Can you find the Radius of the circle? | (Triangle inscribed in a circle) |
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- Опубліковано 19 вер 2024
- 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
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Great work, Professor!❤😀
Thank you so much! 😃❤️
No it's its hard work
To do smart work use properties of circle and chord
Answer expected to come within 6-7 calculations
Area of triangle can be found using Heron's Formula. Then (13 × 15 × 14) ÷ (4 × Area) = R. 🙂
True!
Thanks ❤️🌹
Why does that work?
Properties of Triangles
Area of triangle (Δ)= abc/4R where R is circumradius @@sr2291
@@sr2291 The radius of the circumcircle equals abc/(4*area). For proof, see Wikipedia for "Law of sines".
@@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
Easy solution: Heron's formula gives the area of the triangle.Then the radius of the circle is abc / 4A.
Will you elaborate it please
@@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.
Great!
Thanks ❤️🌹
That's the short way.I think so❤
R=a.b.c/4.∆ which gives R=65/8
Excellent work. Thanks
We can solve it by using formula a*b*c/4*area of triangle 13*14*15/4*84
Sure!
Thanks ❤️🌹
Always helpful videos thanks for doing it @@PreMath
R=ABC/4S...S=(Erone)=4*3*7=84...R=65/8
Wow!
Thanks ❤️🌹
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
Thanks ❤️🌹
Hello sir..I m from India..and watch your videos...I really like it and encourage others to watch
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
How did you derive the measure 84?
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)
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
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.
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 . 🙏❤❤
That is how I did it.
Super!!!
Thanks ❤️🌹
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
The one of the intelligent problems found in u tube videos
@@sudhangshubhattacharya4991
Great geometric problem !!😅😅😅
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.
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
Great job!
Thanks ❤️🌹
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.
Nice solution sir
Thanks and welcome❤️🌹
Cosine rule to find one angle then doing sine rule =2R and you find R
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
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
I did enjoy that. I have not done any Maths for far too long. I hardly know any of it any more.
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
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
Great!
Thanks ❤️🌹
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.
Excelent explanation
Glad you liked it❤️
Thanks ❤️🌹
Thanks mr.
Very well explained. Though different methods r also there, the way u explained in ur own method is superb.
So what? I knew that,. but how would that
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
∆ =84
R=abc/4∆=13.14.15/4.84
=65/8 sq units
Thanks ❤️🌹
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
GREAT WORK PROFESSOR.. YOU ARE MY FAVOURITE MATHS GURUJEE
AFTER A LONG TIME IAM WATCHING YOUR VIDEOS.. BLESS ME SIR.
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.
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
If you calculate this, you get 8.0829. See my more long winded explanation above. Thanks
Beautiful solution , without using the heron's formula & R = abc/ area
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.
Hello Dr Tahir
Very clear. You are a very good teacher
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
2R = 65/4 so R = 65/8
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?
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
S=p*R. S= area using Heron's formula. p= semiperimeter
R=a*b*c/sqrt((a+b+c)*(a+b-c)*(a+c-b)*(b+c-a))
🙂👌🏻🙏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 🤔
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.
Full area 207.3942 & tenagale area 15×11.2=168÷2 =84
Use cos rule to work out angle A. BOC is double that. Use cos rule again to get OC=R in triangle BOC😉
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
Than you I like you
CIRCUMRADIUS = multiplication of sides/ 4 times the area of triangle. 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.
Please elaborate
Please elaborate
Nice explanation but we could use herons formula and area of triangle is equal to abc /4R to find the circumradius
Great work sar❤❤❤❤
Thanks a lot ❤️
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
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
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!
.. very good solution
And intresting 😊😊
ARVINDAR YADAV
New Delhi.. India
16.25÷2=8.125×8.125=66.015625×3.14159268=207.3942 full area sarcol
Excellent presentation sir
Thanks dear ❤️🌹
good explanation
thank you teacher
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.
You can use cosine rule to solve for angle alpha
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.
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.
Un ejercicio muy interesante!
Saludos!
Will this solution hold good if 'O' is outside of triangle ABC?
That seems interesting question...
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.
Someone explain why at 5:59 the triangle 12, 14, 15 is right is 144+196 is not 225.
In triangle BCD , BD isn't equal to 14. BD is 14-X (which is 5) so BD is 9. Now 144+81=225
R (radius of circumcircle)=abc/4∆ where ∆ is the area of triangle
You just found the diameter. To get the radius you must devide the diameter by 2. Thanks
I want to buy the book .where can I get it? and what is its name?
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!
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.😊
Thanks ❤️🌹
why the radius is 65 divided by 8 ?it was divided by 4 ..on previous step ?
There is a far quicker way of solving this, and I'm surprised nobody in the comments mentioned it.
Please share 🙏
@@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.
@@User-jr7vf thanks
@@User-jr7vfno trigonometry
R=8,125
Use cose rule to get the angel. Next a'÷Sina =2r
nice explanation thank you
Used trig to solve it (cosine and then sine law) which in hindsight wasn’t necessary.
Можно найти площадь треугольника по формуле Герона. R=a*b*c/(4*s)=8.125.
Great!
Many thanks ❤️🌹
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!
@@STEAMerBear хочется решить не только правильно, но и быстро. Спасибо.
You are amazing.want to say hi from Ethiopia
nicely explained
Thankyou sir
R=abc/4S R=8.125
Good job
very nice
Beyond words
15×11.2=168÷2=84teangle area
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
Just keep watching.
Pause at the beginning and try to think of an answer.
👍
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.
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.
Too remember. Thank you
Very good video, but the constant uhs do really distract
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
Thanks ❤️🌹
R not mention by pai
I believe r = 65/8. I tried this in my head so I could be way off.
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😢😢.
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.
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!
@@rey-dq3nx You don't need to calculate sine. When you get cosine you just use formula sinx=sqrt(1-(cosx)^2).
8.13
15×11.2=168÷2=84 tgl area
Good morning sir
After long time.
Can u remember me?