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 circ
le formed by the intersection of the plane and the sphere?
1 answer:
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
<span>let's assume that the center of the sphere is at the origin
</span>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
the answer is
the area of the circle is 549.5 m²
we know that
the equation of a sphere is
(x-h)²+(y-k)²+(z-l)²=r²
<span>let's assume that the center of the sphere is at the origin
</span>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
the answer is
the area of the circle is 549.5 m²
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