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Zorluk: OrtaCircles, Arc Lengths, and Sector Areas

A square is inscribed in a circle with center OO. The perimeter of the square is 16216\sqrt{2} units. A sector of this circle has an area equal to the total area of the region inside the circle that lies outside the square. What is the arc length of this sector?

  1. A
    4π−84\pi - 8
  2. 8π−168\pi - 16Cevap
  3. C
    16π−3216\pi - 32
  4. D
    16π−816\pi - 8
  5. E
    8π−328\pi - 32

Cevap

8π−168\pi - 16
The correct answer is 8π−168\pi - 16. The inscribed square has side length 424\sqrt{2} and diagonal 88. Since the diagonal of an inscribed square is the circle's diameter, the circle has radius r=4r = 4 and total area 16π16\pi. The square has area 3232, so the region outside the square has area 16π−3216\pi - 32. For any sector of radius rr, the arc length LL and sector area AA satisfy L=2ArL = \frac{2A}{r}. Substituting A=16π−32A = 16\pi - 32 and r=4r = 4 yields L=2(16π−32)4=8π−16L = \frac{2(16\pi - 32)}{4} = 8\pi - 16.

Adım Adım Çözüm

1
Find the side length and diagonal of the inscribed square.
Side length s=1624=42s = \frac{16\sqrt{2}}{4} = 4\sqrt{2}. Diagonal d=s2=(42)2=8d = s\sqrt{2} = (4\sqrt{2})\sqrt{2} = 8.
The perimeter of a square is 4s4s, and the diagonal of a square with side ss is s2s\sqrt{2}.
2
Determine the radius and total area of the circle.
Diameter equals diagonal d=8d = 8, so radius r=4r = 4. Area of circle Acircle=πr2=16πA_{\text{circle}} = \pi r^2 = 16\pi.
A square inscribed in a circle has its diagonal aligned with the diameter of the circle.
3
Calculate the area of the region inside the circle but outside the square.
Area of square Asquare=(42)2=32A_{\text{square}} = (4\sqrt{2})^2 = 32. Area outside square Aoutside=16π−32A_{\text{outside}} = 16\pi - 32.
Subtract the area of the inscribed square from the total area of the circle.
4
Relate sector area to arc length using L=2AsectorrL = \frac{2 A_{\text{sector}}}{r}.
L=2(16π−32)4=32π−644=8π−16L = \frac{2(16\pi - 32)}{4} = \frac{32\pi - 64}{4} = 8\pi - 16.
Since sector area Asector=θ360∘πr2A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2 and arc length L=θ360∘(2πr)L = \frac{\theta}{360^\circ}(2\pi r), we have L=2AsectorrL = \frac{2A_{\text{sector}}}{r}.

Anahtar Kavram

Relationship between circle area, inscribed figures, sector area, and arc length
Tahmini Süre:1m 30s
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Konular

Circles, Arc Lengths, and Sector Areas

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