Lune of Hippocrates | Can you find area of the Purple shaded region? |
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- Опубліковано 7 лют 2025
- Learn how to find the area of the Purple shaded region. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; area of a triangle formula; Semicircle; Quarter circle; area of a circle formula. Step-by-step tutorial by PreMath.com
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Lune of Hippocrates | Can you find area of the Purple shaded region? | #math #maths | #geometry
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It's the Lune of Hippocrates! Awesome problem!
Thanks for the feedback ❤️🙏
Well explained
It was easy peasy
A=(π*(4*√2)²/2)-((π*8²)/4-(8*8/2))
A=32
Surprenant et amusant. Bonne journée
Too easy to solve
Nice problem sir
I solved it 😊😊😊
R=r√2 --> R²= 2r² , Then :
Semicircle area = Quarter circle area
Then :
Shaded area = Isosceles right triangle area
A = ½bh = ½R² = ½8² = 32cm²
Although not mentioned, I'm assuming the wider arc between A and C is a semicircle - it looks that way visually.
The straight line AC is 8*sqrt(2).
r (OC or OB) = 4*sqrt(2). Area of semicircle is 16pi un^2.
Calculate inner arc of AC and subtract it from 16pi.
Quadrant ABC area = 16pi.
Inner arc AC = 16pi - 32
Sln: 16pi - (16pi - 32)
Purple area = 32 un^2.
Bom dia Mestre
Desejo-lhe um mês abençoado q se inicia
Grato pela aula
Thank you!
S=32 square units
АС=8√2, это диаметр полукруга. Он же - основание ▲АВС. Сперва ищем площадь жёлтого сегмента, она получается вычитанием треугольника из четверти круга. Площадь четверти круга S₁=π8²/4=8π*2=16π. Площадь равностороннего прямоугольного ▲АВС S(ABC)=8²/2=8*4=32. S₂=16π-32. Радиус полукруга r=8√2/2=4√2. Его площадь S₂=(4√2)²π/2=16*2π/2=16π. Таким образом искомая площадь серпа S=16π-(16π-32)=16π-16π+32=32, т. е. точно равна площади треугольника.
Radio púrpura =r=CO=8√2/2=4√2 ---> Área sombreada púrpura =(πr²/2)-(π8²/4)+(8²/2) =16π-16π+64/2=32 u².
Gracias y un saludo cordial.
Semicircle:
A₁= ½π(R/√2)² = ½πR²/2= ¼πR²
Circular Segment :
A₂= ½R²(α-sinα)= ½R²(π/2-1)= ¼πR²-½R²
Shaded area:
A= A₁-A₂= ½R²= ½8²= 32 cm² (Solved √)
We have the triangle with sides 8 units and a rt. Angled one so the Hypotenuse is 8^2+8^2.√ of that. @ √128. This is the diameter. Hence its radius is √128/2. =8/√2.contd after seeing the figure.
A₁ = ¼πR²
A₂ = ½π(R/√2)² = ½πR²/2= ¼πR²
A₁=A₂ ; A = A₁-A₂+A₃
A=A₃= ½bh = ½8² = 32 cm² ( Solved √)
AO=OC=√[8²+8²]/2=4√2
Purple shaded area = 8*8*1/2 + 4√2*4√2*π*1/2 - 8*8*π*1/4 = 32 + 16π - 16π = 32
The Area of the Lune 🌙 is the same as the Area of the Triangle 🔺️
Areas ( Sum of both purpol and Yellow is πr^2./2.So it is Half
1/ Label BC=AB = a --> OA=OC=a sqrt2/2
2/ Let A be the area of the isosceles triangle ABC
We have: Area of the yellow segment = area of the 1/4 yellow circle- A
= pix sqa/4 - A (1)
And Area of the purple region = 1/2 Area purple circle - Area of the yelow segment
= 1/2 xpi x sq (asqrt2/2)-
( pixsqa/4-A)
= pix sqa/4 -pixsqa/4+ A
--> Area of the purple region = A=32 sq units😅😅😅
I’m not a math wizard by any means but last I checked acres is a measurement of area
That is what the problem is
To find the area
The measurements given are linear
It if Half πr^2.. But r is 8/√2. U have π×(8/√2) ^2. So it 64/4. Which is 16. Sq units
Let's find the area:
.
..
...
....
.....
Let R be the radius of the quarter circle and let r be the radius of the semicircle. Since ABC is a right triangle, we can apply the Pythagorean theorem:
AC² = AB² + BC²
(2r)² = R² + R²
4r² = 2R²
⇒ r²/2 = R²/4
Now we are able to calculate the area of the purple region:
A(purple)
= A(semicircle) − A(yellow circle segment)
= A(semicircle) − [A(yellow quarter circle) − A(triangle ABC)]
= A(semicircle) − A(yellow quarter circle) + A(triangle ABC)
= πr²/2 − πR²/4 + A(triangle ABC)
= πR²/4 − πR²/4 + A(triangle ABC)
= A(triangle ABC)
= (1/2)*AB*BC
= R²/2
= 8²/2
= 32
Best regards from Germany
😮😮
😮😮
Yellow quarter circle area = (8)(8)π/4 = 16π
Yellow triangle area = (8)(8)/2 = 32
Semicircle area = (π/2)[(8√2)/2]^2 = 16π
Purple area = 16π + 32 - 16π = 32
Easy. The radius of the big circle is R = OA = OC = (8.sqrt(2))/2 = 4.sqrt(2). The area of the big semi circle is (Pi.(R^2))/2 = (Pi.32)/2 = 16.Pi
The area of the triangle ABC 1s (1/2).8.8 = 32. The area of the small quater circle is (Pi.(r^2))/4 with r = 8, so it is (Pi.64)/4 = 16.Pi
Finally the purple area is 16.Pi + 32 - 16.Pi = 32.
32 sir
32
MY RESOLUTION PROPOSAL :
01) OC = OA = R lin un
02) Semicircle Radius (R) = (4 * sqrt(2)) lin un
03) Semicircle Area (SA) = (R^2 * Pi) / 2 ; SA = 32Pi sq un ; SA ~ 100,53 sq un
04) AB = BC = 8 lin un
05) Circular Segment Area (CSA) = [(8^2/2) * (Pi/2 - sin(Pi/2)] ; CSA = 32 * (Pi - 1) ; CSA = (32Pi - 32) sq un ; CSA ~ 68,53 sq un
06) Purple Shaded Area (PSA) = [32Pi - (32Pi - 32)] sq un ; PSA = 32Pi - 32Pi + 32 ; PSA = 32 sq un
Therefore,
MY ANSWER IS:
Beyond any reasonable doubt the Purple Shaded Area is equal to 32 Square Units.
ua-cam.com/video/NGsuCPFjStI/v-deo.htmlsi=n3If6L01GPvrGK2y
(22÷(7*2)×(4√(2)×4√(2))) -
((22÷(4*7)×(8×8)-(8×4))=
31.99999999999998
=32