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

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

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[yalchingedikgedik]

Thanks Sir
Thanks PreMath
That’s very nice
Wonderful method for solve
I like that
❤❤❤❤
With my respects

[misterenter-izrz]

Seems to be odd and difficult, but actually simple, (r-8)^2+(r-16)^2=r^2, r^2-48r+320=(r-8)(r-40)=0, r=8, impossible, r=40, therefore the answer is 1600/4 pi=400pi.😊

[Stuv]

Formula for a circle
X squared + Y squared = r^2
(r-8)^2+(r-16)^2=r^2
r^2-16r+64 +r^2-32r+256=r^2
r^2-48r+320=0
(r- 40 )(r-8)=0
r=8 or 40
8 does not work
r=40
Yellow area = (pi x 40^2)/4
Answer =400pi
Approx. 1256.637 square units

[ОльгаСоломашенко-ьы]

Продлим отрезок AO до диаметра, а отрезок CP до хорды. ( r-8)*(r-8)=16*(2r -16) по теореме об отрезках пересекающихся хорд. Решим квадратное уравнение и выберем подходящий корень. r=40.

[alscents]

Choose E as origin
Equation of the circle is
(x - r)^2 + (y + r)^2 = r^2
and
(8, -16) and (16, -8) are points on the circle

(8 - r)^ + (-16 + r)^2 = r^2
and
(16 - r)^2 + (-8 + r)^2 = r^2
which are both the same equation

64 + r^2 - 16 r + 256 + r^2 - 32 r = r^2
r^2 - 48 r + 320 = 0
(r - 40)(r - 8) = 0

since we know that r > 8 + 8, r = 8 is an extraneous solution and r = 40
Area of quarter circle = pi r^2 / 4 = 400 pi

[quigonkenny]

As AOBE is a square and quarter circle O is fully inscribed in it, OA = OB = AE = EB = r. As STFE is given as a square as well, then ST is parallel to OA, and FT is parallel to OB.

Draw radius OD, and draw DM, where M is the point on OA where DM is perpendicular to OA and parallel to AE, making ∆DMO a right triangle. As MA is parallel to SD, DM is parallel to AS, and every internal angle of ASDM is 90°, then ASDM is a rectangle and DM = AS = AE-SE = r-16 and MA = SD = 8. As OA = r, then OM is therefore equal to r-8.

Triangle ∆DMO:
DM² + OM² = OD²
(r-16)² + (r-8)² = r²
r² - 32r + 256 + r² - 16r + 64 = r²
r² - 48r + 320 = 0
(r-40)(r-8) = 0
r = 40 | r = 8 ❌ r must be > 16
r = 40 ✓

Yellow Quarter circle O:
Aₒ = πr²/4
Aₒ = π(40)²/4 = 1600π/4 = 400π ≈ 1256.64 sq units

[MrPaulc]

Large square's sides are r
Extend ST rightward toconnect with BD, and call it point
ODJ is a right triangle with sides r-8, r-16 and hypotenusev r
(r-8)^2 + (r-16)^2 = r^2
r^2 - 16r + 64 + r^2 - 32r + 256 = r^2
r^2 - 48r + 320 = 0
(48+or-sqrt(2304 - 4*1*320))/2 = r
48+or-sqrt(1024)/2 = r
(48+or-32)/2 = r
r = 40 or r = 8
r = 40
40^2 = 1600, so 1600pi for full circle
400pi for quadrant
100 * 3.14159 * 4
314.159 * 4
314.16 * 4
1256.64 (rounded) un^2
Yes, I must factorise more to simplify the calculation.

[murdock]

Many thanks, Sir, this is awesome!
fast lane: r = √((r - 16)^2 + (r - 8)^2) = 40
φ = 30° → sin⁡(3φ) = 1; ∎AOBE → AO = AP + PO = AU + PU + PO =
8 + 8 + (r - 16) = BO = BW + WQ + QO = 8 + 8 + (r - 16) = BE = EK + FK + BF = 8 + 8 + (r - 16) =
AE = EV + SV + AS = 8 + 8 + (r - 16) → EO = AB = r√2; ∆ ECF → EF = 16; CF = 8 → CE = 8√5 →
FEC = δ → tan⁡(δ) = 1/2; ∆ CED → CE = DE = 8√5; CD = CM + DM = 8√2 = 4√2 + 4√2 → CEM = MED = β
BEO = 3φ/2 → tan⁡(3φ/2) = 1 → tan(3φ/2 - δ) = tan⁡(β) = 1/3 → EM = 3CM = 12√2 → MO = (r - 12)√2 →
∆ CMO → CM = 4√2; CO = r; MO = (r - 12)√2; sin⁡(CMO) = 1 → r^2 = 32 + 2(r - 12)^2 → r = 24 ± 16 →
r = 40 ↔ r = 8 ≠ solution → πr^2/4 = 400π;
btw: MOC = α → sin⁡(α) = √2/10 → tan⁡(α) = 1/7; COB = θ → tan⁡(θ) = tan⁡(3φ/2 - α) = 3/4

[phungpham]

1/ ST(extended) intersects the circle at D’.
We have: sq SA= SDxSD’
—> sq (r-16) = 8.(2r-8)
—> sqr-48r+320=0
Or (sqr - 2 x24 r + sq24 )- sq24+320= 0
—> sq(r-24) =256=sq16
—> r-24= +/- 16
—> r=24+16=40 (r=8 rejected)
Area = 1/4 pi sqr=400 pi😅

[wackojacko]

@ 5:22 Yes! Grouping and factoring...I use mainly down in the crypt plying my skills to protocols to prevent third party interference. Information is power and I'm a hoarder. 🙂

[santiagoarosam]

Potencia de T respecto a la circunferencia r=40→ Área amarilla =400π ud².
Gracias y un saludo cordial.

[AdemolaAderibigbe-js]

Alternatively we can extend ST to the right to intersect BO at U and use the equation for the product of the intersecting chord components at U as follows: (r - 8)^2 = 16(2r -16) to get r = 40 and the area of the yellow shaded area = (1/4)*pi*r^2 = 400*pi square units.

[unknownidentity]

Let's find the area:
.
..
...
....


Let s be the side length of the square EFTS and let r be the radius of the quarter circle (and the side length of the square AEBO as well). Obviously we have:

s = ST = DS + DT = 8 + 8 = 16

Now let's add point R on AO such that ODR is a right triangle. In this case we can apply the Pythagorean theorem:

DO² = OR² + DR²
DO² = (AO − AR)² + AS²
DO² = (AO − DS)² + (AE − ES)²
r² = (r − DS)² + (r − s)²
r² = (r − 8)² + (r − 16)²
r² = r² − 16*r + 64 + r² − 32*r + 256
0 = r² − 48*r + 320

r = 24 ± √(24² − 320) = 24 ± √(576 − 320) = 24 ± √256 = 24 ± 16
r₁ = 24 + 16 = 40
r₂ = 24 − 16 = 8

Since r>s, only r=r₁=40 is a useful solution. Now we are able to calculate the area of the yellow quartercircle:

A(yellow quartercircle) = πr²/4 = π*40²/4 = 400π

Best regards from Germany

[marcgriselhubert]

Let's use an orthonormal, center O, first axis (OA).
The equation of the circle is x^2 + y^2 = R^2, with R the unknown radius of this circle.
C(R -16; R -8) is on the circle, so we have:
(R - 16)^2 + (R - 8)^2 = R^2
The end is the same as yours...

[johnjones]

Intersecting chords theorem!

16(2r-16)=(r-8)^2
r=8 (rejected)
r=40 ☆

A=400pi

[johnbrennan]

Can be done similarly by joining D to O and dropping a perpendicular from D onto AO

[gelbkehlchen]

Solution:
r = radius of the yellow quarter circle.
The point C must fit into the circle equation x²+y² = r². The point T has the coordinates (-t;t). Then C = (-t;t+8).
C = (-t;t+8) inserted into the circle equation x²+y² = r² means:
(1) (-t)²+(t+8)² = r²
You can also read at point B: (2) t = r-16 | inserted into (1) ⟹
(1a) [-(r-16)]²+(r-16+8)² = r² ⟹
(1b) (16-r)²+(r-8)² = r² ⟹
(1c) 16²-32r+r²+r²-16r+64 = r² |-r² ⟹
(1c) r²-48r+320 = 0 |p-q formula ⟹
(1d) r1/2 = 24±√(24²-320) = 24±16 ⟹
(1e) r1 = 24+16 = 40 and (1f) r2 = 24-16 = 8 ⟹
But you can see from the drawing that the Radius should be greater than 8, so r = 40.
Then the area of ​​the yellow quarter circle = π*40²/4 = 400π ≈ 1256.6371

[professorrogeriocesar]

Usei 4r^2=a^2+b^2+c^2+d^2.

Show, obg!!! 😊

[EPaozi]

r=8 est acceptable.... et A = 16 pi !!!