Awesome, enjoyed the method. Thank you.🌱

@PreMath

4 ай бұрын

Glad it was helpful! You are very welcome! Thanks ❤️

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.

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

@hcgreier6037

3 ай бұрын

@@PreMath No it's not. You only post "Excellent, thank You" on user replies here, even if users have good suggestions about what could be done better in some examples.

If you are going to approximate the angle of 36.86989... as 37, you should not really give you answer looking like 2 dp of accuracy. You cannot really justify better that 2 sig fig. Keeping accuracy of a higher order on a calculator give 18.39 to 2dp not 18.65.

@user-uu2rf8ev7z

4 ай бұрын

Excellent point. This site spends 9 minutes explaining a 2 minute process but cannot understand significant figures.

@armel334

4 ай бұрын

Lo emportante es el procedimiento, lo demás es solo refinamiento en la precisión del resultado.

@waheisel

4 ай бұрын

@@user-uu2rf8ev7z I appreciate PreMath for providing me with a fun daily math puzzle. If you run the speed at 2x and occasionally hit the left arrow, you can get the explanation in a couple minutes. PS nobody's perfect.

@waheisel

4 ай бұрын

@@someonespadre He explained how he came up with 37 degrees by showing that it is the arccosine of 4/5. He should have said that that was an approximation. Of course this is a great channel to get a fun daily math puzzle, not get engineering training.

@PreMath

4 ай бұрын

Thanks ❤️

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

@PreMath

4 ай бұрын

Thanks ❤️

Nice

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

It's awesome!

@PreMath

4 ай бұрын

Glad you think so! Thanks ❤️

Very good!!

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

Love 🎉it

@PreMath

4 ай бұрын

Excellent! Glad to hear that! Thanks ❤️

❤❤❤❤❤

@PreMath

4 ай бұрын

Thanks ❤️

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.

@PreMath

4 ай бұрын

Thanks ❤️

Very clever.

@PreMath

4 ай бұрын

Thanks ❤️

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.

@PreMath

4 ай бұрын

Thanks ❤️

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).

@PreMath

4 ай бұрын

Thanks ❤️

Thanks sir 😊

@PreMath

4 ай бұрын

You are very welcome! Thanks dear ❤️

Thank you!

@PreMath

4 ай бұрын

You are very welcome! Thanks ❤️

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

@PreMath

4 ай бұрын

You are very welcome! Thanks Sonia❤️

😀😁🥂❤️

@PreMath

4 ай бұрын

Excellent! Thanks a lot ❤️

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

@PreMath

4 ай бұрын

Great job You are very welcome! Thanks ❤️

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!)

@PreMath

4 ай бұрын

Thanks ❤️

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

@simpleman283

4 ай бұрын

He divided by 45. He skipped show the multiplication. Kinda weird how he did it, but maybe just to show there are almost always different paths you can take to find the answer.

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...

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

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

@ 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. 🙂

@PreMath

4 ай бұрын

😊 Thanks ❤️

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

@PreMath

4 ай бұрын

Thanks ❤️

I did not know the Sector Area formula, but I did figure the radius before watching the video. 9 x 9 = 81 81/3 = 27 27 + 3 = 30 30/2 = 15 👍

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.

@PreMath

4 ай бұрын

Bravo! Thanks ❤️

Love you sir 🎉🎉🎉🎉🎉🎉

@PreMath

4 ай бұрын

Thanks dear ❤️🌹

r^2=(r-3)^+9^2, 6r=9+81=90, r=15, therefore the area is 225pi arctan(3/4)/(2pi)-1/2×9×12=(225/2)arctan(3/4)-54=18.4 approximately. 😅

@PreMath

4 ай бұрын

Thanks ❤️

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.

@PreMath

4 ай бұрын

Thanks ❤️

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.

@hanswust6972

3 ай бұрын

I found it easier to calculate the radius using the Theoreme of Chords: 9×9 = 3×(2R - 3)

I could use Integral Calculus for this problem but I'm not gonna do it; it's not needed. Finding the Radius of the Semicircle. Let's call it R. 1) 9^2 + (R - 3)^2 = R^2 ; 81 + R^2 - 6R + 9 = R^2 ; (cancelling the two R^2) 81 - 6R + 9 = 0 ; 6R = 90 ; R = 90/6 ; R = 15. 2) Radius equal to 15. 3) tan of angle COA = 9/12 = 3/4. 4) arctan (3/4) ~ 36,9º. 5) Slice [AOC] Area = ([(arctan(3/4)º * Pi) / 180º] * R^2) / 2 ~ 144,8 / 2 sq un = 72,4 sq un. 6) Area of triangle [AOB] = (9*12)/2 = 108/2 = 54 sq un. 7) Area of Green Shaded Region = 72,4 - 54 = 18,4 sq un. Answer: The Green Shade Region Area is equal to 18,4 Square Units.

@PreMath

4 ай бұрын

Thanks ❤️

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)

@PreMath

4 ай бұрын

Thanks ❤️

@cyruschang1904

4 ай бұрын

@@PreMath 🙂🙏

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)

@PreMath

4 ай бұрын

Thanks ❤️

3*(2*r-3)=9^2 => r=15 (to avoid 2nd deg?)

@petrdub8650

4 ай бұрын

Yes, I've done it this way too. Thales theorem and right triangle altitude theorem led me directly to the fact, that rest of the circle's diameter must be 9^2/3=27, so the radius is 15.

@PreMath

4 ай бұрын

Thanks ❤️

Is it possible to make the answer becoming integer?

@joeschmo622

4 ай бұрын

Only if you make pi = 3.

@ybodoN

4 ай бұрын

If we consider that 2∠AOB ≈ 1 radian then _⅔ · c · h_ is a good approximation for the area of the complete circular segment. In our case, the chord _c_ is 18 and the height _h_ is 3. Therefore, half of the circular segment is approximately 18 square units.

(3)^2=9° (9)^2=81° (81°+9°)=90° (180°-90°) =√90° 3^√30 3^√5√^6 √3√5^13^2.√ 3^1 1^1 3^2 √1^√1 √1^√1 3^2 (x+2x-3)

@PreMath

4 ай бұрын

Thanks ❤️🙏

Checking the Area with Integral Calculus!!! Integrate, from 3 to 0, sqrt(-x^2 + 30x) dx Integrate, from x=3 to x=0, (30x - x^2)^(1/2) dx Using the help of Wolfram Alpha to do the Calculations it gives me 18,394 Square Units. QED.

@PreMath

4 ай бұрын

Thanks ❤️

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 .

@PreMath

4 ай бұрын

Thanks ❤️

What's important is that Sohcahtoa helped Lewis And Clark make it to The Pacific Coast through The Louisiana Territory! The more you know...

@christopherlinder7618

4 ай бұрын

Yeah, but they wouldn't have made it rounding like this!

@joeschmo622

4 ай бұрын

@@christopherlinder7618 They just circled around but found their way...

@PreMath

4 ай бұрын

Thanks ❤️

A = 18.3939.. because the angle = sin-1(9/15) = tan-1(9/12) = 36.8699.. deg., not 37 deg.

@user-uu2rf8ev7z

4 ай бұрын

And it did not take anyone 9 minutes to figure that out.

@PreMath

4 ай бұрын

Thanks ❤️

37 is just an approximation though

@PreMath

4 ай бұрын

Yes! Thanks ❤️

شكرا لكم على المجهودات S=AOC/2 ×15^2-54 AOCبالراديان

@PreMath

4 ай бұрын

Thanks ❤️

225π×arctan(0.75)/2π-54

PreMath, I generally LOVE your content, but this result was a bit disappointing. I feel that the problems get nicer as the answers are integer, fraction, radical or at least in some exact notation. It may be nice to remind students that you get APPROXIMATELY 37° and 53° for the famous 3,4,5 triangle, but you should definitely never use it in calculations. The integration over sqrt(225-x^2) from x = 12 to 15 has to yield the same result, and it doesn't. 18.39... is a far cry from 18.65. I heard that the U.S. state of Georgia at one point wanted to "define pi" as 3.14 to make it "easier for the students." Luckily, a few people intervened, and rightly so! We have to abstain from accepting too many rounded values from our students. As a teacher I cannot accept getting 0.7 or even 0.6 for 2/3, and your result in this video is almost as problematic. You could have written arctan(3/4) for the angle and used the old 1/2 * arc * radius formula for the sector, but of course that's not "beautiful" by most standards. I do realize that many international standards want "3 significant figures" (which is a bit crazy in itself), but obviously, as some have mentioned here already: DON'T ROUND UNTIL THE VERY END. Thanks for your wonderful videos!

@PreMath

4 ай бұрын

Thanks for your honest feedback❤️ Kind regards

asnwer=125cm isit

Yep… find that radius (we ALL did it the same)   𝒓² = 9² + (𝒓 - 3)²   𝒓² = 81 + 𝒓² - 6𝒓 ⊕ 9   6𝒓 = 90   𝒓 = 15 Got radius. I went with   △ABC + lens ACO = area ABC So, need lens. Easy, find θ of pie slice of circle   sin θ = 9 ÷ 15   θ = arcsin( 9 ÷ 15 )   θ = 0.643501 radians   Lens = ½θ𝒓² - △ACO   Lens = (½ 0.643501 × 15²) - (½ 15 × 9)   Lens = 72.3939 - 67.5000   Lens = 4.8939 To finish up   area = △ABC + Lens   area = ½ 3 × 9 ⊕ 4.8939   area = 13.5000 ⊕ 4.8939   area = 18.3939 Ta da … ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

4 ай бұрын

Thanks ❤️

Let's find the size of the green area: . .. ... .... ..... The triangle OAB is a right triangle, so we can apply the Pythagorean theorem in order to obtain the radius R of the semicircle: OA² = OB² + AB² OA² = (OC − BC)² + AB² R² = (R − BC)² + AB² R² = (R − 3)² + 9² R² = R² − 6*R + 9 + 81 6*R = 90 ⇒ R = 15 Now we can calculate the angle ∠AOB: tan(∠AOB) = AB/OB = AB/(R − BC) = 9/(15 − 3) = 3/4 ⇒ ∠AOB = arctan(3/4) Finally we are able to get the size of the green area: A(green) = A(circle sector OAC) − A(triangle OAB) = πR²*arctan(3/4)/(2π) − (1/2)*OB*AB = R²*arctan(3/4)/2 − (1/2)*(R − BC)*AB = 15²*arctan(3/4)/2 − (1/2)*(15 − 3)*9 ≈ 18.39 Best regards from Germany

@PreMath

4 ай бұрын

Excellent! Thanks ❤️🌹

Tan-1(0,75)~36,87...

@PreMath

4 ай бұрын

Thanks ❤️

Me, desperately trying to find a 30, 60,90 triangle somewhere.

@PreMath

4 ай бұрын

Thanks ❤️

@richardchristie3203

4 ай бұрын

@@PreMath I will try and commit 37,53,90 to memory for a 3,4,5 triangle, as I haven’t quite got a grasp on trigonometry yet, but enjoy your geometry problems.

@simpleman283

4 ай бұрын

@@richardchristie3203 37, 53, 90 is not the exact answer. He rounded off. Be careful.

Very disappointing problem!!! I spent some time trying to figure how you were going to accurately "calculate" the angle. I have been doing this sort of maths for over 50 years and have never ever committed to memory 37 and 53 because it is only a rough approximation. I have committed many important numbers to memory (3:4:5 , 5:12:13, 7:24:25 60/30, 45/45, Sqrt 2 , Sqrt3 triangles etc) but never any obscure angles. I expected much more from this channel. I thought the whole idea was to calculate and simplify without the use of a calculator. I am an engineer and to solve any of these problems in a real world work situation I just draw the problem in AutoCad or similar package and take a direct measurement. 😞

@PreMath

4 ай бұрын

Thanks for your honest feedback🌹

Kinda different method. First it took the radius as x. Other leg of the triangle=x-3 (x-3)²+81=x² x=15 The traingle was a 4,3,5 triangle as in 12,9,15. cosQ=4/5 Q=37° Sector angle=53° Area of adjacent sector=1/2×225×53/360=104.11 Remaining quadrant area=pi225/4 Area of shaded=pi225/2-pi225/4-(1/2×9×12) -104. 11=pi225/4-54-104.11=18.65 unit²