Given: -Square ABCD with side length 2. -Two quarter circles, one with center A and the other at center B. Call their intersection E. What is the area enclosed by Arc CE, Arc DE, and side CD? I've already done the problem, but I want to verify my answer. Also, my solution does not use calculus, trigonometry, and any other form of high-powered math. The whole solution only required geometry.
ack!!...picture would help
No problem. Give me a minute.
This ought to help. Essentially, finding the area of the shaded.
thats easier :)
we essentially got 4 areas, lets call them a,b,c, and d
When I did that, I formuated the equation a=4-b-c-d
thats good... we also need to realize that: b+d = c+d
b = c
a = 4-2b-d right?
lets try to fill in some values here: lets say b = 5, c = 5, and d = 10 we know that: (5+10) + (5+10) but the area here is exagerated by "an extra 10"
we have: the true area for these overlapping circles is: 2b + 2d - d right?
2b + d jsut gets us right back to where we were .... can we use the circumference of a circle in our solution?
2pi(2) = 4pi for a whole circle; which means that pi is the length for a quarter of it......I think I got it. unless you already know the answer :)
do we know that area of an equilateral triangle?
that middle triangle has an area of: sqrt(3)
Thats what I did too!
each side section has an area of 1 so the total unshaded area is 2sqrt(3)
pi(r^2) is the area of a complete circle the r here =2 4pi is the total area of our representative circle... we only want 30 out of 360 of that circle
30 degrees is 1/6 of a circle right?
so the area of each side should be 4pi/6 = 2pi/3
got something turned around...hold on.
360/30 = 12, so we have 1/12 of a circles area.... 4pi(1/12) = 4pi/12 = pi/3... am i right?
it should be 2pi/3 for the other stuff
a =4-sqrt(3)-(2pi/3) a = 12 - 3sqrt(3) - 2pi ---------------- 3 is my answer
a = .1735 if I did it right :)
I didn't evaluate for that, but the formula for a is what I got too! Awesome, thanks! You just earned a fan.
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