i watch your videos everyday, your problems are really amazing! But if you don't mind, could you please come with more hard problems? It would be more helpful for me to take my skills to the next level. Thank you!
Диагональ квадрата равна радиусу, значит, сторона будет r/√2. Площадь треугольника r²/(2√2)=2√2⇔r²=(2√2)²⇔r=2√2. Тогда сторона а=2, площадь 4, а её половина - снова 2. Площадь сегмента равна четверти полукруга или ¹/₈ круга, т. е. πr²/8=8π/8=π. Отнимаем нашу половинку (совпадает по диагонали с сегментом): π-2.
Overthought this one WAY too much! After mulling it over for an hour and getting nowhere, I finally realized that the diagonal of the square was equal to the radius of the semicircle. After I realized that, it was a piece of cake.
Radius = r and square side length = x r = diagonal of the square, so r = x√2 2√2 cm^2 = (r)(x)/2 = (x^2)(√2)/2 => x = 2 cm, r = 2√2 cm Green area = (r^2)π/8 - (x^2)/2 = ((8)π/8 - 2) cm^2 = (π - 2) cm^2
The side lenght c of the square is R/sqrt(2), with R the radius of the circle. The area of the yellow triangle is (1/2).c.R = (sqrt(2)/4).(R^2) So we have ((sqrt(2)/4).(R^2) = 2.sqrt(2), and then (R^2) = 8 and R = 2.sqrt(2). Angle AOE = 45°, so the area of the sector AOE is Pi.(R^2).(45/360) = Pi. 8.(1/8) = Pi. The area of the triangle A=EFO is (1/2).(c^2) = (1/2).(2^2) = 2. So, finally, the green area is Pi - 2.
As OE is the diagonal of square ODEF and OE = r, then the side length of ODEF is r/√2. As ∠AOE = 45°, then as the green shaded area is equal to the area of sector AOE minus the area of triangle ∆EFD, then: Green shaded area: Aɢ = (45°/360°)πr² - bh/2 Aɢ = πr²/8 - (r/√2)²/2 Aɢ = πr²/8 - r²/4 = r²(π-2)/8 --- [1] Yellow triangle ∆DOB: Aʏ = bh/2 = r(r/√2)/2 2√2 = r²/2√2 r² = 2√2(2√2) r = 2√2 --- [2] Aɢ = (2√2)²(π-2)/8
Let's find the area: . .. ... .... ..... The triangle OEF is a right triangle. With r being the radius of the semicircle and s being the side length of the square we obtain: OF² + EF² = OE² s² + s² = r² 2s² = r² s² = r²/2 ⇒ s = r/√2 From the given area of the yellow right triangle OBD we can conclude: A(OBD) = (1/2)*OB*OD = (1/2)*r*s = (1/2)*r*r/√2 = r²/(2√2) ⇒ r² = 2√2*A(OBD) = 2√2*(2√2)cm² = 8cm² Now we are able to calculate the area of the green region: A(green) = A(circle sector OAE) − A(right triangle OEF) = πr²*(∠AOE/360°) − (1/2)*OF*EF = πr²*(∠FOE/360°) − (1/2)*s*s = πr²*(arctan(EF/OF)/360°) − (1/2)*s² = πr²*(arctan(s/s)/360°) − (1/2)*r²/2 = πr²*(arctan(1)/360°) − r²/4 = πr²*(45°/360°) − r²/4 = πr²/8 − r²/4 = (π/8 − 1/4)*r² = (π/8 − 1/4)*8cm² = (π − 2)cm² Best regards from Germany
Solution: Angle AÔE = θ Angle AÔE = 45°, since DEFO is a square DE = EF = FO = DO = x Green Shaded Area (GSA) = Sector AOE Area - ∆ OFE Area GSA = θ/360° π r² - ½ x² ... ¹ So, we must find "x" and radius EO = x√2, since DEFO is a square and EO is the radius of the semicircle AO = BO = EO = radius = x√2 Yellow Triangle Area = 2√2 = ½ b h = ½ x√2 . x 2√2 = ½ x√2 . x 4√2 = √2 x² x² = 4 x = 2 Therefore, AO = BO = EO = radius = 2√2 Substituting in Equation ¹ GSA = 45°/360° π (2√2)² - ½ (2)² GSA = 1/8 π . 8 - ½ . 4 GSA = (π - 2) cm² ✅ GSA ≈ 1,1415 cm² ✅
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Square Side = a cm 02) Semicircle Radius = R cm 03) (a * R) / 2 = 2 * sqrt(2) sq cm ; (a * R) = 4*sqrt(2) 04) Square Diagonal OE = R ; R = (a * sqrt(2)) 05) Square Side = a = (R * sqrt(2)) / 2 06) a * R = (4 * sqrt(2)) 07) (R * sqrt(2)) / 2 * R = 4*sqrt(2) ; R^2 * sqrt(2) / 2 = 4*sqrt(2) ; R^2 = 8 ; R = sqrt(8) ; R = 2*sqrt(2) cm 08) R = 2*sqrt(2) cm 09) a = 2 cm 10) Sector [OEA] Area = R^2 * (pi/4) / 2 ; Sector Area = Pi sq cm 11) Green Shaded Region = (Sector Area) - (Half Square Area) ; NOTE : Square Area = (2 * 2) = 4 sq cm 12) GSA = (Pi - 2) sq cm 13) GSA ~ 1,1415 sq cm Therefore, OUR BEST ANSWER : The Green Shaded Area is equal to (Pi - 2) Square Centimeters or approx. equal to 1,1415 Square Centimeters.
Awesome breakdown sir so Thanku for these daily exercises!
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Enjoyed a lot by solving & knowing your problems. Thanks a lot sir
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i watch your videos everyday, your problems are really amazing! But if you don't mind, could you please come with more hard problems? It would be more helpful for me to take my skills to the next level. Thank you!
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Bom dia Mestre.
Não acredito q acertei
Aprendi Geometria graças às vossas aulas
Deus lhe Abençoe
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Soo great
This channel is so nice
Can you share other math problems except geometry pls
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Диагональ квадрата равна радиусу, значит, сторона будет r/√2. Площадь треугольника r²/(2√2)=2√2⇔r²=(2√2)²⇔r=2√2. Тогда сторона а=2, площадь 4, а её половина - снова 2. Площадь сегмента равна четверти полукруга или ¹/₈ круга, т. е. πr²/8=8π/8=π. Отнимаем нашу половинку (совпадает по диагонали с сегментом): π-2.
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Nice! ar = 4√2; a = r√2/2 → r = 2√2 → a = 2 → area AFE = π - 2√2 + 2(√2 - 1 ) = π - 2
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Overthought this one WAY too much! After mulling it over for an hour and getting nowhere, I finally realized that the diagonal of the square was equal to the radius of the semicircle. After I realized that, it was a piece of cake.
EO=OB=r EF=FO=OD=DF=r/√2 r*r/√2*1/2=2√2 r=2√2
Green shaded area = 2√2*2√2*π*45/360 - 2*2*1/2 = π - 2(cm²)
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OF=a---> Radio =r=a√2---> Amarillo =a*a√2/2=2√2---> a=2--->r²=8---> Verde =(Sector circular 45º)-(FEDO/2) =(πr²/8)-(a²/2) =(π-2) cm².
Gracias y saludos
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Radius = r and square side length = x
r = diagonal of the square, so r = x√2
2√2 cm^2 = (r)(x)/2 = (x^2)(√2)/2 => x = 2 cm, r = 2√2 cm
Green area = (r^2)π/8 - (x^2)/2 = ((8)π/8 - 2) cm^2 = (π - 2) cm^2
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@ Thank YOU
S=π-2≈1,14 cm²
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Green shaded area=(π-2)cm^2.❤❤❤
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Yellow triangle:
A = ½bh = ½R.s = 2√2 cm²
R.s = 4√2 cm²
Radius of circle:
R = s√2
R²= Rs√2= (4√2)√2= 8 cm²
Green Area: (Half circular segment)
A = ½[½R²(90°-sin90°)] = ¼ R² (π/2-1)
A = π-2 = 1,1416 cm² ( Solved √ )
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lr/2=2√2...(r=√2l)...=> l=2,r=2√2..Agreen=πr^2/2-Asett(135°)-l^2/2=πr^2/2-(3/8)πr^2-4/2=(π/8)r^2-2=π-2
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The side lenght c of the square is R/sqrt(2), with R the radius of the circle. The area of the yellow triangle is (1/2).c.R = (sqrt(2)/4).(R^2)
So we have ((sqrt(2)/4).(R^2) = 2.sqrt(2), and then (R^2) = 8 and R = 2.sqrt(2).
Angle AOE = 45°, so the area of the sector AOE is Pi.(R^2).(45/360) = Pi. 8.(1/8) = Pi. The area of the triangle A=EFO is (1/2).(c^2) = (1/2).(2^2) = 2.
So, finally, the green area is Pi - 2.
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Pi-2
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As OE is the diagonal of square ODEF and OE = r, then the side length of ODEF is r/√2. As ∠AOE = 45°, then as the green shaded area is equal to the area of sector AOE minus the area of triangle ∆EFD, then:
Green shaded area:
Aɢ = (45°/360°)πr² - bh/2
Aɢ = πr²/8 - (r/√2)²/2
Aɢ = πr²/8 - r²/4 = r²(π-2)/8 --- [1]
Yellow triangle ∆DOB:
Aʏ = bh/2 = r(r/√2)/2
2√2 = r²/2√2
r² = 2√2(2√2)
r = 2√2 --- [2]
Aɢ = (2√2)²(π-2)/8
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Shapes, sizes, and drop dead curves... I'm gonna hit the beach when the sun comes up! Practice doin maths in my head. 😊
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Let's find the area:
.
..
...
....
.....
The triangle OEF is a right triangle. With r being the radius of the semicircle and s being the side length of the square we obtain:
OF² + EF² = OE²
s² + s² = r²
2s² = r²
s² = r²/2
⇒ s = r/√2
From the given area of the yellow right triangle OBD we can conclude:
A(OBD) = (1/2)*OB*OD = (1/2)*r*s = (1/2)*r*r/√2 = r²/(2√2)
⇒ r² = 2√2*A(OBD) = 2√2*(2√2)cm² = 8cm²
Now we are able to calculate the area of the green region:
A(green)
= A(circle sector OAE) − A(right triangle OEF)
= πr²*(∠AOE/360°) − (1/2)*OF*EF
= πr²*(∠FOE/360°) − (1/2)*s*s
= πr²*(arctan(EF/OF)/360°) − (1/2)*s²
= πr²*(arctan(s/s)/360°) − (1/2)*r²/2
= πr²*(arctan(1)/360°) − r²/4
= πr²*(45°/360°) − r²/4
= πr²/8 − r²/4
= (π/8 − 1/4)*r²
= (π/8 − 1/4)*8cm²
= (π − 2)cm²
Best regards from Germany
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Solution:
Angle AÔE = θ
Angle AÔE = 45°, since DEFO is a square
DE = EF = FO = DO = x
Green Shaded Area (GSA) = Sector AOE Area - ∆ OFE Area
GSA = θ/360° π r² - ½ x² ... ¹
So, we must find "x" and radius
EO = x√2, since DEFO is a square and EO is the radius of the semicircle
AO = BO = EO = radius = x√2
Yellow Triangle Area = 2√2 = ½ b h = ½ x√2 . x
2√2 = ½ x√2 . x
4√2 = √2 x²
x² = 4
x = 2
Therefore, AO = BO = EO = radius = 2√2
Substituting in Equation ¹
GSA = 45°/360° π (2√2)² - ½ (2)²
GSA = 1/8 π . 8 - ½ . 4
GSA = (π - 2) cm² ✅
GSA ≈ 1,1415 cm² ✅
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Square Side = a cm
02) Semicircle Radius = R cm
03) (a * R) / 2 = 2 * sqrt(2) sq cm ; (a * R) = 4*sqrt(2)
04) Square Diagonal OE = R ; R = (a * sqrt(2))
05) Square Side = a = (R * sqrt(2)) / 2
06) a * R = (4 * sqrt(2))
07) (R * sqrt(2)) / 2 * R = 4*sqrt(2) ; R^2 * sqrt(2) / 2 = 4*sqrt(2) ; R^2 = 8 ; R = sqrt(8) ; R = 2*sqrt(2) cm
08) R = 2*sqrt(2) cm
09) a = 2 cm
10) Sector [OEA] Area = R^2 * (pi/4) / 2 ; Sector Area = Pi sq cm
11) Green Shaded Region = (Sector Area) - (Half Square Area) ; NOTE : Square Area = (2 * 2) = 4 sq cm
12) GSA = (Pi - 2) sq cm
13) GSA ~ 1,1415 sq cm
Therefore,
OUR BEST ANSWER :
The Green Shaded Area is equal to (Pi - 2) Square Centimeters or approx. equal to 1,1415 Square Centimeters.
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