Question
In circle D with m, angle, C, D, E, equals, 130, degreesm∠CDE=130
∘
and C, D, equals, 12CD=12 units, find the length of arc, C, E
CE
⌢
. Round to the nearest hundredth.
∘
and C, D, equals, 12CD=12 units, find the length of arc, C, E
CE
⌢
. Round to the nearest hundredth.
Answers
GPT-4o mini
To find the length of arc \( \overset{\frown}{CE} \) in circle \( D \), we can use the formula for the length of an arc:
\[
L = \frac{m\angle}{360^\circ} \times C
\]
where:
- \( L \) is the length of the arc,
- \( m\angle \) is the measure of the angle in degrees that subtends the arc,
- \( C \) is the circumference of the circle.
First, we need to find the circumference \( C \) of the circle. The circumference is given by the formula:
\[
C = 2\pi r
\]
where \( r \) is the radius of the circle. Since \( C, D = 12 \) units represents the length from point \( C \) to point \( D \) which is not directly the radius; we need to analyze the geometry. In a circle, if we have an angle at the center (and since \( D \) is the center and \( C \) is a point on the circumference), the radius \( r \) to points \( C \) and \( E \) is the same.
From the inscribed angle theorem:
- The inscribed angle \( m\angle CDE \) is half the measure of the central angle that subtends the same arc.
Thus, the central angle that corresponds to \( \overset{\frown}{CE} \) is
\[
m\angle DOE = 2 \times m\angle CDE = 2 \times 130^\circ = 260^\circ
\]
Next, we can calculate the circumference \( C \) of the circle. Since \( CD \) is given:
- Let’s assume the radius \( r \) is equal to \( CD \) since it represents a segment from the center to the circumference. Therefore, \( r = 12 \) units.
Now calculate the circumference \( C \):
\[
C = 2\pi(12) = 24\pi
\]
Next, find the length of arc \( \overset{\frown}{CE} \):
\[
L = \frac{m\angle DOE}{360^\circ} \times C = \frac{260^\circ}{360^\circ} \times 24\pi
\]
This simplifies to:
\[
L = \frac{260}{360} \times 24\pi = \frac{26}{36} \times 24\pi = \frac{13}{18} \times 24\pi
\]
Calculating this fraction gives:
\[
L = \frac{13 \times 24\pi}{18} = \frac{312\pi}{18} = \frac{156\pi}{9}
\]
Now we can evaluate this numerically:
\[
L \approx \frac{156 \times 3.14159}{9} \approx \frac{489.40964}{9} \approx 54.37885
\]
Rounding to the nearest hundredth yields:
\[
L \approx 54.38
\]
Thus, the length of arc \( \overset{\frown}{CE} \) is approximately \( \boxed{54.38} \) units.
\[
L = \frac{m\angle}{360^\circ} \times C
\]
where:
- \( L \) is the length of the arc,
- \( m\angle \) is the measure of the angle in degrees that subtends the arc,
- \( C \) is the circumference of the circle.
First, we need to find the circumference \( C \) of the circle. The circumference is given by the formula:
\[
C = 2\pi r
\]
where \( r \) is the radius of the circle. Since \( C, D = 12 \) units represents the length from point \( C \) to point \( D \) which is not directly the radius; we need to analyze the geometry. In a circle, if we have an angle at the center (and since \( D \) is the center and \( C \) is a point on the circumference), the radius \( r \) to points \( C \) and \( E \) is the same.
From the inscribed angle theorem:
- The inscribed angle \( m\angle CDE \) is half the measure of the central angle that subtends the same arc.
Thus, the central angle that corresponds to \( \overset{\frown}{CE} \) is
\[
m\angle DOE = 2 \times m\angle CDE = 2 \times 130^\circ = 260^\circ
\]
Next, we can calculate the circumference \( C \) of the circle. Since \( CD \) is given:
- Let’s assume the radius \( r \) is equal to \( CD \) since it represents a segment from the center to the circumference. Therefore, \( r = 12 \) units.
Now calculate the circumference \( C \):
\[
C = 2\pi(12) = 24\pi
\]
Next, find the length of arc \( \overset{\frown}{CE} \):
\[
L = \frac{m\angle DOE}{360^\circ} \times C = \frac{260^\circ}{360^\circ} \times 24\pi
\]
This simplifies to:
\[
L = \frac{260}{360} \times 24\pi = \frac{26}{36} \times 24\pi = \frac{13}{18} \times 24\pi
\]
Calculating this fraction gives:
\[
L = \frac{13 \times 24\pi}{18} = \frac{312\pi}{18} = \frac{156\pi}{9}
\]
Now we can evaluate this numerically:
\[
L \approx \frac{156 \times 3.14159}{9} \approx \frac{489.40964}{9} \approx 54.37885
\]
Rounding to the nearest hundredth yields:
\[
L \approx 54.38
\]
Thus, the length of arc \( \overset{\frown}{CE} \) is approximately \( \boxed{54.38} \) units.