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.
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.
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
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.
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.
Great job it's so difficult but will trying more and more
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.
Here’s my version with trigonometry to share with.
Let
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 --> Pytagorean triplet 3-4-5 !!!
R = ½*2,5*5/3 = 25/12 cm
A = πR²= 625π/144 cm² (Solved √)
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.
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 √ )
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 --> Pytagorean triplet !!!
R = ½*2,5*5/3 = 25/12 cm
A = πR²= 625π/144 cm² (Solved √)
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
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.
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.