# Can't understand this solution.

I came across a problem which was already present on the internet.

If an arc with a length of $12\pi$ is $\frac{3}{4}$ of the circumference of the circle, what is the shortest distance between the endpoints of the arc?

According to a certain site the solution is something like this

$$12\pi\left(\frac{4}{3}\right)=\text{circumference}=16\pi=2\cdot\text{radius}\cdot\pi$$ $$\text{radius}=8$$ $$x^2+y^2=64$$ Let $x=0$ for the first endpoint and let $y=0$ for the other, then find the two points $(0,8)$ and $(8,0)$. Now find the distance between these two points: $$d=((0-8)^2+(8-0)^2)^{1/2}=(128)^{1/2}\qquad \text{Ans}$$

I on the other hand decided to take my own approach since I couldnt figure out what happened after the radius

Step 1:

$12\pi = (3/4)$ (Circumference) Cirum $= 16\pi$ so radius of the circle in question is $8$

Step 2: Since $16\pi = 360^{\circ}$ so $12\pi$ is $270^{\circ}$ .

Edited: From the suggestions i got from users here is how i would solve this: Construct a line from the origin that goes to $270^{\circ}$ which is equal to radius and acts as a base and another line goes from origin to $360^{\circ}$ which acts as a perpendicualr then we calculate the hypotenuse (shortest distance). This definitely makes sense. But what if the question changes and angle is not $90^{\circ}$. I would appreciate it if someone could explain how to solve this using the distance formula as done above without the need of calculating $270^{\circ}$

• If $12\pi=\dfrac34(2\pi r)$, you should be able to obtain the radius of the circle easily. You now want the length of the chord corresponding to the $90^\circ$ arc of this circle, which you should find easy to do with a bit of Pythagoras... (hint: the chord is the hypotenuse of an isosceles (why?) right triangle.) – J. M. is a poor mathematician Jun 9 '12 at 14:05

Draw a picture. By your calculation, the angle subtended at the origin by the arc is $270^\circ$. So going from one endpoint of the arc to the other the short way around, the angle is $360^\circ-270^\circ=90^\circ$.
Then by the Pythagorean Theorem the distance between the endpoints is $\sqrt{8^2+8^2}$, or more simply $8\sqrt{2}$.
Added: Let $A$ and $B$ be on a circle with radius $r$ and centre $O$. There are two arcs joining $A$ and $B$, the "short" one and the "long" one. Let $\theta$ be the angle subtended at $O$ by the short arc. So $\theta=\angle AOB$. We want to calculate $AB$.
Drop a perpendicular from $O$ to $AB$, meeting $AB$ at point $P$. Then $\angle AOP=\theta/2$. We have $\frac{PA}{OA}=\frac{PA}{r}=\sin(\theta/2)$, so $PA=r\sin(\theta/2)$ and therefore $$AB=2r\sin(\theta/2).$$
• There are two cases to consider: (i) the angle (your $270^\circ$) is $\le 180^\circ$ and (ii) the angle is $\gt 180^\circ$. In case (i), work with that angle, in case (ii) work with $360^\circ$ minus the angle. Call the smallet angle $\theta$. Draw the triangle $OAB$ with vertices the centre and the two endpoints. Drop a perpendicular from $O$ to $P$ on $AB$. Then $\frac{AP}{OA}=\sin(\theta/2)$. But $OA$ is the radius $r$. So $AP=r\sin(\theta/2)$ and therefore $AB=2r\sin(\theta/2)$. [This is actually how the sine function began, for solving exactly this problem.] – André Nicolas Jun 9 '12 at 14:24