Radius of circle: r = R cos 45° = 8 / √2 = √32 r = 5,657 cm Area circle : A = πr² = 32π cm² Area circular segment: A₂ = ½ R² (α - sin α) A₂ = ½ 8² (π/2- sin π/2 ) A₂ = 32 (π/2- 1)= 16π - 32 A₂ = 18,2655 cm² Blue shaded area : A₁ = ½ A - A₂ A₁ = 16π - (16π - 32) A₁ = 32 cm² ( Solved √ )
another method we have circle with radius 8 so this area is (8X8 pi)/4 remaining two small congruent parts can be calculated as (area of circle of radius 4 root 2 - area of square formed with side 8 unit) X 2.........since that angle is 90 so joining A and C will give us diagonal....and hence we can imagine a square to find the area of segment now we got all the areas then subtract it from the original circle and and is 32
AC is a diameter, so D= r sqrt 2. AB is the radius of the quarter circle inscribed (ABC), so r = 8. A=Area of the full circle (center O, radius D/2). pi D^2/4 = pi r^2/2 A2=Area of the quarter circle is pi r^2/4 = A/2 A3=Area of the half circle AOC is half of full circle, or pi r^2/2 = A/2 The overlap between A2 and A3 is the area of the right triangle of side r. A4 = Area of triangle = r^2/2 BA= Blue area A = A2 + A3 - A4 + BA = A/2 + A/2 - A4 + BA BA = A4 = r^2/2 = 64/2 = 32
I connected A to C and AC ( diameter of large circle) according to Pythagoras theory is 11.313 . so radius is 5.656. The area of large circle= 100.45÷2=50.225. The area of triangle ABC =32==>the area of segmental circle=50.24. So 50.24-32=18.24 and the area of blue shaded region is 50.225-18.24=31.985.
Everytime we have this configuration: Blue shaded area = 1/π Area circle Radius of circle: r = R cos 45° = 8 / √2 = r = √32 cm Area of circle : A = πr² = 32π cm² Blue shaded area : A₁ = 1/π x A = 1/π x 32π A₁ = 32 cm²
Thales Flat Earth floating on water....when Pythagoras showed up , he took Thales "know thyself" to the extreme! Quadricircle BCA is nice and neat with radius = 8. Redo this problem with Inscribed 3 4 5 triangle. It's more fun! 🙂
32 The radius of the circle: sqrt (4^2 + 4^2) [Pythagorean] sqrt ( 32) The area of the circle 32 pi [pi r^2] Draw a right angle to complete an 8 by 8 square The area of this square is 64 The area not covered by the square = 32 pi - 64 Hence the area of each segment (32pi - 64)/4 Since two are unshaded, the area of those two is (32pi -64)/4 *2 = 32 pi - 64)/2 = 16 pi -32 Since the next unshaded area is the quarter circle, its area is 16 pi ( 64 pi /4) Hence the area of the SHADED area is 32 pi - (16 pi - 32) + 16 pi 16 pi - [ 16 pi - 32 ] 16 pi - 16 pi + 32 32 Answer
Diameter of circle: AC = √(8^2 + 8^2) = 8√2 → Radius = 4√2 Radius of quarter circle = 8 Area of blue region = Area of semi-circle (with radius 4√2) − (Area of quarter circle (with radius 8) − Area of △ABC) = Area of semi-circle (with radius 4√2) + Area of △ABC − Area of quarter circle (with radius 8) = 1/2 × π(4√2)² + 1/2 × (8 × 8) − 1/4 × π(8)² = 16π + 32 − 16π = 32
Just my opinion: this is not olimpiad level, not even for a regional olympiad. It is very suitable for an end of semester (or trimester, what have you) test. Depending on the country, grades 7-8 in mid-school, or maybe first year high schoo (not higher). Thanks for the post 👍
I could also solve this math problem, but my way is much more complicated than yours 😅 radius of the big circle = R, radius of the quarter circle = r = 8 draw a line from the midpoint of AB to the midpoint of BC, you will have a triangle with a 90°angle at point B R²=4²+4²=32 --> R²=32 area big circle=32π area quarter circle=π*r²/4=16π draw a square : from point A a vertical line, from point C a horizontal line the blue region is now divided in three parts, two parts are equal to the parts of the circle, below the line AB and at the right of line BC let's these parts be A1 = (area big circle-area square)/4=(32π-64)/4 A1=8π-16 area blue region=area big circle-area quarter circle-2*A1 area blue region=32π-16π-2*(8π-16)=32π-16π-16π+32 area blue region=32
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Nice trick
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Radius of circle:
r = R cos 45° = 8 / √2 = √32
r = 5,657 cm
Area circle :
A = πr² = 32π cm²
Area circular segment:
A₂ = ½ R² (α - sin α)
A₂ = ½ 8² (π/2- sin π/2 )
A₂ = 32 (π/2- 1)= 16π - 32
A₂ = 18,2655 cm²
Blue shaded area :
A₁ = ½ A - A₂
A₁ = 16π - (16π - 32)
A₁ = 32 cm² ( Solved √ )
Nice! 16π - 16(π - 2) - the "Lune of Hippocrates"... 🙂
Thanks for video.Good luck sir!!!!!!!°
So nice of you, dear❤️
another method
we have circle with radius 8 so this area is (8X8 pi)/4
remaining two small congruent parts can be calculated as (area of circle of radius 4 root 2 - area of square formed with side 8 unit) X 2.........since that angle is 90 so joining A and C will give us diagonal....and hence we can imagine a square to find the area of segment
now we got all the areas then subtract it from the original circle and and is 32
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
You could perhaps generalize to the case of a quarter-circle placed anywhere in a circle.
AC is a diameter, so D= r sqrt 2.
AB is the radius of the quarter circle inscribed (ABC), so r = 8.
A=Area of the full circle (center O, radius D/2). pi D^2/4 = pi r^2/2
A2=Area of the quarter circle is pi r^2/4 = A/2
A3=Area of the half circle AOC is half of full circle, or pi r^2/2 = A/2
The overlap between A2 and A3 is the area of the right triangle of side r.
A4 = Area of triangle = r^2/2
BA= Blue area
A = A2 + A3 - A4 + BA
= A/2 + A/2 - A4 + BA
BA = A4 = r^2/2 = 64/2 = 32
Thanks for sharing! Cheers!
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I connected A to C and AC ( diameter of large circle) according to Pythagoras theory is 11.313 . so radius is 5.656. The area of large circle= 100.45÷2=50.225. The area of triangle ABC =32==>the area of segmental circle=50.24. So 50.24-32=18.24 and the area of blue shaded region is 50.225-18.24=31.985.
Everytime we have this configuration:
Blue shaded area = Radius²
A = r²
A = (8 cos45°)² = ( 8 / √2 )²
A = 8² / 2
A = 32 cm² ( Solved √ )
Wow, I solved it in my head!
(I’ve seen the trick over the years)
❤
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Being R the radius of quarter circle.
Everytime we have this configuration:
Blue shaded area = ½ R²
A = ½ 8² = 32 cm² ( Solved √ )
Radius of circle:
r = R cos 45° = 8 / √2 = √32
r = 5,657 cm
Area circle :
A = πr² = 32π cm²
Area quarter circle:
A₁ = ¼ π R² = ¼ π 8²
A₁ = 16π cm²
Area circular segment:
A₂ = ½ r² (α - sin α)
A₂ = ½ 32 (π/2- sin π/2 )
A₂ = 16 (π/2- 1)= 8π - 16
A₂ = 9,1327 cm²
Blue shaded area :
A₃ = A - A₁ - 2. A₂
A₃ = 32π - 16π - 2. 9,1327
A₃ = 32 cm² ( Solved √ )
❤😁👍
Excellent!
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I got it! I feel proud!
Bravo!
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You are awesome. Keep it up 👍
Everytime we have this configuration:
Blue shaded area = 1/π Area circle
Radius of circle:
r = R cos 45° = 8 / √2 =
r = √32 cm
Area of circle :
A = πr² = 32π cm²
Blue shaded area :
A₁ = 1/π x A = 1/π x 32π
A₁ = 32 cm²
Excellent!
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Everytime we have this configuration:
Blue shaded area = Triangle area
A = ½ b.h = ½ 8²
A = 32 cm² ( Solved √ )
S=32
Very nice! And surpising that there is no PI in the solution.
Note that the radius of the oitside circle is 4root2 and AOC is a diameter, therefore the answer is 16pi-(64pi/4-64/2)=32.😊
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@@PreMath respond in deadly painful in my left knee, maynot I can continue my participation in premath in future😪.
@@misterenter-iz7rz
No worries!
Your health is more important. You are always in our thoughts and prayers. 🙏
Thales Flat Earth floating on water....when Pythagoras showed up , he took Thales "know thyself" to the extreme!
Quadricircle BCA is nice and neat with radius = 8. Redo this problem with Inscribed 3 4 5 triangle. It's more fun! 🙂
Ablue=(pi(sqrt32)^2)/2-(pi8^2/4-8*8/2)=32
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32
The radius of the circle: sqrt (4^2 + 4^2) [Pythagorean]
sqrt ( 32)
The area of the circle 32 pi [pi r^2]
Draw a right angle to complete an 8 by 8 square
The area of this square is 64
The area not covered by the square = 32 pi - 64
Hence the area of each segment (32pi - 64)/4
Since two are unshaded, the area of those two is (32pi -64)/4 *2 = 32 pi - 64)/2 = 16 pi -32
Since the next unshaded area is the quarter circle, its area is 16 pi ( 64 pi /4)
Hence the area of the SHADED area is 32 pi - (16 pi - 32) + 16 pi
16 pi - [ 16 pi - 32 ]
16 pi - 16 pi + 32
32 Answer
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Curious to me that a moon shape has arcs of circles for both sides, the area does not involve pi.
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The area of "moony" it's equal with area of triangle! 😀😉
Very true!
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What grade math is it?
It looks like the π disappearing in the round type area...
Yes, you are right
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Diameter of circle: AC = √(8^2 + 8^2) = 8√2 → Radius = 4√2
Radius of quarter circle = 8
Area of blue region
= Area of semi-circle (with radius 4√2) − (Area of quarter circle (with radius 8) − Area of △ABC)
= Area of semi-circle (with radius 4√2) + Area of △ABC − Area of quarter circle (with radius 8)
= 1/2 × π(4√2)² + 1/2 × (8 × 8) − 1/4 × π(8)²
= 16π + 32 − 16π
= 32
This is the lune of Hippocrates, Prof, wellknown solution
Easy peezy
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If I take the length of the angle at 8, Make the diameter 16, square it, then x .7854 = 201 x .25= the area you're looking for.(50.56?)
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@@PreMath thanks. I sold hydraulic cylinders for 35 years. Area of a circles burned into my brain.
Just my opinion: this is not olimpiad level, not even for a regional olympiad. It is very suitable for an end of semester (or trimester, what have you) test. Depending on the country, grades 7-8 in mid-school, or maybe first year high schoo (not higher). Thanks for the post 👍
This problem is over 2400 years old and is known as the *lune of Hippocrates* 😉
Amazing!
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I could also solve this math problem, but my way is much more complicated than yours 😅
radius of the big circle = R, radius of the quarter circle = r = 8
draw a line from the midpoint of AB to the midpoint of BC, you will have a triangle with a 90°angle at point B
R²=4²+4²=32 --> R²=32
area big circle=32π
area quarter circle=π*r²/4=16π
draw a square : from point A a vertical line, from point C a horizontal line
the blue region is now divided in three parts, two parts are equal to the parts of the circle, below the line AB and at the right of line BC
let's these parts be A1 = (area big circle-area square)/4=(32π-64)/4
A1=8π-16
area blue region=area big circle-area quarter circle-2*A1
area blue region=32π-16π-2*(8π-16)=32π-16π-16π+32
area blue region=32
Wait around 4:39, how did he know that the area is quarter circle?
That was the given at the beginning, quarter circle with center at B.
I think the answer is 32
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连接ac就一定过o点？？这题目出的？？？
This was not that hard