Pleasant Challenge. Its how simple it becomes after the explanation .😊👍

@PreMath

Ай бұрын

Glad you liked it Thanks ❤️

El resultado final será el mismo si desplazamos el círculo pequeño hasta que Q coincida con O→ En la figura resultante: r²+(1/2)²=R²→ R²-r²=1/4. El área amarilla es la de la corona circular resultante de esa transformación → Área amarilla =πR²-πr² =π(R²-r²) =π(1/4) = π/4. Gracias y un saludo cordial.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Extremely easy: A = ¼ π c² A = ¼ π 1² A = π/4 cm² ( Solved√ ) Yellow area doesn't depend on the center position of white circle. We can apply the formula for circular ring area, with respect to the chord.

@LuisdeBritoCamacho

Ай бұрын

Faster than the Speed of Light! Thanks for Sharing your knowledge!!

If we move the white circle so that it shares its center with the yellow circle, we can draw a bicentric square of side 1 🤓 Therefore, the diameter of the white circle is 1 and the diameter of the yellow circle is √2. So, their area is in a 1:2 ratio. As we will subtract one circle from the other to get the area of the ring, the shaded/unshaded regions are in a 1:1 ratio. Since the area of a circle is ¼ π d², the area of the withe circle is ¼ π 1² = ¼ π = the area of the yellow region. *Thank you PreMath!* 🙏

@Ddntitmattrwhtuthnk

Ай бұрын

I politely disagree. How do you get πr^2=1/4πd^2? Plus, There are no restrictions on the diameter of either circle. Example: If the larger circle was R=4, the math for the circle difference in area is still π/4 square units, but the radius of the smaller circle is 16 - 1/4 = r^2, r=√63/4. In real life applications, AutoCAD, or Solid Works, any drafting software, etc... the circles would just change size together with the only constraint being the 1 unit on the chord, which ties the circles by the coincident radius of the smaller.

@ybodoN

Ай бұрын

@@Ddntitmattrwhtuthnk To answer your first question: ¼ π d² = ¼ π (2r)² = ¼ π 4r² = π r² where _r_ is the the radius and _d_ is the diameter of a circle. To answer your second question: since there is no restriction on the diameters, assuming there is only one valid answer, we can then choose a special case that is very easy to solve. Of course, we must also assume that OQ and AB are parallel. Obviously, this information is missing from the problem statement.

Yellow region area=πR^2-πr^2=π(R^2-r^2) Chord theorem (1/2)(1/2)=(R+r)(R-r) So R^2-r^2=1/4 Yellow region area=π/4=0.79 square units.❤❤❤ Thanks sir.

Ay=π(R^2-r^2)...R^2=r^2+(1/2)^2...Ay=π/4

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Amazing!

I looked at it and said pi/4. Just shrink the diameter of the small circle to zero in your head and 1 becomes the diameter of a yellow circle (r =1/2).

Draw a diameter through O and through the midpoint of AB. As AB is a chord, a radius through its midpoint is perpendicular to it, and thus parallel to QP, since P is a point of tangemcy between the chord and circle Q. Let QP = x. By Intersecting chords theorem, the products of the segments on either side of the intersection point of two chords are equal for the two chords. (r+x)(r-x) = (1/2)(1/2) r² - x² = 1/4 x² = r² - 1/4 Yellow area: A = πr² - πx² A = πr² - π(r²-1/4) A = π/4

Thank you!

Label the radius of ⊙O as R and the radius of ⊙Q as r. Yellow Region Area = ⊙O Area - ⊙Q Area A = πR² A = πr² Yellow Region Area = πR² - πr² = π(R² - r²) = π(R + r)(R - r) Draw radius PQ. PQ = r. By the Circle Theorem, radius PQ is perpendicular to tangent segment AB. But segment AB is also a chord of ⊙O. Draw a diameter perpendicular to chord AB. By the Perpendicular Chord Bisector Theorem, this diameter bisects chord AB. Apply a translation on ⊙Q to map center Q onto center O. P' = R. Then, OR = r & AR = BR = 1/2. Use the Intersecting Chords Theorem. (R + r)(R - r) = (1/2)(1/2) R² - r² = 1/4 Input into the formula. Yellow Region Area = π(R² - r²) = π(1/4) = π/4 So, the area of the yellow shaded region is π/4 square units (exact), or about 0.79 square units (approximation).

Muito criativo!

@PreMath

Ай бұрын

Excellent! Thanks dear ❤️

Ok. First things, first! 1) Make Point O and Point Q to coincide. Now we have two Concentric Circles: The Outer Circle and the Inner Circle. It's the very same! 2) Now, Point P (vertically raised) is the Middle Point between A and B. 3) The Big Radius, R, is equal to R = OP + PP' ; name the Lines as follow : OP = A and PP' = B ; Conclude that : R = A + B 4) By the Law of Chords, one can easily see that AP * BP = OP * PP' ; 0,5 * 0,5 = OP * OP' ; in other words : A * B = 1/4 Now we have a relationship between A and B : B = 1/4A or A = 1/4B, and (A + B) = R; being OP = r or A = r (Small Radius) 5) R^2 = A^2 + (1/2)^2 ; (A + B)^2 = A^2 + (1/2)^2 ; (A + 1/4A)^2 = A^2 + 1/4 ; A^2 + 1/(16A^2) = A^2 + 1/4 ; 1/(16A^2) = 1/4 ; (16A^2)/4 = 1 ; 4A^2 = 1 ; A^2 = 1/4 ; A = sqrt(1/4) ; A = 1/2 6) So : A = 1/2 7) OA^2 = (1/2)^2 + (1/2)^2 ; R^2 = 1/4 + 1/4 ; R^2 = 1/2 ; R = sqrt(1/2) ; R = sqrt(2)/2 ; R ~ 0,7071 8) R = sqrt(2)/2 and r = 1/2 9) BCA = Pi/2 and SCA = Pi/4 10) Yellow Area = BCA - SCA ; Yellow Area = Pi/2 - Pi/4 ; YA = 2Pi/4 - Pi/4 ; YA = Pi/4 sq un 11) Answer: The Yellow Region Area is equal to Pi/4 Square Units or approx. to 0,785 Square Units.

AO = BO = r; AB = 1; CO = a; AC = √(r^2 - a^2) → BC = 1 - √(r^2 - a^2) → a^2 = r^2 - (1 - √(r^2 - a^2))^2 → √(r^2 - a^2 ) = 1/2 → yellow region area = π/4

👏👏👏👏👏

You didn't say the lines are parallel when explaining the task. They don't have to be. If they are then move Q to O (doesn't change the areas). You get a right triangle with sides R, 1/2 and r, which makes R²-r² = (1/2)² and the yellow area is (R²-r²)­ Pi . So we get (1/2)² ­­­Pi without calculating anything. Are you wondering how I make those ²'s? Hold Alt and type 253 on numpad. 252 for ³.

@LuisdeBritoCamacho

Ай бұрын

That's an Implicit Information that you should know. If the Point P is the Point of Tangency of Line AB to the Inner Circle, it means that Line AB is perpendicular to Line QP. And: 1) If : Line AB is Tangent with Line QP, and 2) If : Line QP is Parallel to Line OC, 3) Then : Line OC is Perpendicular to Line AB. 4) Moreover Distance QP is equal to Distance OP. Best Regards from Portugal.

@ybodoN

Ай бұрын

On a French keyboard, the key just above the tab key is dedicated to "²" and "³" (superscript "2" and "3") 😉

@szkoclaw

Ай бұрын

@@LuisdeBritoCamachodraw the point P anywhere else on the circle and it’s no longer parallel.

@LuisdeBritoCamacho

Ай бұрын

@@szkoclaw, ok but this problem is not only about what is stated, it have a draw showing clearly that Line AB // Line OQ.

@szkoclaw

Ай бұрын

​@@LuisdeBritoCamacho Doesn't matter what it looks like, if it's not stated in the description of the problem, it's not parallel unless you can prove it is from given data. He even said "the picture may not be true to the scale".

Pretend that the white circle has a radius of 0. -> The diameter of the large circle becomes 1, its radius 0.5 and therefor its area 0.25 pi or pi/4.

What I got was the formula for a chord length 1=2√(r²-d²) That means that 1/4=(r²-d²) The yellow circle is πr², the white is πd², so the difference is π(r²-d²) Substitute ¼ for (r²-d²) and get π/4

Solution: r = radius of small circle = QP R = Radius of the great circle = OA = OB Pythagoras: (AB/2)²+r² = R² ⟹ 1/4+r² = R² ⟹ Yellow region = π*R²-π*r² = π*(R²-r²) = π*(1/4+r²-r²) = π/4 ≈ 0.7854

Mind blowing ❤ Love from Pakistan

@PreMath

Ай бұрын

Glad to hear that! Thanks dear ❤️ Love and prayers from the USA! 😀

With pythagorean th. R² - r² =(1/2)²

Obviously, this can work only when line OQ is parallel to the chord, because other chords tangent to the smaller circle but not drawn parallel to OQ will have lengths different from 1, so the value of (R^2 - r^2) will vary. Still, it's a very interesting problem. Thanks!

Let's find the area: . .. ... .... ..... The yellow area corresponds to the difference of the areas of the big circle and the small circle inside the big circle. Therefore the size of the yellow area does not depend on the location of the small circle as long as it is fully inside the big circle. So we can shift the small circle such that the centers of both circles coincide: Q=O. In this case P becomes the midpoint of AB and the triangles OPA and OPB become right triangles, so we can apply the Pythagorean theorem. With R being the radius of the big circle and r being the radius of the small circle we obtain: OA² = OP² + AP² OA² = OP² + (AB/2)² OA² − OP² = (AB/2)² R² − r² = (AB/2)² = (1/2)² = 1/4 Finally we can calculate the size of the yellow area: A(yellow) = A(big circle) − A(small circle) = πR² − πr² = π(R² − r²) = π/4 Best regards from Germany

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Simpler way is (R^2-r^2)pi=(1/2)^2pi=pi/4.😊

1) It doesn't matter where the position of the small circle is: let's move it to the middle of the bigger circle. 2) Chord Theorem: 1/2 * 1/2 (the secant) = (R + r) (R - r) 3) 1/4 = R² - r² 4) A(yellow) = A(big circle) - A(small circle) = R²π - r²π = (R² - r²) π A(yellow) = 1/4 π ≈ 0.785 square units P.S. Your solution is nice, too! Thanks for sharing!

@PreMath

Ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

I would say that it should be stated as a given that the cord of length 1 is parallel to the diameter that passes through the centers of the two circles . I don't think it is a good idea to assume this just because it looks that way in the diagram .

😉

@PreMath

Ай бұрын

Thanks ❤️

Assumption that OC is r is incorrect. It is possible that OC is r only if AB || OQ. But it is not mentioned like that.

@alphacentauri7757

Ай бұрын

I am agree with you

@LuisdeBritoCamacho

Ай бұрын

It's not an Assumption, it's a Mathematical Fact!! The Perpendicular Distance from a Diameter parallel to a Chord is always the same!! Being the Domain the Length of the Chord. Plane and Simple. Nothing is incorrect!!

@laxmikantbondre338

Ай бұрын

@@LuisdeBritoCamacho no it is incorrect. you can draw line AB as tangent but little tilted. I.e. from some other point of tangency

@laxmikantbondre338

Ай бұрын

​@@LuisdeBritoCamacho So in simple words It is not mentioned that line AB is parallel to OQ. It is appearing parallel. It is just mentioned that line AB is a tangent to smaller Circle. So if you move point p little upwards or downwards on the smaller circle's circumference and draw line AB tangent from that point. Will it be parallel. The perpendicular distances between two lines are same if and only if they are parallel.

@laxmikantbondre338

Ай бұрын

So it is impossible to solve this problem with given data unless it is mentioned that line AB is parallel to OQ. OR mention that angle OQP is 90.

I thought a line can only be tangent to a circle at one point at a time. In thus case line AB appears to be tangent at 2 points C and P. Can someone explain?

@waheisel

Ай бұрын

You are right. C is on AB but not (quite) on the circle. So C is not a second point of tangency.

OK, no difficulty.

Aren't O and Q always colinear?

@MegaSuperEnrique

Ай бұрын

Did you mean OQ and AB are parallel lines?