Can you find area of the Purple shaded Rainbow? | (Semicircles) |
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- เผยแพร่เมื่อ 30 ก.ย. 2024
- Learn how to find the Purple shaded Rainbow. Important Geometry and Algebra skills are also explained: Pythagorean theorem; circle theorem; area of the circle formula; perpendicular bisector theorem. Step-by-step tutorial by PreMath.com
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Extremely easy:
A = ½(¼πC²) = ⅛π888² = 98568π cm² ( Solved √ )
WE can make the medium sized semicircle as small as we want to, so let's make it infinitesimally small. Consequently, AB is basically the same as the diameter of the big circle.
So Purple area = (πR²)/2 = (π * 444²)/2 = 98568π
What's the need of the small circle??
Yes, there's no reason
What's the need of all video?
A = ½(¼πC²) = ⅛π888² = 98568π cm² ( Solved √ )
MUCH EASIER !?!
A Red Herring.
to misleading you 😂
Solution:
Chords Theorem
(R + r) . (R - r) = 444 . 444
R² - r² = 197.136 ... ¹
Purple Area = πR²/2 - πr²/2
Purple Area = π/2 (R² - r²)
Purple Area = π/2 197.136
Purple Area = 98.568 π Square Units
Purple Area ~ 309660,5 Square Units
Easiest method. The radius of the small circle is immaterial. Reduce it to zero. Then, the shaded area is semicircle of diameter 444.
Thats cheating
😊 an ordinary candy 🍬 flavor 😋 👌
Excellent Sir
Very nice and easy method.
Thanks PreMath.
Good job .
❤❤❤❤
R^2 - 444^2 = r^2 by BPO.
Therefore, R^2 - r^2 = 444^2
It looks like that might be the answer, but surely it's more complicated than that. Purple area = 444^2/2 = 197,136/2 un^2.= 98,568 un^2.(division by 2 due to semicircle).
Ok there was slightly more to it as I forgot to multiply by pi. Oops!
Remembering the area of an annulus, (full circle), as
Pi ( R^2 - r^2 )..........(1)
Triangle OPB, R^2 - r^2 = 444^2.
Substituting for R^2 - r^2 = 444^2 in (1).
Area annulus (full circle) = Pi x 444^2 = Pi x 197136.
Then area of purple (half annulus) = Pi x 98568.
309660.5.
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OP = r
02) OA = OB = R
03) AB = 888
04) AP = PB = 444
05) Big Semicircle Area = (Pi * R^2) / 2
06) Medium Semicircle Area = (Pi * r^2) / 2
07) Purple Shaded Area = Big Semicircle Area - Medium Semicircle Area = (Pi * R^2) / 2 - (Pi * r^2) / 2 = (Pi * (R^2 - r^2)) / 2
08) R^2 - r^2 = 444^2
09) R^2 - r^2 = 197.136
10) Purple Shaded Area = Pi * (197.136) / 2
11) Purple Shaded Area = Pi * 98.568
Thus,
OUR BEST ANSWER :
Purple Shaded Area is equal to 98.568Pi Square Units.
What an arduous way to do this problem! Let’s clean this up. First, realize that the smaller semicircle COULD have a radius equal to ZERO. Next, realize that the outer semicircle, by implication, would have a diameter equal to the cord length of 888. The radius would likewise be 444. The task suddenly becomes…
• compute the area of a circle with a radius of 444
• divide the computed area by 2
•the result is then
309,660.5093
(expressed to 4 decimal places)
Let the radius of the outer semicircle O (containing A and B) be R and the radius of the middle semicircle O (containing P) is r. From here on the semicircles will be referred to as outer and middle, and the smaller one will be ognored as it does not figure into the calculations.
The purple area we are looking for is the area of the outer semicircle minus the area of the middle, or πR²/2-πr²/2 = (R²-r²)π/2.
Draw OA, OP, and OB. As AB is tangent to the middle semicircle at P and OP is a radius of the middle semicircle, then ∠OPB = 90°. As AB is a chord of the outer semicircle and both semicircles share O as their center then that means P must be the midpoint of AB, as OP and AB are perpendicular. This AP = PB = 888/2 = 444 and ∆APO and ∆OPB are congruent.
Triangle ∆OPB:
OP² + PB² = OB²
r² + 444² = R²
R² - r² = 444²
The area we are looking for, as established above, is (R²-r²)π/2. Therefore vwr now have all we need.
Purple shaded area:
A = (R²-r²)π/2
A = (444²)π/2
A = 197136π/2
A = 98568π ≈ 309660.50 sq units
My way of solution ▶
The area of the purple shaded ring:
Apurple= r₃²*π/2 - r₂²*π/2
Apurple= π/2(r₃² - r₂²)
Let's consider the right triangle ΔOBP
[PO]= r₂
[OB]= r₃
[BP]= 888/2
[BP]= 444 length units
By applying the Pythagorean theorem for this triangle:
[PO]²+[BP]²= [OB]²
⇒
r₂²+444²= r₃²
r₃²-r₂²= 444²
r₃²-r₂²= 197.136
Apurple= π/2(r₃² - r₂²)
Apurple= π/2*197.136
Apurple= 98.568 π
Apurple ≈ 309.660,5 square units
It looks as though you don't have enough givens at first. However you dont need to work out the absolute values of the radii. Then its quite easy
This seems to be one of the largest areas ever calculated on this channel. 🙂By the way: I used the same method shown in the video.
Best regards from Germany
I operate on a need too know basis. The smallest of the 3 concentric semi circles is more than I need too know. 🙂
OB²-OP²=(AB/2)² → Área de la corona circular =π(OB²-OP²)=π(AB/2)² → Área púrpura =π(888/2)²/2 =π444²/2 =309660,5047..ud².
Gracias y un saludo.
R^2-r^2=444^2...Aviolet=πR^2/2-πr^2/2=(π/2)444^2=309660,5..
(888＊（sqrt(2)-1）)^2＊pi/4/2
Thank you!
The answer is 98568pi units square. Looks like this is an example of geometry being the equivalent of algebraic equivalency. And I really appreciate the theorems cited and I shall use that for practicing rapid fire citing of basic circle theorems so that I can justify the relevant equivalency. I hope that this means that I am paying attention. And I hope that there is a playlist of problems that cite either a circle theorem or basic theorem. Or both.
It's a simple application of Pythagoras and the circle area formula. Pose the problem, define the variables, write down some equations, solve. I wouldn't overcomplicate it.
@@pwmiles56 I agree and that is the fist of that video!!!
He forgot the main theorem, the formula of circular ring area
A = ¼πC²
Easiest solution:
A = ½(¼πC²) = ⅛π888² = 98568π cm² ( Solved √ )
@@marioalb9726 I think that he wanted a way of solving that does NOT require the main theorem. And his presentation is somehow easier to understand. I could be wrong.
@@michaeldoerr5810
Then, He should explain both methods.
He forgot the easier method
Small circle???