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

EDIT:

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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    C$i$rcumference is the perimeter of a full circle. If you don't have a full circle, then you don't have a circumference.2012-07-25

2 Answers 2

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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.

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    Not sure why you unaccepted my answer. As the question stands, it is asking about the arc length $AQB$ which is $5\pi$ as I said earlier.2012-07-25
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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

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