Can you find area of the largest semicircle? | (Semicircles) |

[@jamestalbott4499]

Thank you!

[@PreMath]

You're welcome! Thanks for the feedback ❤️

[@danmike2305]

You're definitely a knowledgeable person. Thanks 😊

[@allanflippin2453]

Around 7:20 where you find the length of FB = 4+8+6 = 18, I found a simpler way to calculate R. Consider right triangle OFE. OF = 18-R, EF = 6 and EO = R. R can be found using Pythagorean theorem. (18-R)^2 + 6^2 = R^2. Multiplying out, 324 - 36R + R^2 + 36 = R^2. Eliminating R^2, and combining numbers we get 36R = 360 or R = 10.

[@KenW-kb4uk]

Could also extend big semi-circle to full circle and use intersecting chords to deduce AF length: 6 x 6 = (6+8+4) x AF => AF = 2

[@gabrielmizrachi9826]

Thank you

[@alexnikola7520]

super

[@SUPERTUTORALGEBRA]

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[@prossvay8744]

Green semicircle area=18cm^2 1/2(π)r1^2=18π r1=6cm Yellow semicircle area=8cm^2 1/2(π)r2^2=8 So r2=4cm Connect P to Q and C to Q In CPQ CQ^2+CP^2=PQ^2 6^2+CP^2=10^2 So CP=8cm CB=CP+BP=8+4=12cm CF=OE=6cm FB=12+6=18cm AF=2R-18 Connect O to F and E to F AF=2R-18 OF=R-(2R-18)=18-R In∆ OEF EF^2+OF^2=OE^2 6^2+(18-R)^2=R^2 So R=10cm BIg sermicle area=1/2(π)(10^2)=50π cm^2=157.08 cm^2.❤❤❤

[@sergioaiex3966]

Good job, congrats !

[@PreMath]

Excellent! Thanks for sharing ❤️

[@리사두]

❤

[@PreMath]

Excellent! Thanks ❤️🙏

[@himadrikhanra7463]

R1= 6 , R2 =4 Length of transverse common tangent is radius = R ( out) Area = pi R ^2 /2

[@marcgriselhubert3915]

The radius of the green semi circle is 6 and the radius of the yellow semi circle is 4. We use an orthonormal center P and first axis (PB). We have B(4; 0) and E(-14; 6) on the big semi circle whose equation is (x -a)^2 + y^2 = R^2 or x^2 +y^2 - 2.a.x + a^2 - R^2 = 0 (with a the abscissa of its center O and R its radius). B is on it, so: 16 - 8.a + a^2 - R^2 = 0, E is on it, so: 196 + 36 + 28.a +a^2 - R^2 = 0. We substract and obtain: 36.a + 216 = 0, and so a = -6. We can now have a^2 - R^2 and then R, but simpler: abs(a) = 6 = OP, and then R = 6 + 4 = 10. Finally the area of the big semi circle is then (1/2).Pi.(10^2) = 50.Pi.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@misterenter-iz7rz]

(4+8-r+6)^2+36=r^2, (18-r)^2+36=r^2, 18^2-36r+36=0, r=(18^2+36)/36=10, thus the answer is 50pi.😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Los radios verde y amarillo son 6 y 4---> CP²=(6+4)²-6²=64---> CP=8. --->. Potencia de F =6²=(6+8+4)*AF---> AF=2---> AO=(2+6+8+4)/2=10---> Área semicírculo grande =10²π/2=50π. Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

My first attempt is almost the same as yours😅 So it is more fun to suggest another solution. 1/ Extend EQ to build the diameter EE’ and BP to build the diameter BB’. We have: EE’=12 and BB’=8 2/ Focus on the tangent point D: We have: the three points P,D,andQ are collinear( tangent theorem) EE’//BB’-> angle E’QD=angle B’PD Because the two angles mentioned above are the exterior angles of the two isiosceles triangles EQD and BPD So, angle QDE=angle BDP -> E,D,and B are collinear --> similarly E’D and B’ are collinear -> E’B’//=AE ( both are perpendicular to EB) -> AB’=12 and AB=20 Radius of the big semicircle=10 Area=1/2 pix100= 50 pi sq units😊😅😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@s900991]

R = 5+4 = 9 4= r of the yellow, 5 = po

[@LucasBritoBJJ]

I solved almost the same way…. Just got h²=mn h=6 m=18 n=x And it is easy to find x = 2

[@PreMath]

Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let's use the labels R, R(g) and R(y) for the radii of the big semicircle, the green semicircle and the yellow semicircle, respectively. From the known areas of the green and the yellow semicircle we obtain: A(g) = πR²(g)/2 (18π)cm² = πR²(g)/2 36cm² = R²(g) ⇒ R(g) = √(36cm²) = 6cm A(y) = πR²(y)/2 (8π)cm² = πR²(y)/2 16cm² = R²(y) ⇒ R(y) = √(16cm²) = 4cm The yellow and the green semicircle have exactly one point of intersection. Therefore the distance PQ of their centers is equal to the sum of their radii: PQ = PD + QD = R(y) + R(g) = 4cm + 6cm = 10cm Since AB is a tangent to the green semicircle, the triangle PQC is a right triangle and we can apply the Pythagorean theorem: PQ² = QC² + PC² PQ² − QC² = PC² PQ² − R²(g) = PC² (10cm)² − (6cm)² = PC² 100cm² − 36cm² = PC² 64cm² = PC² ⇒ PC = √(64cm²) = 8cm Now let's add point F on AB such that CFEQ is a square. Then we obtain: BF = PB + PC + CF = R(y) + PC + R(g) = 4cm + 8cm + 6cm = 18cm According to the theorem of Thales the triangle ABE is a right triangle, so we can apply the right triangle altitude theorem: AF*BF = EF² (AB − BF)*BF = EF² (2*R − BF)*BF = R²(g) (2*R − 18cm)*(18cm) = (6cm)² (36cm)*R − 324cm² = 36cm² (36cm)*R = 360cm² ⇒ R = (360cm²)/(36cm) = 10cm Now we are able to calculate the area of the big semicircle: A = πR²/2 = π*(10cm)²/2 = (50π)cm²

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@philipkudrna5643]

I managed to get to the point where the last piece was missing (by only calculating in my head), but got stuck then, as I failed to see that you can solve it by proportion via constructing a „Thales triangle“. Once you realize that, you can also calculate the pretty solution in your head (18/6=6/2!). The radius of the large semi circle is 10 and thus the area is 50pi.

[@rey-dq3nx]

QC=6 QP=10 so CP=8 Let CO=x Draw a tangent line from O to D then DO=CO=x From triangle ODP x²+4² =(8-x)² x²+16=64-16x+x² 16x=48 x=3 OP=CP-CO OP=8-3=5 Therefore OB=OP+BP OB=5+4=9 OB=9=radius of big semi circle where did I go wrong, professor or anyone ?

[@sergioaiex3966]

Solution: A = π a² 18π = π a²/2 π a² = 36π a² = 36 a = 6 A = π b² 8π = π b²/2 π b² = 16π b² = 16 b = 4 Triangle CPQ CQ = 6 PQ = 4 + 6 = 10 CP = 8 (Pythagorean Triple) Then, CB = 12 CF = 6 EF = h = 6 AF = m = ? FB = n = 6 + 12 = 18 h² = m . n 6² = m . 18 36 = m . 18 m = 2 Diameter = 2 + 18 Diameter = 20 Radius = 10 A = π r² A = π 10²/2 A = 50π Square Units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@michaeldoerr5810]

The radius of the larger circle is 10 units. Also at the 9:25, I think that I have noticed a fact regarding AA similarity and right angles: apart from aides different sides being similar, there also has to be a comparison between different triangles that are are formed. The first triangle is formed by beta-alpha and the second triangle is formed by alpha beta. I am also wondering the way that I actually did that is just good as yours. I set the original sides as AF/EF=FE/FB. I am hoping that my construction is too far off.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@aljawad]

I connected OD, then used a variety of triangles similarities coupled with trigonometry to reach the same result.

[@PreMath]

Thanks for the feedback ❤️

[@quigonkenny]

Let Rɢ be the radius of green semicircle Q, Rʏ the radius of yellow semicircle P, and R the radius of big semicircle O. Green semicircle Q: Aɢ = πRɢ²/2 18π = πRɢ²/2 Rɢ² = 18(2) = 36 Rɢ = √36 = 6 Yellow semicircle P: Aʏ = πRʏ²/2 8π = πRʏ²/2 Rʏ² = 8(2) = 16 Rʏ = √16 = 4 As the centers of two circles are collinear with their point of tangency, D is on PQ. As PD = Rʏ = 4 and QD = Rɢ = 6, PQ = 6+4 = 10. Triangle ∆QCP: QC² + PC² = PQ² 6² + PC² = 10² PC² = 100 - 36 = 64 PC = √64 = 8 Drop a perpendicular from E to AB at M. As EQ is parallel to AB and QC is perpendicular to AB as C is the tangent point between AB and semicircle Q, EM = QC = 6 and CM = QE = 6. MB = CM + PC + PB = 6 + 8 + 4 = 18 Triangle ∆EMO: EM² + OM² = OE² 6² + (18-R)² = R² 36 + 324 - 36R + R² = R² 36R = 360 R = 10 Big semicircle O: Aꜱ = πR²/2 = π10²/2 = 100π/2 Aꜱ = 50π sq units ≈ 157.08 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ 1) first way: radius of the largest semicircle: R radius of the small semi circle: r₁ radius of the middle semicircle: r₂ A₁= π*r₁²/2 A₁= 8π cm² ⇒ 8π= πr₁²/2 16= r₁² r₁= 4 cm A₂= π*r₂²/2 A₂= 18π cm² ⇒ 18π= πr₂²/2 36= r₂² r₂= 6 cm Let's consider the right triangle ΔCPQ [QC]= r₂= 6 cm [PQ]= r₁ + r₂ [PQ]= 4+6 [PQ]= 10 cm By applying the Pythagorean theorem: [QC]²+[CP]²= [PQ]² 6²+[CP]²= 10² ⇒ [CP]= 8 cm Let's say: [CO]= x [OP]= 8-x ⇒ [OP]+[PB]= R 8-x+r₁ = R 8-x+4= R 12-x= R..........Equation-1 Let's consider the right triangle ΔOSE S is a point on the radius r₂ so that [ES] ⊥ [SO] By applying the Pythagorean theorem we get: [OS]²+[SE]²= [EO]² [OS]= r₂ = 6 cm [SE]= x+r₂ = 6+x [EO]= R ⇒ R²= 6²+(6+x)² R²= 36+36+12x+x² R²= 72+12x+x² R= 12-x........Eqaution-1 ⇒ (12-x)²= 72+12x+x² 144-24x+x²= 72+12x+x² 72= 36x x= 2 cm ⇒ [CO]= 2 cm R= 12-x R= 10 cm Alarge-semicircle= πR²/2 R= 10 cm ⇒ Alarge-semicircle= π*10²/2 Alarge-semicircle= 50π Alarge-semicircle≈ 157,08 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

Second way of solution ▶ radius of the largest semicircle: R radius of the small semi circle: r₁ radius of the middle semicircle: r₂ By making an accurate drawing, I found out that the line [PQ] is parallel to [EO] (also equal: = R) ❗ [PQ] // [EO] A₁= π*r₁²/2 A₁= 8π cm² ⇒ 8π= πr₁²/2 16= r₁² r₁= 4 cm A₂= π*r₂²/2 A₂= 18π cm² ⇒ 18π= πr₂²/2 36= r₂² r₂= 6 cm Let's consider the right triangle ΔCPQ [QC]= r₂= 6 cm [PQ]= r₁ + r₂ [PQ]= 4+6 [PQ]= 10 cm By applying the Pythagorean theorem: [QC]²+[CP]²= [PQ]² 6²+[CP]²= 10² ⇒ [CP]= 8 cm If we move the [QP] to the left side of r₂ cm , we get the line [EO] ⇒ [PO] would be equall to [QE] [PO]= [QE] [PO]= 6 cm❗ ⇒ [CO]= 8-6 [CO]= 2 cm Let's consider the right triangle ΔOSE S is a point on the radius r₂ so that [ES] ⊥ [SO] By applying the Pythagorean theorem we get: [OS]²+[SE]²= [EO]² [OS]= 6 cm [SE]= [EQ]+[QS] [QS]= [CO] [QS]= 2 cm [SE]= 6+2 [SE]= 8 cm ⇒ 6²+8²= R² R= 10 cm Alarge-semicircle= π*10²/2 Alarge-semicircle= 50π Alarge-semicircle≈ 157,08 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

The Answer is that Big Semicircle Area is equal to 50Pi Square Cm. 01) Big Diameter Diameter (D) : D = 4 + 4 + 6 + 6 = 8 + 12 ; D = 20 02) Moving the Green Circle to left until Line AE is orthogonal to AB reduces 2 cm, that is the projected intersection of Green and Yellow Circle, in Line AB. 03) CP = 8. CP^2 = 10^2 - 6^2 ; CP^2 = 100 - 36 ; CP^2 = 64 ; CP = 8. 04) Moving Point C, 2 cm to the Left there will be no Diagonal Tangent. Now the two Circles are Tangent in a Vertical Line. 05) So : The Big Circle Diameter must be equal to the Sum of the Green Diameter and the Yellow Diameter. D = 12 + 8 = 20 cm 06) So : The Big Circle Radius = 10 cm 07) Semicircle Area = 100*Pi / 2 = 50Pi sq cm ANSWER: Big Semicircle area equal 50Pi Square Centimeters.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@fakeplaystore7991]

A = 36π + some change. NEXT!

[@wackojacko3962]

Anytime I call on paranormal entities too help with solving math problems, ...nothin! Thales couldn't figure it out either. So he didn't rely on the paranormal too define his existence...All Hail Thales! 🙂

[@PreMath]

Well said! Thanks for the feedback ❤️

[@sergeyvinns931]

50pi.

[@Nothingx303]

Είναι μια γεύση μπισκότου σοκολάτας 😋 😜