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... .
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😊
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
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
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.
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
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
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 √ )
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.
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.
Ok, well, BAO is clearly a 45 degree angle, which tells us the length of the double-stroked segments is 8. Now imagine the remainder of the circle (let A', B', and C' be the mirror image points of A, B, and C). Then CD and DC' are both 16, AD is 8, and DA' is 2*DO+8. So we can use the intersecting chords theorem: 16*16 = 8*(2*DO+8) 256 = 16*DO + 64 DO = 192/16 = 12 So, our radius is 12+8 = 20. Therefore, the quarter circle area is pi*20^2/4 and the unshaded triangle area is 20^2/2, so the yellow area is Yellow_Area = pi*100 - 200 Yellow area = 100*(pi-2) Q.E.D.
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
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.
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.
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.
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).
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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...
.
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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😊
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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
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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
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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.
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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
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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
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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)
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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
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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 √ )
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R^2=(2*8)^2+(R-8)^2...R=20...Ay=400π/4-20*20/2=100π-200
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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.
Gostei muito dessa questão e de sua explicação! Obrigada
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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
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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.
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merci , excellent exercice
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Ok, well, BAO is clearly a 45 degree angle, which tells us the length of the double-stroked segments is 8. Now imagine the remainder of the circle (let A', B', and C' be the mirror image points of A, B, and C). Then CD and DC' are both 16, AD is 8, and DA' is 2*DO+8. So we can use the intersecting chords theorem:
16*16 = 8*(2*DO+8)
256 = 16*DO + 64
DO = 192/16 = 12
So, our radius is 12+8 = 20. Therefore, the quarter circle area is pi*20^2/4 and the unshaded triangle area is 20^2/2, so the yellow area is
Yellow_Area = pi*100 - 200
Yellow area = 100*(pi-2)
Q.E.D.
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
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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.
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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)
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Good method of solution
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Mashallah very nice sharing sir❤❤
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(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).😊
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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.
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Достроим окружность и продлим отрезки CD и AO до хорды и диаметра. 16*16=8*(2* r-8). r=20.
A partir desse ponto o resto é Pura Rotina Algébrica.
@@LuisdeBritoCamacho é tudo matemática. obrigado pelo comentário
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I got this one!
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.
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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).
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114.29 square units
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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.
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