Try this

Toast · 2007-02-11 01:13:01

This is a little problem I thought up while I was bored, I can't seem to solve it, but you have a go:

mathsyperson · 2007-02-11 01:41:06

I think I've got it. I've got an answer that sounds feasible anyway.
I probably used an inefficient and long-winded way of working it out, so I don't want to put up all of my lengthy workings, but here's my answer in case anyone else wants to check:

luca-deltodesco · 2007-02-11 02:00:21

(looking at pic)

Last edited by luca-deltodesco (2007-02-11 02:12:28)

mathsyperson · 2007-02-11 02:17:01

Huh. That's pretty much the same method I used, and yet it somehow took up 3 pages of scribbles. Ah well.

Toast · 2007-02-11 23:03:33

Could you please explain how you got x and y? I don't really understand . Thanks.

mathsyperson · 2007-02-12 01:43:19

x is a sector of the circle. You get its area by finding the angle to see what fraction of the circle it is, and then taking that fraction of the circle's area.

luca has worked out angle b by using cos θ = adj/hyp = 5/6, and the angle of the sector is double angle b. To find the area as a fraction of the circle's area, you divide the angle by 2π.

So then we have 2b/2π = b/π. Multiplying this by the circle's area will get the area of the sector. b/π * π*6² = b*6² = 36b.

y is a triangle. luca has used Pythagoras to find that half of the triangle's base is √11, and then multiplied by 2 to get the total base. Because the square has side length 10, it's fairly easy to see that the triangle's height is 5.

Therefore, it's area is (2√11*5)/2 = 5√11.

And then z is the triangle subtracted from the sector, and the shaded regions are the quadruple of z.

Math Is Fun Forum

#1 2007-02-11 01:13:01

Try this

#2 2007-02-11 01:41:06

Re: Try this

#3 2007-02-11 02:00:21

Re: Try this

#4 2007-02-11 02:17:01

Re: Try this

#5 2007-02-11 23:03:33

Re: Try this

#6 2007-02-12 01:43:19

Re: Try this

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