Can you find area of the Brown Semicircle? | (Circle) |

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Learn how to find the area of the Brown Semicircle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Circle formula; circle theorem. Step-by-step tutorial by PreMath.com.
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3 окт 2024

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Комментарии : 43

@michaeldoerr5810 День назад

This is one great example of collinearity being useful as substitution!!! And the radius is 2!!!

@PreMath День назад

Glad you liked it! Thanks for the feedback ❤️

@jamestalbott4499 5 часов назад

Thank you!

@PreMath 58 минут назад

You are very welcome! Thanks for the feedback ❤️

@sorourhashemi3249 День назад

Easy.thanks

@PreMath 57 минут назад

Excellent! You are very welcome! Thanks for the feedback ❤️

@marioalb9726 День назад

A₁= π cm² = π.r₁² - > r₁= 1cm R = 2r Pytagorean theorem, twice: a² = (r+r₁)²-(r-r₁)² =(R-r₁)²-r₁² (r+1)²-(r-1)² =(2r-1)²-1² (r²+2r+1)-(r²-2r+1)=(4r²-4r+1)-1 4r = 4r² - 4r r = 2 cm Area of semicircle: A = ½πr² = 2π cm² (Solved √ )

@PreMath 52 минуты назад

Excellent! Thanks for sharing ❤️

@santiagoarosam430 День назад

Área del círculo pequeño =π cm² → Radio =1 cm. Radio del cuarto de círculo =2r = Diámetro del semicírculo. r+1)²-(r-1)²=OA²=(2r-1)²-1²→ r=2→ Área semicírculo =2π. Gracias y un saludo.

@PreMath 52 минуты назад

Excellent! Thanks for sharing ❤️

@imetroangola4943 День назад

*Outro método:* Na construção de OB inicialmente na figura, podemos usar potência no ponto O: AO²=(2R - 2)×2R = 4R² - 4R Por Pitágoras no ∆PDQ: PQ²=DQ²+PD², temos: PQ=1+R, DQ=OA (DQAO é um retângulo) e PD=R - 1. Assim, (R+1)²= 4R² - 4R + (R - 1)² R² + 2R +1=4R² - 4R + R² - 2R +1 2R = 4R² - 4R - 2R →4R² -8R=0 como R>0 então R=2. Área é π×2²/2 = *2πcm²*

@PreMath 55 минут назад

Excellent! Thanks for sharing ❤️

@quigonkenny День назад

Green corcle G: A = πR² π = πR² R² = 1 R = 1 Let the radius of semicircle P be r. By observation, the radius of quarter circle O is the diameter of semicircle P, or 2r. Let the point of tangency between semicircle P amd circle G be T. Draw OB and GA. As a radius of circle G, GA = 1, and is perpendicular to OA as A is the point of tangency between circle G and segment OA. As the centers of two tangent circles and their point of tangency are always collinear, G is collinear with OB, so OG = 2r-1. Triangle ∆OAG: OA² + GA² = OG² OA² + 1² = (2r-1)² OA² + 1 = 4r² - 4r + 1 OA² = 4r² - 4r = 4(r²-r) OA = √(4(r²-r)) = 2√(r²-r) Draw GM, where m is the point on OC where GM is parallel to to OA and perpendicular to OC. As GM and OA are parallel and OM and GA are both perpendicular to OA and GM, then all internal angles of OAGM are equalt to 90°, and OAGM is a rectangle. Therefore, OM = GA = 1 and GM = OA. As OP = r, PM = r-1. As the centers of two circles and their point of tangency are always collinear, T is collinear with PG, so PG = PT+GT = r+1. Triangle ∆PMG: PM² + GM² = PG² (r-1)² + (4r²-4r) = (r+1)² r² - 2r + 1 + 4r² - 4r = r² + 2r + 1 4r² - 6r = 2r 4r² - 8r = 0 4r(r-2) = 0 r = 0 ❌ | r = 2 Semicircle P: A = πr²/2 = π(2²)/2 = 2π ≈ 6.28 cm²

@PreMath 56 минут назад

Excellent! Thanks for sharing ❤️

@sventheviking4635 День назад

A line from C to bottom of quadrant E makes angle 45˚OCE. A radius from O bisects this line at K with angle 90˚ at J at half of OE from O = R/2, but OKJ is 45˚, so JK = OJ = R/2 =1, so R =2

@PreMath 57 минут назад

Thanks for the feedback ❤️

@SkinnerRobot День назад

πr² = π r = 1 (2R - 1)² - 1² = (R + 1)² - (R - 1)² 4R² - 4R + 1 - 1 = R² + 2R + 1 - R² + 2R -1 4R² - 8R = 0 4R(R - 2) =0 R = 2 Area = 1/2π2² = 2π cm²

@phungpham1725 21 час назад

@SkinnerRobot Simple and elegant! Thank you so much!

@PreMath 54 минуты назад

Excellent! Thanks for sharing ❤️

@giuseppemalaguti435 День назад

(r+1)^2-((r-1)^2=(R-1)^2-1^2...R=2r..4r=4r^2-4r...8=4r.r=2

@PreMath 52 минуты назад

Excellent! Thanks for sharing ❤️

@prossvay8744 День назад

Area of the small circle=π cm^2 πr^2=π So r=1cm Let Radius of semicircle=R Connect O to B and A to Q So OQ=2R-1 OA=x and AQ=1cm I ∆ OAQ OQ^2=AQ^2+OA^2 (2R--1)^2=1^2+x^2 So x^2=(2R-1)^2-1=4R^2-4R (1) Connect P to Q and Q to M (M on OP and right OP) In∆MPQ MP=R-1; MQ=OA=x ; PQ=R+1 (R-1)^2+x^2=(R+1)^2 R^2-2R+1+x^2=R^2+2R+1 So x^2=4R (2) (1)&(2) 4R^2-4R=4R R(R-2)=0 So R=2 cm So Brown semicircle area=1/2(π)(2^2)=2π cm^2=6.28 cm^2.❤❤❤

@PreMath 53 минуты назад

Excellent! Thanks for sharing ❤️

@andreasproteus1465 День назад

Since it is obvious that PQ = OQ => R +1 = 2R - 1, hence R = 2.

@ganeshkambli7050 День назад

How is obvious that PQ = OQ? can you please explain?

@PreMath 55 минут назад

Thanks for the feedback ❤️

@MukeshYadav-gl6sy День назад

Most of his videos are based on circle😂😂❤❤❤

@PreMath 54 минуты назад

Thanks for the feedback ❤️

@himadrikhanra7463 День назад

( r - 1)^2 + 2r = h^2 r^2 + 2r = c^2 Pi R^2 2 root 2 R Pi r^2 /2 Root 2 4pi r^2: Pi r^2 : Pi 8 Sq. Unit ?

@misterenter-iz7rz День назад

(r+1)^2-(r-1)^2=(2r-1)^2-1, 4r=4r^2-4r, as r non zero, r=2 and the answer 2pi.😊

@PreMath 50 минут назад

Excellent! Thanks for sharing ❤️

@JinnirasFlair День назад

1st comment 🙋🏻‍♀️

@PreMath 53 минуты назад

Excellent! Glad to hear that! Thanks for the feedback ❤️

@LuisdeBritoCamacho 23 минуты назад

I am going to give my Rational about this Problem. 01) OC = R cm 02) OP = (R / 2) cm = r cm 03) R = 2r 04) Big Quarter of Circle Area = (Pi * R^2) / 4 05) Semicircle Area = (Pi * r^2) / 2 06) Big Quarter of Circle Area give in terms of r = (Pi * (2r)^2) / 4 = (Pi * 4r^2) / 4 = (Pi * r^2) sq cm 07) As we can see the Semicircle Area is Half of Big Quarter Area. 08) So, the White Area (without the Green Circle) is equal to Brown Semicircle Area; as stated before. 09) So we can see a Radius Pattern here : 1 ; 2 ; 4 ; 8 ; 16 ; ... ; 2^n 10) So I must conclude that the Brown Semicircle is 2Pi sq cm. 11) That's all. ANSWER : Brown Semicircle Area equal 2Pi Square Centimeters.

@Birol731 21 час назад

My way of solution ▶ The area of the green circle, Agreen Agreen= π*r₁² Agreen= π cm² ⇒ π= π*r₁² r₁²= 1 r₁= 1 cm K ∈ [PO] and [KQ] ⊥ [KO] I) [OB]= R [QB]= r₁ [OQ]= R-r₁ Let's consider the triangle ΔOQK: [KO]= r₁ [QK]= d [OQ]= R-r₁ r₁= 1 cm by applying the Pythagorean theorem we get: [KO]²+[QK]²= [OQ]² r₁²+d²= (R-r₁)² 1²+d²= (R-1)² 1+d²= R²-2R+1 d²= R²-2R R= 2r₂ ⇒ d²= 4r₂²-4r₂ II) Let's consider the triangle ΔKQP: [PK]= r₂-r₁ [KQ]= d [QP]= r₂+r₁ r₁= 1 cm by applying the Pythagorean theorem we get: [PK]²+[KQ]²= [QP]² (r₂-r₁)²+d²= (r₂+r₁)² (r₂-1)²+d²= (r₂+1)² d²= r₂²+2r₂+1-(r₂-1)² d²= r₂²+2r₂+1-(r₂²-2r₂+1) d²= r₂²+2r₂+1-r₂²+2r₂-1 d²= 4r₂ III) d²= 4r₂²-4r₂ and d²= 4r₂ ⇒ 4r₂²-4r₂ = 4r₂ 4r₂²-8r₂ = 0 4r₂(r₂-2)=0 4r₂= 0 r₂= 0 cm ❌ r₂ > 0 r₂-2=0 r₂= 2 cm Abrown= πr₂²/2 Abrown= π*2²/2 Abrown= 2π cm² is the answer !

@PreMath 57 минут назад

Excellent! Thanks for sharing ❤️

@nenetstree914 День назад

2PI

@PreMath 53 минуты назад

Excellent! Thanks for sharing ❤️

@turtle752436 День назад

If you don't like things to scale, why the cm^2 and not m^2 Never mind. You are good enough for me. Do they understand the scale thing, I wonder. Why it may not be to scale. Be well, Thank you for these.

@turtle752436 День назад

Still just for clarity. Before we get to arcs and such. 1^2=1 but 0.01^2=0.0001. Or also { [10^(-2)]^2}= 10^(-4) It is perhaps when someone would say, maths and perhaps geometry, don't have units, baby

@PreMath 50 минут назад

Thanks for the feedback ❤️