A sphere with radius 16 m is cut by a plane that is 9 m below its center. what is the best approximation of the area of the circle formed by the intersection of the plane and the sphere?
Accepted Solution
A:
the complete question in the attached figure
we know that the equation of a sphere is (x-h)²+(y-k)²+(z-l)²=r²
let's assume that the center of the sphere is at the origin so (h,k,l)-------> (0,0,0) for r=16 m (x-h)²+(y-k)²+(z-l)²=r²--------> (x)²+(y)²+(z)²=16²
for z=9 m (x)²+(y)²+(9)²=16²--------> (x)²+(y)²=256-81--------> (x)²+(y)²=175 the radius of the circle is r=√175 m so the area of the circle A=pi*r²-----> A=pi*175-----> A=549.5 m²