Can you find area of the Yellow shaded Circle?

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Can you find area of the Yellow shaded Circle? | (Quarter Circle) |

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Learn how to find the area of the Yellow shaded Circle. Important Geometry skills are also explained: Pythagorean Theorem; circle area formula; tangent secant theorem; circle theorem; Perpendicular bisector theorem. Step-by-step tutorial by PreMath.com
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Пікірлер: 41

@user-ie1ey1rv2r Ай бұрын

Fantastique

@PreMath Ай бұрын

Merci chéri❤️

@thinker821 Ай бұрын

We can use intersecting secants theorem for secants to the yellow circle originating from O: 2*(2 + r - 2 + r - 2) = 14*(14 - 2r) which gives r = 25/4 readily.

@angeluomo Ай бұрын

The simplest solution. Well done.

@phungpham1725 Ай бұрын

I did it the same way!😊

@PreMath Ай бұрын

Thanks for the feedback ❤️

@unknownidentity2846 Ай бұрын

Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the quarter circle: A = πR²/4 49π = πR²/4 4*49 = R² ⇒ R = √(4*49) = 2*7 = 14 The circumference of the quarter circle and the circumference of the yellow circle have exactly one point of intersection. Therefore this point and the centers of the two circles are located on the same line. So with r being the radius of the yellow circle we obtain: OC + BC = OB OC + r = R ⇒ OC = R − r A is a point of tangency. Therefore AC is parallel to OQ and the triangle OAC is a right triangle. By applying the Pythagorean theorem we obtain: OA² + AC² = OC² OA² + r² = (R − r)² ⇒ OA² = (R − r)² − r² Now let's add point R such that AR is the diameter of the yellow circle. According to the theorem of Thales the triangle APR is a right triangle. Let's also add point S such that PS is the height of the triangle APR according to its base AR. Since APR is a right triangle, we can apply the right triangle altitude theorem: AS*RS = PS² AS*(AR − AS) = PS² OP*(2*r − OP) = OA² OP*(2*r − OP) = (R − r)² − r² 2*(2*r − 2) = (14 − r)² − r² 4*r − 4 = 196 − 28*r + r² − r² 32*r = 200 ⇒ r = 200/32 = 25/4 Now we are able to calculate the area of the yellow circle: A(yellow circle) = πr² = (625/16)π ≈ 122.72 Best regards from Germany

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@angeluomo Ай бұрын

After figuring out that the radius of the quarter circle is 14, you can solve for the radius of the yellow circle with 2 right triangles: 1) OA^2+r^2=(14-r)^2 and 2) OA^2+(r-2)^2=r^2. With triangle 1, you get OA^2=196-28r. With triangle 2, you get OA^2= 4r-4. Setting these two equations equal to each other results in: 4r-4 = 196-28r. Solving for r, we get r= 25/4.

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@jamestalbott4499 Ай бұрын

Thank you! Good introduction to the tangent secant theorem.

@PreMath Ай бұрын

Glad you enjoyed it! Thanks for the feedback ❤️

@PrithwirajSen-nj6qq Ай бұрын

Join OCB( these points are collinear according to tangent / kissing circle property) . Suppose it cuts yellow circle at point R . In respect of yellow circle OB & OQ are secants. Hence by intersecting secants theorem we may get OP*OQ = OR * OB---(1) OP=2 (given) OQ = 2+2(r -2) OR= R (radius of the quarter circle) = 14 OR= OB - BOR =14 -2r Hence from equation (1) we may write 2[2+2(r - 2) = (14 - 2r) 14 > 4r - 4 = 196 - 28r > r = 200/32=25/4 Area of yellow circle = 22/7*25/4*25/4 =122 . 77 sq units (approx.)

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@kd._.star. Ай бұрын

intresting

@PreMath Ай бұрын

Glad to hear that! Thanks for the feedback ❤️

@PrithwirajSen-nj6qq Ай бұрын

In triangle OCA Hypotenuse (OC) =14 - r (circles touches internally and distance between the centres is difference of their radii) (14 - r) ^2 = r^2 + [2√(r - 1)]^2 >196 - 28r +r^2 =r^2+4r -4 > r = 200/32

@PreMath Ай бұрын

Thanks for the feedback ❤️

@murdock5537 Ай бұрын

This is really awesome,many thanks, Sir, very challenging 🙂 fast lane: AO = 2√(r - 1); BO = 14 = BT + TO = 2r + (14 - 2r) → 4(r - 1) = 196 - 28r → r = 25/4 btw: AC = AN + = r + (r - 2) → CPN = α → sin⁡(α) = 17/25; COA = δ → sin⁡(δ) = 25/31

@PreMath Ай бұрын

You are very welcome! Thanks for the feedback ❤️

@sergioaiex3966 Ай бұрын

Nice ... Congrats

@PreMath Ай бұрын

Glad to hear that! Thanks for the feedback ❤️

@quigonkenny Ай бұрын

Let R be the radius of quarter circle O and r be the radius of circle C. Quarter circle O: Aₒ = πR²/4 49π = πR²/4 R² = 49π(4/π) = 196 R = √196 = 14 Draw radius OB. When two circles are tangent to each other their point of tangency is collinear with their centers. As B is the point of tangency between circle C and quarter circle O, then C is collinear with OB. As CB = r and OB = R = 14, OC = 14-r. Draw CA. As CA is a radius and OA is tangent to circle C at A, ∠OAC = 90°. Triangle ∆OAC: OA² + CA² = OC² OA² + r² = (14-r)² OA² = 196 - 28r + r² - r² OA = √(196-28r) Draw CM, where M is the point on PQ where CM is perpendicular to PQ. As PQ is a chord of circle C, CM bisects PQ, so PM = MQ. As ∠CMP = ∠MOA = ∠OAC = 90°, then ∠ACM = 90° and OACM is a rectangle. As CA = OM = r and OP = 2, PM = r-2. Triangle ∆CMP: CM² + PM² = CP² (196-28r) + (r-2)² = r² 196 - 28r + r² - 4r + 4 = r² 32r = 200 r = 200/32 = 25/4 Circle C: Aᴄ = πr² = π(25/4)² = 625π/16 ≈ 122.72 sq units

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@giuseppemalaguti435 Ай бұрын

OA=a..raggio(big)=14...le equazioni sono R+√(R^2+a^2)=14..R=2+√(R^2-a^2)... risulta a^2=4R-4..e la sostituisco nella prima..32R=200...R=25/4

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@marcgriselhubert3915 Ай бұрын

Let's use an orthonormal center O, first axis (OA) The equation of the yellow circle is: x^2 + y^2 +a.x + b.y +c = 0 P(0; 2) is on this circle, so 4 + 2.b + c = 0 and c = -2.b -4 The equation is now x^2 + y^2 +a.x +b.y -2.b -4 = 0 At the intersection with (OA) (which equation is y = 0) we have: x^2 + a.x -2.b -4 = 0. This equation must have a double solution (point of tangency) so its delta is 0 Delta = a^2 +4.(2.b +4) = a^2 + 8.b +16 = 0 So we have b = (-(a^2)/8) -2 Now the equation of the yellow circle is: x^2 + y^2 +a.x - ((a^2 +16)/8).y +(a^2)/4 = 0 or (x +a/2)^2 + (y - (a^2 +16)/16)^2 = ((a^2 + 16)/16)^2 The radius of the yellox circle is then r = (a^2 +16)/16 (*) The center C of this circle has for abscissa -a/2, so OA = -a/2 and OC^2 = ((a^2) /4 ) + r^2 in OAC Then OC = sqrt((a^2)/4) + r^2) OC + CB = radius of the big quater circle = 14 So we have:sqrt((a^2)/4 +r^2) + r =14 (a^2)/4 + r^2 = (14 - r)^2. , then (a^2)/4 = -28.(r^2) + 196 We replace (a^2)/4 by 4.(r^2) -4 (by *) and have: 32.r = 200 and r = 200/32 = 25/4 The yellow area is Pi.((25/4)^2) = (625/16).Pi Another method to find r at the end: When the coordinates of C are known, we can write the equation of (OC), find the intersection B with the big quater circle and then r = CB. (This problem was not so evident.)

@PreMath Ай бұрын

Excellent! Thanks for the feedback ❤️

@professorrogeriocesar Ай бұрын

Essa questão foi maravilhosa!

@PreMath Ай бұрын

Gracias, querido❤️

@christopherellis2663 Ай бұрын

r²=x²+(r-2)²

@PreMath Ай бұрын

Thanks for the feedback ❤️

@misterenter-iz7rz Ай бұрын

14 is the radius of the quarter circle, ....it is difficult puzzle😅

@PreMath Ай бұрын

Thanks for the feedback ❤️

@WernHerr Ай бұрын

Sorry, no idea: Why are O, C and B collinear, particulary O ??

@MegaSuperEnrique Ай бұрын

They are both tangent at point B. So by def, the line from their center to B is perpendicular (90 angle) to the tangent line. If they both have a perpendicular line at B going to their center, then the lines are colinear, and OCB is a straight line.

@WernHerr Ай бұрын

@@MegaSuperEnrique Thank you very much, I had an obvious mistake in thinking