Find area of the Blue shaded region | Three circles given | Important Geometry skills explained
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- เผยแพร่เมื่อ 30 ก.ย. 2024
- Learn how to find the area of the Blue shaded region. Important Geometry and Algebra skills are also explained: Similar triangles, area of the circle, Thales's Theorem. Step-by-step tutorial by PreMath.com
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Find area of the Blue shaded region | Three circles given | Important Geometry skills explained
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Good evening sir
Same to you, dear
So nice of you.
Thank you! Cheers! 😀
When you got to ①, couldn't you just use the chord theorem to say that CE•ED = AE•EB? Since AE = EB, it yields the same result of course.
I came here to say this, you beat me to it.
Many approaches are possible to find the solution to this problem!
Thank you! Cheers! 😀
You absolutely could, but PreMath also quickly showed *_why_* that chord theorem is true.
PreMath is solving with the Geometry basics method without using theorems. Well done.
@@madhusudangupta3661 Several theorems are used in this solution.
@8:33 once you got A(blue shaded) = pi/2 (CE)(ED) , you can get (CE)(ED) more directly by using the Intersecting Chord Theorem:
(CE)(ED) = (AE)(EB) = (10)(10) = 100. So that A(blue shaded) = (pi/2) (100) = 50 pi.
The position of line AB is not defined, so can be moved to the right to be the diameter of the large circle equal to 20 or 10 radius.
The yellow and green circles will then be congruent, with diameters equal to10 or 5 radii.
Then area of large circle = Pi x 10^2 = 100 Pi.
Area of the 2 small circles = 2 x Pi x 5^2 = 50 Pi.
Then area of shaded area = 100Pi - 50Pi = 50 Pi.
Excellent!
Thank you! Cheers! 😀
Brilliant
Now suppose this is a trick question and the area of the blue region is not constant🤔
@@ybodoN If the area of the blue region were not constant, then the problem, as formulated, could not be solved at all. That's the point. There are many such problems (in geometry especially) where the absence of certain data (in this case, the exact positioning of AB relative to the center of the larger circle) just "tells" you that the answer is constant within a set of specified parameters. Here, so long as AB=20 and intersects the diameter of the larger circle at a 90 degree angle, the area of the blue region will remain constant. You can safely and reliably infer that from the way the problem is set up. It's a good trick to know when you just need to quickly get to the answer, and it comes up fairly often.
Yes , that's the way I did it. Took me about 20 seconds with no pen and paper 😊
(Chords Theorem)
CE＊ED=AE＊EB=10＊10=100
Blue region :
π/2＊100=50π=157
Excellent!
Thank you! Cheers! 😀
absolutely i also solved your way easily
I solved this with no triangles used in any way. I defined the radius of the big circle in terms of the radii of the small circles, so I could create a formula for the area of the blue region in terms of pi and the two smaller radii. Conveniently, this formula reduced to a single term in which the two smaller radii were multiplied: 2pi*Ry*Rg. I calculated that product using the chord theorem: since AB = 20, then 10 x 10 = 100, so 2Ry × 2Rg also = 100. 4RyRg = 100, so RyRg = 25. Finally, 2piRyRg = 50pi.
I personally started with the intersecting chord theorem, giving me CE.ED = 100 hence xy = 25, where x and y are the radii of the inner circles. I then applied the area formula to all 3 circles and thereby arrived at the answer.
Since the only measurement given is the 20-unit chord, the specific dinensions of the circles may be irrelevant.
Move the vertical line AB to the right, so that it crosses CD at O, changing the various circle sizes as necessary. This makes the diameter of the big circle 20 units, and the two small circles are now identical with diameters of 10 units each.
The area of the big circle is 100 pi and the smaller ones are 25 pi each for a total of 50 pi.
The blue shaded area is the difference, or 100 pi -- 2(25 pi) = 50 pi, the same as your answer.
Carpe Diem! 🤠
Blue Shaded Area:
A = ⅛ π c²
A = ⅛ π 20²
A = 50π cm² ( Solved √ )
Why similar triangles theory . Why not intersecting chords theory . CE.ED = AE.EB (=AE²)
Now I'm closer too figuring out the meaning of crop circles.🙂
Excellent!
Thank you! Cheers! 😀
Sir inspite of construction of lines , we can also use intersecting chords theorem. Anyways the solution is awesome
you could solve it much more it easily using the power theorem on a circle
Important Geometry skill here! Love it.
I also prefer intersecting chords theorem.
I came up with the solution using a data plot. y=radius of yellow circle. g = radius of green circle. b = radius of blue circle. In the diagram 2y + 2g = 2b. dividing both sides of the equation by 2 gives y + g = b. With the intersecting chord theorem, 2y * 2g = 10 * 10. 4yg = 100 and gy = 25. I picked 3 points where g < 25. g = 10 means y = 2.5 and b = 12.5. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. The same with g = 8 which means y = 25/8 and b = 89/8. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. It is exactly the same answer with g = 25/4 which means y = 4 and b = 1681/16. Calculating all the areas and subtracting out the yellow and green gives 50*pi for the answer. I wonder if this ties back into Thales Theorem for a circle and right angles. The answer is consistently 50*pi. 50*pi is the answer.
Blue Shaded Area:
A = ⅛ π c²
A = ⅛ π 20²
A = 50π cm² ( Solved √ )
Easier !!!
A=Ac-(A₁+A₂)
A= ¼πD²-¼π(D₁²+D₂²)
Intersecting chords theorem:
(AB/2)²=D₁.D₂=10²=100
(D₁+D₂)²=D₁²+2D₁D₂+D₂²=D²
D₁²+D₂²=D²-2D₁D₂ = D²-2.100
A= ¼πD² - ¼π(D²-200)
A= ¼πD² - ¼πD² +50.π
A= 50π =157,08cm² ( Solved √ )
You are my boss...
Mahin from Bangladesh..
So nice of you, Mahin dear
Thank you! Cheers! 😀
Thank you so much Proffesor for your sharing
You are very welcome
Here's an interesting observation! AS LONG AS THE CHORD AB = 20 UNITS, then it doesn't matter what the sizes of the circles are - the blue shaded area will always be 50pi square units!
Question: Is that independant how big is CE is, we get always 50Pi? (Only that ED and CD change)
It seems to be a very challenge puzzle.😊First y,g,b be the radii of yellow, green and blue circles, then OA=b, b=y+g, OE=b-2y=g-y, thus 100=AE^2=b^2-(g-y)^2=(g+y)^2-(g-y)^2=4gy, so gy=25. Our answer is (b^2-y^2-g^2)pi=((y+g)^2-y^2-g^2)pi=2gyxpi=2x25pi=50pi=157.1.😃
Excellent!
Thank you! Cheers! 😀
I got snagged trying to figure out OE. Anyone know what that is in terms of the radius of the big circle?
Once you got the equation for area of blue shaded area, could have used intersecting cords theorem.
Is there a way to solve for the radius of the two smaller circles? I can see great responses here but i don't know how to solcve the smaller circle diameters.
Is there some intuitive way to understand that the difference of the area of the outer circle and the two inner circles remains constant as the diameter of the larger inner circle goes from equal to the smaller inner circle to equal to the outer circle?
Bro if u = √(x-√(x-√(x-√...)))
Then
[(u²(u² -2u +1))/2 ]=x²/2
Prof, your proportions on ACD triangle is the second Euclid theorem
moreover I think your videos are 100% truely brilliant
I believe the following solution is simpler: note that R=r1 + r2 (where R , r1, r2 are the radii of the big, small, medium circles).
Then note that OE = R-2r1 = r2-r1
Apply the Pythagorean theorem to OEA : (r1+r2)^2 = (r2-r1)^2 = 10^ 2
The terms of r1^2 and r2^2 on both sided cancel each other, yielding 4 r1 r2 = 100, so r1 r2 = 25
Now, the blue area is Pie (R^2 - r1^2 - r2^2) , and applying R= r1 + r2, again the square
Terms disappear, yielding Area = pie * 2 r1 r2 = 50 pie
Exactly , my solution was the same, too.
Blue Shaded Area:
A = ⅛ π c²
A = ⅛ π 20²
A = 50π cm² ( Solved √ )
Easier !!!
Archimedes' Book of Lemmas, proposition 4
fantastic sum purely based on basic theorems
Very complicated problem. Beautifully explained bringing clarity
Thanks PreMath Guru Ji
See above comments , it's not complicated at all
Thanks for video.Good luck sir!!!!!!!!!
I was nowhere on that one... much tougher than some of the others
At first, we have to make the right construction of the three circles. Then we can recognize that the little circle has a diameter of 46,32, the middle circle has 132,72, and the big one 334,58. For calculating the blue shaded area, we have to make the subtraction of the area of the little circle and the middle circle, from the area of the big circle. That means: 334,58 - (132,72+46,32)= 155,5. The calculation can be fulfilled with the known calculation-type: A=(π•d^2)÷4. The blue shaded area has 155,5 area units.
Let the radii of the yellow and green circles be a and b respectively.
Radius of the largest circle = a+b
Area of blue shaded region = pi(a+b)^2 - pi•a^2 - pi•b^2
=pi•2ab
Using intersecting chord theorem,
2a•2b=20/2•20/2
4ab= 100
2ab=50
Therefore, area of blue region
.=50pi
I have a short method Sir.
Briliant! I just love how you solve these geometry problems. Geometry has always been a favorite subject and I can see the wisdom on why you sometimes take the longer method - so that the student/observer gets to know the applications of the various theorems and methods. Cheers!
So, the formula for the area of the blue region is _⅛ π c²_ where _c_ is the length of the line segment AB.
Congratulations !!!
Blue Shaded Area:
A = ⅛ π c²
A = ⅛ π 20²
A = 50π cm² ( Solved √ )
Easier and Faster !!!
You were the most cunning of all !!!
Strange way to solution.
Thank you!
Very good
S=314
50pi
I used two methods, one specific and one more general.
In both cases, we make use of the information you used in video:
- Centres and points of tangency are collinear, i.e. the centers of all 3 circles lie on CD.
Therefore, CD is diameter of larger circle, and CE and ED are diameters of the smaller circles.
- If a radius of a circle is perpendicular to a chord, then the radius bisects the chord.
Therefore, AE = EB = AB/2 = 10
Specific solution:
Since the position of chord AB seems irrelevant to solution, we assume AB is a diameter of larger circle.
So larger circle has radius = 10, and the two smaller circles have radius = 5
Area of blue region = π(10)² − 2 × π(5)² = 100π − 50π = 50π
General solution:
Let r and R be radius of yellow and green circles respectively:
CE = 2r, ED = 2R → CD = 2r+2R → radius of larger circle = r+R
By intersecting chord theorem, we get:
CE × ED = AE × EB
2r × 2R = 10 × 10
4rR = 100
rR = 25
Area of blue region = Area of larger circle − Areas of the 2 smaller circles
= π(r+R)² − πr² − πR²
= π(r² + 2rR + R²) − πr² − πR²
= (r² + 2rR + R² − r² − R²)π
= (2rR)π [ but rR = 25 from above ]
= 50π
Blue Shaded Area:
A = ⅛ π c²
A = ⅛ π 20²
A = 50π cm² ( Solved √ )
Easier !!!