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$ABCD$ is a square having each side = $6$ cm. Two arcs are drawn within the square as shown in the figure. Find the radius of the circle (centre $P$ in the figure) which touches both the arcs and the side of the square.
enter image description here

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  • $\begingroup$ What you have tried? $\endgroup$ Apr 17 '17 at 14:58
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It is only a matter of pythagorean theorem: enter image description here

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  • $\begingroup$ In fact, if the side length of the square is $s$, then $r=s/6$. $\endgroup$
    – amd
    Apr 17 '17 at 17:40
  • $\begingroup$ @amd. Yes exactly. $\endgroup$
    – Seyed
    Apr 17 '17 at 18:27
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let the radius of the little circle be $r$. Let X be the point on BC where this little circle is tangential to BC. Let the angle PBX be $\phi$. Consider the triangle PBX \begin{eqnarray*} sin \phi = \frac{r}{6-r} \end{eqnarray*} Now condiser the triangle ABP ... Its sides are of length $6,6+r , 6-r$ and the angle at $B$ satisfies $cos \theta = \frac{r}{6-r}$. Now apply the cosine rule to this triangle. \begin{eqnarray*} (6+r)^2=6^2+(6-r)^2-2 \times 6 (6-r) \frac{r}{6-r} \end{eqnarray*} Solving this we have $\color{red}{r=1}$. Diagram

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  • $\begingroup$ Thank you for edited my answer. You have corrected an error & added a diagram (This is fantastic). I need to learn how to do diagrams like these ... can you suggest a nice peice of software for this ? ... thank you again. $\ddot \smile$ $\endgroup$ Apr 18 '17 at 18:59
  • $\begingroup$ I used Desmos for graphing. $\endgroup$ May 3 '17 at 11:02
  • $\begingroup$ Right ... Desmos ... math.stackexchange.com/questions/2254969/… ... levap gives another really nice example of this software here. Thank you, just what I was looking for. $\endgroup$ May 3 '17 at 19:47

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