Angle BPQ encloses a quarter of the circle when we complete the circle and hence its measurement is 45° and hence triangle BQB is right and isosceles so QB=5 we extend AP until we get square PEBQ and from it AB²=AE²+EB²=146 so the radius of the circle R=√146/√2=√73 we have [AOBE]=[AOB]+[ABE]=(√73*√73)/2+(11*5)/2=64 and from it [OAPQB]=[AOBE]-[PEBQ]=64-25=39 and from it the area of the red region is equal to (π*(√73)²)/4-39=(73π)/4-39
Angle APB is 135 degree, so angle QPB=135-90=45. QPB is right isosceles then QB=5 Now we can find Rsqrt2 with cosine law on APB knowing that Cos (135)=cos(90+45)=-cos(90-45)=-cos45=-sqrt2/2 And that PB=5sqrt2. Once found the radius (sqrt73) we have to subtract the areas of a trapezoid and a right triangle to the quarter circle. Tracing the radius that splits the chord AP in two equal parts and extending BQ untill intersects AO we get two similar right triangles and with proportions we can find all the sides we need
I drew the diagonal AB and labeled angle ABP as θ. I calculated cos(45 + θ) and found that cos(θ) - sin(θ) = 3sqrt(2)/R. I then translated side PQ so that point Q coincided with point B. This results in sin(θ) = 5/(Rsqrt(2)). Using cos(θ) = sin(θ) + 3sqrt(2) gives cos(θ) = 11/(Rsqrt(2)). So we have a right triangle with sides Rsqrt(2), 5 and 11 from which R = sqrt(73). Also BQ is 5. Now angle AOP is (90 - 2θ) and angle POB is 2θ. So the required area can be calculated by: A = 73π/4 - (1/2)R^2sin(90 - 2θ)-(1/2)R^2sin(2θ) + 25/2 = 73π/4 - 39.
AB=√[5²+(6+5)²]=√146→ Radio del círculo =r=√146/√2→ r²=146/2=73→ Área círculo =73π → Si copiamos la figura propuesta y la giramos 90º tres veces sucesivas en el mismo sentido obtenemos un círculo completo con una cruz griega inscrita; ésta se compone de un cuadrado central de 6*6 y cuatro brazos rectangulares de 5*6→ Área de la cruz inscrita = 6²+4*5*6 =156 → Área sombreada =(73π-156)/4 =(73π/4)-39. Gracias y un saludo cordial.
@@AzouzNacir There were the four pieces of the jigsaw. one fitted exactly in each quadrant. Twelve sides, Angles: 270 degrees times four and 90 degrees times eight. Total 1800 degrees. Any length but 5 would not have fitted in. ----- I am sorry that this is not as rigorous as Math Booster's solutions. -------- You could also check all this with a drawing of the cross with those measurements and circumscribing a circle. The shaded area would change in shape if the quadrant were drawn at any other angle, but the area between 1/4 of a circle and 1/4 of a Greek Cross inscribed would measure up to the same answer. Please thank Math Booster for regular rigorous solutions.❤
I know this but you should have given the justification before calculating the length of AB and it is clear that angle BPQ encloses an arc of the circle measuring 90° so its measure is 45°@@kateknowles8055
@@AzouzNacir The arc PB shown has centre at O and not at Q . I could see that QB would measure 5, but I was not able to give a reason . You might use t = tan theta = 3/8 then OBQ is arctan (3/8) so QB measures 5, using Math Booster's video at 9:00 minutes, instead.
Best Wishes to you, and all fellow viewers who have had a go at this before watching the video. Extending PQ to cross OB at R is making AQ the diameter of another circle, so that angle OAP will equal QRB. Now extending OR to A' we have a semicircle with AA' as diameter.Rotating the pentagonal area OAPQB around by 90 , 180, 270 degrees to make three more pentagons will form a Greek Cross. Dividing it into a central square of 6.6 and four reclangular arms of 6.5 there is a total area of Cross which will be 120+36 =156, The shaded area is 1/4 ( large circle - Cross). Think about the length of AA', the diameter because finding the radius OA = OA' is the way to finisihing the problem. PA' = 5 + 6 + 5 = 16 In triangle APA' , Pythagoras' theorem gives ( AA' )^2 = 6.6 + 16.16 = 256+36 = 292 = 4.73 OA.OA will be 73 so the large circle area is 73.pi 1/4(73.pi - 156) will equal the shaded area This is approximately 18.334 square units.
Cosine formula Ptolemy's theorem PB Angle subtend 135 degrees, Should find the radius EDIT: I think that PB is sqrt(50) because it's 135- 90 degree. The rest is wrong
Angle BPQ encloses a quarter of the circle when we complete the circle and hence its measurement is 45° and hence triangle BQB is right and isosceles so QB=5 we extend AP until we get square PEBQ and from it AB²=AE²+EB²=146 so the radius of the circle R=√146/√2=√73 we have [AOBE]=[AOB]+[ABE]=(√73*√73)/2+(11*5)/2=64 and from it [OAPQB]=[AOBE]-[PEBQ]=64-25=39 and from it the area of the red region is equal to (π*(√73)²)/4-39=(73π)/4-39
Two chords:
c₁ = 6cm. ; c₂ = 5/cos45° = 5√2 cm
Pytagorean theorem:
(R√2)² = 5² + (6+5)² --> R= √73 cm
Isósceles inscribed triangles:
sin ½α₁ = ½6/R --> α₁ = 41,112°
sin ½α₂= ½5√2/R --> α₂ = 48,888°
Shaded area:
A₁= ½R²(α₁-sinα₁) = 2,1902 cm²
A₂ = ½R²(α₂-sinα₂) = 3,6438 cm²
A₃ = ½b.h = ½5² = 12,5 cm²
A = A₁+A₂+A₃= 18.334 cm² ( Solved √ )
Eu fiz diferente ! Gosto muito da resolução do professor ! Todos os dias eu aprendo uma nova resolução.
Parabéns pelos vídeos! Questões interessantes e instrutivas!
You used a Good idea to solve this question. Congratulations.
Pytagorean theorem:
R²=(½6)²+(½6+5)²= 3²+8² --> R=√73 cm
Internal areas :
A₀ = ¼πR² = 73/4 π cm²
A₁ = ½b.h = ½(5+½6)*½6= 12 cm²
A₂ = b.h = 5*½6 = 15 cm²
Shaded area:
A = A₀- 2.A₁-A₂= 18.334 cm² ( Solved √ )
Angle APB is 135 degree, so angle QPB=135-90=45. QPB is right isosceles then QB=5
Now we can find Rsqrt2 with cosine law on APB knowing that Cos (135)=cos(90+45)=-cos(90-45)=-cos45=-sqrt2/2
And that PB=5sqrt2.
Once found the radius (sqrt73) we have to subtract the areas of a trapezoid and a right triangle to the quarter circle. Tracing the radius that splits the chord AP in two equal parts and extending BQ untill intersects AO we get two similar right triangles and with proportions we can find all the sides we need
I drew the diagonal AB and labeled angle ABP as θ.
I calculated cos(45 + θ) and found that cos(θ) - sin(θ) = 3sqrt(2)/R.
I then translated side PQ so that point Q coincided with point B.
This results in sin(θ) = 5/(Rsqrt(2)).
Using cos(θ) = sin(θ) + 3sqrt(2) gives cos(θ) = 11/(Rsqrt(2)).
So we have a right triangle with sides Rsqrt(2), 5 and 11 from which R = sqrt(73). Also BQ is 5.
Now angle AOP is (90 - 2θ) and angle POB is 2θ.
So the required area can be calculated by:
A = 73π/4 - (1/2)R^2sin(90 - 2θ)-(1/2)R^2sin(2θ) + 25/2 = 73π/4 - 39.
AB=√[5²+(6+5)²]=√146→ Radio del círculo =r=√146/√2→ r²=146/2=73→ Área círculo =73π → Si copiamos la figura propuesta y la giramos 90º tres veces sucesivas en el mismo sentido obtenemos un círculo completo con una cruz griega inscrita; ésta se compone de un cuadrado central de 6*6 y cuatro brazos rectangulares de 5*6→ Área de la cruz inscrita = 6²+4*5*6 =156 → Área sombreada =(73π-156)/4 =(73π/4)-39.
Gracias y un saludo cordial.
Same way that I saw it. ✅ y saludo cordial para usted!
How did you know that BQ = 5?
@@AzouzNacir
There were the four pieces of the jigsaw. one fitted exactly in each quadrant. Twelve sides, Angles: 270 degrees times four and 90 degrees times eight. Total 1800 degrees.
Any length but 5 would not have fitted in. -----
I am sorry that this is not as rigorous as Math Booster's solutions. --------
You could also check all this with a drawing of the cross with those measurements and circumscribing a circle.
The shaded area would change in shape if the quadrant were drawn at any other angle, but the area between 1/4 of a circle and 1/4 of a Greek Cross inscribed would measure up to the same answer.
Please thank Math Booster for regular rigorous solutions.❤
I know this but you should have given the justification before calculating the length of AB and it is clear that angle BPQ encloses an arc of the circle measuring 90° so its measure is 45°@@kateknowles8055
@@AzouzNacir
The arc PB shown has centre at O and not at Q .
I could see that QB would measure 5, but I was not able to give a reason .
You might use t = tan theta = 3/8 then OBQ is arctan (3/8) so QB measures 5, using Math Booster's video at 9:00 minutes, instead.
شكرا
Best Wishes to you, and all fellow viewers who have had a go at this before watching the video.
Extending PQ to cross OB at R is making AQ the diameter of another circle, so that angle OAP will equal QRB. Now extending OR to A' we have a semicircle with AA' as diameter.Rotating the pentagonal area OAPQB around by 90 , 180, 270 degrees to make three more pentagons will form a Greek Cross.
Dividing it into a central square of 6.6 and four reclangular arms of 6.5 there is a total area of Cross which will be 120+36 =156,
The shaded area is 1/4 ( large circle - Cross).
Think about the length of AA', the diameter because finding the radius OA = OA' is the way to finisihing the problem.
PA' = 5 + 6 + 5 = 16 In triangle APA' , Pythagoras' theorem gives ( AA' )^2 = 6.6 + 16.16 = 256+36 = 292 = 4.73
OA.OA will be 73 so the large circle area is 73.pi
1/4(73.pi - 156) will equal the shaded area This is approximately 18.334 square units.
Cosine formula
Ptolemy's theorem
PB Angle subtend 135 degrees, Should find the radius
EDIT: I think that PB is sqrt(50) because it's 135- 90 degree. The rest is wrong
(6)^2 (5)^2={36+25}=61 90°APBO/61=1.29APBO 1.13^16 3^6 3^3^3 1^1^3 1^3 (APBO ➖ 3APBO+1).