There is no reason to have points C, O and B on the same line, this should be one of the initial hypothesis of this problem if you want your proof is true.
Actually, the green rectangle is not needed. It is there to confuse us. The question can be rephrased as following: You draw a vertical diameter CB, then a horizontal tangent line on the bottom of the circle. Finally you draw a red line from C to A, such that its length is 25, 16 inside the circle and 9 outside. What is the area of the circle?
Would it not have been easier to say that the total hypotenuse length is 25. Then from similar triangle theory it's a scaled (5*) 3:4:5 triangle making the other lengths 15 and 20. Thus the radius is 10 and then it's just πr^2.
Подходит пифагорова тройка 7;24;25 и, следовательно, R=12. Вот только гипотенуза 7 никак не может быть меньше малого катета 9. Ищем дальше. ▲ABD й очень похож на утроенный Египет - (3;4;5)×3=(9;12;15). В этом случае хорда BD разбивает его на два подобных. Если ▲BCD тоже египетский, тогда DB=12, а CB=20, т. е. R=10. Проверим гипотезу. Если ▲ABD действительно Египет, тогда АВ=15. 15;20;25 - это пятикратный Египет, (3;4;5)×5. Значит догадка верна. Тогда искомая площадь будет 100π.
Solution: Tangent Secant Theorem AB² = AD × AC ... ¹ AD = 9 AC = 9 + 16 = 25 Substituting in ¹ AB² = 9 × 25 AB² = 225 AB = 15 B is the tangency point, so AB is the tangent line and C, O and B are collinear. Therefore we have a right triangle ABC, such that angle B is 90° and COB is the diameter, like that, let's applying the Pythagorean Theorem AB² + BC² = AC² (15)² + (2r)² = (25)² 225 + 4r² = 625 4r² = 625 - 225 4r² = 400 r² = 100 r = 10 Final Step Yellow Circle Area = π r² Yellow Circle Area = π (10)² Yellow Circle Area = 100 π Square Units ✅ Yellow Circle Area ≈ 314,1592 Square Units ✅
AB/25 = 9/AB Therefore AB square = 225. CB square = AC square - AB square = 400. Square root = 20 = Diameter. No need to say how to get the circle area.
You could solve your problem by only applying Euclid's First Theorem to the right triangle ABC ( note that BD is perpendicular to AC in fact the triangle BDC is inscribed in a semicircle ). 4r² = BC² = AC⋅DC = 25⋅16 = 400 Consequently, the area of the yellow circle is 100π.
Sol is too short Pythagorean triplet (3, 4,5) Hypotenuse is 16+9=25 We get another triplet multiplying by 5 (15 , 20,25) Hence 2r =20 >r=10 Area = 100π sq units
It feels like there is a flaw in the diagram and it possibly confused me. If r = 10, surely two-tangent theorem would give the height of the rectangle as 10 due to the lower-right vertex being equidistant from the two tangents? However, the rectangle's diagonal is shown as 9 when it should be >10. The alternative would be that the lower-right vertex is not making a right angle with the tangent line. Ah well, it is Christmas LOL.
Alternative solution: BD is the height from B to the hypotenuse AC so, SqBD= ADxDC= 9x16 -> BD= 12=3x4 Focus on the triangle BDC, CD = 16=4x4->BC=5x4=20 ( 3-4-5 triples) Area of the yellow circle= pi x sq BC/4 =100 pi sq units😅😅😅
Let's find the area: . .. ... .... ..... Let E be the bottom right corner of the green rectangle, A be the center of the coordinate system and B on the x-axis. With r being the radius of the circle we obtain the fo llowing coordinates: A: ( 0 ; 0 ) B: ( xB ; 0 ) C: ( xB ; 2r ) D: ( xD ; yD ) E: ( xD ; 0 ) O: ( xB ; r ) From the known length of the chord CD we can conclude: (xC − xD)² + (yC − yD)² = CD² (xB − xD)² + (2r − yD)² = 16² (xB − xD)² + (2r − yD)² = 256 Since D is located on the circle, we also can conclude: (xO − xD)² + (yO − yD)² = r² (xB − xD)² + (r − yD)² = r² The right triangles ADE and ABC are obviously similar. Therefore we obtain: BC/AC = DE/AD (yC − yB)/AC = (yD − yE)/AD (yC − yB)/(AD + CD) = (yD − yE)/AD (2r − 0)/(9 + 16)= (yD − 0)/9 2r/25 = yD/9 ⇒ yD = 18r/25 (xB − xD)² + (2r − yD)² = 256 (xB − xD)² + (r − yD)² = r² (2r − yD)² − (r − yD)² = 256 − r² 4r² − 4*yD*r + yD² − (r² − 2*yD*r + yD²) = 256 − r² 4r² − 4*yD*r + yD² − r² + 2*yD*r − yD² = 256 − r² 4r² − 2*yD*r = 256 r² − yD*r/2 = 64 r² − (18r/25)*r/2 = 64 r² − 9r²/25 = 64 25r² − 9r² = 64*25 16r² = 64*25 r² = 4*25 = 100 ⇒ r = √100 = 10 Now we are able to calculate the area of the yellow circle: A(circle) = πr² = π*10² = 100π Best regards from Germany
Thank you!
You are very welcome!
Thanks for the feedback ❤️
There is no reason to have points C, O and B on the same line, this should be one of the initial hypothesis of this problem if you want your proof is true.
Premath just said that the 3 points are collinear😊
It is a given fact.
@@phungpham1725 Yes I heard that at a moment, but not in the initial hypothesis.
@@marcgriselhubert3915: Premath said: “ and furthermore these points B, O and C are collinear as well”.😊
@@marcgriselhubert3915 he said that at 42"
Actually, the green rectangle is not needed. It is there to confuse us. The question can be rephrased as following:
You draw a vertical diameter CB, then a horizontal tangent line on the bottom of the circle. Finally you draw a red line from C to A, such that its length is 25, 16 inside the circle and 9 outside. What is the area of the circle?
Join BD & BC.
Note angle BDC = 90 degrees in semicircle.
Triangles BDA & BDC are similar.
9 / BD = BD / 16.
BD^2 = 9 x 16 = 144.
BD = 12.
Triangle BDC is of 3,4,5 ratio.
Therefore sides 12,16, 20.
BC = 20.
r = 10.
Therefore, the area is 10^2 *pi = 314.16 sq. units.
Let me know how BC is treated as diameter for assuming angle BDC is 90 degree.
Nice 👍 I was a bit confused because the miniature did not show that COB are colinear.
Why is COB colinear, which thoeram
@ I think this is part of the initial « given » data otherwise I don’t know why.
Tangent Secant theorem:
a² = 9*(9+16) ---> a² = 225cm²
Pytagorean theorem:
(2R)² = c² - a²= (9+16)² - 225
R = 10 cm
Area of yellow circle:
A = πR² = 100π cm² ( Solved √)
9*(9+16)=AB²=225→ AB=15→ (2r)²=(9+16)²-15²→ r²=100→ Área círculo =100π u².
Gracias y saludos.
c = chord BD
Similarity of triangles:
c/9 = 16/c --> c²= 144cm²
Pytagorean theorem:
(2R)² = 16² + c² --> R=10cm
Area of yellow circle :
A = πR² = 100π cm² ( Solved √ )
Thanks for sharing sir❤❤
That’s very nice
Thanks Sir
Thanks PreMath
❤❤❤❤❤
Would it not have been easier to say that the total hypotenuse length is 25. Then from similar triangle theory it's a scaled (5*) 3:4:5 triangle making the other lengths 15 and 20. Thus the radius is 10 and then it's just πr^2.
Actually no - just because the hypotenuse is 25, it doesn't mean that the triangle must be 3:4:5
Подходит пифагорова тройка 7;24;25 и, следовательно, R=12. Вот только гипотенуза 7 никак не может быть меньше малого катета 9. Ищем дальше. ▲ABD й очень похож на утроенный Египет - (3;4;5)×3=(9;12;15). В этом случае хорда BD разбивает его на два подобных. Если ▲BCD тоже египетский, тогда DB=12, а CB=20, т. е. R=10. Проверим гипотезу. Если ▲ABD действительно Египет, тогда АВ=15. 15;20;25 - это пятикратный Египет, (3;4;5)×5. Значит догадка верна. Тогда искомая площадь будет 100π.
Most of these can be solved with similarity 😢
wow
How can we assume BOC is collinear?
Bom dia Mestre
Muitíssimo obrigado pelos ensinamentos
Deus Lhe Abençoe
25cosα=2r...cosα=(8/r)...r=10
Solution:
Tangent Secant Theorem
AB² = AD × AC ... ¹
AD = 9
AC = 9 + 16 = 25
Substituting in ¹
AB² = 9 × 25
AB² = 225
AB = 15
B is the tangency point, so AB is the tangent line and C, O and B are collinear.
Therefore we have a right triangle ABC, such that angle B is 90° and COB is the diameter, like that, let's applying the Pythagorean Theorem
AB² + BC² = AC²
(15)² + (2r)² = (25)²
225 + 4r² = 625
4r² = 625 - 225
4r² = 400
r² = 100
r = 10
Final Step
Yellow Circle Area = π r²
Yellow Circle Area = π (10)²
Yellow Circle Area = 100 π Square Units ✅
Yellow Circle Area ≈ 314,1592 Square Units ✅
Triangles ABC and BCD are Egyptian, so from any given triangle:
2r= 20; r= 10.
Thanks sir😊
Qual a necessidade do retângulo verde?
Easier solution:
cos angle(ACB) = 2*r / 9+16 = 16/(2*r)
4*r² = 16*25 = 400
r² = 100
r = 10
A = pi*r² = pi*10 = 314,16 unit²
Its a 3-4-5 ratio to the sides. When the hypotenuse is 25 the rest is easy to find.
AB/25 = 9/AB Therefore AB square = 225. CB square = AC square - AB square = 400. Square root = 20 = Diameter. No need to say how to get the circle area.
You could solve your problem by only applying Euclid's First Theorem to the right triangle ABC ( note that BD is perpendicular to AC in fact the triangle BDC is inscribed in a semicircle ).
4r² = BC² = AC⋅DC = 25⋅16 = 400
Consequently, the area of the yellow circle is 100π.
100.pai when pai=22/7
Sol is too short
Pythagorean triplet
(3, 4,5)
Hypotenuse is 16+9=25
We get another triplet multiplying by 5
(15 , 20,25)
Hence 2r =20
>r=10
Area = 100π sq units
It feels like there is a flaw in the diagram and it possibly confused me.
If r = 10, surely two-tangent theorem would give the height of the rectangle as 10 due to the lower-right vertex being equidistant from the two tangents? However, the rectangle's diagonal is shown as 9 when it should be >10. The alternative would be that the lower-right vertex is not making a right angle with the tangent line.
Ah well, it is Christmas LOL.
The right side of the rectangle is not a tangent to the circle.
@@unknownidentity2846 Too many Christmas festivities here, I think :)
368.08
368.08
😄😄😄
100pi
Alternative solution:
BD is the height from B to the hypotenuse AC so,
SqBD= ADxDC= 9x16
-> BD= 12=3x4
Focus on the triangle BDC,
CD = 16=4x4->BC=5x4=20 ( 3-4-5 triples)
Area of the yellow circle= pi x sq BC/4 =100 pi sq units😅😅😅
Let's find the area:
.
..
...
....
.....
Let E be the bottom right corner of the green rectangle, A be the center of the coordinate system and B on the x-axis. With r being the radius of the circle we obtain the fo
llowing coordinates:
A: ( 0 ; 0 )
B: ( xB ; 0 )
C: ( xB ; 2r )
D: ( xD ; yD )
E: ( xD ; 0 )
O: ( xB ; r )
From the known length of the chord CD we can conclude:
(xC − xD)² + (yC − yD)² = CD²
(xB − xD)² + (2r − yD)² = 16²
(xB − xD)² + (2r − yD)² = 256
Since D is located on the circle, we also can conclude:
(xO − xD)² + (yO − yD)² = r²
(xB − xD)² + (r − yD)² = r²
The right triangles ADE and ABC are obviously similar. Therefore we obtain:
BC/AC = DE/AD
(yC − yB)/AC = (yD − yE)/AD
(yC − yB)/(AD + CD) = (yD − yE)/AD
(2r − 0)/(9 + 16)= (yD − 0)/9
2r/25 = yD/9
⇒ yD = 18r/25
(xB − xD)² + (2r − yD)² = 256
(xB − xD)² + (r − yD)² = r²
(2r − yD)² − (r − yD)² = 256 − r²
4r² − 4*yD*r + yD² − (r² − 2*yD*r + yD²) = 256 − r²
4r² − 4*yD*r + yD² − r² + 2*yD*r − yD² = 256 − r²
4r² − 2*yD*r = 256
r² − yD*r/2 = 64
r² − (18r/25)*r/2 = 64
r² − 9r²/25 = 64
25r² − 9r² = 64*25
16r² = 64*25
r² = 4*25 = 100
⇒ r = √100 = 10
Now we are able to calculate the area of the yellow circle:
A(circle) = πr² = π*10² = 100π
Best regards from Germany
Thank you!