Find the arc length of a circle

Question

Question: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

My Workout: Chord of length 40 cm has arc length of semicircle curcumference i.e = π × 20 = 20π CM

So chord of length 10 CM has (proportionately) length = 10π cm

Answer (given): 20π / 3

Why am I wrong

Answer 1

Why am I wrong

I don't know because your work seems like guesswork and possibly not related to the problem, since the chord in the problem and your chord ($20 \not = 10$) are of different length.

Regardless, we continue. Our circle has a diameter of $40$ and a radius of $20$ by definition. We may then realize that our chord is of equal length to the radius of our circle, meaning our chord can be formed by an angle of $60^o$ or $\frac{\pi}{3}$. There are two ways you can see this,

• first as suggested in the comments you can form an equilateral triangle with the two intersections of the chord and the circle with the center, and thus you have $\frac{\pi}{3}$ radians for an angle.
• Second, we can do this formulaically. The formula for a chord length is $$2r \sin(\frac{\theta}{2}) = L$$ Where $r$ is the radius of the circle, $\theta$ is the angle, and $L$ is the length of the chord. Then, we can find through manipulation that $\theta = 2 \sin^{-1}(\frac{1}{2}) = 2 \frac{\pi}{6} = \frac{\pi}{3}$.

Now that we have our angle, we can easily find the arc length using the formula: $$A = \theta r = \frac{pi}{3} \cdot 20 = \frac{20 \pi}{3}$$ And thus we have obtained our answer.

Answer 2

In order to find a length of an arc of a given chord, you should determine what the angle $\angle AOB$ formed by the center $O$ of the circle with radius 20 cm and the two end points of the chord $AB$. Here lies the real proportionality, that is, the length of this arc $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}$ to the circumference of the circle is equal to the angle $\angle AOB$ to the measure of the whole circle, i.e., $360^{\circ}$ or $2\pi$. Hence, \begin{align*} \frac{\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}}{2\pi r}=\frac{\angle AOB}{2\pi}. \end{align*} Simplifying and solving for $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}$ we get \begin{align*} \newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}=r\angle AOB_{\text{rad}}. \end{align*} Since $AB=AO=OB=r=20$ cm, then $\triangle AOB$ is equilateral, and thus $\angle AOB=\frac{\pi}{3}$. Substituting gives the result \begin{align*} \newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}=\frac{20\pi}{3}. \end{align*} A circle with radius 20 cm