Can you find the radius of the circle? | Triangle inscribed in a circle | 3 Different Methods 

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After finding AD=3 and DC=2, we can find the 3rd the length of side AC =√(3²+2²)=√13 (using Pythagoras theorem) We can use abc=4R△ (an identity/formula) to find R, the circumcircle radius Plugging in a=6, b=√13, c=5, △=area of triangle=9 in this formula,we get: i.e., (6)(√13)(5)=4(R)(9)=> R=(6)(√13)(5)/[(4)(9)]=(5√13)/6 units

I used Heron's formula and came up with two solutions to line length AC. AC = √(13) which gives radius about 3.004626 or AC=√(109) which gives radius about 8.700255. I put this in FreeCAD sketcher and it shows to work.

AC can be equal to the root of 109.

Let CA = b, AB = c = 5, BC = a = 6, ∠A = α, ∠B = β, and ∠C = γ. Let O be the center point of the circle. First method: Triangle ∆ABC: [ABC] = acsin(β)/2 9 = 6(5)sin(β)/2 9 = 15sin(β) sin(β) = 9/15 = 3/5 = O/H cos(β) = A/H = √(H²-O²)/H cos(β) = √(5²-3²)/5 cos(β) = √(25-9)/5 cos(β) = √16/5 = 4/5 b² = a² + c² - 2accos(β) CA² = 6² + 5² - 2(6)5(4/5) CA² = 36 + 25 - 48 CA² = 13 CA = √13 2r = b/sin(β) = √13/(3/5) r = 5√13/6 ≈ 3.0046 cm Method 2: Drop a perpendicular from A to D on BC. Let DA = h. [ABC] = bh/2 = 6h/2 = 3h 9 = 3h h = 3 As AB = 5 and DA = 3, ∆BDA is a 3-4-5 Pythagorean triple triangle and BD = 4. As DC = BC-BD, DC = 6-4 = 2. Triangle ∆CDA: CD² + DA² = CA² 2² + 3² = CA² CA² = 4 + 9 = 13 CA = √13 Draw a diameter from C to E, passing through O, and draw EA. As CE is a diameter, by Thales' Theorem, ∠EAC = 90°. Additionally, as ∠E and ∠B both cover the arc AC and E and B are both on the circumference of circle O, ∠E = ∠B = β. As ∠E = ∠B = β and ∠EAC = ∠ADB = 90°, ∆EAC and ∆ADB are similar. CE/CA = AB/DA 2r/√13 = 5/3 6r = 5√13 r = 5√13/6 = 3.0046 cm

1. tiangle's area via (AB * BC * sinB) => We find sinB 2. Find AC using cosine theorem 3. Find R with sin theorem

I believe, it's wrong as angle B can be either acute (90 degrees), so point D may be either inside BC or on the continuation of BC. This means that there are actually two solutions. Let me show another method giving both of them. We have: Area = 1/2 * 5 * 6 sin B = 9. Therefore sin B = 3/5, cos B = ±4/5. Let b = length of AC. According to cosine theorem: b² = 5² + 6² ∓ 2*5*6*4/5, so b²=61∓48. That gives two values of b: b₁ =√13₁ and b₂ = √109. According to sine theorem R = b/(2sin B). Since sin B = 3/5, R = 5b/6, so R₁ = 5√13 /6 (from the video) and R₂ = 5√109 /6 (missed by the video). Actually methods of the video can be corrected to give both solutions, by considering two different positions of point D.

Another way to solve is to construct perpendicular bisector to BC at point O'. Construct line through O parallel to BC intersecting AD at O". Let X = OO'. 2 right triangles can be formed. One OO'B (sides R, 3, X) & OO"A (sides R, 3 - X, 1). Using Pythagorean formula on both gives X = 1/6, & from this & triangle OO'B you get R = sqrt(325/36) which simplifies to (5/6)√13 .

r=(5×root13)/6, or (5×root109)/6 May be the second one is not accepted

There is a simple formula for the radius of a circumscribing circle which is abc/4K where K is the area of the triangle. I taught this to Mathletes 40+ years ago.

(axbxc)/4R=Area, so AC=(13)^1/2, so R is easy

2R = a/sinA =b/sinB =c/sinC

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Fun problem.

I always seem to do things 'the harder way'. Basically, I found the 2 functions-of-a-line for line A and line C. Then inverted them, choosing the 𝒃 in [ f(𝒙) = 𝒎𝒙 + 𝒃 ] for each inversion to be the center point of the A and C lines. This way, 'mathematically crossed' (equated), I found the 𝒙 on line B to be perpendicular to the 𝒐 origin; Using either inverted equation with that gives a 𝒚 Then [1.1]  𝒓² = 𝒙² + 𝒚² Lemme tell you though, it sounds good, but the algebra is tiresome. First start with finding an imaginary height to the triangle: [1.1]  area = ½𝒉𝒃 [1.2]  2 × 9 = 6 𝒉 [1.3]  𝒉 = 3; Also need the 𝒌 part, which is the left-part of the baseline 𝒃 [2.1]  𝒌 = √(𝒂² - 𝒉²) [2.2]  𝒌 = √(25 - 9) [2.3]  𝒌 = 4 OK, now the formulæ for the A(𝒙) and C(𝒙) sides [3.1]  A(𝒙) = (𝒉/𝒌)𝒙 ⊕ 0 [3.2]  A(𝒙) = ¾𝒙 ⊕ 0 [4.1]  C(𝒙) = (-𝒉/(𝒃 - 𝒌))𝒙 + 𝒃𝒉/(𝒃 - 𝒌) [4.2]  C(𝒙) = -1.5𝒙 ⊕ 9 Now need the inversions that have trailing constants that make 'em intercept the ½𝒉 Inversion of A(𝒙) is [5.1]  iA(𝒙) = (-1/𝒎)𝒙 + something [5.2]  iA(𝒙) = -4⁄3𝒙 + (3² ⊕ 4²)/(2×3) [5.3]  iA(𝒙) = -⁴⁄3𝒙 + 25⁄6 And the inversion of C(𝒙) is [6.1]  iC(𝒙) = (-1/𝒎)𝒙 + something [6.2]  iC(𝒙) = ⅔𝒙 + (3² ⊕ 4² - 6²)/(2×3) [6.3]  iC(𝒙) = ⅔𝒙 - 11⁄6 Now set [5.3] and [6.3] to be equal, finding where these lines cross [7.1]  -⁴⁄3𝒙 + 25⁄6 = ⅔𝒙 - ¹¹⁄6 … rearrange [7.2]  36⁄6 = 6⁄3𝒙 [7.3]  6 = 2𝒙 [7.4]  𝒙 = 3 then use either [5.3] or [6.3] equation to find 𝒚 [8.1]  𝒚 = -4⁄3 × 3 + 25⁄6 [8.2]  𝒚 = -4 + 25⁄6 [8.3]  𝒚 = ⅙ And of course using Pythagoras, now we find radius [9.1]  𝒓 = √(𝒙² + 𝒚²) [9.2]  𝒓 = √(3² + (⅙)²) [9.3]  𝒓 = ⅚√13; [9.4]  𝒓 = 3.00462 BUT … rather remarkably (which I didn't remember beforehand) [9.5]  𝒓 = ⅚√13 = 𝒂𝒄/𝒃 How about that!!! I've derived that time, and time again, and it always surprises me when it pops back up. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅