Can you find area of the Green shaded region? | (Semicircle) |

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@manuelantoniobahamondesa.3252]

Excelente!

[@quigonkenny]

Let AB = x. As ∠CAB = 60°, ∆ABC is a 30-60-90 special right triangle, ∠BCA = 30°, CA = 2x, and BC = √3x. Draw DE. As ∠CDE = ∠ABC = 90° and ∠ECD is common, ∆CDE is similar to ∆ABC. As EC = 2r, DE = EC/2 = r and CD = √3r. BC/CD = CA/EC (√3+2r)/√3r = (6+√3r)/2r 2r(√3+2r) = √3r(6+√3r) 2√3r + 4r² = 6√3r + 3r² r² - 4√3r = 0 r(r-4√3) = 0 r = 4√3 As ∠ECD = 40°, ∠EOD = 2(30°) = 60°. Therefore ∠COD = 180°-60° = 120°. The green area will be equal to the area of the sector subtended by arc DC minus the area of triangle ∆COD. Geeen area: Aɢ = (120/360)πr² - r²sin(120°)/2 Aɢ = πr²/3 - √3r²/4 Aɢ = π(4√3)²/3 - √3(4√3)²/4 Aɢ = 48π/3 - 48√3/4 Aɢ = 16π - 12√3 ≈ 29.48 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Right triangle ABC: (6+2rcos30°).cos30°=√3+2r 6√3/2 + 3/2 r = √3 + 2r ½ r = 6√3/2 - √3 r = 6√3 - 2√3 = 4√3 cm Círcular segment area: A = ½ . r² . ( α - sinα ) A = ½ . 48 . (120°-sin120°) A = 29,48 cm² ( Solved √ )

[@marioalb9726]

Pytagorean theorem, right triangle ABC: (6+2rcos30)²=(√3+2r)²+((√3+2r)/tan60)² (6+√3r)²=(√3+2r)²+(1+2/√3 r)² 6²+12√3r+3r²=(3+4√3r+4r²)+(1+4/√3 r+4/3 r²) 7/3 r² - 11,547r - 32 = 0 r² - 4,9487 r - 13,7143 = 0 r = 6,9282 cm = 4√3 cm Círcular segment area: A = ½ . r² . (α-sinα) A = ½ . 48 . (120°-sin120°) A = 29,48 cm² ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Right triangle ABC: (√3 + 2r)/cos30° = 6+2r.cos30° 2 + 4/√3 r = 6 + √3 r 4/√3 r - 3/√3 r = 6 - 2 = 4 r = 4√3 cm Círcular segment area: A = ½ . r² . ( α - sinα ) A = ½ . 48 . (120°-sin120°) A = 29,48 cm² ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing ❤️

[@KipIngram]

Below D will denote the y coordinate of the point D in the diagram. Reflect the entire figure in the horizontal axis. Then we can use the law of sines to write 2*D/sin(60) = 2*R Note also that the triangle inscribed in our full circle is an equilateral triangle, so DC = 2*D. Now we can use standard triangle trig to write (6 + 2*D)*cos(30) = sqrt(3) + 2*R (6 + 2*R*sin(60))*cos(30) = sqrt(3) + 2*R 6*sin(60) + 2*R*sin^2(60) = sqrt(3) + 2*R This is easily solved for R to yield R = 4*sqrt(3). Note that R^2 = 48. Now, the area of an equilateral triangle inscribed in a circle is given by Area = 3*R^2*sqrt(3)/4 Area = 48*3*sqrt(3)/4 = 36*sqrt(3) We can subtract this from the circle area and divide by three to get our desired result: Green_Area = R^2*(pi - 3*sqrt(3)/4)/3 Green_Area = 29.4808 Q.E.D.

[@sergeyvinns931]

R=4\/3; Area=1/2R^2(pi*a/180-sina)=1/2*48(2pi/3-\/3/2)=24(4pi-3\/3)/6=4(4*3,1415-3*1,732)=29,48.

[@JSSTyger]

You're a great teacher. It takes alot of work to go through these and until you've made sure its right. I just did it but found two dumb errors (one was π(d/2)²(1/3) = πd²/6) but ended up ironing it all out to get it right.

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️🙏

[@juanalfaro7522]

I obtained the same result, but I did not use the cosine rule to determine X, though I drew the triangle to find the angle COD (OD=OC=R -> ODC = OCD = 30 -> COD = 180-2*30 = 120 -> EOD = 180-120 = 60). I used (6+X) * cos (30) = √3 + 2R = 3√3 + √3/2 * X and then X = 2R * cos (30) = √3R. Then √3 + 2R = 3√3 + 3/2 * R --> r/2 = 2√3 -> R = 4√3; then X=R√3 = 12 and did everything else as in the video.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@prossvay8744]

Angle BAC=60° Angle ACB=30° Connect E to D So ∆ CDE is right triangle Angle DCE=30° Angle CDE=60° So (30° ; 60°; 90°) BC=2r Hence DE=r and CD=r√3 AC=6+r√3 ; BC=√3+2r In∆ ABC Siin,(60°)=(√3+2r)/6+r√3) So r=4√3 Green shaded area=1/2(π)(4√3)^2-(60°/360°)(π)(4√3)^2-1/2(4√3)^2sin(120°=16π-12√3=29.48 square units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@montynorth3009]

Joining E to D, EDC = 90 degrees in semicircle. Triangle EDC, sin 60 = DC / 2r. DC = 2r x sin 60 = 1.732r. Triangles EDC & ABC are similar. So DC / EC = BC / AC. 1.732r / 2r = (1.732 + 2r) / (6 + 1.732r). Cross multiply. 4r^2 + 3.464r = 3r^2 +10.392r. Divide both sides by r. 4r + 3.464 = 3r +10.392. r = 6.928. Area triangle COD = 0.5 x 6.928^2 x sin 120. 20.783. Area sector = Pi x 6.928 x 6.928 / 3. 50.263. Green segment = 50.263 - 20.783. 29.48.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Celanish]

Grazie !

[@josephsalinas6725]

O professor resolvendo parece tão fácil !

[@PreMath]

Gracias, querido❤️

[@santiagoarosam430]

Las proyecciones ortogonales de D sobre AB y BC son F y G, respectivamente.FD=BG=AD*√3/2=3√3---> EG=BG-BE=2√3 ---> Ángulo BCA=30º--->BOD=60º y. DOCr120º---> OD=r--->GO=r/2 y GD=r√3/2---> r=2*EG=2*2√3=4√3 ---> Área verde =(πr²/3)-(OC*GD/2) =16π-12√3 ud². Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

The triangle COD is isosceles (OC = OD = R the radius of the semi circle), so angle ODC = angle OCD = 30° and the angle COD = 120°. In triangle COD: CD^2 = OD^2 + OC^2 - 2.OC.OD.cos(120°) =R^2 + R^2 - 2.R.R.(-1/2) = 3.R^2, so CD = sqrt(3).R and CA = 6 + sqrt(3).R In triangle ABC: cos(angle BCA) = cos(30°) = sqrt(3)/2 = CB/CA = (sqrt(3) +2.R)/(6 + sqrt(3).R), giving that R = 4.sqrt(3). The area of thiangle COD is (1/2).OC.OD.sin(angle COD) = (1/2).(4.sqrt(3)).(4.sqrt(3).(sqrt(3)/2) = 12.sqrt(3) The area of sector EOD is Pi.((4.sqrt(3))^2).(60°/360°) = 8.Pi The area of the whole semi circle is Pi.((4.sqrt(3))^2).(1/2) = 24.Pi So the green area is 24.Pi - 8.Pi - 12.sqrt(3) = 16.Pi -12.sqrt(3). Finally, I see that I did it just like you did...

[@PreMath]

Excellent! Thanks for sharing ❤️

[@satrajitghosh8162]

Radius of this circle be r ∆ EDC is right angular. DC = EC * cos (π/6) = √ 3 * r BC = AC * sin (π/3) = ( 6 + √3 * r)* ( √3 /2) 2 ( √ 3 + 2 r) = 6 √3 + 3 r r = 4 √ 3 Arc of Green region sweeps π - 2 * π/6 = 2π/3 radian with centre of circle. If r be the radius of this circle Green Area = r ^2 * ( π /3 - √3 /4) = 48 ( π /3 - √3 /4) = 16 π - 12 √3

[@phungpham1725]

1/ The triangle CDE is a 30/90/60 special one, so CE=2r, ED=r and CD=rsqrt3. 2/ The quadrilateral ABED is cyclic, so CDxCA=CExCB -> rsqrt3(rsqrt3+6)=2r(2r+sqrt3) -> r=4sqrt3 -> CD=12 3/ Area of the triangle DOC: Drop the height OH to CD. Note that angle DOC= 120 degrees so angle DOH=60 degrees->OH=r/2=2sqrt3 Area of triangle DOC= 1/2x2sqrt3x12=12sqrt3 Area of sector DOC= pi. sqr/3= 16 pi Area of the green segment=16 pi-12sqrt3=29.48 sq units😅😅😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

Interesting question, thank you 🙂🙏 Here my solution ▶ By drawing the triangle in the given semicircle: ΔOCD [OC]= [DO]= r ⇒ ∠CDO= ∠OCD = 30° ⇒ ∠DOC= 120° By applying the cosine theorem for the triangle ΔOCD we can find [CD] : [CD]²= [DO]²+[OC]² - 2*[DO]*[OC]*cos(∠DOC) [DO)= r [OC]= r cos(120°)= -0,5 [CD]²= r²+r² +2r²*(0,5) [CD]²= 2r²+r² [CD]= √3r ⇒ [CA]= [CD]+[DA] [CA]= √3r + 6 II) By considering the right triangle ΔBCA, we can write ∠BCA= 30° tan(30°)= 1/√3 1/√3 = [AB]/[BC] [BC]= 2r+√3 ⇒ 1/√3 = [AB]/(2r+√3) 2r+√3 = √3 [AB] [AB]= √3(2r+√3)/3 By applyig the Pythagorean theorem for the triangle ΔBCA, we can write: [AB]²+[BC]²= [CA]² [√3(2r+√3)/3]² + (2r+√3)²= (6+√3r)² (1/3)*(4r²+4√3r+3) + 4r²+4√3r+3 = 36+12√3r+3r² 4r²/3 + 4√3r/3 + 1+ 4r²+4√3r+3 = 36+12√3r+3r² by multiplying both sides with 3 we get 4r²+4√3r+3+12r²+12√3r+9 = 108+36√3r+9r² ⇒ 16r²- 9r²+16√3r -36√3r+12-108=0 ⇒ 7r² - 20√3r - 96 = 0 Δ= 20²*3 - 4*7*(-96) Δ= 3888 3888 = 2⁴*3⁴*3 √Δ = √2⁴*3⁴*3 √Δ = 2²*3²√3 √Δ = 36√3 r₁= (20√3+36√3)/14 r₁= 56√3/14 r₁= 4√3 r₂= (20√3 - 36√3)/14 r₂= -16√3/14 r₂= -8√3/7 < 0 ❌ ⇒ r= 4√3 III) The area of the semicircle: Asc= πr²/2 r= 4√3 ⇒ Asc= π*(4√3)²/2 Asc= 24π By subtracting the tringle ΔOCD from this semicircle, plus the 60° segment of the circle, we can find the green area: Acircle segment = πr²*(60°)/360° Acircle segment = π*( 4√3)²*(1/6) Acircle segment = 8π A(ΔOCD)= 1/2 * [DO]*[OC]*sin(120°) A(ΔOCD)= 1/2*r²*sin(120°) A(ΔOCD)= 1/2*(4√3)²*√3/2 A(ΔOCD)= 48√3/4 A(ΔOCD)= 12√3 Agreen= Asc - Acircle segment - A(ΔOCD) Agreen= 24π - 8π - 12√3 Agreen= 16π - 12√3 Agreen= 4(4π - 3√3) Agreen≈ 29,48 squre units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@zdrastvutye]

there are 2 solutions, just try this: 10 vdu5:for a=0 to 15:gcola:print a;:next a 20 print "premath-can you find area of the green shaded region" 30 @zoom%=1.4*@zoom%:dim x(3),y(3):rem print"test";sin(rad(45)),"%",cos(rad(45)) 40 l1=sqr(3):l2=6:w=60:sw=l1^2/(l1+l2)/10:r=sw:gosub 80:goto 130 50 h=(r*2+l1)/tan(rad(w)):xm=l1+r:dgu1=(l2*sin(rad(w))-xm)^2/l2^2 60 dgu2=(h-l2*cos(rad(w)))^2/l2^2:dgu3=r*r/l2^2 70 dg=dgu1+dgu2-dgu3:return 80 gosub 50 90 dg1=dg:r1=r:r=r+sw:if r>100*l1 then stop 100 r2=r:gosub 50:if dg1*dg>0 then 90 110 r=(r1+r2)/2:gosub 50:if dg1*dg>0 then r1=r else r2=r 120 if abs(dg)>1E-10 then 110 else print "r=";r:return 130 print "Loesung1":gosub 140:goto 200 140 xd=l2*sin(rad(w)):yd=h-l2*cos(rad(w)):l3=sqr((xd-l1-2*r)^2+yd^2) 150 l3n=h/cos(rad(w))-l2:fe=l3/l3n:fe=(1-fe)*100:print "der fehler=";fe;"%" 160 ym=0:xc=l1+2*r:yc=0:rem das skalarprodukt berechnen *** 170 dx1=xd-xm:dy1=yd-ym:dx2=xc-xm:dy2=yc-ym 180 zx=dx1*dx2:zy=dy1*dy2:n=sqr(dx1^2+dy1^2)*sqr(dx2^2+dy2^2) 190 we=acs((zx+zy)/n):ar=r*r*(we/2-sin(we/2)*cos(we/2)):print "die flaeche="; ar:return 200 print "loesung 2":r=r+sw:gosub 80:gosub 140:masx=1200/(l1+2*r):masy=850/h 210 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

[@robertlynch7520]

Basically ... used a different identity to get the same answer. For ALL right triangles, the interior height [H] is [AB/C] from that, and from the congruence of 30-60-90 triangles through the diagram, all the rest of the measurements are easy.

[@PreMath]

😀 Thanks for the feedback ❤️

[@-wx-78-]

Draw EF || AC and EG || AB, △ABC is split in several similar triangles (and parallelogram): △ABC ~ △EDC ~ △FBE ~ △GEC ~ △GDE. From △FBE EF = AG = 2, therefore DG = AD−AG = 4. From △CDE DE = r; on the other hand, r² = DE² = CD·DG = r√3·4 as an altitude in right triangle, hence r = 4√3. Now is the time for circular segment area formula: a = ½r²(θ−sin θ), where θ is obviously ⅔π.

[@PreMath]

Thanks for sharing ❤️

[@giuseppemalaguti435]

Calcolo r..(6+2rcos30)^2=(2r+√3)^2+((2r+√3)tg30)^2...risulta r=4√3...Agreen=πr^2/2-πr^2/6-(rcos30rsin30)=(1/3)πr^2-√3r^2/4=(π/3-√3/4)r^2=29,48..

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johnbrennan3372]

Triangle abc having angles 30,60 and 90 ,ac=(4r/root3)+2 so dc = (4r/root3)-4. In the triangle dec with angles 30,60 and 90 , de=r,ec=2r and dc = (4r/root3)-4.. Using pythag. You get r=4root 3 etc. It involved tedious use of quadratic formula.

[@PreMath]

Thanks for sharing ❤️

[@kunalchakraborty9735]

Draw a perpendicular from o to DC At F. Then FC/ r = cos 30 FC = sqrt(3)*r /2 so DC=r * sqrt(3) The remaining calculation are same

[@PreMath]

Thanks for sharing ❤️

[@murdock5537]

Nice! φ = 30° → cos⁡(φ) = √3/2 = (2r + √3)/(6 + r√3) → r = 4√3 → DOC = 4φ → sin⁡(4φ) = sin⁡(6φ - 4φ) = sin⁡(2φ) = cos⁡(φ) = √3/2 → area ∆ DOC = (1/2)sin⁡(4φ)48 = 12√3 → area of green shaded area = 4(4π - 3√3); btw: EAB = α → sin⁡(α) = √7/14 → AE = 2√21 → AB = 9

[@matthewmarston5149]

Explain in Russian.

[@matthewmarston5149]

Explain how in Russian, not Greek Geek

[@matthewmarston5149]

No. I didn't get to these classes. I dropped out. But went to the United States Navy.

[@PreMath]

Thanks for sharing ❤️

[@scottdort7197]

I solved by taking the 30-60-90 proportions of the 2 triangles. That quickly yield the diameter of the circle as 8*sqrt3. Divide that by 2 to get the radius. Then do was premath did to get the resultant areas.

[@PreMath]

Thanks for the feedback ❤️

[@cyruschang1904]

sin60° = (✓3)/2 = (✓3 + 2r)/(6 + x) = x/2r 2(✓3 + 2r) = (6 + x)✓3 2✓3 + 4r = 6✓3 + x✓3 x = (2✓3 - 6✓3 + 4r)/✓3 x = 4r/✓3 - 4 x = r✓3 4r/✓3 - 4 = r✓3 r/✓3 = 4 r = 4✓3, x = r✓3 = 12 green area = (4✓3)(4✓3)π/3 - (6)(2✓3) = 16π - 12✓3

[@michaeldoerr5810]

The area is 4[4pi-3*sqrt(3)]. I did a sanity check on my work twice. Also regarding yesterday's video, I got called out for stating the wrong answer. I put down a "I stand corrected" response. I am just letting other people know.

[@PreMath]

Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... First of all we observe that the triangle ABC is not only a right triangle, but it is also a 30°-60°-90° triangle. According to the theorem of Thales the triangle CDE is another right triangle and since ∠ACB=∠DCE=30°, this triangle is also a 30°-60°-90° triangle. Therefore with R being the radius of the semicircle we can conclude: DE:CD:CE = AB:BC:AC = 1:√3:2 CE = 2*R DE = R CD = √3*R AC = AD + CD = 6 + √3*R AB = AC/2 = 3 + √3*R/2 BC = BE + CE = √3 + 2*R AB:BC = 1:√3 BC = √3*AB √3 + 2*R = √3*(3 + √3*R/2) √3 + 2*R = 3√3 + 3*R/2 2√3 + 4*R = 6√3 + 3*R ⇒ R = 4√3 The triangle OCD is an isosceles triangle (OC=OD=R). So we obtain: ∠ODC=∠OCD=∠ACB=30° ∠COD = 180° − ∠ODC − ∠OCD = 180° − 30° − 30° = 120° Now we are able to calculate the area of the green region: A(green) = A(circle sector OCD) − A(triangle OCD) = π*R²*(∠COD/360°) − (1/2)*OC*OD*sin(∠COD) = π*R²*(∠COD/360°) − (1/2)*R²*sin(∠COD) = π*(4√3)²*(120°/360°) − (1/2)*(4√3)²*sin(120°) = π*48*(1/3) − (1/2)*48*(√3/2) = 16π − 12√3 ≈ 29.48 Best regards from Germany

[@PreMath]

Excellent!😀 Thanks for sharing ❤️

[@MrPaulc222]

A couple of needlessly complicated dead ends before I finally got there.

[@LuisdeBritoCamacho]

STEP-BY STEP RESOLUTION PROPOSAL : 01) Departing from a Unit Circle, anyone can see that the Distance from the Center of the Circle to the Side of any Equilateral Triangle inscribed in a Unit Circle is half of the Radius. The Inner Angle in this case is equal to 120º. 02) Now, draw 2 Orthogonal Lines from Point D, one parallel to AB e the other Parallel to BC. Find the Point F and Point G. 03) The Right Triangle [ADG] have Sides: a) AG = 3 b) DG = 3sqrt(3) c) AD = 6 This can be verified by Trigonometry : Sine and Cosine of 60º. 04) GD = BF = 3sqrt(3) 05) OF = (EF - BE) = (3sqrt(3) - sqrt(3)) ; OF = 2sqrt(3) 06) Radius (R) = 2 * 2sqrt(3) ; R = 4sqrt(3) 07) Puting this data in a Segment Area Calculator with the following data : a) R = 4sqrt(3) b) Inner Angle = 120º c) Height of Segment = 2sqrt(3) d) Chord Length = 12 e) It returns me the Area = 29,5 sq un Therefore, As far as my best knowledge can lead me, the Area of The Green Shaded Region is equal to 29,5 Square Units.

[@PreMath]

Excellent!😀 Thanks for sharing ❤️

[@misterenter-iz7rz]

(2r+sqrt(3))/(6+2r×sqrt(3)/2)=sqrt(3)/2......

[@PreMath]

Thanks for sharing ❤️

[@matthewmarston5149]

Can you please explain in Russian 😊

[@PreMath]

Good question. We are looking into translating into multiple languages... Thanks for the feedback ❤️