Property of any inscribed triangle: A = a.b.c/4R --> R = a.b.c/4A A² = s(s-a)(s-b)(s-c) = 84² A = 84cm² R = (13*14*15)/(4*84) R = 8,125 cm² ( Solved √ )
a = BC = 15, b = AC = 13, c = AB = 14; p = (1/2).(a + b + c) = 21. The erea of ABC is: A = sqrt(p.(p-a).(p-b).(p-c)) = sqrt(21.6.8.7) = sqrt(7056) = 84 The radius of the circle is R = (a.b.c)/(4.A) = (15.13.14)/(4.84) = 2730/336 = 65/8.
*Outro Método:* Lei dos cossenos no ∆ABC: 15² = 14² + 13² - 2×14×13 × cos A 225 = 365 - 364 cos A cos A = (365 - 225)/364 = 140/264 cos A = 5/13. Assim, cos² A + sen² A = 1, daí sen² A = 1 - 25/169 = (169 -25)/169 sen² A = 144/169 → sen A=12/13. Usando a relação: 2r = BC/sen A, temos: 2r = 15 ÷ 12/13 = 15 ×13/12 2r = 5 × 13/4 = 65/4 *r = 65/8 unidades.*
Cosine law. 14^2 = 13^2 + 15^2 - 2 x 13 x 15 x cos C. cos C = 0.5077. C = 59.49 degrees. Dropping perpendicular from O to mid point of AB at D. Angle AOD = 59.49 degrees. Sin 59.49 = 7 / r. r = 7 / sin 59.49. r = 8.125.
Let's find the radius: . .. ... .... ..... First of all we calculate the area of the triangle ABC according to the formula of Heron: s = (AB + AC + BC)/2 = (14 + 13 + 15)/2 = 42/2 = 21 A(ABC) = √[s*(s − AB)*(s − AC)*(s − BC)] = √[21*(21 − 14)*(21 − 13)*(21 − 15)] = √(21*7*8*6) = √(2⁴*3²*7²) = 2²*3*7 = 84 Now we can calculate the radius of the circumscribed circle by using the following formula: R = AB*AC*BC/[4*A(ABC)] = 14*13*15/(4*84) = 65/8 Best regards from Germany
a/sinA=b/sinB=c/sinC=2R Hence R =a/2sinA= abc /2bcsinA =abc/ 4 🔺 [ as 🔺 =1/2 be sinA] 🔺 = √(21*8*7*6) sq units = 3*7*4 sq units R= (13*14*15)/(4*3*7*4) units
About 4 months ago, PreMath also solved a problem where a triangle was inscribed in a circle and we needed to find the radius. The triangle side lengths were different: 9, 10 and 11, but the problem is otherwise identical. The solution method used in that video can be applied to this problem, with the side lengths changed to 13, 14, and 15. Reviewing the 2 videos, I find that the solution methods presented are basically the same. However, the steps are presented in a different order.
I wished that I have noticed that. And I better compare that video to the video from 4 months ago!!! Also this comment makes a ton of sense. And the radius is 65/8 units.
Use heron's law. Find the angles of the triangle. Use formula: circumradius = side of triangle/2(sine of angie opposite). Using angle at C (59.49 degrees) and line AB, Circumradius is 8.125.
I know the triangle area is 84 because I remember it from previous videos (using Heron's Formula), so that was a shortcut for me :) I did it slightly differently, but the main contents were the same.
The Central Angle Theorem always makes me think of the revered Delta Star Trek Insignia . Sometimes at Warp speed the Delta gets a bit skewed as C moves about the circle but Central Angle Theorem maintains it's integrity. 🙂
1) Here circumradius (r) And Sin theta (sin ACB ) has been derived in terms of r and derived relation is as under--- Sin theta (sinACB) =7/r (7 is half of 14 i e the length of AB which is opposite to angle ACB) ---------------------------------- 2)Now I like to prove the above relation in a different way.. ----------------------- 3) PROOF ********************* Construction ঃ --- a) Join AO and extend AO to D( a point on circumference) .AD =2r(r is circumradius). b) Join BD. c) proof +++----++++ I) Now 🔺 ABD is a rt 🔺 where angle ABD is a rt angle as it is an angle on semicircle. II) Hence sin angle ADB =AB/AD=14/2r=7/r III) Now angle ADB =.angle ACB( as both are inscribed on the same arc AB) iv ) Hence sin angle ACB=sin angle ADB=7/r Comment please
STEP-BY-STEP RESOLUTION PROPOSAL : 01) O = (X ; Y) ; A = (0 ; 0) ; B = (5 ; 12) ; C = (0 ; 14) 02) Distances : d(OA) = d(OB) = d(OC) = R 03) d(OA) = X^2 + Y^2 04) d(OB) = (X - 14)^2 + Y^2 05) d(OC) = (X - 5)^2 + (Y - 12)^2 06) Solutions : X = 7 and Y = 33/8 07) R^2 = 7^2 + (33/8)^2 08) R^2 = 4.225/64 09) R = sqrt(4.225/64) 10) R = 65/8 lin un or R = 8,125 lin un OUR BEST ANSWER : R = 65/8 Linear Units or R = 8,125 Linear Units
c² = a² + b² - 2.a.b.cosα cosα = (13²+15²-14²)/(2*13*15) α = 59,49° ( Angle ACB & angle AOD ) sin α = ½c / R R = ½14/ sinα = 8,125 cm (Solved √) There's no need of Heron's formula, neither area of triangle If we calculate area of triangle, has to be any reason !!! : A = 84 cm² = a.b.c/4R R = a.b.c/4A = 13*14*15/(4*84) R = 8,125 cm ( Solved √ ) Why was calculated "sinα=7/R" ???? , if then this formula was not used It doesn't make sense !!! Very confused this vídeo, propierties of inscribed triangles, all mixed up and messy !!
Beautiful work!
I'm glad you enjoyed it! 😊❤️
Thanks for the feedback ❤️🙏
@@PreMath 😀😃
Property of any inscribed triangle:
A = a.b.c/4R --> R = a.b.c/4A
A² = s(s-a)(s-b)(s-c) = 84²
A = 84cm²
R = (13*14*15)/(4*84)
R = 8,125 cm² ( Solved √ )
Excellent!
Thanks for sharing ❤️
a = BC = 15, b = AC = 13, c = AB = 14; p = (1/2).(a + b + c) = 21. The erea of ABC is: A = sqrt(p.(p-a).(p-b).(p-c)) = sqrt(21.6.8.7) = sqrt(7056) = 84
The radius of the circle is R = (a.b.c)/(4.A) = (15.13.14)/(4.84) = 2730/336 = 65/8.
Excellent!
Thanks for sharing ❤️
*Outro Método:*
Lei dos cossenos no ∆ABC:
15² = 14² + 13² - 2×14×13 × cos A
225 = 365 - 364 cos A
cos A = (365 - 225)/364 = 140/264
cos A = 5/13. Assim,
cos² A + sen² A = 1, daí
sen² A = 1 - 25/169 = (169 -25)/169
sen² A = 144/169 → sen A=12/13.
Usando a relação:
2r = BC/sen A, temos:
2r = 15 ÷ 12/13 = 15 ×13/12
2r = 5 × 13/4 = 65/4
*r = 65/8 unidades.*
Excellent!
Thanks for sharing ❤️
Cosine law.
14^2 = 13^2 + 15^2 - 2 x 13 x 15 x cos C.
cos C = 0.5077.
C = 59.49 degrees.
Dropping perpendicular from O to mid point of AB at D.
Angle AOD = 59.49 degrees.
Sin 59.49 = 7 / r.
r = 7 / sin 59.49.
r = 8.125.
Excellent!
Thanks for sharing ❤️
area of triangle ABC = 84 CD = 12
Radius : CO=OE=r CE=2r ⊿ACD∞⊿ECB 13/12 = 2r/15 2r = 65/4 r = 65/8
Excellent!
Thanks for sharing ❤️
CBA=α ..cos(α/2)=√(21*8/15*14)=√(4/5)...S=(1/2)*14*15*sinα=105sin(2arccos√(4/5))=84...r=13*14*15/4*84=13*15/24=195/24
Thanks for sharing ❤️
Let's find the radius:
.
..
...
....
.....
First of all we calculate the area of the triangle ABC according to the formula of Heron:
s = (AB + AC + BC)/2 = (14 + 13 + 15)/2 = 42/2 = 21
A(ABC) = √[s*(s − AB)*(s − AC)*(s − BC)] = √[21*(21 − 14)*(21 − 13)*(21 − 15)] = √(21*7*8*6) = √(2⁴*3²*7²) = 2²*3*7 = 84
Now we can calculate the radius of the circumscribed circle by using the following formula:
R = AB*AC*BC/[4*A(ABC)] = 14*13*15/(4*84) = 65/8
Best regards from Germany
Excellent!
Thanks for sharing ❤️
a/sinA=b/sinB=c/sinC=2R
Hence
R =a/2sinA= abc /2bcsinA
=abc/ 4 🔺 [ as 🔺 =1/2 be sinA]
🔺 = √(21*8*7*6) sq units
= 3*7*4 sq units
R= (13*14*15)/(4*3*7*4) units
Thanks for sharing ❤️
❤❤❤
Excellent!
Thanks dear❤️
About 4 months ago, PreMath also solved a problem where a triangle was inscribed in a circle and we needed to find the radius. The triangle side lengths were different: 9, 10 and 11, but the problem is otherwise identical. The solution method used in that video can be applied to this problem, with the side lengths changed to 13, 14, and 15.
Reviewing the 2 videos, I find that the solution methods presented are basically the same. However, the steps are presented in a different order.
I'm glad you recognized the similarity in the problems and how the solution method applies to both scenarios!
Thanks for the feedback ❤️
I wished that I have noticed that. And I better compare that video to the video from 4 months ago!!! Also this comment makes a ton of sense. And the radius is 65/8 units.
Use heron's law. Find the angles of the triangle. Use formula: circumradius = side of triangle/2(sine of angie opposite). Using angle at C (59.49 degrees) and line AB, Circumradius is 8.125.
Excellent!
Thanks for sharing ❤️
✨Magic!✨
I would've had to have thought about that one quite a while to get it.
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
Best Wishes for DIWALI. Happy Diwali to Premath team.
Thanks dear🙏
Happy Diwali❤️
I know the triangle area is 84 because I remember it from previous videos (using Heron's Formula), so that was a shortcut for me :) I did it slightly differently, but the main contents were the same.
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
The Central Angle Theorem always makes me think of the revered Delta Star Trek Insignia . Sometimes at Warp speed the Delta gets a bit skewed as C moves about the
circle but Central Angle Theorem maintains it's integrity. 🙂
Per William Shatner: Get a life. 😂😂
Right on!😀
Thanks for the feedback ❤️
R=a*b*c/(4* S), S по формуле Герона.
Excellent!
Thanks for the feedback ❤️
Just use Cosine rule for theta (or any angle) the Sine Rule says that 2R = sin theta/ opp side. End
Thanks for the feedback ❤️
1) Here circumradius (r)
And Sin theta (sin ACB ) has been derived in terms of r and derived relation is as under---
Sin theta (sinACB) =7/r
(7 is half of 14 i e the length of AB which is opposite to angle ACB)
----------------------------------
2)Now I like to prove the above relation in a different way..
-----------------------
3) PROOF
*********************
Construction ঃ ---
a) Join AO and extend AO to D( a point on circumference) .AD =2r(r is circumradius).
b) Join BD.
c) proof
+++----++++
I) Now 🔺 ABD is a rt 🔺 where angle ABD is a rt angle as it is an angle on semicircle.
II) Hence sin angle ADB =AB/AD=14/2r=7/r
III) Now angle ADB =.angle ACB( as both are inscribed on the same arc AB)
iv ) Hence sin angle ACB=sin angle ADB=7/r
Comment please
Excellent!
Thanks for sharing ❤️
21R=84=>R=4 unit.ans
Using cos(theta) and cos(2theta) for solution.
Thanks for the feedback ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) O = (X ; Y) ; A = (0 ; 0) ; B = (5 ; 12) ; C = (0 ; 14)
02) Distances : d(OA) = d(OB) = d(OC) = R
03) d(OA) = X^2 + Y^2
04) d(OB) = (X - 14)^2 + Y^2
05) d(OC) = (X - 5)^2 + (Y - 12)^2
06) Solutions : X = 7 and Y = 33/8
07) R^2 = 7^2 + (33/8)^2
08) R^2 = 4.225/64
09) R = sqrt(4.225/64)
10) R = 65/8 lin un or R = 8,125 lin un
OUR BEST ANSWER :
R = 65/8 Linear Units or R = 8,125 Linear Units
Excellent!
Thanks for sharing ❤️
R=65/8units.❤
Excellent!
Thanks for sharing ❤️
16.25÷2=8.125
Thanks for sharing ❤️
8:40
why you say (2r) not one r
He multiplies the 2 and r those were in denominator and he writes 2r in the denominator and pronounces 2r
@PrithwirajSen-nj6qq thanks 💚🌹
c² = a² + b² - 2.a.b.cosα
cosα = (13²+15²-14²)/(2*13*15)
α = 59,49° ( Angle ACB & angle AOD )
sin α = ½c / R
R = ½14/ sinα = 8,125 cm (Solved √)
There's no need of Heron's formula, neither area of triangle
If we calculate area of triangle, has to be any reason !!! :
A = 84 cm² = a.b.c/4R
R = a.b.c/4A = 13*14*15/(4*84)
R = 8,125 cm ( Solved √ )
Why was calculated "sinα=7/R" ???? , if then this formula was not used
It doesn't make sense !!!
Very confused this vídeo, propierties of inscribed triangles, all mixed up and messy !!
A = a.b.c/4R --> R = a.b.c/4A
A² = s(s-a)(s-b)(s-c) = 84²
A = 84cm²
R = (13*14*15)/(4*84)
R = 8,125 cm² ( Solved √ )
A/sinA=b/sinb=2r
Thanks for the feedback ❤️