Cool ! My first method : Same method like you from 0:00 to 12:35 But, with : a+c=b+c=32, i find a=b (i won't calculate c, i choose to calculate a) Blue area = a+b=2*a a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3)) a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3)) Blue area=2*a=64*(2-sqrt(3))
Wow, this is really nice, it skipped my mind to calculate a instead of c, just like you did in this method, because honestly even I found calculating c Quite boring
Second method : Same method like you from 0:00 to 6:40 Blue area = a+b a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3)) a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3)) b=area(BST)=BS*H/2 and BS=AS-AB=8-8*(2-sqrt(3))=8*(sqrt(3)-1) H=PT*sin(15°)=8*sqrt(2)*(sqrt(6)-sqrt(2))/4=4*(sqrt(3)-1) Then, b=BS*H/2=8*(sqrt(3)-1)*4*(sqrt(3)-1)/2=16*(sqrt(3)-1)^2=16*(4-2*sqrt(3))=32*(2-sqrt(3)) Blue area=a+b=32*(2-sqrt(3))+32*(2-sqrt(3))=64*(2-sqrt(3))
1/ We have: the triangle PAQ is a right isosceles and PTQ is an equilateral one. -> QA= 8 and PA=8sqrt2 2/ AT intersects PQ at point I. Note that IT is the perpendicular bisector of PQ so, point I is the midpoint of PQ and IT//QS 3/ Demonstrating that a=b Label the area of triangle ABT=c We have: Area of QAT=Area of BAT ( same base, same height) -> a+c= b+c --> a=b --> Area of the blue are= 2 times area of the triangle QAB= 8 x AB😅 4/ Calculating the base AB Focus on the triangle QBS The angle BQS= 30 degrees, QSB=45 degrees --> angle QBS= 105 degrees. Drop the height BH = h to QS. We have two special triangles: BHS is a 45-90-45 triangle and BHA is a 60-90-30 triangle So, SH= h and HA=h. sqrt3 -> h + hsqrt3 = QS =8 sqrt2 --> h = 8sqrt2/(sqrt3+1) --> BS= hsqrt2=16/(sqrt3+1) AB= AS-BS= 8- 16/(sqrt3+1) AB= 8(sqrt3-1)/(sqrt3+1) --> Area of the blue= 8 AB =64(sqrt3-1)/(sqrt3+1) = 64(2-sqrt3) 😅😅😅
Lado del cuadrado AQRS: 32/4=8 → QPA=45º→ AP=AQ→ PQ=8√2=QS. QPT=45º+15º=60º y PQ=PT=QT=QS→ PQT es triángulo equilátero con ángulos 60º/60º/60º→ AQB=60º-45º=15º→ ABQ=90º-15º=75º=TBS. QP y QS son diagonales de cuadrados de 8*8 adosados y por tanto son perpendiculares→ TQS=90º-60º=30º→ Cómo QT=QS→QTS=QST=75º→ Los triángulos SBT y SQT son isósceles y semejantes → Si M es punto medio de BT y QS=8√2→ MS=4√2→ QM=4√2√3→ MT=MB=QT-QM=8√2-4√2√3=4√2(2-√3) → Razón de semejanza entre MBS y BAQ =s=MS/AQ=4√2/8=√2/2→ Razón entre áreas =s² =2/4=1/2→ Áreas SBT=QTS → Área azul =BAQ+SBT =2*MT*MS*2/2=2MT*MS =2[4√2(2-√3)]*4√2 =128-64√3 m². Sencillo pero farragoso. Gracias y un saludo cordial.
Thank you so much for sharing this amazing method of yours. Please keep sharing with us your method for the Question as I find them interesting and Educating, I remember that day, I wanted to solve one of the Questions I posted here, I didn't use my methodology, I used the one you shared and he understood it clearly thanks
Cool !
My first method :
Same method like you from 0:00 to 12:35
But, with : a+c=b+c=32, i find a=b (i won't calculate c, i choose to calculate a)
Blue area = a+b=2*a
a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3))
a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3))
Blue area=2*a=64*(2-sqrt(3))
Wow, this is really nice, it skipped my mind to calculate a instead of c, just like you did in this method, because honestly even I found calculating c Quite boring
Second method :
Same method like you from 0:00 to 6:40
Blue area = a+b
a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3))
a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3))
b=area(BST)=BS*H/2 and BS=AS-AB=8-8*(2-sqrt(3))=8*(sqrt(3)-1)
H=PT*sin(15°)=8*sqrt(2)*(sqrt(6)-sqrt(2))/4=4*(sqrt(3)-1)
Then, b=BS*H/2=8*(sqrt(3)-1)*4*(sqrt(3)-1)/2=16*(sqrt(3)-1)^2=16*(4-2*sqrt(3))=32*(2-sqrt(3))
Blue area=a+b=32*(2-sqrt(3))+32*(2-sqrt(3))=64*(2-sqrt(3))
Thanks for sharing this as well. I prefer the other method your shared though, but this is also a perfect one, thanks 👍
I think I could solve it . a and b are equal as
a=64-32√3 and b=64-32√3 so a+b=128-64√3
Thank you sir for your great efforts.
Perfect, thanks for sharing
You"re very welcome sir.
1/ We have: the triangle PAQ is a right isosceles and PTQ is an equilateral one.
-> QA= 8 and PA=8sqrt2
2/ AT intersects PQ at point I. Note that IT is the perpendicular bisector of PQ so, point I is the midpoint of PQ and IT//QS
3/ Demonstrating that a=b
Label the area of triangle ABT=c
We have: Area of QAT=Area of BAT ( same base, same height)
-> a+c= b+c
--> a=b
--> Area of the blue are= 2 times area of the triangle QAB= 8 x AB😅
4/ Calculating the base AB
Focus on the triangle QBS
The angle BQS= 30 degrees, QSB=45 degrees --> angle QBS= 105 degrees.
Drop the height BH = h to QS.
We have two special triangles: BHS is a 45-90-45 triangle and BHA is a 60-90-30 triangle
So, SH= h and HA=h. sqrt3
-> h + hsqrt3 = QS =8 sqrt2
--> h = 8sqrt2/(sqrt3+1)
--> BS= hsqrt2=16/(sqrt3+1)
AB= AS-BS= 8- 16/(sqrt3+1)
AB= 8(sqrt3-1)/(sqrt3+1)
--> Area of the blue= 8 AB
=64(sqrt3-1)/(sqrt3+1)
= 64(2-sqrt3) 😅😅😅
Wow so simple and straightforward, thanks for sharing.
Lado del cuadrado AQRS: 32/4=8 → QPA=45º→ AP=AQ→ PQ=8√2=QS.
QPT=45º+15º=60º y PQ=PT=QT=QS→ PQT es triángulo equilátero con ángulos 60º/60º/60º→ AQB=60º-45º=15º→ ABQ=90º-15º=75º=TBS.
QP y QS son diagonales de cuadrados de 8*8 adosados y por tanto son perpendiculares→ TQS=90º-60º=30º→ Cómo QT=QS→QTS=QST=75º→ Los triángulos SBT y SQT son isósceles y semejantes → Si M es punto medio de BT y QS=8√2→ MS=4√2→ QM=4√2√3→ MT=MB=QT-QM=8√2-4√2√3=4√2(2-√3) → Razón de semejanza entre MBS y BAQ =s=MS/AQ=4√2/8=√2/2→ Razón entre áreas =s² =2/4=1/2→ Áreas SBT=QTS → Área azul =BAQ+SBT =2*MT*MS*2/2=2MT*MS =2[4√2(2-√3)]*4√2 =128-64√3 m².
Sencillo pero farragoso. Gracias y un saludo cordial.
Thank you so much for sharing this amazing method of yours. Please keep sharing with us your method for the Question as I find them interesting and Educating, I remember that day, I wanted to solve one of the Questions I posted here, I didn't use my methodology, I used the one you shared and he understood it clearly thanks