Love it

@PreMath

14 күн бұрын

Glad to hear that! 🌹 Thanks for the feedback ❤️

Nice solution❤

@PreMath

13 күн бұрын

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R^2=(2*8)^2+(R-8)^2...R=20...Ay=400π/4-20*20/2=100π-200

@PreMath

13 күн бұрын

Excellent!👍🌹 Thanks for sharing ❤️

Gostei muito dessa questão e de sua explicação! Obrigada

@PreMath

13 күн бұрын

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Intersecting chords theorem: (2R-8).8 = 16² R = 20 cm Area of circular segment: A = ½R²(α - sin α) A = ½ 20² (π/2 - sin π/2 ) A = 114,16 cm² ( Solved √ )

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

merci , excellent exercice

@PreMath

13 күн бұрын

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Complete the circle such that Radius AO meets circle at point X.So AX is a diameter with AO = 8 and OX = 2r -8. Similarly extend CD to meet Circle in point Y. So CD= 16 and DY = 16. Use intersecting chords to find r. 16*16 = 8*(2r-8) Divide by 8 32=2r-8 2r=40 r=20

@PreMath

13 күн бұрын

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Thanks Sir That’s very nice Good method of solution Good luck with glades ❤❤❤❤

@PreMath

13 күн бұрын

Many many thanks, dear🌹🙏

For fun, a more convoluted use of intersecting chords: 1. AE = hypo of △ ADE = 8√2. 2. Consider whole circle and reflect quarter circle across AO with CD continuing down to pt F. CF and and AB are intersecting chords 3. AE * EB = CE * EF 8√2 * EB = 8 * 24 = 192 EB = 192 / 8√2 = 24/√2 = 12√2 4. AB = AE + EB = 8√2 + 12√2 = 20√2 5. AB = hypo of △ ABO so AO = 20√2 / √2 = *20 = r* Proceed as in the video... .

@PreMath

13 күн бұрын

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AD = 8 => DE = 8 => CD = 16 r² = 16² + (r - 8)² r² = 256 + r² - 16r + 64 = r² - 16r + 320 16r = 320 r = 20 A(yellow) = A(quarter circle) - A(triangle AOB) A(yellow) = 1/4 * r² π - 1/2 * r² = 1/4 * 20² π - 1/2 * 20² A(yellow) = 1/4 * 20² (π - 2) = 100 (π - 2) ≈ 114.16 square units

@PreMath

13 күн бұрын

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Mashallah very nice sharing sir❤❤

@PreMath

13 күн бұрын

Thanks for liking

The yellow area is a circular segment, and the area is the difference between the sector subtended by arc AB and the triangle formed by the quarter circle center O and chord AB. We need to determine the radius r. Aₛ = (90°/360°)πr² - r(r)/2 Aₛ = πr²/4 - r²/2 Aₛ = (r²/2)((π/2)-1) As OA = OB = r, ∆AOB is an isosceles tight triangle and ∠BAO = ∠OBA = (180°-90°)/2 = 45°. As ∠ADE = ∠AOB = 90° and ∠EAO is common, ∆ADE amd ∆AOB are similar triangles. As OA = OB = r, AD = DE = 8. As DE = 8, EC = 8. Draw OC. Triangle ∆CDO: OD² + DC² = OC² (r-8)² + 16² = r² r² - 16r + 64 + 256 = r² 16r = 320 r = 320/16 = 20 Aₛ = ((20)²/2)((π/2)-1) Aₛ = (400/2)((π/2)-1) Aₛ = 200((π/2)-1) Aₛ = 100π - 200 ≈ 114.16 sq units

@PreMath

13 күн бұрын

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I got this one!

Triangle ABO is an isosceles triangle with angles OAB and OBA both at 45 degrees. Triangles AED and ABO are similar so length DE is the same as length AD = 8. So length CD=16. In right-angle triangle CDO, side OD=r-8, side DE=16 and hypotenuse OC=r. Therefore, 16^2 + (r-8)^2=r^2 which simplifies to 320/16=r so r=20. Area of full circle = pi r^2 Area of quarter circle = (pi r^2) / 4 = 100 pi Area of triangle ABO = 1/2 Base x Height = 1/2 . 20 . 20 = 200 Area of yellow segment = (100 pi) - 200 = 114.159 units^2

@PreMath

13 күн бұрын

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ABO is an isosceles 🔺 angle OAB = angle OBA ( two sides are radius of the same circle) 🔺 DAE angle DAE = angle DEA (angle DEA = corresponding angle OBA as OB II DE) Hence DE=8 DC =16 Now complete the semicircle .Name the other end point of diameter as M Hence 8*DM=16*16 ( 16 is the geometric mean of 8 & DM) DM =32 Hence diameter = 32 +8=40 Radius =20 Area of shaded portion =400π/4 - 1/2*20*20 = 100π - 200 =114.159 sq units (approx.) Sir I only derived radius with the help geometric mean theorem.

@PreMath

13 күн бұрын

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Wow! Not often I'm first here. Let's see if I can solve it though :) As ADE is a right isosceles, AD = DE = CE = 8. Several ways to find radius. I choose to imagine a full circle and going for intersecting chords. 16*16 = 8(2r-8) 256 = 16r-64 320=16r r = 20 Area of full circle is 400pi, so the quadrant is 100pi Area of AOB is (20*20)/2=200 Yellow area is 100pi-200 = 314.16-200=114.16 un^2 EDIT: I see you chose the calculate the radius another way.

@PreMath

14 күн бұрын

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AD=8=DE=EC→ Potencia de D respecto a la circunferencia de radio OA=r: 8(2r-8)=(8+8)*(8+8)→ r=20→ Área amarilla =(20²π/4)-(20*20/2) =100(π-2)=114,15926....ud². Gracias y un saludo cordial.

@PreMath

14 күн бұрын

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Think out of the box: expand it to semicircle, F point is at the opposite side on the semicircle. ACD and ACF triangles are similar (ACD angle common, DAC and CAF are both 90 degrees). AC=√(AD^2*5)=√(64*5)=√320 => AF=√(AC^2*5)=√(320*5)=√1600=40 => r=AF/2=20

@PreMath

13 күн бұрын

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Thank you!

@PreMath

13 күн бұрын

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due to the linear increase from 0 to r, it must be that l2=l1 see line 20, and then the endpoint must be on the circle: 10 print "premath-can you find area of the yellow shaded region" 20 l1=8:l2=l1:r=(l1^2+4*l2^2)/2/l1 30 print r:ages=r*r*(pi/4-1/2):print "die gesuchte flaeche=";ages 40 mass=1100/r:gcol 11:move r*mass,0:move 0,0:plot165,r,r 50 gcol8:line l1*mass,0,0,0:line l1*mass,0,l1*mass,2*l1*mass:gcol 11:line 0,0,r*mass,r*mass 60 premath-can you find area of the yellow shaded region 20 die gesuchte flaeche=114.159265 > run in bbc basic sdl and hit ctrl tab to copy from the results window.

By using the chord theorem or the height altitude triangle theorem ( in a semi circle) We have: sq CD= 8.(2r-8) -> r= 20 Area of the yellow segment = 100 pi - 200 = 100. ( pi-2)= 114.16 sq units😊

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

Angle OAB=45 degrees, CD=16, OD=R-8, OC=R, 16^2+(R-8)^2=R^2, 256+R^2-16R+64=R^2, 16R=320, R=20. Area of a sector of a circle AOB=pi*R^2/4, area of a friangle ABO=R^2/2, areaof the Yellow shaded region = piR^2/4-R^2/2=114,16.

@PreMath

13 күн бұрын

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Nice, many thanks, Sir! AD = 8 = DE = CE → AQ = 2r = AD + (2r - 8) → 16(16) = 8(2r - 8) → r = 20 → yellow area = 100(π - 2)

@PreMath

13 күн бұрын

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CE=ED=8 8+8=16 16＊16=8＊(2r-8) 256=16r-64 16r=320 r=20 AO=BO=r=20 Yellow Area = 20＊20＊π＊1/4 - 20＊20＊1/2 = 100π - 200 = 314 - 200 = 114

@PreMath

14 күн бұрын

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without looking scale 8=1 then rescale solution DC² =4 OC²=R² OD²=(R-1)² = R²-2R +1 So, R²-2R+5 =R² 2R=5 Rescale 2R=5×8 R=20 segment = πR²/4 - R²/2 = (R²/4)(π-2) Answer100(π-2)

Достроим окружность и продлим отрезки CD и AO до хорды и диаметра. 16*16=8*(2* r-8). r=20.

@LuisdeBritoCamacho

14 күн бұрын

A partir desse ponto o resto é Pura Rotina Algébrica.

@user-sk9oi9jl2g

14 күн бұрын

@@LuisdeBritoCamacho é tudo matemática. obrigado pelo comentário

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

(r-8)^2+16^2=r^2, 16r=64+256=320, r=20, therefore the area is 1/4 20^2 pi-1/2 20^2=20^2(1/4 pi-1/2)=100(pi-2).😊

@PreMath

13 күн бұрын

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Let's find the area: . .. ... .... ..... The right triangles ADE and ABO are obviously similar. So with R being the radius of the quarter circle we can conclude: DE/AD = BO/AO = R/R = 1 ⇒ DE = AD = 8 ⇒ CD = CE + DE = 2*DE = 2*8 = 16 The triangle CDO is also a right triangle. So we can apply the Pythagorean theorem: CO² = CD² + DO² CO² = CD² + (AO − AD)² R² = 16² + (R − 8)² R² = 256 + R² − 16*R + 64 0 = 320 − 16*R ⇒ R = 320/16 = 20 Now we are able to calculate the area of the yellow region: A(yellow) = A(quarter circle) − A(right triangle ABO) = π*R²/4 − (1/2)*AO*BO = π*R²/4 − R²/2 = π*400/4 − 400/2 = 100π − 200 ≈ 114.16 Best regards from Germany

@PreMath

13 күн бұрын

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Let's use an orthonormal, center O, first axis (OA). We have A(R;0) B(0; R) D(R -8; 0) The equation of the circle is x^2 + y^2 = R^2 C is on the circle and its abscissa is R -8, its ordinate is y with (R -8)^2 + y^2 = R^2, so R^2 - 16.R +64 +y^2 = R^2, y^2 = 16.R -64, So we have C(R -8; 4.sqrt(R -4)) and E(R -8; 2.sqrt(R -4)). The equqtion of (AB) is x + y - R = 0 and point E is on (AB), so R - 8 +2.sqrt(R -4) - R = 0. That gives: 2.sqrt(R -4) = 8, and R -4 = (8 -2)^2, and then R = 20. The area of the quater circle is (Pi/4).(20^2) = 100.Pi. The area of the triangle OAB is (1/2).(20^2 )= 200. Finally the yellow area is 100.Pi - 200 or 100.(Pi -2).

@PreMath

13 күн бұрын

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STEP-BY-STEP RESOLUTION PROPOSAL : 01) Consider a Quarter of a Circle with Radius equal to 1. 02) Consider a Function f(X) that is a Straight Line with the folowing Equation : f(X) = - X + 1. This is the Line AB. 03) Now, consider another Function that is the Perimeter of a Quarter of a Circle defined by the Equation : g(X) = sqrt(1 - X^2) 04) As the Line CD is divided into two equal parts, one can say that : g(X) = 2 * f(X) 05) So : sqrt((1 - X^2) = 2 * (- X + 1) 06) Solving this Equality between Functions we get the following Solutions : X = 1 and X = 3/5 (or X = 0,6) 07) As AD = 8 and is equal to 2/5 of the Radius : 8 = 2R / 5 ; 2R = 40 ; R = 20. 08) So : AD = 8 and OD = AO - AD : OD = 20 - 8 = 12 09) Quarter of a Semicircle Area (QCA) with Radius equal to 20. 10) QCA = 400 * Pi / 4 ; QCA = 100Pi sq un 11) Area of Triangle AOB = (20 * 20) / 2 ; Triangle AOB = 400 / 2 ; Traingle AOB = 200 sq un 12) Yellow Area = Quarter of a Semicircle Area - Area of Triangle AOB 13) Yellow Area = (100*Pi - 200) sq un ; Yellow Area ~ 114,16 sq un ANSWER : The Yellow Area is approx. equal to 114,16 Square Units. Best Regards from The Universal Islamic Institute for the Study of Ancient Mathematical Thinking, Knowledge and Wisdom. Cordoba Caliphate.

@PreMath

13 күн бұрын

Excellent! 👍 Thanks for sharing ❤️🙏

114.29 square units

@PreMath

13 күн бұрын

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You don't need a calculator to get to 5 digits of the answer if you have that many digits of pi memorized.

@PreMath

13 күн бұрын

Thanks for the feedback ❤️