Originally posted by clandarkfire
I just got to a question on my homework tonight that I can't make heads or tails of. Its probably quite simple, but I'm not getting it. Here it is:
What is the area of the space trapped between a circle radius 10 and a equalateral triangle inscribed inside it?
Obviously, the circle has an area of 100pi, but how do you find a side lenght of the triangle? could somebody help me out?
I solved it in a very un-elegant way...It hinges on you understanding triginometry, and Cartesian graphing...so it may or may not be the method your looking for...
On a cartesian graph start with the equation of a circle of raduis 10 which is defined by the following equation:
x^2 + y^2 = 10^2
now place the vertex of the equallateral triangle at an arbitrary point "p" on the circumference of the circle. (chosing the point on one of the axises simplifies the math but is not necessary
I chose the point (0,10) on the y-axis
Now, the triangle in question has 3 interior angles of 60 degrees each.
next find the equation of the line that passes through point "p" at a 30 degree angle refrenced from the y-axis.
here is where some trig comes in handy
give the sides of your triangle the variable lenght "L"
slope of the line in question = (y-y1)/(x-x1)
slope = (L*cos 30 deg.)/(L*sin30deg.) = (sqrt(3)/2) / (1/2) = sqrt(3)
the equation of my line is
y = -sqrt(3)*x + 10
we are trying to find the the side lenght "L", so first we need to find the points of intersection of the circle equation and the line equation just developed.
so solve the system of equations
(1) x^2 + y^2 = 10^2
(2) y = -sqrt(3)*x +10
and i get
x= 5*sqrt(3) & y = -5
from the distance formula
d(0,10),(5*sqrt(3) , -5) = SQRT [(5*sqrt(3) - 0)^2 + (-5-10)^2] = 10*sqrt(3) = L
now that we have "L"
we calculate the area to be
100*pi - (1/2)*(10*sqrt(3))*(10*sqrt(3)*(cos 30 deg.)
A = 100*pi - 75*sqrt(3)
approx = 184.255 units^2
like i said, not elegant, but a solution.
hope this helps, but if not, im sure others will simplify the solution...