# Word Problem (Circles/Rectangles)

Question: The diagram shows a $400m$ running track. The "Bends" (semi cricles) are $100m$ and the "straights" are $100m$. Find the distance run by an athlete who runs $x$ metres out from the kerb. (Had to make the diagram in paint because the picture is not really clear so apologies)

The part where I am stuck is what it means by runs $x$ metres out from the kerb.

The answer is $400 + 2\pi x$ but not too sure how they got it

• '$x$ meters out from the kerb' means 'following a path around the track, at a distance of $x$ meters outside it'. Mar 1, 2016 at 4:11
• @Jared: I agree the figure you drew is what the problem intends, but OP's figure was an ellipse. I suspect that is where the problem lies. I think we should not replace it. OP did say that drawing in paint was hard, but the straights could be straighter easily Mar 1, 2016 at 4:15
• @RossMillikan hmm, not sure how to undo my picture...I don't know whether or not the OP thought the shape was an ellipse or not to be frank or if they just didn't know how to create the figure "correctly" (I assumed they just didn't know how to correctly create the figure--perhaps that was wrong). Mar 1, 2016 at 4:29
• @Jared: you can roll back the edit to the original post, then edit the rest as you see fit. Mar 1, 2016 at 4:30

Your picture is misleading. The intent is where the semicircles have length $100$ meters and the straights have length $100$ meters. Clearly running out from the kerb on the straights has no effect on the path length. Running out from the kerb by $x$on the circles increases the radius of the circle by $x$, so the circumference by $2 \pi x$.

• Thank you very much for the explanation and sorry for the misleading diagram (Where program did you use for your picture?) Mar 1, 2016 at 4:14
• @dydxx: I used GeoGebra, which is free. I makes nice figures for many applications. Mar 1, 2016 at 4:16

I'm not sure if you're picture is "wrong" or just "the best you could do". The picture should look like the following:

Assuming that this is the "interior" then the radius of the two semi-circles is ($180^\circ = \pi$):

$$\pi r = 100 \rightarrow r = \frac{100}{\pi}$$

Then adding to $x$ meters to the radius would give:

\begin{align} d =&\ 200 + 2(r + x)\pi \\ =&\ 200 + 2 r\pi +2x\pi \\ =&\ 200 + 2\cdot 100 + 2x\pi \\ =&\ 400 + 2x\pi \end{align}

This is using the fact that $r\pi=100$ because the two semi-circle loops are $100$ meters.