Can you find area of the Yellow circle? | (Secant) | #math #maths | #geometry

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Can you find area of the Yellow circle? | (Secant) | #math #maths | #geometry

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#FindYellowCircleArea #Rectangle #TangentSecantTheorem #Circle #GeometryMath #PythagoreanTheorem
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[MrPaulc]

Thank you. Tangent secant is new for me.

[jamestalbott]

Thank you! Appreciate the focus on the Tangent Secant Theorem.

[michaelkouzmin]

I used a bit more cumbersome method applying Pythagorean theorem thrice:
1. Let us connect T and C, angle BTC is a right one as it leans on diameter.
2. Let x= AC; y=TC;
3. Tr. ABC:
(27+48)^2 = x^2 +(2r)^2 => x^2 = 75^2 - (2r)^2 (1)
4. Tr. BCT:
(2r)^2 = 48^2 + y^2 ; (2)
5. Tr. ACT:
x^2 = 27^2 + y^2 => y^2 = x^2 - 27^2 (3)
6. Let us put (3) into (2)
(2r)^2 = 48^2 + (x^2 - 27^2) (4)
7. Let us put (1) into 4:
(2r)^2 = 48^2 + (75^2-(2r)^2) - (2r)^2;
(2r)^2 + (2r)^2 = 75^2+48^2-27^2;
2(4r^2) = (75^2+48^2-27^2);
r^2 = (75^2+48^2-27^2)/8 = 900;
8. Ayellow = 900*pi.

[AllmondISP]

Man.I didn't think about the tangent Secant Theorem. It could simplify my solution so much.

[yalchingedikgedik]

Thanks Sir
Thanks PreMath
That’s very nice
We are learning more from your explain
❤❤❤❤❤
Good luck with glades

[devondevon]

900 pi
Let R = the point of tangency on the lower part of the yellow circle, then according to the tangent secant theorem
AB * BT = (AR)^2
75 * 27 =(AR)^2
sqrt 2025 = AR
AR =45
Using Pythag, BR =30
Hence, the radius of yellow = 30

[toninhorosa]

I solved a little different:
From point "A" I drew the diameter of the small circle (parallel to diameter BC) and formed the line AD.
Then I joined point "D" to point "C" and formed 2 ∆s. ∆TBC and ∆TAD. They are similar (3 equal angles)
r = radius of small Circle
R = radius of big Circle.
AT = 27 ( side of small ∆)
TB = 48 (side of big ∆)
Then
27/2r = 48/2R
R/r= 96/54 = 16/9
16r = 9R
r = 9R/16
We join point "P" to point "Q" with a straight line, forming the line PQ = r + R.
From point "P" we draw a parallel to line AC until we meet line BC and mark point "E".
We now have ∆PEQ.
PQ = r + R = 9R/16 + 16R/16
PQ = 25R/16
QE = R - r= 16R/16 - 9R/16
QE = 7R/16
Applying Pythagoras:
(25R/16)^2=(7R/16)^2+ PE^2
PE^2 =(625R^2+49R^2)/256
PE^2 = 576R^2/256
PD = 24R/16 = (3/2)R
PD = AC = (3/2)R
Now: ∆ ABC
Applying Pythagoras:
AB^2 = BC^2 + AC^2
75^2 = (2R)^2 + (3R/2)^2
5625 = 4R^2 + (9/4)R^2
22500 = 16R^2 + 9R^2
25R^2 = 22500
R^2 = 22500/25
R^2 = 900 and R = 30
Area of yellow Circle=
900π units^2

[cyruschang]

R ÷ r = 48/27 = 16/9
r = 9R/16
(R + r)^2 - (R - r)^2 + (2R)^2 = (48 + 27)^2
(25R/16)^2 - (7R/16)^2 + (2R)^2 = (48 + 27)^2
(25^2 - 7^2 + 32^2)(1/16^2)(R^2) = (75)^2
R^2 = (75^2)(16^2)/(25^2 - 7^2 + 32^2) = (75^2)(16^2)/1600 = (75^2)(16)/100 = (75^2)(4)/25 = 75(3)(4) = 300(3) = 900
area = π(R^2) = 900π

[makermaker]

Sir can you make visualozation of these circlw proofs tangent secant and more pls

[awcampbell]

Upon recognizing the hypotenuse (line segment AB) is 75 and divisible by 5 this becomes a special triangle (3, 4, 5). The duameter of the yellow circle is 4 * 15 or 60. This makes the radius 30 and the area (30^2)π or 900π.

However the secant is new to me and i appreciate seeing it here

[santiagoarosam]

27*(27+48)=d²=2025---> (27+48)²-d²=(2r)²---> r²=900---> Área amarilla =900π.
Gracias y saludos.

[wackojacko]

Too solve for this problem one needs to know what the internal and external segment of a secant and the Intersecting tangent segment are. Simple as pi ...🙂

[marcgriselhubert]

Fine.
PA^2 = PB.PC is called the power of P for the circle. Old geometry, probably no longer much studied.

[michaelavishay]

16r=9R 2*srR*r=L sr=squre root 2R^2+L^2=75^2

[giuseppemalaguti]

(r+R)^2=(R-r)^2+((27+48)^2-(2R)^2)...4rR=75^2-4R^2..13, 5=rcosθ..24=Rcosθ...13, 5/r=24/R..R/r=24/13, 5=8/4, 5=16/9...9R=16r...r=9R/16...la metto nella 1 equazione, risulta

[phungpham]

Alternative solution:
The angle BTC= 90 degrees so:
Focus on the right triangle ABC:
sq CT=27x48
—> sqBC=sq48+sqCT=sq 48+ 27x48=3600
Area of the yellow circle= 3600/4 xpi= 900 pi sq units😅

[juanalfaro]

I solved the problem by the same method and obtained same answer of 900π sq. units.

[devondevon]

radius = 30, Hence, area 900 pi square

[jimlocke]

Note that the small circle, the vertical sides of the rectangle, and the points P and Q are not needed to solve the problem, as long as the horizontal sides remain and are shown as being parallel. The solution method presented may be more obvious with these items removed!

[jarikosonen]

How can I check that the point T is actually the point of contact between these two circles?