ASB is quarter circle. PQRS is a rectangle with side PQ=8 and PS=6 . What is length of ARC AQB ? Ans $5\pi$

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Here is how I am solving it:

Radius of Quarter circle = diagonal of rectangle = $\sqrt {100} = 10$

For a Full Circle : $Length_{Arc}=\frac{Arc_{Angle}}{360} \times Circumference$--->A

Now Circumference of complete circle will be : $2\pi10 = 20\pi$

Circumference of $\frac{1}{4}$ of circle = $5\pi + 10 + 10 =5\pi + 20$

Now For Quarter Circle $Length_{Arc}= \frac{90}{90} \times (5\pi + 20)$


I realize that I could have gotten the answer by simply not adding the 20 in the circumference. But according to this link. Circumference is the same as parameter and we should add the radius twice when determining the perimeter/circumference of quarter circle (refer to example 2 on bottom of the page) , since the formula above (Equ. A) requires the circumference I wanted to know why we shouldn't be adding the 20 since after all isnt it a part of the circumference of circle portion ?

For convenience I have posted an image of the example from that site

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  • $\begingroup$ Circumference and perimeter are only the same for a full circle. The question is asking for arc length AQB, which is only the curved part. $\endgroup$ – Chris Cudmore Jul 25 '12 at 21:02
  • $\begingroup$ Does that mean in case the circle is not full they will have different definitions ? $\endgroup$ – Rajeshwar Jul 25 '12 at 21:03
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    $\begingroup$ Circumference is the perimeter of a full circle. If you don't have a full circle, then you don't have a circumference. $\endgroup$ – Chris Cudmore Jul 25 '12 at 21:07

Once you get circumference $ = 20\pi$, just divide by $4$ to get the quarter circumference.

Answer is $20\pi / 4 = 5\pi$. simple.

However, the title of the question seems to be misleading. The question in the body seems to have nothing to do with the area and the circumference (perimeter?) of the quadrant.

  • $\begingroup$ But I thought that when we need to find the circumference we need to make up for the horizontal and vertical bars? $\endgroup$ – Rajeshwar Jul 25 '12 at 20:26
  • $\begingroup$ @Rajeshwar But the question only seems to be asking about the length of the arc $AQB$. $\endgroup$ – Old John Jul 25 '12 at 20:28
  • $\begingroup$ YES I agree and that is for the parameter of the quarter-circle thanks.. $\endgroup$ – Rajeshwar Jul 25 '12 at 20:31
  • $\begingroup$ Also I was initially doing $\frac{90}{360}$ which is suppose to be $\frac{90}{90}$ for a quarter circle $\endgroup$ – Rajeshwar Jul 25 '12 at 20:33
  • $\begingroup$ Hi , I was having some doubts about the question so I re-edited it and re-opened the question. Could you kindly look at the changes and post your opinion. $\endgroup$ – Rajeshwar Jul 25 '12 at 20:58

Note that the radius is $r=SQ=\sqrt{(SR)^2+(SP)^2}=\sqrt{8^2+6^2}=\sqrt{100}=10$ by the Pythagorean theorem.

Then, we know that the perimeter of a whole circle is $2\pi r$, so a quarter of a circle has an arc length of

$$L=\frac{1}{4} 2\pi r = \frac{\pi r}{2}$$

Plugging in the radius of the circle that we found above

$$L=\frac{\pi r}{2}=\frac{\pi \cdot 10}{2}=5\pi$$


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