Can you find the Radius of the circumscribed circle? | (In-depth Proof) | #math #maths | #geometry

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Can you find the Radius of the circumscribed circle? | (In-depth Proof) | #math #maths | #geometry

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[Abby-hisf]

Great skill! Love to see how easily you are showing proof theory

[marioalb]

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 √ )

[jimlocke]

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.

[RAG]

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

[PrithwirajSen-njqq]

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 )

[zehradiyab]

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)

[hongningsuen]

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).

[murdock]

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

[lasalleman]

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

[unknownidentity]

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

[prossvay]

Area of triangle ABC
S=(9+10+11)/2=15

units ❤❤❤

[davew]

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

[marcgriselhubert]

R = (2.A)/p, where p is the perimeter and A the area given by the Heron formula.

[AndreasPfizenmaier-yw]

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

[MrPaulc]

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

[luigipirandello]

Belíssimo problema de geometria. Obrigado mestre. O senhor é um professor nota dez.

[ОльгаСоломашенко-ьы]

Формула Героина для определения площади и R=a*b*c/(4*S)

[misterenter-izrz]

It has a well-known formula to compute circumradius.😮

[LuisdeBritoCamacho]

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.

[jamestalbott]

Thank you! Appreciated the problem using the formula for a triangle inscribed in a circle.