Monday

March 30, 2015

March 30, 2015

Posted by **me** on Monday, March 18, 2013 at 7:18am.

- urgent maths pls -
**Steve**, Monday, March 18, 2013 at 10:09amAssuming A is on the edge of the field (circle F), let the radius of the field be R, and the rope be length L.

Let the circle with radius L and center A intersect the field at points B and C.

Then we want the area enclosed by the arcs BC to be 1/2 pi R^2.

The sector of circle A subtended by arc BC has area 1/2 L^2 θ where θ subtends BC.

The rest of the area comprises two segments of circle F, where each has area

1/2 R^2 (φ - sinφ)

where φ subtends arc AB or AC.

So, we want

1/2 L^2 θ + R^2 (φ - sinφ) = pi/2 R^2

So, what are θ and φ?

In triangle FAB, let FB be s.

s^2 = L^2 + R^2 - 2RL cos θ/2

s/sin θ/2 = L/sinφ

That should yield the result you need.

- urgent maths pls -
**me**, Monday, March 18, 2013 at 10:45amPls d answer isn't really clear. And where did d triangle come from??

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