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The circle is given by $x^2+y^2=25$.

FGHI are midpoints on the rhombus

Calculate the area of FGLMHIJK (taking into account the curved lines)

Calculate the area of FGLMHIJK (taking into account the curved lines)

rschwieb
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Mike Ashfield
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3 Answers3

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No calculus needed here. We need to deduct the area of the four circular segments. We have $r=5$. To get the coordinates of $L$, we have $x^2+y^2=25; y=\frac 43(6-x)$, which gives $(x,y)=(\frac {117}{44},\frac {25}{44})$. Now compute $c=\sqrt{(\frac {117}{44}-3)^2+(4-\frac {25}{44})^2}=\frac {\sqrt{14930}}{44}$ and get the area.

Ross Millikan
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  • Interesting approach... although it still seems pretty hard to retrieve the area. It looks like it involves double angle identities and inverse trig functions in an overall nasty computation. (That or there is some easy identity I'm not finding on the wiki page :) ) For being independent of integration it's OK. – rschwieb Oct 17 '13 at 17:16
  • You can get the cosine of the angle from the equation for chord length without any double angle identities-just clear the square root. Then get the sine from that and you have the area. – Ross Millikan Oct 17 '13 at 17:38
  • Sure, but that's equivalently nasty :) As the integral I had in mind also requires the inverse trig function too, I guess even using heavy machinery runs into the complication. – rschwieb Oct 17 '13 at 19:13
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My first instinct would be to take these steps:

  1. Compute the radius of the circle using $G$. (Oh, you were given the radius. I didn't even see that until just a moment ago.)
  2. Use the radius to find $L$'s location.
  3. Write an equation for the circle and the rhombus side in the first quadrant, and then use integral calculus to find out the area of the sliver between $G$ and $L$.
  4. Then, I would compute the area of the circle sector in the first quadrant, then subtract the sliver area from the last bullet point.
  5. Finally, I would appeal to symmetry and multiply by 4.
rschwieb
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draw $EG=5$ , $EL=5$ and $EH \bot GL$. first note that $EH=(6.8)/10=4.8$ by the area. in $EGH$ using pyth. theo. $GH=7/5$. $\tan (GEH)=7/24$ so $\tan(GEL)=\tan(2.(GEH))$ now you can find $\angle GEL$. finding area between arc and trinagle is easy.

mert
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