Can you find area of the Green shaded region? | (Semicircle) | #math #maths #geometry

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Can you find area of the Green shaded region? | (Semicircle) | #math #maths #geometry

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[mohabatkhanmalak]

Awesome, enjoyed the method. Thank you.🌱

[boneistt]

I got 18.39 units^2 (to 2 decimal places), because I was always taught never to round until the very end.

Using the angles of 37° and 53° in a 3, 4, 5 triangle is a rough approximation, and I’m surprised (and disappointed) you’re advocating the usage of this mid-calculation.

[SergeCeyral]

A slightly faster way to get r (circle radius) is to consider the rectangle triangle ACD. (AD is the horizontal diameter of circle) : AB² = CB.BD (as AB perpendicular to CD) so BD = 27 and r = (27+3)/2 = 15

[quigonkenny]

By observation, OA is the radius r of semicircle O, and BO = r-3

Triangle ∆ABO:
a² + b² = c²
9² + (r-3)² = r²
81 + r² - 6r + 9 - r² = 0
6r = 90
r = 90/6 = 15

sin(θ) = O/H
sin(θ) = 9/15 = 3/5
θ ≈ 36.87° *

Green area:
A = (θ/360)πr² - bh/2
A = 0.10π(15²) - 12(9)/2
A = 23.04π - 54 ≈ 18.39*

* - note: values and further calculations resultant from inverse sine of 3/5 are shown to two places but were calculated with all precision allowed by scientific calculator (45 significant digits).

[hcgreier]

With (r-3)² + 9² = r² we get r = 15. The area A = sector ABC - triangle ABO = (α/360)·π·r² - 9·(15 - 3)/2, where tan α = 9/(15-3) = 9/12 => α ≈ 36.87°. So A ≈ (36.87/360)·π·15² - 54 = 18.39 square units.

[marcgriselhubert]

Easy to find the radius of the circle (just as you did): 15. In an adapted orthonormal the equation of the semi circle is y = sqrt(225 - (x^2))
So the unknown area is the integral from -15 to -12 of sqrt(225 - (x^2)) dx. Or (by symetry) this integral from 12 to 15.
It is not difficult to calculate but difficult to write here with the computer with no special characters. We note x/15 = sin(t), dx = 15.cos(t)dt
Finally the area is (225/2).Arccos(4/5) - 54 just as you found. (I know this method is too complicated here!)

[prossvay]

Area of the green shaded region=π(15)^2× (36.87/360)-1/2(9)(12)=18.4 square units. ❤❤❤ Thanks teacher.

[jimlocke]

Several viewers have effectively noted that the first answer given in your videos has been an exact answer and it was followed by a decimal approximation. In this problem, (185/8)π - 54 is not an exact answer. For an exact answer, you could have left Θ as arccos(4/5) or arccos(0.8), so the sector area would be ((arccos(0.8))/(360°))πr² = (0.625)(arccos(0.8))π. Perhaps better is express the arccos() in radians. Note that 360° = 2π radians. So, the sector area is (112.5)(arccos(0.8)) where arccos() is in radians and the green region area is exactly (112.5)(arccos(0.8)) - 54, when the arccos() is expressed in radians.

[giuseppemalaguti]

r^2=9^2+(r-3)^2..r=15...Ag=π15^2/2-Asettorecirc(180-arctg9/12)-9*12/2=225π(1/2-0, 39758)-54=18, 38...

[soniamariadasilveira]

Tks teacher, I did not know how to solve this question!

[tombufford]

At a quick glance: To find the green shaded area, the sector angle and radius are calculated. R^2=9^2+(R-3)^2. 6*R=81+9. R=90/6=15. The sector angle OCA is tan^-1 (9/12) = 36.87 degrees. The sector area = PI *R^2 * 36.87/360 = 72.4. The triangle OBA area = 0.5 * 12*9=54. The green shaded area = 72.4 - 54 = 18.6 .

[ybodoN]

Let's mirror the diagram so that we get a complete circle and a circular segment.
Let _c_ be the chord length and _h_ the sagitta (height) of the circular segment. So, in our case, _c = 18_ and _h = 3._
In terms of _c_ and _h, _ the area of a circular segment is _((c² + 4h²) / 8h)² cos⁻¹((c² − 4h²) / (c² + 4h²)) − (c / 16h) (c² − 4h²)._
The area of the green shaded region is half of our circular segment, that is approximately 36.7877 / 2 or 18.39 square units.

[Krestor]

I used calculus to solve this
Using the formula where if 2 chords perpendicular bisect each other then the multiples of the length of a chord of each side will be equal to be multiple of the length of the other chord of their sides I found radius. Here one of the chord is the diameter so I knew that if the other chord was extended it would lead to a length 9 below. So
9 * 9 = 3 * x
After that i solved for x and then added a 3 and got the diameter.
The formula for circle is x^2 + y^2 = r^2
here 'r' is 15 so I used
∫ (12, 15) √(225 - x^2) dx
and with this i got the result.

Also if you're gonna get both the decimal places wrong then you should probably just stick to writing A = 18 instead of 18.65 because it's really 18.19

[wackojacko]

@ 6:07, I got right on the phone for this one. When I dialed the number, operator came on and said Press One for English Two for Dipthongs Three for Ululating and Four for Mnenomics . I pressed Four and requested SOH CAH Toa immediately! Welcome to America. 🙂

[murdock]

φ = 30°; 81 = 3(2r - 3) → r = 15 → BO = r - 3 = 12 → AO = 15 → AC = 3√10 → sin⁡(ϑ) = 3/5 →
225π(ϑ/12φ) → green area = 225π(ϑ/12φ) - 54 = (3/4)(25πϑ/φ - 72)

[fjb]

How did you come up with 185 x? Pi / 8? Please tell me how you did that. What did I overlook?

[hanswust]

I think that if you finally use a calculator, you could take advantage of its power and get the exact figures, later you can round them at will.

Let it be θ = arc sin (0.6) in radians.

The area of the sector is 225*θ/2π = 72, 39387474

Now you substract 54 and the green area is 18, 39387474 or 18.39 with just two decimals.

[MrPaulc]

I would probably begin with calculating the radius.
OBA is a right triangle with sides of r-3, 9, and r
(r-3)^2 + 9^2 = r^2
r^2 - 6r + 9 + 81 = r^2
Simplify to 90=6r, so r=15
Triangle AOB is (9*12)/2 = 54

Inverse sine of 9/15 = 36.87 (rounded) degrees. Ouch.
need a better way.
OAC is (36.87/360)*225pi
which comes out at 72.4 (rounded).
72.4-54=18.4 sq units, subject to small rounding errors.
I imagine the video shows a simpler way.
Indeed, I now see that the video used this method, but simplification appears to be at the expense of accuracy.
Is it usual to have 37 and 53 degrees as the angles in a 3, 4, 5 triangle? I had them coming up as 36.87 and 53.13. I imagine this accounts for my answer being slightly lower than on the video.

[cyruschang]

9^2 + (r - 3)^2 = r^2
9^2 - 6r + 9= 0
6r = 9^2 + 9 = 90
r = 15
green area = (πr^2)(θ/2π) - 9(r - 3)/2 = (θ/2)(15)^2 - 9(6) = 225θ/2 - 54
(θ is the smallest inner angle of a 3 - 4 - 5 triangle. sinθ = 3/5, cosθ = 4/5, θ is about π/4.882)

[mathswan]

r^2=9^2+(r-3)^2
r=15
sin a=9/15=3/5
a=arcsin 3/5
Area=1/2*12^2xarcsin 3/5-1/2x9x12
=72arcsin3/5-54