Can you find area of the Blue Portion? | (Fun Geometry Problem) |

[@matthieudutriaux]

1/2*200*100*sqrt(3)/2-1/2*200*100*sqrt(3)/4=2500*sqrt(3) mm²

[@MathandEngineering]

Thanks it's accurate

[@juanalfaro7522]

I got the same answer. Let F be midpoint CE -> CF=100/82=50 --> AD = √ (150^2 + (50√3) ^2) = 100√3 --> tan (DAE) = 50√3/150 = √3/3 --> ADE=30 --> FDA = 90-30 = 60. But CDE equilateral ---> ,FDE = 60/2 = 30 --> ADE = 60+30 = 90 ---> ADP] = AD*DP/2. Now DP = DE/2 = 100/2 = 50 ---> [ADP] = 100√3 * 50/2 = 2500√3 = 4330 mm^2.

[@MathandEngineering]

@@juanalfaro7522 perfect 👍, thanks for sharing

[@marioalb9726]

Angle ADE = 90°
b = DP = 200/4 = 50 mm
h = AD = 200 cos30° = 100√3 mm
A = ½b.h = ½ 50 100√3
A = 2500√3 mm² ( Solved √ )

[@MathandEngineering]

Perfect 👍 thanks

[@ediosantos1420]

Brasil

[@MathandEngineering]

Brasil?

[@murdock5537]

way too complicated: same base, same height = same area 🙂
φ = 30° → sin⁡(3φ) = 1; AE = 4a = 200mm → a = 50mm
BAC = ACB = DCE = CED = 2φ; DE = 2a = DP + EP = a + a
sin⁡(EDA) = 1 → DAE = φ → DA = 2a√3 = h →
area ∆ AEP = area ∆ APD = (1/2)area ∆ AED = a^2√3 ≈ 4330(mm)^2
btw: PAE = α → sin⁡(α) = √39/26; DAP = β → sin⁡(β) = √13/13

[@MathandEngineering]

@murdock5537 hmmmmm. So you chose to find an answer whose percentage error is 0. This is good an perfect. Thanks for sharing