Can you calculate area of the Green shaded region? | (Circles) |

Learn how to find the area of the Green shaded region. Important Geometry and Algebra skills are also explained: Trigonometry; Pythagorean theorem; circle theorem. Step-by-step tutorial by PreMath.com.
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Can you calculate area of the Green shaded region? | (Circles) | #math #maths | #geometry
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[@imetroangola17]

Parabéns pelos vídeos e por trazer excelentes questões! ❤

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️🙏

[@jaimeyomayuza6140]

Gracias por su apoyo. DIOS. BENDIGA NUESTROS PAISES!! From BOGOTA D. C. COLOMBIA

[@johankotze42]

Reflect the semi circle. Six small circles fitting like that into a larger circle implies a seventh small circle on the center of the large circle, all being tangent. Therefore the diameter of the large circle is: 24 = 6*r => r = 4. No trigonometry, no Pythagoras.

[@egillandersson1780]

If you mirror all to get the big full circle, the centers of the small circles make an hexagon. So, triangle OPQ is equilateral ; therefore 2r = r - 12 => r = 4

[@SkinnerRobot]

2r = 12 - r

[@PreMath]

Thanks for the feedback ❤️🙏

[@Noval5s]

Is there a theorem that proves them equilateral?

[@quigonkenny]

​​​@@Noval5s Don't know if there's a specific theorem per se, but a regular hexagon is made up of six equilateral triangles. Since each vertex in a regular hexagon is 60° from the next and the distances from the vertices to the center are all the same, then you have six 60° isosceles triangles. A 60° isosceles triangle is necessarily equilateral since the other two angles must also be 60°, so there you go.

[@egillandersson1780]

@ Yes ! Lapsus calami. Thank you for notice.

[@alanthayer8797]

VERY NIVE breakdown ! Thanks for different Solutions Daily sir !

[@gaylespencer6188]

Another way. Draw a half circle equal to the r of the smaller circles. This half circle will be tangent to those three small circles. Now the distance from the center of the big half circle to the top of the big half circle is three r. If 3r = 12 then r is 4.

[@professorrogeriocesar]

Muito bom! Eu fiz sem precisar calcular o ângulo. Excelente, obrigado.

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏

[@jacquespictet5363]

Making a horizontal mirror and adding the 7th small circle shows directly that R=3r.

[@jimlocke9320]

You haven't proved that the 7th circle has the same radius as the 6 other circles. Any other number of circles (e. g. 4, 5, or 6) and the middle circle will not have the same radius as the other circles.

[@Claudio_Bruzzone]

@@jimlocke9320 It is easily demonstrated. If you join the 6 centers of the small triangles, you get a regular hexagon of side 2r. Each of the 6 triangles is equilateral, therefore OP=OQ=OT=2r...

[@PreMath]

Thanks for the feedback ❤️🙏

[@jimlocke9320]

@@Claudio_Bruzzone Yes, that is correct, thanks for the proof. jacquespictet5363 made the assumption that the 7th circle would have the same radius as the other circles, but did not prove it. With 6 circles in the large full circle, the added 7th circle in the middle will have the same radius. With more circles, you won't get a regular hexagon and the added middle circle will not have the same radius.

[@marcgriselhubert3915]

Fine.

[@quigonkenny]

By mirroring the figure about AB it's clear to see that the centers of the internal circles form a regular hexagon. Therefore each of the centers is 60° from the next. As the figure is symmetrical about OD, then ∠AOT = ∠POB = (180°-120°)/2 = 30°. As AB is tangent to circles T and P at G and F respectively, and TG and PF are radii (length r) of the congruent circles T and P, then ∠TGO = ∠OFP = 90° and thus ∆TGO and ∆OFP are congruent 30-60-90 special tight triangles. As OP = 2PF and PF = PC = r, then OC = R = OP+PC = 2r+r = 3r. AB = 2R 24 = 2R R = 24/2 = 12 3r = 12 r = 12/3 = 4 Green area: A = πR²/2 - 3πr² A = π(12)²/2 - 3π(4)² A = 144π/2 - 3(16)π A = 72π - 48π [ A = 24π ≈ 75.398 sq units ]

[@marioalb9726]

sin30°=r/(12-r) = 1/2 2r = 12-r --> r = 4 cm A = ½πR² - 3πr²= ½π12² -3π4² A = 72π - 48π = 24π cm² (Solved √)

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️🙏

[@sventheviking4635]

OTQP form a rhombus of sides of equal length so OP=PQ=2r, so radius =3r

[@santiagoarosam430]

En la figura propuesta es fácil visualizar dos hexágonos regulares concéntricos; el exterior tiene lado ED=DO =24/2=12 y el interior TQ=2r=DO-r=12-r---> r=12/3=4 ---> Área sombreada verde= (π12²/2)-(3*π4²) =24π u². Gracias y un saludo cordial.

[@himo3485]

AO=OB=24/2=12 r/(12-r)=1/2 2r=12-r 3r=12 r=4 Green shaded area = 12*12*π*1/2 - 4*4*π*3 = 24π

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@sabinafacondo1638]

Si calcola l'area del semicerchio,(conoscendo la diagonale). Si calcolano le aree dei 3 cerchi inscritti(deducendo le diagonali). Si sottraggono dall'area del semicerchio 226,08,le aree dei 3 cerchi 150,72. Risultato 75,36.

[@cyruschang1904]

R = 12 = 3r Green area = (R^2)(π/2) - (r^2)(3π) = (144/2 - 16 x 3)π = (72 - 48)π = 24π

[@lwels49]

The line OD must be 3 times radius of small circle. So r equals 4. Etc

[@AmirgabYT2185]

S=24π≈75,43

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@Claudio_Bruzzone]

No calculation is needed. Given the symmetry of the configuration, if the circumference is completed with the other 3 smaller circles, the only possible and compatible configuration is that of 7 small tangent circles (6 external and one internal with center in the center of the large circumference). Therefore trivially: r = D/6 = 24/6 = 4 The green area follows: 72π - 3*16π = 24π u² In fact, if you join all the centers of the small circles, you obtain a regular hexagon which, having a side equal to the radius of the circumscribed circle, such radius can only be 2r, and therefore r=1/3R.

[@PreMath]

Thanks for the feedback ❤️🙏

[@michaelkouzmin281]

Just another solution: 1. Let us draw 2 auxiliary lines: OD and horizontal TV through point T where V is crosspoint of OD and TV; 2. Let r= TG, x = GO Then OT = 12 -r; QV = 12 -2r; TQ = 2r 3. So we have 2 right triangles: OGT: (12-r)^2 = x^2 +r^2 => x^2 = (12-r)^2 - r^2 = 144-24r (1) TVQ: x^2+(12-2r)^2 = (2r)^2 => x^2 = (2r)^2 - (12-2r)^2 = 48r - 144 (2) 4. Let us compare (1)& (2) : left sides are equal: 48r -144 = 144-24r: 72r =288 r = 4. /* x= 4sqrt(3) just for reference */ 5. A(semicirc) = pi*D^2(4*2) = 72*pi; A(white) = 3*pi*r^2 = 2*pi*3^2 = 48*pi; A(green) = 72*pi-48*pi = 24*pi sq units.

[@peterkrauliz5400]

That's the best approach Man!

[@sergioaiex3966]

Solution: We must find the radius "r" to solve this question Applying Pythagorean Theorem in ∆FOP, we will have: OF² + r² = (12 - r)² OF² + r² = 144 - 24r + r² OF = √(144 - 24r) Applying, once again, Pythagorean Theorem in ∆PQW, such that W is the midpoint in the line PT [√(144 - 24r)]² + (12 - 2r)² = (2r)² 144 - 24r + 144 - 48r + 4r² = 4r² 144 - 24r + 144 - 48r = 0 288 - 72r = 0 72r = 288 r = 4 White Region Area = 3 × π (4)² WRA = 48π Green Shaded Area = ½ π (12)² - 48π GSA = 72π - 48π GSA = 24π Square Units ✅ GSA ≈ 75.3982 Square Units ✅

[@giuseppemalaguti435]

OP=b,R=12...risultano due equazioni 12=2r+√((2r)^2-b^2)...12=r+√(r^2+b^2)...(12-2r)^2=4r^2-b^2,(12-r)^2=r^2+b^2...[sommo le equazioni]...288=72r=>r=4...Agreen=π(12)^2/2-3π4^2=π(72-48)=24π

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The white circle in the middle has exactly one point of intersection with the white circle on the left and with the white circle on the right. So with r being the radius of the white circles we can conclude: PQ = QT = 2r The semicircle has exactly one point of intersection with each of the white circles. Therefore with R being the radius of the semicircle we obtain: OP = OC − PC = R − r OQ = OD − QD = R − r OT = OE − TE = R − r The triangles OPT and PQT are isosceles triangles (OP=OT and PQ=QT). So with M being the midpoint of PT we obtain two pairs of congruent right triangles: OMP/OMT and MPQ/MQT. By applying the Pythagorean theorem to the right triangles OMP and MPQ we obtain: OP² = OM² + PM² PQ² = QM² + PM² OP² − PQ² = OM² − QM² OP² − PQ² = OM² − (OD − OM − QD)² OP² − PQ² = PF² − (OD − PF − QD)² (R − r)² − (2r)² = r² − (R − r − r)² (R − r)² − (2r)² = r² − (R − 2r)² R² − 2Rr + r² − 4r² = r² − (R² − 4Rr + 4r²) R² − 2Rr + r² − 4r² = r² − R² + 4Rr − 4r² 2R² = 6Rr R = 3r ⇒ r = R/3 = (AB/2)/3 = (24/2)/3 = 12/3 = 4 Now we are able to calculate the area of the green region: A(green) = A(semicirle) − A(white circles) = πR²/2 − 3πr² = π*12²/2 − 3π*4² = 72π − 48π = 24π Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@johankotze42]

This looks like an AI generated solution. 😀

[@unknownidentity2846]

@@johankotze42 I can assure you that my solutions are always HI generated (HI = human intelligence).🙂 Best regards from Germany

[@unknownidentity2846]

@@johankotze42 I can assure you that my solutions are HI generated (HI = human intelligence).🙂 Best regards from Germany

[@AzouzNacir]

Let r be the radius of the circles, from which we have (12-r)²-r²=(2r)²-(12-2r)², so r=4, and the green area is equal to π*(12)²/2-3*(π*4²)=24π

[@jimlocke9320]

You beat me to it! To provide missing details: To derive that equation, construct OD and PT. Label the intersection as point G. Construct CD. Let PG have length x. Then, from ΔCDG, (2r)² = (12-2r)² + x² and, from ΔCGO, (12-r)² = r² + x². Solve for x² in each equation and equate the values of x² to produce the equation (12-r)²-r²=(2r)²-(12-2r)². Expand to (12)² - 24r + r² = 4r² - ((12)² - 48r + 4r²) and simplify to find r = 4.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@AzouzNacir]

Are these simple things worth writing a newspaper about? They are repetitive and not new. I think that users know this.​@@jimlocke9320

[@AzouzNacir]

Are these simple things worth writing a newspaper about? They are repetitive and not new. I think that users know this.​@@jimlocke9320

[@AzouzNacir]

Are these simple things worth writing a newspaper about? They are repetitive and not new. I think that users know this.

[@SaurabhYadav-hr9nk]

please keep

[@PreMath]

Thanks for the feedback ❤️🙏

[@wackojacko3962]

When I call customer service, I don't press one for English, I don't press two for diphthongs, and I don't press three for ululation. I press four to SOHCAHTOA! 😊

[@PreMath]

😀 Thanks for sharing ❤️🙏

[@Christopher-e7o]

X,2x+5=8

[@MohamedMd-l5d]

Why angle equal 60 degrés please with explain

[@dantallman5345]

Since the small circles are all equal, he drew tangent lines to them from the midpoint of the semicircle diameter, thus dividing the semicircle into three equal wedges (180 degrees /3=60 degrees).

[@nenetstree914]

24PI

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@brettgbarnes]

OG² = OT² - TG² OG² = (12 - r)² - r² OG² = (144 - 24r + r²) - r² ------------------------------------- OG² = 144 - 24r ------------------------------------- OG² = TQ² - (12 - QD - TG)² OG² = (r + r)² - (12 - r - r)² OG² = (2r)² - (12 - 2r)² OG² = 4r² - (144 - 48r + 4r²) ------------------------------------- OG² = 48r - 144 ------------------------------------- 48r - 144 = 144 - 24r 72r = 288 ------------------------------------- r = 4 -------------------------------------

[@texitaliano64]

Il semicerchio di diametro D=24 ha raggio R=D/2=12 R=12 chiamiamo O il centro dellla semicirconferenza chiamiamo O1 il centro del primo cerchio inscritto a sinistra chiamiamo O2 il centro del secondo cerchio inscritto a destra chiamiamo O3 il centro del terzo cerchio inscritto in alto chiamo 2*a la distanza O1O2 con pitagora considero il triangolo rettangolo O1KO3 dove K è il punto medio di O1O2 a=sqrt((r+r)^2-(R-r-r)^2 a=sqrt((2*r)^2-(R-2*r)^2) a=sqrt((2*r)^2-(R^2+(2*r)^2-2*R*2*r)) a=sqrt(4*R*r-R^2) a=sqrt(4*12*r-12^2) a=sqrt(48*r-144) a=sqrt(48*(r-3)) con pitagora considero il triangolo rettangolo O1KO dove K è il punto medio di O1O2 a=sqrt((R-r)^2-r^2) a=sqrt(R^2+r^2-2*R*r-r^2) a=sqrt(12^2-24*r) a=sqrt(144-24*r) a=sqrt(24*(6-r)) uguagliando a=a otteniamo sqrt(24*(6-r))=sqrt(48*(r-3)) sqrt(144-24*r)=sqrt(48*r-144) eleviamo al quadrato entrambi 144-24*r=48*r-144 144+144-24*r-48*r=0 288-72*r=0 288=72*r r=288/72 r=4 La superficie della semicirconferenza è Ss=(pi*D^2)/4/2 Ss=72*pi La superficie di ciascuna circonferenza inscritta è So=pi*r^2 So=pi*4^2 So=16*pi Ora facciamo la differenza per trovare l'area richiesta S=Ss-3*So S=72*pi-3*16*pi S=72*pi-48*pi S=(72-48)*pi S=24*pi

[@LuisdeBritoCamacho]

MY RESOLUTION PROPOSAL : 01) TG = r 02) TQ = 2r 03) Let's drop a Line between Point O and Point D. OD = R = 12 04) Let's draw an Isoscles Triangle [PQT) with Sides (TQ ; PQ ; TP) and with TQ = PQ 05) Let's divide Line OD in 3 different segments : OD (R) = OQ'(r) + Q'Q + QD (r). Q' is the Middle Point between Side TP. 06) R = r + Q'Q + r 07) 12 = 2r + Q'Q 08) Q'Q = (12 - 2r) 09) Let's give a call to Mr. Pythagoras !! 10) Let's draw a Rectangle [OFPQ'] 11) OF^2 = OP^2 - FP^2 ; OF^2 = (12 - r)^2 - r^2 ; OF^2 = 144 - 24r + r^2 - r^2 ; OF^2 = (144 - 24r) 12) OF'^2 = Q'P^2 13) (12 - 2r)^2 + (144 - 24r) = 4r^2 14) Only one Solution r = 4 15) Green Shaded Area = 72 * Pi - (3 * (16 * Pi)) 16) GSA =72 * Pi - 48 * Pi 17) GSA = 24 * Pi MY BEST ANSWER IS : Green Shaded Area equal 24Pi Square Units or approx. equal to 75,4 Square Units.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏