Can you find area of circle? | (Fun Geometry Problem) | #math #maths | #geometry

[Abdelfattah-hrtt]

Great job it's so difficult but will trying more and more

[santiagoarosam]

Prolongamos BO y corta la circunferencia en E---> DE=2, 5---> EDB es simétrico de ECB respecto al diámetro BE---> BE y CD se cortan en F---> CF=FD=4/2=2---> FE=√(2, 5²-2²)=3/2 ---> Potencia de F respecto a la circunferencia =2²=(3/2)BF---> BF=8/3---> BE=(8/3)+(3/2)=25/6---> Radio =r=25/12---> Área del círculo =(25/12)²π =625π/144 u².
Gracias y saludos.

[xualain]

Here’s my version with trigonometry to share with.
Let <BAD=<BCD=a and AD=2*r where r is the radius of the circle.
BD=2*r*sin(<BAD=2*r*sin a —>BD=2*r*sin a ….(1)
cos(<BAD)=cos a=AB/AD=2.5/(2*r)—>cos a=2.5/(2*r) ….(2)
CBD is isosceles triangle, hence BC*cos(<BCD)=CD/2 —>BD*cos a=4/2=2 …(3)
Where BC=BD and<BCD=<BAD=a
Replacing (1) and (2) to (3)
2*r*sin a*2.5/(2*r)=2 —>sin a=4/5
hence cos a=3/5 replacing to (2)
cos a=3/5=2.5/(2*r) —>r=25/12
Area=pi*(25/12)^2=625/144*pi that’s our answer.

[AdemolaAderibigbe-js]

Let radius of circle; AO = OD = r. Let BC = BD = c. Then since AO is a diameter, angle ABD = 90 degrees. Also angle BAD = angle BAC (we'll name the angle theta) since they are subtended on the circumference by the (same) arc BD. For the right angle triangle ABD, we get cos(theta) = (2.5/2r). For isosceles triangle CBD we get cos(theta) = 2/c (explanation draw the perpendicular (also the median and the altitude) from B to the line CD to intersect the line at E; that gives us the triangle BCE with right angle BEC from which we get the cosine of theta). These cosine values are equal so (2.5/2r) = (2/c) or c = 8r/5. Next apply Pythagoras' theorem to triangle ABD to get: (2r)^2 = (2.5)^2 + c^2. Now substitute for c in terms of r and simplify the expression to get r^2 = 625/144. The area requested is pi*r^2 = (625/144)*pi.

[solimana-soli]

Angles BAD and BCD are equal because they tend the same chord BD. Setting BAD = BCD = alpha and BD = BC = x and applying sines law on BCD:
x/sin alpha = 4/sin(180-2alpha)
and being sin(180-2alpha)=sin 2alpha
x/sin alpha = 4/2sin alpha cos alpha
cos alpha = 2/x
Considering ABD
5/2 = 2R*cos alpha
2R = 5/4x
applying pythagorean theorem on ABD
(5/2)² + x² = (5/4x)²
x = 10/3
2R = 5/4*10/3= 25/6
R = 25/12
area = 625/144 pi

[MrPaulc]

I've bookmarked this one for later as it's a bit advanced for my current level.
I tried comparing two right triangles formed by Thales Theorem which both had 4r^2 as d^2, and then I looked into the formed cyclic quadrilateral which had 4 and 2.5 as opposite sides 2r as a diagonal, and two right angles opposite from each other.
It all became too confusing. A few more months and I reckon I will be figuring it out better.

[marioalb]

α + 2β = 90°
sinβ = 2, 5/2R = 5/4R
cosα=4/2R =2/R=cos(90°-2β)=sin2β
Dividing and cancelling "R" :
sinβ/sin(2β) = 5/8 = 1/(2cosβ)
cosβ = 4/5 --> sinβ = 3/5
Returning above:
2R = 2, 5/sinβ = 2, 5*5/3 =25/6
R = 25/12 = 2, 0833 cm
Area of circle :
A = πR² = 625π/144 = 13, 635 cm² ( Solved √ )

[marioalb]

α + 2β = 90°
sinβ = 2, 5/2R = 5/4R

Dividing and cancelling "R" :
sinβ/sin(2β) = 5/8 = 1/(2cosβ)
cosβ= 4/5 --> Pytagorean triplet 3-4-5 !!!
R = ½*2, 5*5/3 = 25/12 cm
A = πR²= 625π/144 cm² (Solved √)

[chintamanimoghe]

Is this reasoning ok? The perpendicular bisector OE of the chord CD must pass through B since BC=BD. We have now 3 rt angle triangles namely ABD, BED, and OED. Apply Pythagoras relation to each of them and solve for the value of the radius.