s=½(a+b+c) = ½(9+10+11) =15 cm Heron's formula: A² = s (s-a)(s-b)(s-c) A²= 15(15-9)(15-10)(15-11) A = 42,43 cm² A = a . b . c / 4R R = a. b. c / 4A R = 9 . 10 . 11 / (4 . 42,43) R = 5,83 cm ( Solved √ )
Drop a perpendicular from A to BC and label the intersection as point D. Let BD = x, then CD = 10 - x. Let AD = y. ΔABC has been divided into two right triangles, ΔABD and ΔACD. Applying the Pythagorean theorem to ΔABD, x² + y² = 9² = 81, and to ΔACD, (10 - x)² + y² = (11)² = 121, or 100 - 20x + x² + y² = 121. Replace x² + y² by 81: 100 - 20x + 81 = 121. Solve for x and find x = 3, so y = 6√2. Extend AD into a chord, labelling the other intersection with the circle as point E. Note that BD = 3, CD = 7, AD = 6√2 and DE is unknown, but found to be 7(√2)/4 from the intersection chords theorem, (AD)(DE) = (BD)(CD) in this case. So length AE = 6√2 + 7(√2)/4 = 31(√2)/4. Drop a perpendicular from O to BC and label the intersection as point F, noting that OF bisects BC, so BF = 5. Construct OB. Note that ΔOBF is a right triangle, OB is a radius r, BF is 5 and OF is the distance from D to the midpoint of AE, which is length AE/2 - length DE = 31(√2)/8 - 7(√2)/4 = 17(√2)/8. So r² = (17(√2)/8)² + (5)² = 578/64 + 25 = 2178/64. r = √(2178)/√(64) = (√(1089))(√2)/8 = 33(√2)/8, as PreMath also found.
I dropped a perpendicular from A, through BC to the circumference. Worked out the height of the triangle and then used the intersecting chord theorum 4r^2=a^2+b^2+c^2+d^2 to find radius
The formula circumradius = abc/(4 x area of trianlge) holds even the circumcentre is outside the triangle. A more general proof uses Thales theorem: 1. Start with triangle ABC with the usual notations for angles and sides circumscribed by circumcircle with centre O and circumradius R. 2. Draw a diameter BD from B (or other vertex of the triangle) through O to point D on the other side of the circumference. It does not matter if O is inside or outside the triangle. 3. Triangle BAD is a right-angled triangle by Thales theorem as triangle is in semicircle. 4. With chord AB, the angles in same segment theorem gives angle ADB = angle C of triangle. 5. sinC = sin(ADB) = AB/BD (sine value from sides of right-angled triangle) = c/2R. 6. Area of triangle ABC = (1/2) a b sinC = (1/2) a b (c/2R) = (abc)/4R. Hence R = abc/(4 x area of triangle).
Step 4 is for acute angle C. For obtuse angle C, angle C is supplementary to angle ADB as opposite angles of cyclic quadrilateral. By trigonometric identity, sin(180 - theta) = sin(theta). Hence steps 5 and 6 still hold true.
Herons formula tells us an Area of 42,42. Base 10* half of height, Hence perpendicular is 8,48. Prolong this height to the Circle. 2,47 is length of this Part of the chord. 4r^2=a^2+b^2+c^2+d^2. Hence r= 5,83
Or: circumradius = a/2sin(A) Here, a is a side of the triangle, A is the angle opposite of side a. Using 9 as the side and twice the sin of 50.479 degrees at ACB, Circumradius: 5.83
I solved it by using cos rules Suppose A is angle in a segment opposite the chord which equal 9 9²= 11²+10²-2×11×10 cosA cosA= 140/220= 7/11 cos(2A)= 2cos²A-1= 2(49/121)-1 = -23/121 9²= r²+r²-2r×rcos(2A) 81= 2r²-2r²(-23/121) 81= 2r²(144/121) 9= sqrt(2)×r(12/11) r= (33/8) sqrt(2)
Utilizei, exatamente, o mesmo MODUS OPERANDI. Calcular a Área do Triângulo (A) com a Fórmula de Heron e depois R = (a * b * c) / (4 * A). Sendo a ; b ; c ; os respetivos lados de Triângulo Escaleno. Obrigado.
I did it different, but was wrong by a small margin. I went for 30*sqrt(2) = 15h, so (average)h = 2*sqrt(2). I ended up with r = sqrt(33)> This would have been ok if it was equilateral with side lengths of the average 10. I got r = 5.74 rather than your 5.83
STEP-BY-STEP RESOLUTION PROPOSAL : 1) Triangle [ABC] Area = A = sqrt(15 * 6 * 5 * 4) ; A = sqrt(1800) ; A = (30 * sqrt(2)) sq un 2) R = (9 * 10 * 11) / (4 * A) 3) R = 990 / (4 * 30 * sqrt(2)) 4) R = 990 / (120 * sqrt(2)) 5) R = 33 / (4 * sqrt(2)) 6) R = (33 * sqrt(2)) / (4 * 2) 7) R = [(33 * sqrt(2)) / 8] lin un 8) R ~ 5,834 lin un 9) ANSWER : The Radius Length is approx. equal to 5,834 Linear Units. Greetings from Cordoba Caliphate University, the Center of Ancient Greek, Persian, Indian and Arabic Mathematical Knowledge and Wisdom.
Not Need Finding: sin 8 = 5/r ?? ❌❌ Straight to the Point: Heron's Formula then R = a•b•c / 4•A = (33/8)•√2 = 5,834 You can Cut this video 4' shorter. ⏳ 😉
Great skill! Love to see how easily you are showing proof theory
I appreciate that!
Thanks for the feedback ❤️
s=½(a+b+c) = ½(9+10+11) =15 cm
Heron's formula:
A² = s (s-a)(s-b)(s-c)
A²= 15(15-9)(15-10)(15-11)
A = 42,43 cm²
A = a . b . c / 4R
R = a. b. c / 4A
R = 9 . 10 . 11 / (4 . 42,43)
R = 5,83 cm ( Solved √ )
Drop a perpendicular from A to BC and label the intersection as point D. Let BD = x, then CD = 10 - x. Let AD = y. ΔABC has been divided into two right triangles, ΔABD and ΔACD. Applying the Pythagorean theorem to ΔABD, x² + y² = 9² = 81, and to ΔACD, (10 - x)² + y² = (11)² = 121, or 100 - 20x + x² + y² = 121. Replace x² + y² by 81: 100 - 20x + 81 = 121. Solve for x and find x = 3, so y = 6√2. Extend AD into a chord, labelling the other intersection with the circle as point E. Note that BD = 3, CD = 7, AD = 6√2 and DE is unknown, but found to be 7(√2)/4 from the intersection chords theorem, (AD)(DE) = (BD)(CD) in this case. So length AE = 6√2 + 7(√2)/4 = 31(√2)/4. Drop a perpendicular from O to BC and label the intersection as point F, noting that OF bisects BC, so BF = 5. Construct OB. Note that ΔOBF is a right triangle, OB is a radius r, BF is 5 and OF is the distance from D to the midpoint of AE, which is length AE/2 - length DE = 31(√2)/8 - 7(√2)/4 = 17(√2)/8. So r² = (17(√2)/8)² + (5)² = 578/64 + 25 = 2178/64. r = √(2178)/√(64) = (√(1089))(√2)/8 = 33(√2)/8, as PreMath also found.
Let's find the radius:
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First of all we calculate the area of the triangle using Heron's formula:
s = (a + b + c)/2 = (9 + 10 + 11)/2 = 30/2 = 15
A(ABC) = √[s*(s − a)*(s − b)*(s − c)] = √[15*(15 − 9)*(15 − 10)*(15 − 11)] = √(15*6*5*4) = 30√2
For the height of the triangle according to its base BC we obtain:
A(ABC) = (1/2)*BC*h(BC)
⇒ h(BC) = 2*A(ABC)/BC = 2*30√2/10 = 6√2
Now let's assume that O is the center of the coordinate system and that BC is parallel to the x-axis. Then we obtain the following coordinates:
O: ( 0 ; 0 )
A: ( xA ; yA )
B: ( −5 ; yB )
C: ( +5 ; yB )
Now we can try to calculate the radius R of the circumscribed circle:
(xB − xA)² + (yB − yA)² = AB²
(xB − xA)² + h(BC)² = AB²
(−5 − xA)² + (6√2)² = 9²
(−5 − xA)² + 72 = 81
(−5 − xA)² = 9
−5 − xA = −3
⇒ xA = −2
xA² + yA² = R² ∧ xB² + yB² = R²
xB² + yB² = xA² + yA²
yB² − yA² = xA² − xB²
(yB − yA)(yB + yA) = (−2)² − (−5)² = 4 − 25 = −21
⇒ yB + yA = −21/(yB − yA) = 21/h(BC) = 21/6√2 = (7/4)√2
yA − yB = 6√2
yA + yB = (7/4)√2
⇒ yA = (7/4 + 6)√2/2 = (7/8 + 3)√2 = (+31/8)√2
∧ yB = (7/4 − 6)√2/2 = (7/8 − 3)√2 = (−17/8)√2
Let's check these results:
AC² = (xC − xA)² + (yC − yA)² = (5 + 2)² + h(BC)² = 7² + (6√2)² = 49 + 72 = 121 = 11² ✓
Now we are able to calculate the radius R of the circumscribed circle:
R² = xA² + yA² = (−2)² + (+31√2/8)² = 4 + 961/32 = 128/32 + 961/32 = 1089/32
R² = xB² + yB² = (−5)² + (−17√2/8)² = 25 + 289/32 = 800/32 + 289/32 = 1089/32 ✓
⇒ R = √(1089/32) = 33/(4√2) = (33/8)√2
Best regards from Germany
Now I know an easy formula to calculate the radius of the circumscribed circle.🙂 Thanks for that.👍
I dropped a perpendicular from A, through BC to the circumference. Worked out the height of the triangle and then used the intersecting chord theorum 4r^2=a^2+b^2+c^2+d^2 to find radius
The formula circumradius = abc/(4 x area of trianlge) holds even the circumcentre is outside the triangle.
A more general proof uses Thales theorem:
1. Start with triangle ABC with the usual notations for angles and sides circumscribed by circumcircle with centre O and circumradius R.
2. Draw a diameter BD from B (or other vertex of the triangle) through O to point D on the other side of the circumference.
It does not matter if O is inside or outside the triangle.
3. Triangle BAD is a right-angled triangle by Thales theorem as triangle is in semicircle.
4. With chord AB, the angles in same segment theorem gives angle ADB = angle C of triangle.
5. sinC = sin(ADB) = AB/BD (sine value from sides of right-angled triangle) = c/2R.
6. Area of triangle ABC = (1/2) a b sinC = (1/2) a b (c/2R) = (abc)/4R. Hence R = abc/(4 x area of triangle).
Step 4 is for acute angle C.
For obtuse angle C, angle C is supplementary to angle ADB as opposite angles of cyclic quadrilateral.
By trigonometric identity, sin(180 - theta) = sin(theta). Hence steps 5 and 6 still hold true.
Herons formula tells us an Area of 42,42. Base 10* half of height, Hence perpendicular is 8,48. Prolong this height to the Circle. 2,47 is length of this Part of the chord. 4r^2=a^2+b^2+c^2+d^2. Hence r= 5,83
You beat me to it but I'll not delete mine as I used a slightly different approach (pythagorus)
Or: circumradius = a/2sin(A)
Here, a is a side of the triangle, A is the angle opposite of side a. Using 9 as the side and twice the sin of 50.479 degrees at ACB, Circumradius: 5.83
R = (2.A)/p, where p is the perimeter and A the area given by the Heron formula.
R=abc/4 🔺
🔺 =√(15*4*5*6)=30√2 sq units
R=9*10*11/4*30√2= 8.25/√2=5.834 units (approx )
Belíssimo problema de geometria. Obrigado mestre. O senhor é um professor nota dez.
Great! Again..!
I solved it by using cos rules
Suppose A is angle in a segment opposite the chord which equal 9
9²= 11²+10²-2×11×10 cosA
cosA= 140/220= 7/11
cos(2A)= 2cos²A-1= 2(49/121)-1 = -23/121
9²= r²+r²-2r×rcos(2A)
81= 2r²-2r²(-23/121)
81= 2r²(144/121)
9= sqrt(2)×r(12/11)
r= (33/8) sqrt(2)
Area of triangle ABC
S=(9+10+11)/2=15
Area=√15(15-9)(15-10)(15-11)=30√2
R=(9)(10)(11)/(4)30√2=33√2/8=5.83 units ❤❤❤
Excellent!
Thanks for sharing ❤️
Circumradius = abc/4*area
r=33√2/8≈5,838
Excellent!
Thanks for sharing ❤️
Формула Героина для определения площади и R=a*b*c/(4*S)
Utilizei, exatamente, o mesmo MODUS OPERANDI. Calcular a Área do Triângulo (A) com a Fórmula de Heron e depois R = (a * b * c) / (4 * A). Sendo a ; b ; c ; os respetivos lados de Triângulo Escaleno. Obrigado.
@@LuisdeBritoCamacho pensemos da mesma maneira
This is awesome, many thanks, Sir!
φ = 30°; ∆ ABC → BC = a = 10; AC = b = 11; AB = c = 9; AO = BO = CO = r = ? BCA = ϑ → BOA = 2ϑ
81 = 100 + 121 - 2(10)11cos(ϑ) → cos(ϑ) = 7/11 → sin(ϑ) = √(1 - cos^2(ϑ)) = 6√2/11 →
cos(2ϑ) = cos^2(ϑ) - sin^2(ϑ) = -23/121 → 2ϑ > 3φ → 81 = 2r^2(1 - cos(2ϑ)) → r = 33√2/8
r=ABC/4A=9*10*11/4√(15*6*5*4)=990/4*√1800=990/120√2=99/12√2
Using law of cosines
I did it different, but was wrong by a small margin.
I went for 30*sqrt(2) = 15h, so (average)h = 2*sqrt(2).
I ended up with r = sqrt(33)> This would have been ok if it was equilateral with side lengths of the average 10.
I got r = 5.74 rather than your 5.83
4.12× root 2 ( may be2)
STEP-BY-STEP RESOLUTION PROPOSAL :
1) Triangle [ABC] Area = A = sqrt(15 * 6 * 5 * 4) ; A = sqrt(1800) ; A = (30 * sqrt(2)) sq un
2) R = (9 * 10 * 11) / (4 * A)
3) R = 990 / (4 * 30 * sqrt(2))
4) R = 990 / (120 * sqrt(2))
5) R = 33 / (4 * sqrt(2))
6) R = (33 * sqrt(2)) / (4 * 2)
7) R = [(33 * sqrt(2)) / 8] lin un
8) R ~ 5,834 lin un
9) ANSWER : The Radius Length is approx. equal to 5,834 Linear Units.
Greetings from Cordoba Caliphate University, the Center of Ancient Greek, Persian, Indian and Arabic Mathematical Knowledge and Wisdom.
Thank you! Appreciated the problem using the formula for a triangle inscribed in a circle.
11/10 = 1.1 unit ?
Not Need Finding: sin 8 = 5/r ?? ❌❌
Straight to the Point:
Heron's Formula then R = a•b•c / 4•A = (33/8)•√2 = 5,834
You can Cut this video 4' shorter. ⏳ 😉