Question

In each circle below, a 50° angle with a vertex at the center of the circle is drawn. How are minor arc lengths CD and EF related?

Circle A and B are shown. Line segments D A and C A are radii with lengths of 8 centimeters. Angle D A C is 50 degrees. Line segments F B and E B are radii with lengths of 2 centimeters. Angle F B E is 50 degrees.

They are the same because the central angle measure is the same.
The arc lengths are proportional: Arc C D = 2 arc E F.
The arc lengths are proportional: Arc C D = 4 arc E F.
The arc lengths are proportional: Arc C D = 6 arc E F.

Answer 1

Answer:C

Step-by-step explanation:

Answer 2

The arc lengths are proportional: Arc CD = 4 arc EF.

Given that,

Circle A and B are shown.

Line segments DA and CA are radii with lengths of 8 centimeters.

Angle DAC is 50 degrees.

Line segments FB and EB are radii with lengths of 2 centimeters.

Angle FBE is 50 degrees.

In each circle below, a 50° angle with a vertex at the center of the circle is drawn.

We have to determine,

How are minor arc lengths CD and EF related?

According to the question,

The length of an arc that subtends an angle of θ degrees at the center of the circle is given by,

$\rm Length = \dfrac{\theta}{360} \times 2\pi r$

Line segments DA and CA are radii with lengths of 8 centimeters.

$\rm Length \ of \ CD= \dfrac{50}{360} \times 2\pi \times 8$

Line segments FB and EB are radii with lengths of 2 centimeters

$\rm Length \ of \ EF= \dfrac{50}{360} \times 2\pi \times 2$

On comparing both the lengths

$\rm\dfrac{ Length \ of \ CD }{Length \ of \ EF}= \dfrac{\dfrac{50}{360} \times 8\pi \times 2}{ \dfrac{50}{360} \times 2\pi \times 2\\\\}\\\\\dfrac{ Length \ of \ CD }{Length \ of \ EF} = \dfrac{4}{1}\\\\Length \ of \ CD }{}=4\times Length \ of \ EF$

Hence, The arc lengths are proportional: Arc CD = 4 arc EF.

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