The exact position of D left to right is not pinned down by this problem. That means the answer doesn't depend on that, and we are free to manipulate it to simplify the solution process. So, I choose to make D coincident with E. That means the circle radius is 2.5, and our sought after area is 2.5*5 = 12.5. Q.E.D. Note, as a check on our assumption of invariance, we can also make point F coincident with the top of the circle. Then R = 5/sqrt(2) and the rectangle becomes a square; it's area is A = R^2 = 25/2 = 12.5.
Overlapping rectangle and semicircle. Neither is entirely inside the other. (0.00) (8.15) I Got this answer 25/2 from the similar right-angled triangles with theta in them. This is a very nice bit of geometry. Thank you.
Alternative method using Thales theorem, Pythagoras theorem and similar triangles: 1. Triangle CEF is right-angled triangle by Thales theorem. 2. Let radius of semicircle be R. Triangles CEF and FED are similar triangles by AAA. Hence ED/EF = EF/EC by proportional corresponding sides. ED = EF^2/EC = EF^2/2R While ED = EC - DC = 2R - DC Hence EF^2/2R = 2R - DC 3. In triangle CEF, EF^2 = EC^2 - FC^2 = 4R^2 - 25 by Pythagoras theorem. Substitue EF^2 in (2), (4R^2 - 25)/2R = 2R - DC 4R^2 - 25 = 4R^2 - 2R(DC) 25 = 2R(DC) R(DC) = 25/2 5. Height of rectangle ABCD = R Area of rectangle ABCD = R(DC) = 25/2
Respetando las premisas del trazado propuesto, podemos desplazar el punto A hasta el extremo del radio vertical y el rectángulo original se transforma en un cuadrado de superficie equivalente ---> Área ABCD =5²/2=25/2=12,5 ud. Gracias y saludos.
O centro del semicerchio,DO=a..risulta arccos(a/r)=180-(180-2arccos(2,5/r))=2arccos(2,5/r)...applico cos..a/r=2(2,5/r)^2-1..(*r^2)...ar=2*2,5^2-r^2..Arett=ar+r^2=2*6,25=12,5
En un triángulo rectángulo con hipotenusa=5 u., catetos menor= 3 u. y cateto mayor=4 u. En triángulo rectángulo FDC: FC= 5; FD=3 y DC=4. ∆EDF~∆FDC. Entonces: 3/EF=4/5; EF=3.75 u. En ∆EDF: ED²=EF²-FD²= 3.75²-3²= 5.0625 u². ED= 2.25 u. ED+DC=Diámetro de semicírculo= D. D= 4+2.25= 6.25 u. Radio= D/2= 3.125 u. Radio= AD=BC. DC= lado mayor del rectángulo=4 u. AB= lado menor del rectángulo= 3.125 u. Área del rectángulo= 4×3.125= 12.5 u².
Solution using ratios of sides of similar triangles instead of trigonometry: Construct FE. Note that
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The exact position of D left to right is not pinned down by this problem. That means the answer doesn't depend on that, and we are free to manipulate it to simplify the solution process. So, I choose to make D coincident with E. That means the circle radius is 2.5, and our sought after area is 2.5*5 = 12.5. Q.E.D.
Note, as a check on our assumption of invariance, we can also make point F coincident with the top of the circle. Then R = 5/sqrt(2) and the rectangle becomes a square; it's area is A = R^2 = 25/2 = 12.5.
The area is 25/2. And this is another example of how to solve the using only the properties of right triangles. I shall use that as practice!!!
Overlapping rectangle and semicircle. Neither is entirely inside the other. (0.00) (8.15) I Got this answer 25/2 from the similar right-angled triangles with theta in them.
This is a very nice bit of geometry. Thank you.
5^2 = a *2r
Consider right triangle EFC: CF*CF=CD*CE=CD*2R=25 ==> S= CD*R=25/2
Alternative method using Thales theorem, Pythagoras theorem and similar triangles:
1. Triangle CEF is right-angled triangle by Thales theorem.
2. Let radius of semicircle be R.
Triangles CEF and FED are similar triangles by AAA.
Hence ED/EF = EF/EC by proportional corresponding sides.
ED = EF^2/EC = EF^2/2R
While ED = EC - DC = 2R - DC
Hence EF^2/2R = 2R - DC
3. In triangle CEF, EF^2 = EC^2 - FC^2 = 4R^2 - 25 by Pythagoras theorem.
Substitue EF^2 in (2), (4R^2 - 25)/2R = 2R - DC
4R^2 - 25 = 4R^2 - 2R(DC)
25 = 2R(DC)
R(DC) = 25/2
5. Height of rectangle ABCD = R
Area of rectangle ABCD = R(DC) = 25/2
Respetando las premisas del trazado propuesto, podemos desplazar el punto A hasta el extremo del radio vertical y el rectángulo original se transforma en un cuadrado de superficie equivalente ---> Área ABCD =5²/2=25/2=12,5 ud.
Gracias y saludos.
O centro del semicerchio,DO=a..risulta arccos(a/r)=180-(180-2arccos(2,5/r))=2arccos(2,5/r)...applico cos..a/r=2(2,5/r)^2-1..(*r^2)...ar=2*2,5^2-r^2..Arett=ar+r^2=2*6,25=12,5
area'(abcd) = (EC/2).DC
Triangle (efc) : FC^2 =EC .DC =25
area =25/2 !
En un triángulo rectángulo con hipotenusa=5 u., catetos menor= 3 u. y cateto mayor=4 u.
En triángulo rectángulo FDC:
FC= 5; FD=3 y DC=4.
∆EDF~∆FDC.
Entonces: 3/EF=4/5; EF=3.75 u.
En ∆EDF: ED²=EF²-FD²= 3.75²-3²= 5.0625 u².
ED= 2.25 u.
ED+DC=Diámetro de semicírculo= D. D= 4+2.25= 6.25 u.
Radio= D/2= 3.125 u.
Radio= AD=BC.
DC= lado mayor del rectángulo=4 u.
AB= lado menor del rectángulo= 3.125 u.
Área del rectángulo= 4×3.125= 12.5 u².
CD/CF = (CF/2)/r
CD/5 = (5/2)/r
CDr = 5(5/2) = 25/2 = [ABCD]
2:33 - 2:43: It should be "radius of a semicircle," not "radius of the circle" (because a semicircle is half of a circle).
(5)^2 =25 180°ABCDEO/25=7.5ABCDEO. (ABCDEO ➖ 7ABCDEO +5).
Linda!
12.5
Nice problem and solution, but ABCD ISN’T inside the semi circle. Quite clearly from the diagram, part of ABCD is OUTSIDE the semi circle.
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If I could remember my calculus....areas are easy to solve
ABCD is not within the semi circle
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