@AnkitKumar-sy7pg
Nice question Sir 🙏❤❤❤.
@jamestalbott4499
Thank you!
@AnkitKumar-sy7pg
Sir like same method i had solve it . Thanks Sir
@DadiBob-y7p
Calculating the area of the triangle OCD does not require calculating the length of the side of the square, directly [OCD]=(8*4)/2=16
@sorourhashemi3249
Thanks. Easy. Connect o tod. The side of square =√32. Angle doc= 45-27=18===>1/20.circle area of triangle ocd= 16 +8^2π/0-8^2π/4=24.24
@zawatsky
Итак, диагональ квадрата 8, значит сторона 8/√2=(2√2)²/√2=4(√2)²/√2=4√2.✔Площадь - градусная мера сектора минус полуквадрат. От четверти нужно отнять разницу с диагональным углом, т. е. 90-45+27=45+27=60+5+7=60+5+5+2=72°. Множитель сектора 72/360=1/5.✔Тогда площадь сектора πr²/5=64π/5=12⁴/₅π.✔Полуквадрат (4√2)²/2=16*2/2=16.✔Итого S=12⁴/₅π-16.
@santiagoarosam430
CODº=45º ; DOBº=45º-27º=18º---> Área sombreada verde =π8²/4 -8²/2² -18π8²/360 =(64π/5)-16 ≈24,21238... u².
@anatoliy3323
Thanks for a curious task, Sir. But its interesting if that figure happens in the real life?💯👍
@MrPaulc222
I eventually got to 64pi/5 - 16, but I lost the calculations asking this on a phone. About 24.22, but phone's are fiddly so I may have fat thumbed something.
@marioalb9726
α = 90°+27°-45° = 72° = 2/5 π A₁= ½αR² = 40,212 cm² <-- R=8cm A₂= ½bh= ½R½R= 16 cm² A = A₁ -A₂= 24,212 cm² ( Solved √ )
@marioalb9726
α = 90°+27°-45° = 72°= 2π/5 A₁=½αR² ; A₂=½bh=¼R² <-- R=8cm A= A₁-A₂= ½R²(α-0,5)= ½8²(0,4π-0,5) A= 24,212 cm² ( Solved √ )
@cyruschang1904
Square diagonal = radius = 8 Square side length = 8/√2 = 4√2 Shaded area = quarter circle - half square - circle sector of 45° - 27° = 18° So (8)(8)(π)(1/4) - (4√2)(4√2)(1/2) - (8)(8)(π)(18/360) = 16π - 16 - (16/5)π = (64/5)π - 16
@LuisdeBritoCamacho
Area of Quarter of Circle : R = 8 ; R^2 = 64 ; (R^2 / 4) = 16 Area (C / 4) = 16 * Pi Area (C / 4) ~ 50,26548... Blue Square Diagonal = (sqrt(2) * Square Side) Diagonal = R = 8 8 = sqrt(2) * S ; S = (8 / sqrt(2)) ; S = (8 * sqrt(2)) / 2 Blue Square Side = 4 * sqrt(2) Blue Square Area = 32 ; Blue Square Half Area = 16 Angle (DOB) = 45º - 27º = 18º Area of Sector [ODBO] = ((Pi/10) * 64) / 2 Area of Sector [ODBO] = (16Pi / 5) Green Area = 16Pi - (16Pi/5 + 16) Green Area = (80Pi - 16Pi - 80) / 5 Green Area = (64Pi - 80) / 5 Green Area = (64Pi/5 - 16) Green Area ~ 24,2123859...
@zdrastvutye
the title is misleading. the solution is not the green area:
10 print "premath-can you find the green shaded area may2025":nu=60:dim x(3),y(3)
20 r=8:w=27:l=r/sqr(2):x(0)=0:y(0)=0:x(3)=l*cos(rad(90-w)):y(3)=l*sin(rad(90-w))
30 x(2)=x(3)+l*cos(rad(w)):y(2)=y(3)-l*sin(rad(w)):x(1)=x(2)+x(0)-x(3):y(1)=y(2)+y(0)-y(3)
40 rc=sqr(x(2)^2+y(2)^2):print rc
50 ymin=0:for a=0 to 3:if y(a)<ymin then ymin=y(a)
60 next a:xmin=0:masx=1200/r:masy=850/(r-ymin):if masx<masy then mass=masx else mass=masy
70 gcol5:goto 90
80 xbu=(x-xmin)*mass:ybu=(y-ymin)*mass:return
90 xl=0:yul=0:yol=r:x=xl:y=yul:gosub 80:xba1=xbu:yba1=ybu:y=yol:gosub 80:xba2=xbu:yba2=ybu
100 dx=x(2)/nu:ar=0:dy23=y(2)-y(3):for a=1 to nu:dxa=dx*a:xr=dxa
110 yur=y(3)*xr/x(3):if xr>x(3) then else 130
120 dyur=dy23*(xr-x(3))/(x(2)-x(3)):yur=y(3)+dyur
130 yor=sqr(r*r-xr^2):da=(yol-yul+yor-yur)/2*dx:ar=ar+da
140 x=xr:y=yur:gosub 80:xbn1=xbu:ybn1=ybu:y=yor:gosub 80:xbn2=xbu:ybn2=ybu:goto 170
150 line xba1,yba1,xbn1,ybn1:line xbn1,ybn1,xbn2,ybn2:line xbn2,ybn2,xba2,yba2
160 line xba2,yba2,xba1,yba1:xba1=xbn1:yba1=ybn1:xba2=xbn2:yba2=ybn2:return
170 gosub 150:next a:x=r:y=0:gosub 80:xba=xbu:yba=ybu:gcol8:for a=1 to nu
180 wa=pi/2*a/nu:x=r*cos(wa):y=r*sin(wa):gosub 80:xbn=xbu:ybn=ybu:goto 200
190 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
200 gosub 190:next a:x=x(0):y=y(0):gosub 80:xba=xbu:yba=ybu:for a=1 to 4:b=a:if b=4 then b=0
210 x=x(b):y=y(b):gosub 80:xbn=xbu:ybn=ybu:gosub 190:next a:avk=pi*r*r/4
220 print "die gestreifte flaeche=";ar
premath-can you find the green shaded area may2025
8
die gestreifte flaeche=42.2862025
>
i have found out that pi*r*r/4-42.286+l^2*tan(27)=about 24
@wackojacko3962
🤔 ...See how you are? ...I'm the same and think it's a good thing we're on the same page! 😎
@phungpham1725