I did angle ACO also = 15 degrees, since AO = OC = 8 = isosceles triangle, and also angle AOC = 150 degrees. Then its area = 1/2 x 8 x 8 x sin 150 degrees = 32 x 0.5 = 16. Area of sector including this triangle = Pi x 8^2 x 150 / 360 = 83.7758. Thus Purple area = 83.7758 - 16 = 67.7758.
The formula for the area of a circular segment is ½ R² (θ − sin θ) with the angle θ in radians. If R = 8 and θ = 150° = ⅚ π rad so sin θ = ½, then the purple shaded area is 16/3 (5π − 3) u².
Connect o and c OA=OC=8 triangle AOC is isolated Angle OAc=OCA=15 Angle AOC=150 Angle. BOC=180-150=30 Area of the semicircle=1/2(π)(8^2)=32π Area of the triangle AOC=1/2(8^2)sin(150)=(32)(1/2)=16 square units Area of the BOC=30/360(π)(8^2)=16π/3 Area of the purple shaded region=32π-(16+16π/3)=80π/3-16=67.78 square units. Thanks ❤❤❤
The purple area is a circular segment, which is a circular sector subtended by a chord, minus the triangle formed by that chord and the center of the circle. In this case, as ∠A = 15° and A is on the circumference, the angle of arc CB is 30°, so the sector and chord AC cover 180°-30° = 150°. Purple area: A = (θ/360)πr² - absin(θ)/2 A = (150/360)π(8²) - 8(8)sin(150°)/2 A = (5/12)π(64) = 64(1/2)/2 A = 80π/3 - 16 ≈ 67.78 sq units
I found it easier to draw a line from O to meet line A-C perpendicularly. The length of this line segment is the "height" of the isosceles triangle A-O-C. This length is 8 * sin (15º). The "base" of this triangle is length A - C which is 2 * 8 * cos (15º). This makes the surface of this triangle 8 * sin (15º) * 8 * cos (15º) = 16. The area of circle segment AOC is 80/3 * π. The surface area thus becomes 80/3 π - 16.
@@ybodoN I think you are mistaken. None of the calculations by the professor would be valid if it were not as I posit. Certainly AB would not be a diameter of a semicircle and AO would not the radius. Hopefully, the professor will respond.
@@markscher1968 the circular sector of center A and radius AB is 15/360 of the area of a circle of radius 16 and its area is about 33.51 square units. But AC is √(16² − 4²) = √240 ≈ 15.49 ≠ 16. In fact, the area of ABC (made of a circular sector and an isosceles triangle) is about 32.76 square units.
I understand and agree. The arc BC is from a circle of radius 8. It can’t simultaneously be the arc from a circle of radius 16. Excellent point and thanks for your insight.
This problem can be a lot easier - find the area of the semicircle with the 8 units radius it will be 1/2 pi R^2 where the r is 8 which we got by 16/2 thus becoming 100.5714286 - (i) after then find the area of sector abc where theta is 15 and radius is r = 16 with the circle centre O it becomes - 15/360 pi r^2 thus it becomes 33.52380952 -(ii) now subtracting (ii) from (i) we get 100.5714286 - 33.52380952 = 67.04761908 around the same answer though of course correct me if I am wrong.
Area of semicircle: A = ½πR² = ½π8² A = 100,53 cm² Area of isosceles triangle: A = ½b.h A = ½ 16 cos15°. 8 sin15° A = 16 cm² Area of angular sector: A = ½αR² A = ½ 30° (π/180°). 8² A = 16,775 cm² Purple area: A = A₁ -A₂ - A₃ A = 100,53 - 16 - 16,775 A = 67,756 cm² ( Solved √ )
Well done. My solution was very similar. I just calculated the 30 degree angle using the central angle theorem and then got the 150 degree angle from the straight angle property. Also I calculated the area of the isosceles triangle using the area formula with sin(150) = sin(180 -30) = sin(30) = 0.5.
We know that sin(15°) = (sqrt(6) _ sqrt(2)) /4, and cos(15°) = (sqrt(6) + sqrt(2)) / 4. So, if H is the orthogonal projection of O on (AC), we have OH = OA. sin(15°) = 2. (sqrt(6) - sqrt (2)) and AH = OA. cos(15°) = 2. (sqrt(6) + sqrt(2)). The area of triangle AOH is (1/2).OH.OA = (1/2).2.2. (6-2) = 8, and the area of triangle AOC is the double, so it is 16. The area of sector BOC is (30/360). Pi. (8)^2 = 16 Pi /3 (as angle BOC is 30°, the double of angle BAC in the circle) The area of the bog semi circle is (1/2). Pi. 8^2 = 32 Pi Finally the unknown area is 32 Pi - 16 (Pi/3) - 16 = 80 (Pi /3) - 16.
@@PreMath I now realize that it was not necessary to know the values of sin(15°) and cos(15°) as in the area of triangle AOH we multiply these two values, the area of the triangle AOH is: (1/2). OH.AH = (1/2). (1/2). (OA.sin(15°)). (OA.cos(15°)) = (1/2). (OA^2). ((1/2).sin(30°)) = (1/2).64. (1/4) = 8 (using the formula giving sin(2x))
My answer is A = 16(5π/3-1) = 67.78. The critical part is to realize that angle COB is double that of angle CAB. After that, you find the areas of sector COB and triangle CAO. The purple area is found by subtracting the sector and triangle from the semicircle.
I could not get how you did the area of Triangle AOC. Triangle AOC is an isosceles triangle with base AC and the other two sides of 8 each. The base AC = 2 (8 x cos(15)) i.e 16 x Cos (15). The height is 8 x Sin(15). The area of Triangle AOC is therefore 1/2 x (16 x cos (15) x 8 x Sin (15)) i.e 64 x Cos(15)Sin(15) . From trigonometric identity this will be 64 x 1/2 x sin(30) = 16.
There is an abbreviation: After drawing the isosceles triangle, we have 150 degrees angle. Calculate now the area of the sector and subtract isosceles triangle. Finished.
Whoa yall made that complicated The area of ABC as a sector? I made a mistake But as a triangle ABCarea is 32 Area of purple is 68 which I believe is close enough (opposite = 4.28) I used 4 just to be quick so the discrepancy
Comparing the previous puzzles, it is too easy. Let us to relax? The answer is 1/2 8^2 pi- 1/12 8^2 pi-8 cos 15 8 sin 15=80/3 pi - 16= 67.7758 approximately.😊
isn't the area simply the area of half the circle with diameter 16 (radius 8) - the area of a 15° circular segment of a circle with radius 16? I got the same result, why did we have to solve it the long way?
Thanks Professor, another great problem!😀🥂
Glad you think so!
Thanks ❤️
Happy, safe, and prosperous New Year!
Not really a Problem…
I calculated area of semicircle and then subtracted are of triangle AOC and subtracted area of sector BOC.
Great job!
Thanks ❤
That is how I did it too.
I did angle ACO also = 15 degrees, since AO = OC = 8 = isosceles triangle, and also angle AOC = 150 degrees.
Then its area = 1/2 x 8 x 8 x sin 150 degrees = 32 x 0.5 = 16.
Area of sector including this triangle = Pi x 8^2 x 150 / 360 = 83.7758.
Thus Purple area = 83.7758 - 16 = 67.7758.
Thanks ❤️
Area of triangle=1/2×8×8×sin150
Thanks ❤
S=(16(5π-3))/3≈67,7
The formula for the area of a circular segment is ½ R² (θ − sin θ) with the angle θ in radians.
If R = 8 and θ = 150° = ⅚ π rad so sin θ = ½, then the purple shaded area is 16/3 (5π − 3) u².
Thanks ❤️
Circular segment area:
A = ½.R².(α - sin α)
R = 16/2 = 8 cm
α = 180° - 2 . 15° = 150°
A = ½ 8² (150° - sin 150°)
A = 67.776 cm² ( Solved √ )
Thanks ❤️
Connect o and c
OA=OC=8
triangle AOC is isolated
Angle OAc=OCA=15
Angle AOC=150
Angle. BOC=180-150=30
Area of the semicircle=1/2(π)(8^2)=32π
Area of the triangle AOC=1/2(8^2)sin(150)=(32)(1/2)=16 square units
Area of the BOC=30/360(π)(8^2)=16π/3
Area of the purple shaded region=32π-(16+16π/3)=80π/3-16=67.78 square units. Thanks ❤❤❤
Awesome!
Thanks ❤🌹
The purple area is a circular segment, which is a circular sector subtended by a chord, minus the triangle formed by that chord and the center of the circle. In this case, as ∠A = 15° and A is on the circumference, the angle of arc CB is 30°, so the sector and chord AC cover 180°-30° = 150°.
Purple area:
A = (θ/360)πr² - absin(θ)/2
A = (150/360)π(8²) - 8(8)sin(150°)/2
A = (5/12)π(64) = 64(1/2)/2
A = 80π/3 - 16 ≈ 67.78 sq units
A = A₁ - A₂ - A₃
A = ½πR² - ½R²sinα - ½αR²
A = ½R² (π - sin30°- 30°π/180°)
A = 67,756 cm² ( Solved √ )
Thanks ❤️
I like that the solution does not need any trigonometry, because it makes use of the properties of the special 30-60-90 Triangle!
I found it easier to draw a line from O to meet line A-C perpendicularly. The length of this line segment is the "height" of the isosceles triangle A-O-C. This length is 8 * sin (15º). The "base" of this triangle is length A - C which is 2 * 8 * cos (15º). This makes the surface of this triangle 8 * sin (15º) * 8 * cos (15º) = 16. The area of circle segment AOC is 80/3 * π. The surface area thus becomes 80/3 π - 16.
Thanks ❤️
Very nice, many thanks, Sir!
φ = 30°; AO = BO = CO = r = 8 → CAB = φ/2 → COB = φ → 2π(30/360) = 16π/3
∆ AOC → COB = φ → AOC = 6φ - φ = 5φ → sin(φ) = sin(6φ - φ) = sin(5φ) = 1/2 → (1/2)sin(5φ)64 = 16 →
purple shaded area = 32π - (16π/3 + 16) = (16/3)(5π - 3)
Thanks ❤️
Why not just subtract 15/360 of the area of a circle of radius 16 from the area of a semicircle of radius 8?
Because ABC is not a circular sector (the center of the circular arc BC is O, not A)
@@ybodoN
I think you are mistaken. None of the calculations by the professor would be valid if it were not as I posit. Certainly AB would not be a diameter of a semicircle and AO would not the radius. Hopefully, the professor will respond.
@@markscher1968 the circular sector of center A and radius AB is 15/360 of the area of a circle of radius 16 and its area is about 33.51 square units.
But AC is √(16² − 4²) = √240 ≈ 15.49 ≠ 16. In fact, the area of ABC (made of a circular sector and an isosceles triangle) is about 32.76 square units.
I understand and agree. The arc BC is from a circle of radius 8. It can’t simultaneously be the arc from a circle of radius 16. Excellent point and thanks for your insight.
@@markscher1968 I went down the same path. Special thanks to @yboNob and @markscher1968
This problem can be a lot easier -
find the area of the semicircle with the 8 units radius it will be
1/2 pi R^2 where the r is 8 which we got by 16/2
thus becoming 100.5714286 - (i)
after then find the area of sector abc where theta is 15 and radius is r = 16 with the circle centre O
it becomes -
15/360 pi r^2 thus it becomes
33.52380952 -(ii)
now subtracting (ii) from (i)
we get
100.5714286 - 33.52380952 = 67.04761908
around the same answer
though of course correct me if I am wrong.
Area of semicircle:
A = ½πR² = ½π8²
A = 100,53 cm²
Area of isosceles triangle:
A = ½b.h
A = ½ 16 cos15°. 8 sin15°
A = 16 cm²
Area of angular sector:
A = ½αR²
A = ½ 30° (π/180°). 8²
A = 16,775 cm²
Purple area:
A = A₁ -A₂ - A₃
A = 100,53 - 16 - 16,775
A = 67,756 cm² ( Solved √ )
Thanks ❤️
Well done. My solution was very similar. I just calculated the 30 degree angle using the central angle theorem and then got the 150 degree angle from the straight angle property. Also I calculated the area of the isosceles triangle using the area formula with sin(150) = sin(180 -30) = sin(30) = 0.5.
Thanks ❤️
Ap=1/2π8^2-(1/12π8^2+1/2*8*8sin150)=80/3π-16
Bravo!
Thanks ❤️
شكرا لكم
مساحة نصف الدائرة ذات المركز O والشعاع r هي a
مساحة القطاع الزاوي BOC هوb
X=a-b
=80pi/3 -16
😁I got it (for once). Thanks.
Awesome!
Thanks ❤️
Awesome video sir ❤❤❤❤❤
Thanks dear ❤
We know that sin(15°) = (sqrt(6) _ sqrt(2)) /4, and cos(15°) = (sqrt(6) + sqrt(2)) / 4.
So, if H is the orthogonal projection of O on (AC), we have OH = OA. sin(15°) = 2. (sqrt(6) - sqrt (2)) and AH = OA. cos(15°) = 2. (sqrt(6) + sqrt(2)).
The area of triangle AOH is (1/2).OH.OA = (1/2).2.2. (6-2) = 8, and the area of triangle AOC is the double, so it is 16.
The area of sector BOC is (30/360). Pi. (8)^2 = 16 Pi /3 (as angle BOC is 30°, the double of angle BAC in the circle)
The area of the bog semi circle is (1/2). Pi. 8^2 = 32 Pi
Finally the unknown area is 32 Pi - 16 (Pi/3) - 16 = 80 (Pi /3) - 16.
Thanks ❤️
@@PreMath I now realize that it was not necessary to know the values of sin(15°) and cos(15°) as in the area of triangle AOH we multiply these two values, the area of the triangle AOH is: (1/2). OH.AH = (1/2). (1/2). (OA.sin(15°)). (OA.cos(15°)) = (1/2). (OA^2). ((1/2).sin(30°)) =
(1/2).64. (1/4) = 8 (using the formula giving sin(2x))
i found the area of the semi circle and deduct the area of the sector OBC and area of triangle AOC
Semi circle area - sector BOC = Sector AOC. So it is no different from what the Prof did.
Love it!!!!!!!!!!!!!
Thanks ❤🌹
I like your way too, especially finding the height of Triangle AOC, and Area of Sector AOC. 🙂
Thanks ❤️
My answer is A = 16(5π/3-1) = 67.78. The critical part is to realize that angle COB is double that of angle CAB. After that, you find the areas of sector COB and triangle CAO. The purple area is found by subtracting the sector and triangle from the semicircle.
Thanks ❤️
I could not get how you did the area of Triangle AOC. Triangle AOC is an isosceles triangle with base AC and the other two sides of 8 each. The base AC = 2 (8 x cos(15)) i.e 16 x Cos (15). The height is 8 x Sin(15). The area of Triangle AOC is therefore 1/2 x (16 x cos (15) x 8 x Sin (15)) i.e 64 x Cos(15)Sin(15) . From trigonometric identity this will be 64 x 1/2 x sin(30) = 16.
He declared AO to be the base and determined the height from CD.
it can be solved easier
sin150 = sin30 = 0.5, so [AOC] = 1/2*8*8*0.5 = 16
Purple S = 8*8*n*150/360 - 16
Thanks ❤️
There is an abbreviation: After drawing the isosceles triangle, we have 150 degrees angle. Calculate now the area of the sector and subtract isosceles triangle. Finished.
Whoa yall made that complicated
The area of ABC as a sector? I made a mistake
But as a triangle ABCarea is 32
Area of purple is 68 which I believe is close enough (opposite = 4.28) I used 4 just to be quick so the discrepancy
I'm I'm thinkin using SAS Theorem and Area of Sector Formula, then Area of semi circle - ( area of isosceles + area of sector COB ). 🙂
Thanks ❤️
1/2*8*8*sin(150°)...
Thanks ❤️
Can anyone explain, can this method work?:
Area of the semicircle- the area of the sector which gives about 67.02 units^2
ABC is not a circular sector (the center of the circular arc BC is O, not A)
Comparing the previous puzzles, it is too easy. Let us to relax? The answer is 1/2 8^2 pi- 1/12 8^2 pi-8 cos 15 8 sin 15=80/3 pi - 16= 67.7758 approximately.😊
Thanks ❤️
isn't the area simply the area of half the circle with diameter 16 (radius 8) - the area of a 15° circular segment of a circle with radius 16? I got the same result, why did we have to solve it the long way?