Can you find area of the Yellow shaded region? | (Intersecting Chords) |
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2 ай бұрынLearn how to find the area of the Yellow shaded region in the Circle. Important Geometry and Algebra skills are also explained: Intersecting Chords theorem; perpendicular bisector theorem; Pythagorean theorem. Step-by-step tutorial by PreMath.com.
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Can you find area of the Yellow shaded region? | (Intersecting Chords) | #math #maths | #geometry
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@johnbrennan3372 2 ай бұрын
Excellent Very interesting.
@PreMath 2 ай бұрын
Glad you think so! Thanks ❤️
@reinymichel 2 ай бұрын
What a great question !!!!
@anoopshivhare7223 2 ай бұрын
🎉🎉
@PreMath 2 ай бұрын
Thanks dear❤️
@dirklutz2818 2 ай бұрын
Fantastic!
@RAG981 2 ай бұрын
Very nice solution for the area at the end.
@PreMath 2 ай бұрын
Excellent! Glad to hear that! Thanks ❤️
@yapadek3098 2 ай бұрын
Wow !!
@sergioaiex3966 17 күн бұрын
Nice
@leehowell7734 2 ай бұрын
Nice logic and solution method
@PreMath 2 ай бұрын
Glad you think so! Thanks ❤️
@waheisel 2 ай бұрын
Nice elegant solution! I brute forced it (again) by showing that the area of arcs OCB plus OAD is half a circle (125*pi). Then adding 2 yellow triangles to that, then subtracting 2 white triangles from the arcs gives the answer. As usual PreMath's solution is much better! Thanks again for the exciting daily puzzle.
@nihatarter8750 2 ай бұрын
Thank you sir.
@guptapr 2 ай бұрын
Awesome
@jamestalbott4499 2 ай бұрын
Thank you!
@PreMath 2 ай бұрын
You are very welcome! Thanks ❤️
@michaelstahl1515 2 ай бұрын
Very clever. You divide the circle area into 9 parts and find a solution on this way . Great ! I didn`t see that at first .
@PreMath 2 ай бұрын
Glad it was helpful! Thanks ❤️
@timc5768 2 ай бұрын
Eventually, I came up with a method similar to yours: After 'r^2 = 5^2 + 15^2 = 250', ( or '9^2 + 13^2') Let (P1) and (P2) be the areas of the 2 segments (CAA'D) and (ABC'C). Then req'd area , say 'Y = (P1) + (P2) - 2(A2)' , where (A2) is your representation and = (A4), (Eq.1) Then construct 2 segments (respectively congruent to those given) as you have done , (ADD'B') and (DB'BC). Then the straight lines form a rectangle of area (P3), say , with sides = 18 and 10, so 2[(P1) + (P2)] + (P3) = Area(circle) + 4(A2), (Eq.2) so from Eq.1: Y = (P1) + (P2) - 2(A2) = (1/2) [area(circle) - (P3)] = (1/2)[ 250(pi) - 180]
@MMmaths8800 2 ай бұрын
Nice sir
@PreMath 2 ай бұрын
Glad to hear that! Thanks ❤️
@rooker56 2 ай бұрын
Quite tricky!! Excellent!
@PreMath 2 ай бұрын
Yes, it is! Thanks ❤️
@marcelowanderleycorreia8876 2 ай бұрын
Very interesting!!!!
@PreMath 2 ай бұрын
Glad you think so! Thanks ❤️
@Waldlaeufer70 2 ай бұрын
Very nice problem! :)
@unknownidentity2846 2 ай бұрын
And a very clever solution, not as complicated as mine. Best regards to Switzerland
@PreMath 2 ай бұрын
Excellent! Glad to hear that! Thanks ❤️
@Waldlaeufer70 2 ай бұрын
I thought about solving it for two triangles + their circle segments. But it is just A = (full circle - inner rectangle) / 2. Best regards and a nice weekend to Germany.
@DB-lg5sq 2 ай бұрын
شكرا لكم على المجهودات يمكن استعمال فيثاغورس AD^2=360 AC^2=100 sinADC /AC =1/2R R=5(racine10) S=1/2 R^2(2BDC+2ABD -sinBDC -sinABD) + 1/2. (EC.EB +EA.ED) =125(pi-24/25-24/25) +150 =125 pi -90
@wackojacko3962 2 ай бұрын
Sick! 🙂
@PreMath 2 ай бұрын
Thanks ❤️
@santiagoarosam430 2 ай бұрын
F es la proyección ortogonal de O sobre AB. El diámetro horizontal de la circunferencia es GH. Potencia de E respecto a la circunferencia =8*18=6*EB→ EB=24 → FA=(6+24)/2=15 y FO=(CD/2)-8=5 → r²=OA²=FO²+FA² =5²+15²=250 =r². La figura amarilla CEB la transformamos en su simétrica respecto al eje GH. En la figura amarilla AED, aplicamos un giro de 180º en torno al centro "O" a la parte que queda por encima de GH, de forma que pasa a la derecha de la circunferencia. La figura amarilla resultante es un semicírculo con un entrante rectangular en su parte alta central, de dimensiones [2(FA-6)]x(FO)=18x5. Área amarilla =(πr²/2)-(18*5) =125π-90. Interesante puzle. Gracias y un saludo cordial.
@PreMath 2 ай бұрын
Excellent! Thanks for sharing ❤️
@ZUBAIR457 2 ай бұрын
Millions of likes from Pakistan ❤
@PreMath 2 ай бұрын
Thanks dear❤️ Stay blessed 🌹
@TurquoizeGoldscraper 2 ай бұрын
When I saw the circle split into 9 pieces, I was wondering why A2 and A4 map in that direction, instead of the other direction. Then I realized it maps in both directions so you could have 4 A2, but the way the problem is constructed, it's based on pairs, so it was easier to have two pairs instead of 4.
@PreMath 2 ай бұрын
Thanks for sharing ❤️
@stephenhall5694 2 ай бұрын
After 10:05, the lower left yellow area switched from A4 to A1. Looked like a drawing glitch. Could cause confusion.
@unknownidentity2846 2 ай бұрын
Let's find the area: . .. ... .... ..... According to the intersecting chords theorem we obtain: AE*BE = CE*DE 6*BE = 8*18 ⇒ BE = 8*18/6 = 24 Let's assume that O is the center of the coordinate system. Since AB and CD are perpendicular to each other, we can additionally assume that AB and CD are parallel to the x-axis and y-axis, respectively. Then we get the following coordinates: −xA = xB = AB/2 = (AE + BE)/2 = (6 + 24)/2 = 15 −yD = yC = CD/2 = (CE + DE)/2 = (8 + 18)/2 = 13 xE = xC = xD = xB − BE = 15 − 24 = −9 yE = yA = yB = yC − CE = 13 − 8 = +5 O: ( 0 ; 0 ) A: ( −15 ; +5 ) B: ( +15 ; +5 ) C: ( −9 ; +13 ) D: ( −9 ; −13 ) E: ( −9 ; +5 ) So the radius R of the circle turns out to be: R² = xA² + yA² = (−15)² + (+5)² = 225 + 25 = 250 R² = xC² + yC² = (−9)² + (+13)² = 81 + 169 = 250 ✓ The areas A(AB) above the chord AB (yellow + white) and A(CD) on the left side of the chord CD (yellow + white) can be calculated in the following way: A(AB) = A(circle sector OAB) − A(triangle OAB) = πR²*[∠AOB/(2π)] − (1/2)*AB*h(AB) = R²*(∠AOB/2) − (1/2)*AB*h(AB) = R²*arctan(xB/yB) − xB*yB A(CD) = A(circle sector OCD) − A(triangle OCD) = πR²*[∠COD/(2π)] − (1/2)*CD*h(CD) = R²*(∠COD/2) − (1/2)*CD*h(CD) = R²*arctan(yC/|xC|) − yC*|xC| The upper left white area is the overlap of the two areas A(AB) and A(CD). Since ACE is a right triangle, we can apply the Pythagorean theorem to calculate the length AC. Then we can use this quantity to calculate the angle ∠AOC: AC² = AE² + CE² = 6² + 8² = 36 + 64 = 100 AC² = OA² + OC² − 2*OA*OC*cos(∠AOC) AC² = R² + R² − 2*R²*cos(∠AOC) 100 = 250 + 250 − 2*250*cos(∠AOC) −400 = −500*cos(∠AOC) ⇒ ∠AOC = arccos(4/5) Additionally we need the areas of the triangles OAE and OCE: A(triangle OAE) + A(triangle OCE) = (1/2)*AE*h(AE) + (1/2)*CE*h(CE) = (1/2)*AE*yA + (1/2)*CE*|xC| Now we can calculate the size A(overlap) of the upper left white area: A(overlap) = A(circle sector OAC) − A(triangle OAE) − A(triangle OCE) = πR²*[∠AOC/(2π)] − (1/2)*AE*yA − (1/2)*CE*|xC| = R²*arccos(4/5)/2 − (1/2)*AE*yA − (1/2)*CE*|xC| Finally we are able to obtain the size of the yellow area: A(yellow) = A(AB) + A(CD) − 2*A(overlap) = R²*arctan(xB/yB) − xB*yB + R²*arctan(yC/|xC|) − yC*|xC| − R²*arccos(4/5) + AE*yA + CE*|xC| = 250*arctan(15/5) − 15*5 + 250*arctan(13/9) − 13*9 − 250*arccos(4/5) + 6*5 + 8*9 ≈ 302.70 Best regards from Germany
@unknownidentity2846 2 ай бұрын
To make it complete: tan(x±y) = [tan(x) ± tan(y)]/[1 ∓ tan(x)tan(y)] tan[arctan(15/5) + arctan(13/9)] = tan[arctan(3) + arctan(13/9)] = (3 + 13/9)/[1 − 3*(13/9)] = (27 + 13)/(9 − 3*13) = 40/(−30) = −4/3 tan[arccos(4/5)] = sin[arccos(4/5)]/cos[arccos(4/5)] = sin[arccos(4/5)]/(4/5) = √{1 − cos²[arccos(4/5)]}/(4/5) = √[1 − (4/5)²]/(4/5) = √(1 − 16/25)/(4/5) = √(9/25)/(4/5) = (3/5)/(4/5) = 3/4 tan[arctan(−4/3) − arctan(3/4)] = (−4/3 − 3/4)/[1 + (−4/3)*(3/4)] = (−4/3 − 3/4)/(1 − 1) = (−4/3 − 3/4)/0 The values −4/3 and 3/4 can be regarded as slopes of two intersecting lines. Since the product of their slopes is −1, the lines are perpendicular to each other. So it can be concluded: arctan(−4/3) − arctan(3/4) = π/2 A(yellow) = 250*arctan(15/5) − 15*5 + 250*arctan(13/9) − 13*9 − 250*arccos(4/5) + 6*5 + 8*9 = 250*[arctan(15/5) + arctan(13/9) − arccos(4/5)] − 90 = 250*(π/2) − 90 = 125*π − 90 🙂
@PreMath 2 ай бұрын
Excellent! Thanks for sharing ❤️
@prossvay8744 2 ай бұрын
AE.BE=CE.DE 6.BE=8.18 So AE=144/6=24 units 4R^2=6^2+8^2+18°2+24^2 R=5√10units. Area of the circle=π(5√10)^2=250π In ∆ BED BD^2=BE^2+DE^2=24^2+18^2=900 BD=30 area of ∆ BED=1/2(24)(18)=216 Cos(x)={(5√10)°2+(5√10)^2-30^2}/2(5√10) x=143.13° Area of the segment=π(5√10)^2(143.13/360=312.26 So yellow of the shaded region=250π-312.26=473.14 square units.❤❤❤ Thanks sir.
@PreMath 2 ай бұрын
You are very welcome! Thanks ❤️
@susennath6035 2 ай бұрын
After long time. Can you remember me ?
@MichaelMantion Ай бұрын
at 10 minutes the bottom left a4 becomes a1. lol
@giuseppemalaguti435 2 ай бұрын
326,6976
@PreMath 2 ай бұрын
Thanks ❤️
@LuisdeBritoCamacho 2 ай бұрын
PART I 1) CE * DE = AE * BE 2) 8 * 18 = 6 * BE 3) 144 = 6 * BE 4) BE = 144 / 6 5) BE = 24 PART II 1) tan (ABD) = 18/24 ; tan (EBD) = 3/4 2) tan (CDB) = 24/18 ; tan (EDB) = 4/3 3) Angle (ABD) = arctan (3/4) ~ 37º 4) Angle (CDB) = arctan (4/3) ~ 53º 5) Center Angle (COB) ~ 106º ; In Radians ~ 1,85 Rad 6) Center Angle (COD) ~ 74º , In Radians ~ 1,29 Rad 7) Minor Segment Area of COB = 125 * (1,85 - 0,96) ~ 111,25 sq un 8) Minor Segment Area of AOD = 125 * (1,29 - 0,96) ~ 41,25 sq un PART III 1) Yellow Area CEB = 111,25 + 96 = 207,25 2) Yellow Area AED = 41,25 + 54 = 95,25 3) Total Yellow area = 302,5 sq un ANSWER The Yellow Region Area is equal to Approx. 302,5 Square Units. NOTE: Radius of the Circle is equal to sqrt(250) ~ 15,8 Linear Units; and Area of the Circle is equal to (250*Pi) Square Units ~ 785,4 square Units.
@PreMath 2 ай бұрын
Thanks for sharing ❤️
@misterenter-iz7rz 2 ай бұрын
Clearly EB=24 by intersecting chords' theorem, not difficult to compute the radius =sqrt(5^2+15^2)=sqrt(9^2+13^2)=sqrt(250)=5sqrt(10), but it is clumsy to calculate the area of sectors😢😢😢😢😢😢
@PreMath 2 ай бұрын
Thanks for sharing ❤️
@misterenter-iz7rz 2 ай бұрын
@@PreMath By brute force, to compute the area of two segment, 1/2r^2(t1+t2-sin t1+sin t2), note that t1+t2=pi, so it is 1/2 250(pi -2 24/25)=125pi-240, where sin t1=24/25.😅😅😅
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