Can you find area of the Yellow shaded region? | (Circles) |
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- เผยแพร่เมื่อ 3 ธ.ค. 2024
- Learn how to find the area of the Yellow shaded region. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the circle formula; area of the rectangle formula; circle theorem. Step-by-step tutorial by PreMath.com
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Geometry was baffling for me until I reached the "aha!" moment when the logic train pulled into the station. Now I love geometry.
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Stay blessed 😀
We may get an equilateral 🔺 on joining the centres of the circles and side of the triangle will be 2 units.
Height of triangle is √3*2/2=√3 units
The breadth and length of the rectangle will be
1+1+√3=(2+√3) units
and 4 units
Area of rectangle
=4(2+√3) sq units
Area of yellow portion
=[4(2+√3) - 3π] sq units
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Thanks. Easy.
A₁ = 3π cm²
A₂ = b.h = 4R.(2R+2Rcos30°)
A₂= 14,928 cm²
A= A₂ - A₁ = 14,928 - 3π = 5,5 cm² ( Solved √ )
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r = 1
rectangle length = 4r
rectangle width = (2 + √3)r
area = 4(2 + √3)r^2 - 3π = 4(2 + √3) - 3π
This one was so easy that I had the problem figured out in my head in 5 to 10 seconds. This time, I used the sqrt of three rather than figuring the decimal form of that length of a 30-60 right triangle. I prefer the convertion to decimals rather than leaving the answer in terms of square roots or PI
r*r*π=π r=1
√[2^2-1^2]=√3
Yellow shaded area
= 4(2+√3) - 3π = 8 + 4√3 - 3π
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This is an example of easier than it looks. I must apologize for missing out on PreMath videos. Looks like it is a matter of mentally justifying what geometric observation can be cited as long as that observation respects the fact that nothing is drawn to scale. That requires either memorization or practice. Also I followed along and I almost got every step correct. The yellow shaded area is 8+4*sqrt(3)-3pi units square.
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Got it after obtaining the rectangle's sides
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Sir, in solving the problem u got ur favourite special type of 30 -60-90 triangle.
We may use the knowledge that side opposite to 60 degree angle is √3 unit.
The radius of each circle is R =1, so EF = 4 and it is the big side length of the rectangle ABCD. EFG is an equilateral triangle whose side length is 4, so its height is 4.(sqrt(3)/2) = 2.sqrt(3) and the little side length of the rectangle ABCD is this height plus R, so it is 2.sqrt(3) + 1.
Finally the area of the rectangle ABCD is 4.(2.sqrt(3) + 1) = 8.sqrt(3) + 4, and the yellow area is 8.sqrt(3) + 4 - 3.Pi. (That was easy.)
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r=1---> AB=4r=4 ; AD=r+r√3+r =2+√3 ---> Área sombreada =4(2+√3) -3π =5,5034...
Gracias y saludos
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Note that ABCD is a rectangle and NOT specifically a square.
A = πr²
π = πr²
r² = 1
r = 1
All three circles are known to be identical and therefore each one has a radius of 1.
Connect centers O, P, & Q together. We get an equilateral △OPQ with side length 2.
Label the point of tangency between ⊙O & ⊙P as R. Connect this point to center Q. Segment QR is an altitude of △OPQ. This forms right △ORQ & △PRQ. They are both special 30°-60°-90° right triangles.
b = a√3
QR = 1 * √3
= √3
Draw radii EO & FP. By the Tangent Line to Circle Theorem, the angles these radii form are right.
We get a rectangle ABFE. If we use the radius length, we know that EF = 4.
Therefore, by the Parallelogram Opposite Sides Theorem, AB = 4.
But there is also a point of tangency between ⊙O and side AB. Draw the radius. By the Tangent Line to Circle Theorem, it should also form right angles.
Label the point of tangency as H. Also draw radius GQ.
So, GQ = HO = 1.
So, AD = 2 + √3. This sounds bold, but ∠EOH is a right angle by the Polygon Interior Angle Sum Theorem. ∠ORQ is already a right angle.
Yellow region area = ABCD Area - Combined Area of Three Circles
Combined Area = π + π + π = 3π
A = lw
= 4 * (2 + √3)
= 8 + 4√3
Yellow region area = (8 + 4√3) - 3π
= 8 - 3π + 4√3
So, the area of the yellow shaded region is 8 - 3π + 4√3 square units (exact), or about 5.50 square units (approximation).
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S=8+4√3-3π≈5,508
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The radii are all 1 and the total area of circles is 3pi.
Width of rectangle is 4 due to 4r.
Yellow area is 4y - 3pi, where y is the height of the rectangle.
First task: calculate y
Make a point S so that S is to the left of Q and below O, forming right triangle SQO.
QO = 2 as it is 2r and is the hypotenuse.
S is actually 1 from the left of the rectangle, meaning that QS = 1.
2^2 - 1^2 = 3, so QS = sqrt(3).
Yellow area = (4*(2+sqrt(3)) - 3pi un^2
Simplify to 8 + 4*sqrt(3) - 3pi.
14.9282 - 9.4248 = 5.5. This value is rounded because the final value is the result of two irrationals, sqrt(3) and pi.
Yes, I used pretty much the same method but with points in different places.
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Let the square be named ABCD. Each side is 4R. So the area is 4R^2.But the area of the three circles is 3πR^2.So Yellow shaped figure ara is R^2(16-3π).
Instant answer is 4×(2+sqrt(3))-3pi.😊
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It was easy to work this out in my head, but I still needed a calculator to get a numerical approximation. I could have done it on paper, of course, but who has paper to hand these days? 😊
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Yellow shaded area= 4(2+√3)-3π=5.50 square units.❤❤❤
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Im so happy being able to solve Premath, I'm gonna go sell popcorn at the carnival. 😊
I'm just grateful to the guy for helping to keep my ageing brain working: that becomes more important with the passing of the years.
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14.93- 3pi or 5.51
Answer 5.5034
Since the area of each is pi, then the radius = 1 (r= sqrt (pie/pie) = sqrt 1 =1)
Using OPQ, draw an equilateral triangle. Its sides are 2, 2, 2
Its hypotenuse = 2 and its given base = 1
Hence, the length of the other base (Pythoagrean) = 1.73205
Hence, the rectangle's width is the length of the two radii (2) + 1.73205 = 3.73205
The length of the rectangle is four radii (4)
Hence, the area of the rectangle = 4 * 3.73205 = 14.9282
Since the area of the three circles = 3 pi (given) or 9.4248, then
the area of the shaded region = 14.9282 - 9.4248 = 5.5034
R= 1 -> EF=4
The width of the rectangle= 1+sqrt3+1=2+sqrt3
Area of the yellow region=4x(2+sqrt3)-3pi=5.50 sq unnits😅😅😅
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Seeing the diagram i thought that it was a square and it messed up my result but i checked the intro again and it was a rectangle thus i tried again and found the correct answer!
No worries😀
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It is a square! ... only special. 😊
{{90°A+90°B+90°C+90°D+}=360°ABCD/27pi=10.10pi 5^5.5^5pi 2^3^2^3.2^3^2^3pi 1^1^1^1.2^1^1^3pi 23pi (x ➖ 3pix+2).
Thanks for the feedback ❤️
Alien math! Nobody explains this! 🤣🤣🤣 This guy is on all the math channels, spreading his Mars math!
Let's find the area:
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..
...
....
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First of all we calculate the radius r of the white circles:
A = πr²
π = πr²
1 = r²
⇒ r = 1
Each pair of circles has exactly one point of intersection. Therefore any distance between two of the three centers is equal to the sum of the corresponding radii:
OP = OQ = PQ = 2r = 2
So we can conclude that OPQ is an equilateral triangle. The height h of such a triangle can be calculated by applying the Pythagorean theorem:
h² = 2² − (2/2)² = 2² − 1² = 4 − 1 = 3 ⇒ h = √3
Now we can calculate the side lengths of the rectangle ABCD:
AB = OE + OP + PF = r + 2r + r = 4r = 4
AD = AE + h + QG = r + √3 + r = 2r + √3 = 2 + √3
Finally we are able to calculate the area of the yellow region:
A(yellow) = A(ABCD) − 3*A(circle) = AB*AD − 3π = 4*(2 + √3) − 3π = 8 + 4√3 − 3π ≈ 5.503
Best regards from Germany
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My way of solution ▶
Let's calculate the radius of these same circles :
A= πr²
π= πr²
r= 1 length unit
[AB]= [EF]
[EF]= 4r
⇒
[AB]= 4 length units
step-2) Let's calculate the length [AD]
By considering the equilateral triangle ΔOQP :
[OQ]= [QP]= [PQ]= a
a= 2r
a= 2
The height of this triangle divides the length [PO] into two equal parts; therefore, by applying the Pythagorean theorem, we can write :
(a/2)²+h²= a²
a= 2
⇒
1+h²= 2²
h= √3
[AD] = [AE] + h + [QG]
[AE] = r
[QG] = r
h= √3
⇒
[AD]= 1+1+√3
[AD]= 2+√3
step-3) Let's calculate the yellow shaded region :
A(ABCD)= [AB]*[AD]
A(ABCD)= 4*(2+√3)
A(ABCD)= 8+ 4√3
Ayellow= 8+ 4√3 - 3π
Ayellow≈ 5,50 square units
80 /7 (16 -3π)..... May be
Thanks for the feedback ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB = EF = DC = 4 lin un
02) Let M be the Middle Point between Point E and Poit F.
03) QM^2 + MP^2 = PQ^2
04) QM^2 = PQ^2 - MP^2
05) QM^2 = 4 - 1 ; QM^2 = 3
06) QM = sqrt(3)
07) AD = BC = [2 + sqrt(3)] lin un
08) Rectangle [ABCD] Area (RA) = 4 * 2 + 4 * sqrt(3) ; RA = 4 * (2 + sqrt(3)) sq un ; RA ~ 15 sq un
09) Yellow Shaded Area (YSA) = RA - 3Pi
10) YSA = [4 * (2 + sqrt(3)) - 3Pi] sq un
11) 3Pi ~ 9,425 sq un
11) YSA ~ 5,5 sq un
Thus,
OUR BEST ANSWER :
The Yellow Shaded Area equal to approx. 5,5 Square Units.
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It cracks me up how so many of the commenters here act like they're Matt Damon in Good Will Hunting. I'm just your standard idiot that enjoys watching intelligent logic.
I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility!😀
We are all lifelong learners!
Thanks for the feedback ❤️
That's not how I'd have solved it. I think my way is easier. For all I know may be a new way as I have iq 193. This is all new to me. I'm in college at 47 learning Calc. I'm finding my thinking is needed in math. I'm discovering how I visualize things are easier than how it's done. Lol