Note that point E is not at a fixed position on the circumference, the only requirement is that AE =18. This means that both the radius & dimensions of the rectangle are not fixed. If Point E coincides with point F, then we have a semicircle of radius 9, and a rectangle of height 9 & width 18. Hence 9 x 18 = 162.
Agreed! This is the special case where right ΔABE has collapsed into a straight line and is the limiting case as EF approaches 0. The problem statement implies that the area of rectangle ABCD is the same for all valid lengths EF. One could argue that EF = 0 is not valid because the lines connecting A, B and E not longer form a triangle. However, EF can be infinitesimal and we still have a triangle, as long it is not zero.
@@alexshelest1937 No, if EB = r, then B coincides with O and E coincides with C (at top of circle) and we get a square with side length r (which is no longer 9) and diagonal 18. We could calculate r, but we'll use area formula for square with diagonal d. Area = (d^2)/2 = (18^2)/2 = 162
Just solved it via Pythagoras: 18^2= (r+OB)^2 + BE^2 and r^2= OB^2 + BE^2 and A= r*(r+OB) => A=r^2+r*OB and 18^2=r^2 +2*r*OB +OB^2 + BE^2; substitute OB^2+BE^2 by r^2 results in 18°2=2(r^2+rOB) and finally substituting A=r^2+r*OB results in 18^2=2*A => A=9*18=162. Makes sense?
It is interesting that the radius cancels in the final equation which means the area is independent the radius and any rectangle with the similar configuration (ncluding a square will have the same area.
To me, this appears to be a "minimal information" problem. We cant solve for the radius or the height of the triangle but we know they are dependent on eachother. If I choose h = 0, it will make r = 9. The area of the rectangle would the be 9(18) = 162.
The tricky part here is that the lengths of the sides of rectangle ABCD are variable but their product is a constant. We can not solve for the value of r because there is a range of values for r which are valid.
Any value of r greater or equal to 9 is valid. When r < 9, then diameter < 18, so there can be no chord = 18, since the diameter is the longest chord in a circle.
Method using Thales theorem, Pythagoras theorem and similar triangles: 1. Let radius be R. 2. Triangle AEF is right-angled triangle by Thales theorem. Hence EF^2 = (2R)^2 - 18^2 = 4R^2 - 324 by Pythagoras theorem. (equation 1) 3. Triangles AEF and EBF are similar (AAA) Hence EF/BF = AF/EF (EF)^2 = (AF)(BF) = (2R)(BF) (equation 2) 4. From equations (1) and (2) 4R^2 - 324 = (2R)(BF) 4R^2 - 2(R)(BF) = 324 2R^2 - (R)(BF) = 162 4. When DC extends to DC' such that DC' = AF, area of rectangle ADC'F = 2R^2 and Area of rectangle BCC'F = (R)(BF) Area of ABCD = area ADC'F - area BCC'F = 2R^2 - (R)(BF) = 162
Again, since there are no variables, it means that the blue area is constant, take the special case where the semi-circle is inscribed in the rectangle meaning B, F, and E are all the same points and [AE] is now a diameter of the circle, meaning [AE] = 2 r = 18 This gives r = 9 and the sides of the rectangle as r and 2 r Area of rectangle = 2 r^2 = 2 * 9^2 = 2 * 81 = 162 square units.
two alternative methods first one: similarity tracing perpendicular to chord AE we have right triangle OAH (H is the midpoint of AE) similar to AEB, then 9 : AB = R : 18 area = AB*R = 9*18 = 162 second one: logical extending AE untill it lies on diameter AF we have that area = radius*diameter=18*9=162
the result of the area does not depend on the radius, see line 30: 10 l1=18:l4=13:if l40 then 70 80 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs 90 if abs(dg)>1E-10 then 80 100 print xs;"%";"die gesuchte flaeche=";xs*r:masx=1200/2/r:masy=850/r 110 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
Let R be the radius OA and (x,y) be the coordinates of B. So B is the intersection of the semi-circle with the circle of radius 18 centered at A. Thus (x,y) must satisfy 1) ( x+R)^2 + y^2 = 18^2 and 2) x^2 + y^2 = R^2 Subtracting 2) from 1) eliminates the x^2 and y^2 terms and simplifies to (X+R)R = 18^2/2 = 162. But (x+R)R is the blue area!
Third method: Let r = radius of semi-circle. Then OA = OE = OF = r. Also AD = BC = r Let AB = x, BE = y. Then OB = AB − OA = x − r In right △ABE, we use Pythagorean Theorem: AB² + BE² = AE² x² + y² = 18² = 324 In right △OBE, we use Pythagorean Theorem: OB² + BE² = OE2 (x−r)² + y² = r² (x²−2rx+r²) + y² = r² x² − 2rx + y² = 0 2rx = x² + y² = 324 x = 162/r Area of square = AB * AD = x * r = (162/r) * r = 162
Took me a while but I found the solution by the second method eventually. I was curious to see what the other method was but it seems to me they are basically both the same method, the trig in the first solution is just sort of a bookkeeping detail, it really is just using similar triangles. I thought there might be some way to solve it only using pythagorean but I couldn't find such a solution and I suspect there isn't such a solution because professor would have probably found that and included it as one of the methods if such a method were possible.
You are too suspicious )))) Apply our beloved Pythagorean theorem to triangles ABE and OBE. Observe that BE^2 is the same in both. Apply some algebra and get doubled product of r and r+OB. Divide it by 2 and that is it.
The horizontal location of B is not specified. This implies that the solution is INDEPENDENT of that parameter. Therefore, we can choose it to make solving the problem easier. I choose to put it at O. This makes the area of the square R^2, where R is the radius of the semicircle. This also implies that sqrt(2)*R = 18. Square this, and we get 2*R^2 = 18^2 = 324. Therefore, the area R^2 is 162. Q.E.D.
Llamaremos G a la proyección ortogonal de O sobreDC. Respetando las condiciones del trazado de la figura propuesta, desplazando E hasta G, el rectángulo original se transforma en un cuadrado de dimensiones r*r y diagonal de longitud 18 ud cuya superficie es igual a la buscada- ---> Área ABCD =18²/2=162 ud². Gracias y saludos.
Let's use an orthonormal, center O, first axis(OB). Be OB = c, and R the radius of the circle. The equation of the circle is x^2 + y^2 = R^2 At point E we have x = c and c^2 + y^2 = R^2, so E(c, sqrt(R^2 - c^2) Then VectorAE(c + R, sqrt(R^2 - c^2) AE^2 = (c + R)^2 + (R^2 - c^2) = 2.R^2 + 2.c.R = 18^2 Meaning that: R.[R + c] = (18^2)/2 = 162 Or that the blue area is 162.
Let's find the area: . .. ... .... ..... According to the theorem of Thales the triangle AEF is a right triangle. Therefore we can apply the right triangle altitude theorem: BE² = AB*BF The triangle ABE is also a right triangle, so we can apply the Pythagorean theorem. With R being the radius of the semicircle we obtain: AE² = AB² + BE² = AB² + AB*BF = AB² + AB*(2*R − AB) = AB² + 2*AB*R − AB² = 2*AB*R Since AD=R, we can now calculate the area of the blue rectangle: A(ABCD) = AB*AD = AB*R = (1/2)*2*AB*R = (1/2)*AE² = (1/2)*18² = (1/2)*324 = 162 Best regards from Germany
I have another method, not necessarily better, but I found the solution. That's the most important 😆 Let x=BF en y=BE 18²=y²+(2r-x)² --> 324=y²+4r²-4rx+x² --> 324=x²+y²+4r²-4rx Draw a line from o to E, that's the radius r r²=OB²+y² --> r²=(r-x)²+y² --> r²=r²-2rx+x²+y² --> 2rx=x²+y² 324=x²+y²+4r²-4rx 324=2rx+4r²-4rx 324=4r²-2rx 162=2r²-rx (divide by 2) area rectangle=(2r-x)*r=2r²-rx so, area rectangle=162 square units
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Note that point E is not at a fixed position on the circumference, the only requirement is that AE =18.
This means that both the radius & dimensions of the rectangle are not fixed.
If Point E coincides with point F, then we have a semicircle of radius 9, and a rectangle of height 9 & width 18.
Hence 9 x 18 = 162.
Agreed! This is the special case where right ΔABE has collapsed into a straight line and is the limiting case as EF approaches 0. The problem statement implies that the area of rectangle ABCD is the same for all valid lengths EF. One could argue that EF = 0 is not valid because the lines connecting A, B and E not longer form a triangle. However, EF can be infinitesimal and we still have a triangle, as long it is not zero.
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@@jimlocke9320 So if EB tends to r? Then our rectangle will have area about 81 sq.un.
@@alexshelest1937 No, if EB = r, then B coincides with O and E coincides with C (at top of circle) and we get a square with side length r (which is no longer 9) and diagonal 18. We could calculate r, but we'll use area formula for square with diagonal d.
Area = (d^2)/2 = (18^2)/2 = 162
Just solved it via Pythagoras: 18^2= (r+OB)^2 + BE^2 and r^2= OB^2 + BE^2 and A= r*(r+OB) => A=r^2+r*OB and 18^2=r^2 +2*r*OB +OB^2 + BE^2; substitute OB^2+BE^2 by r^2 results in 18°2=2(r^2+rOB) and finally substituting A=r^2+r*OB results in 18^2=2*A => A=9*18=162. Makes sense?
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EOB=2arccos(9/r)...Ablue=2r*r-r(r-rcosEOB)=2r^2-r^2+r^2(2*(9/r)^2-1)=r^2+r^2(162/r^2-1)=162
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It is interesting that the radius cancels in the final equation which means the area is independent the radius and any rectangle with the similar configuration (ncluding a square will have the same area.
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Given that the area of ABDC = AD*(AO+OB) ∴ ABDC=(r)*(r+OB). ∴ (AO+AB)^2+(BE)^2=AE^2 & (OE)^2=(OB)^2+(BE)^2 ∴ 2*(r^2+(r+OB))=324 ∴ ABCD = 162.
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To me, this appears to be a "minimal information" problem. We cant solve for the radius or the height of the triangle but we know they are dependent on eachother. If I choose h = 0, it will make r = 9. The area of the rectangle would the be 9(18) = 162.
Let l= length
r= width
Area=lr
EB²=r²-(l-r)²
EB²=18²-l²
r²-l²+2lr-r²=324-l²
2lr=324
lr=162 sq units
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The tricky part here is that the lengths of the sides of rectangle ABCD are variable but their product is a constant. We can not solve for the value of r because there is a range of values for r which are valid.
Right on!
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Any value of r greater or equal to 9 is valid. When r < 9, then diameter < 18, so there can be no chord = 18, since the diameter is the longest chord in a circle.
Method using Thales theorem, Pythagoras theorem and similar triangles:
1. Let radius be R.
2. Triangle AEF is right-angled triangle by Thales theorem.
Hence EF^2 = (2R)^2 - 18^2 = 4R^2 - 324 by Pythagoras theorem. (equation 1)
3. Triangles AEF and EBF are similar (AAA)
Hence EF/BF = AF/EF
(EF)^2 = (AF)(BF) = (2R)(BF) (equation 2)
4. From equations (1) and (2)
4R^2 - 324 = (2R)(BF)
4R^2 - 2(R)(BF) = 324
2R^2 - (R)(BF) = 162
4. When DC extends to DC' such that DC' = AF, area of rectangle ADC'F = 2R^2 and
Area of rectangle BCC'F = (R)(BF)
Area of ABCD = area ADC'F - area BCC'F = 2R^2 - (R)(BF) = 162
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Again, since there are no variables, it means that the blue area is constant, take the special case where the semi-circle is inscribed in the rectangle meaning B, F, and E are all the same points and [AE] is now a diameter of the circle, meaning [AE] = 2 r = 18
This gives r = 9 and the sides of the rectangle as r and 2 r
Area of rectangle = 2 r^2 = 2 * 9^2 = 2 * 81 = 162 square units.
Good Method
From Gujrat
Only one evident datum but how nice is this task! I'm enjoyed! Thank you so much , Professor 💯👍
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two alternative methods
first one: similarity
tracing perpendicular to chord AE we have right triangle OAH (H is the midpoint of AE) similar to AEB, then
9 : AB = R : 18
area = AB*R = 9*18 = 162
second one: logical
extending AE untill it lies on diameter AF we have that area = radius*diameter=18*9=162
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the result of the area does not depend on the radius, see line 30:
10 l1=18:l4=13:if l40 then 70
80 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs
90 if abs(dg)>1E-10 then 80
100 print xs;"%";"die gesuchte flaeche=";xs*r:masx=1200/2/r:masy=850/r
110 if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
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3rd method:
1) Let r=AO, x=OB, h=OE; y=EF;
2) triangle ABE: 18^2=h^2+(r+x)^2 => h^2 = 18^2-(r+x)^2; (1)
triangle OBE : r^2=h^2+x^2; => h^2 = r^2-x^2; (2)
3) compare (1) and (2) => r^2-x^2 = 18^2-(r+x)^2;
r^2-x^2 = 324 - (r^2+2rx+x^2);
r^2-x^2 = 324 - r^2-2rx-x^2;
2r^2+2rx=324;
2*(r^2+rx)=324;
r^2+rx=162
4) Let us observe that Ablue = r^2+rx= 162sq units.
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Let R be the radius OA and (x,y) be the coordinates of B. So B is the intersection of the semi-circle with the circle of radius 18 centered at A. Thus (x,y) must satisfy
1) ( x+R)^2 + y^2 = 18^2 and
2) x^2 + y^2 = R^2
Subtracting 2) from 1) eliminates the x^2 and y^2 terms and simplifies to
(X+R)R = 18^2/2 = 162.
But (x+R)R is the blue area!
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Good morning sir
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Good morning❤️
Trigonometry made it easier
Third method:
Let r = radius of semi-circle. Then OA = OE = OF = r. Also AD = BC = r
Let AB = x, BE = y. Then OB = AB − OA = x − r
In right △ABE, we use Pythagorean Theorem:
AB² + BE² = AE²
x² + y² = 18² = 324
In right △OBE, we use Pythagorean Theorem:
OB² + BE² = OE2
(x−r)² + y² = r²
(x²−2rx+r²) + y² = r²
x² − 2rx + y² = 0
2rx = x² + y² = 324
x = 162/r
Area of square = AB * AD = x * r = (162/r) * r = 162
Took me a while but I found the solution by the second method eventually. I was curious to see what the other method was but it seems to me they are basically both the same method, the trig in the first solution is just sort of a bookkeeping detail, it really is just using similar triangles.
I thought there might be some way to solve it only using pythagorean but I couldn't find such a solution and I suspect there isn't such a solution because professor would have probably found that and included it as one of the methods if such a method were possible.
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You are too suspicious ))))
Apply our beloved Pythagorean theorem to triangles ABE and OBE.
Observe that BE^2 is the same in both.
Apply some algebra and get doubled product of r and r+OB.
Divide it by 2 and that is it.
I solved it by the 2nd method.
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The horizontal location of B is not specified. This implies that the solution is INDEPENDENT of that parameter. Therefore, we can choose it to make solving the problem easier. I choose to put it at O. This makes the area of the square R^2, where R is the radius of the semicircle. This also implies that sqrt(2)*R = 18. Square this, and we get 2*R^2 = 18^2 = 324. Therefore, the area R^2 is 162. Q.E.D.
|OB|=x,|BE|=y
(r+x)^2+y^2=18^2
x^2+y^2=r^2
2r(r+x)=18^2
A=r(r+x)=9*18=162
Llamaremos G a la proyección ortogonal de O sobreDC. Respetando las condiciones del trazado de la figura propuesta, desplazando E hasta G, el rectángulo original se transforma en un cuadrado de dimensiones r*r y diagonal de longitud 18 ud cuya superficie es igual a la buscada- ---> Área ABCD =18²/2=162 ud².
Gracias y saludos.
Area=AB . Radius
Radius=AF/2,
Therefore;
Area= AB . AF/2,
Now, AF= AB + BF
So Area= AB . (AB + BF)/2
Therefore;
Area= (AB^2)/2 + (BF . AB)/2
Therefore;
Twice Area= AB^2 + (BF . AB)
By intersecting chord theory;
EB^2 = AB . BF
So Twice Area= AB^2 + EB^2
By Pythag,
AE^2= EB^2 + AB^2
So AB^2= AE^2 - EB^2
By substitution;
Twice Area=AE^2 - EB^2 + EB^2
Simplifies to;
Twice Area= AE^2
So, Area= (AE^2)/2 = 18^2 / 2
Area= 324 / 2 = 162 units^2
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Second method is mo bettah!...cuz i take great care to avoid enclosed spaces. Can't think inside a box. 🙂
😀wow
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Let's use an orthonormal, center O, first axis(OB).
Be OB = c, and R the radius of the circle.
The equation of the circle is x^2 + y^2 = R^2
At point E we have x = c and c^2 + y^2 = R^2,
so E(c, sqrt(R^2 - c^2)
Then VectorAE(c + R, sqrt(R^2 - c^2)
AE^2 = (c + R)^2 + (R^2 - c^2) = 2.R^2 + 2.c.R = 18^2
Meaning that: R.[R + c] = (18^2)/2 = 162
Or that the blue area is 162.
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Let's find the area:
.
..
...
....
.....
According to the theorem of Thales the triangle AEF is a right triangle. Therefore we can apply the right triangle altitude theorem:
BE² = AB*BF
The triangle ABE is also a right triangle, so we can apply the Pythagorean theorem. With R being the radius of the semicircle we obtain:
AE² = AB² + BE² = AB² + AB*BF = AB² + AB*(2*R − AB) = AB² + 2*AB*R − AB² = 2*AB*R
Since AD=R, we can now calculate the area of the blue rectangle:
A(ABCD) = AB*AD = AB*R = (1/2)*2*AB*R = (1/2)*AE² = (1/2)*18² = (1/2)*324 = 162
Best regards from Germany
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A = 162 square units?
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My thoughts:
1) Move C to the top of the circle, so that C = E => AC = 18/2 * √2 = 9√2
2) A(blue) = (9√2)² = 81 * 2 = 162 square units
162
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In trick, we can take rectangle as square and the answer is 162
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However, you have to solve the problem generally to prove it works for all cases.
@@spafon7799You are absolutely right.
My original solution is
18cosA × 18sinA /sin2A=162
I have another method, not necessarily better, but I found the solution.
That's the most important 😆
Let x=BF en y=BE
18²=y²+(2r-x)² --> 324=y²+4r²-4rx+x² --> 324=x²+y²+4r²-4rx
Draw a line from o to E, that's the radius r
r²=OB²+y² --> r²=(r-x)²+y² --> r²=r²-2rx+x²+y² --> 2rx=x²+y²
324=x²+y²+4r²-4rx
324=2rx+4r²-4rx
324=4r²-2rx
162=2r²-rx (divide by 2)
area rectangle=(2r-x)*r=2r²-rx
so, area rectangle=162 square units
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