. Can u remember me?

8.13

Very good video, but the constant uhs do really distract

I was expecting to find a general answer.

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.

abc/4∆ =R

8.125

16.25

that was a lot of work! hahaha

65/8=8.125, maybe

65/8?

R=8,125

diameter is 16.25

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

13² = x² + h² h² = 13² - x² 15² = (14 - x)² + h² 15² = 14² - 28x + x² + h² h² = 15² - 14² - x² + 28x h² = (15 + 14)(15 - 14) - x² + 28x h² = 29 - x² + 28x h² = h² 13² - x² = 29 - x² + 28x 13² = 29 + 28x 28x = 13² - 29 x = (13² - 29)/28 = 140/28 = 5 h² = 13² - x² h² = 169 - 25 = 144 h = 12 Chord theorem: 5 * 9 = 12 * y 45 = 12y y = 45/12 = 3.75 12 + 3.75 = 15.75 14 - 2 * 5 = 4 d² = 15.75² + 4² = (63/4)² + 4² = (63/4)² + (16/4)² = (63² + 16²)/16 = 4225/16 d = 65/4 = 16.25 r = 65/8 = 8.125

@PreMath

5 ай бұрын

Thanks ❤️🌹

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.

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE . Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

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

24 күн 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

23 күн бұрын

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

And with this video the circumscribed triangles will no longer have secrets for us!!! 😂

@PreMath

5 ай бұрын

Awesome! Thanks ❤️🌹

@soli9mana-soli4953

5 ай бұрын

@@PreMath ❤

the radius is half the diameter!

there are some geometry gurus in the comment section

You took too long way I got it in very short way its short cut. Just draw divider line to 14 line and at the same time is triangle with 90 degree angle and you have one radius and the divider line is radius/2. And then just solve the problem as 90 degree angle.

I'm

Error! The assumption that B is 90 degrees is wrong.

@axelbehr43

20 күн бұрын

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

@user-pu2fw2jx6t

4 күн бұрын

It isn't an error he used a law in circle theorem where a diameter makes a 90 degree at the other end of the circle. That's in the video at 7:00 which is referred to as Thales theorem

Very hard problem, not trivial

Digital presentation is not appealing well. Instead using a pen is easily understandable.

Heron's formula is simple than your solution

Треугольник (13,14,15) склеен из (5, 12, 13) и (9, 12, 15) по стороне 12. То есть высота к стороне 14 равна 12. Дальше все элементарно.

In English, when a number is multiplied by itself that number is said to have been SQUARED. You keep saying "square." Maybe your enunciation needs to be improved. Mathematics needs to be precise. Thank you for your attention.

Presh Talwalkar does a better job.