Great skill! Love to see how easily you are showing proof theory
@PreMath
4 ай бұрын
I appreciate that! Thanks for the feedback ❤️
@RAG981
4 ай бұрын
Cos rule to find cos C = 7/11. Sine C = 6rt2/11. Then use Sine rule using 2R = 9/sinC to give R = 33rt2/8
@marioalb9726
4 ай бұрын
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 √ )
@jimlocke9320
4 ай бұрын
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.
@lasalleman6792
4 ай бұрын
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
@PrithwirajSen-nj6qq
4 ай бұрын
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 )
@hongningsuen1348
4 ай бұрын
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).
@hongningsuen1348
4 ай бұрын
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.
@zehradiyab3439
4 ай бұрын
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)
@davew9652
4 ай бұрын
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
@unknownidentity2846
4 ай бұрын
Let's find the radius: . .. ... .... ..... 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
@unknownidentity2846
4 ай бұрын
Now I know an easy formula to calculate the radius of the circumscribed circle.🙂 Thanks for that.👍
@murdock5537
4 ай бұрын
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
@marcgriselhubert3915
4 ай бұрын
R = (2.A)/p, where p is the perimeter and A the area given by the Heron formula.
@prossvay8744
4 ай бұрын
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 ❤❤❤
@PreMath
4 ай бұрын
Excellent! Thanks for sharing ❤️
@MrPaulc222
25 күн бұрын
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
@AndreasPfizenmaier-y7w
4 ай бұрын
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
@davew9652
4 ай бұрын
You beat me to it but I'll not delete mine as I used a slightly different approach (pythagorus)
@AmirgabYT2185
4 ай бұрын
r=33√2/8≈5,838
@PreMath
4 ай бұрын
Excellent! Thanks for sharing ❤️
@luigipirandello5919
4 ай бұрын
Belíssimo problema de geometria. Obrigado mestre. O senhor é um professor nota dez.
Формула Героина для определения площади и R=a*b*c/(4*S)
@LuisdeBritoCamacho
4 ай бұрын
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.
@ОльгаСоломашенко-ь6ы
4 ай бұрын
@@LuisdeBritoCamacho pensemos da mesma maneira
@nenetstree914
4 ай бұрын
Using law of cosines
@ManojkantSamal
2 ай бұрын
4.12× root 2 ( may be2)
@misterenter-iz7rz
4 ай бұрын
It has a well-known formula to compute circumradius.😮
@LuisdeBritoCamacho
4 ай бұрын
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.
@jamestalbott4499
4 ай бұрын
Thank you! Appreciated the problem using the formula for a triangle inscribed in a circle.
@rudychan8792
4 ай бұрын
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. ⏳ 😉
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