A sector of a circle with center O has a central angle of 150^\circ and an area of 15\pi. … — SAT

A sector of a circle with center $O$ has a central angle of $150^\circ$ and an area of $15\pi$ . What is the length of the minor arc that bounds this sector?

$5\pi$ Cevap
B
$\frac{5\pi}{2}$
C
$12\pi$
D
$30\pi$

Cevap

5\pi

To find the arc length, we first determine the radius of the circle using the sector area. The area of a sector with a central angle of

150^\circ

\frac{150}{360} = \frac{5}{12}

of the total circle area. Setting up the equation

15\pi = \frac{5}{12}\pi r^2

allows us to solve for

r^2 = 36

, which gives a radius of

r = 6

. Using the radius, we find the length of the minor arc by taking the same fraction of the total circumference:

\frac{5}{12} \times 2\pi(6) = 5\pi

. This matches the correct option.

Adım Adım Çözüm

Set up the equation for the area of the sector to solve for the radius

r

15\pi = \frac{150}{360} \pi r^2

The area of a sector is given by the formula

A = \frac{\theta}{360} \pi r^2

, where

\theta

is the central angle in degrees.

Simplify the fraction and solve for

r^2

and

r

15\pi = \frac{5}{12} \pi r^2 \implies 15 = \frac{5}{12} r^2 \implies r^2 = 36 \implies r = 6

Dividing both sides by

\pi

and multiplying by

\frac{12}{5}

isolates

r^2

, and taking the square root gives the radius

r

Calculate the length of the minor arc using the radius and central angle.

Arc Length

= \frac{150}{360} \times 2\pi(6) = \frac{5}{12} \times 12\pi = 5\pi

The arc length formula is

L = \frac{\theta}{360} \times 2\pi r

, representing the fraction of the total circumference.

Anahtar Kavram

The relationship between a circle's sector area, central angle, radius, and arc length.

Alternatif Yöntem

Instead of solving for the radius first, note that the ratio of the sector area to the total area is equal to the ratio of the arc length to the total circumference. Since Sector Area

= \frac{1}{2} r L

(where

L

is the arc length and

r

is the radius), we have

15\pi = \frac{1}{2} r L

. Since we also know that the area

15\pi = \frac{5}{12} \pi r^2 \implies r = 6

, substituting this directly into the area-arc relation gives

15\pi = \frac{1}{2} (6) L \implies 15\pi = 3L \implies L = 5\pi

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Konular

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