A sphere has a radius of 16 cm and a plane passes thru it at 9 cm below the center, what is the area of the intersection between the plane and the sphere?
Visualize slicing a nice round object such as an apple in the way you described.
Try to sketch a 3-D diagram.
Your slice will have a circular surface.
Sketch a right-angled triangle with hypotenuse of 16 and height of 9
The third side will the the radius of you circular slice.
r^2 + 9^2 = 16^
r^2 = 175
area of that circular slice = πr2
= 175π cm^2
To find the area of the intersection between the plane and the sphere, we first need to determine the shape of the intersection.
Given that the plane passes through the sphere at a certain distance below the center, it means the plane intersects the sphere and creates a circle on the sphere's surface. The circle is the intersection between the plane and the sphere.
To calculate the area of the intersection circle, we can use the formula for the area of a circle:
Area = π * r^2
where r is the radius of the circle.
In this case, the radius of the circle formed by the intersection is the same as the radius of the sphere, which is 16 cm. Therefore, the area of the intersection circle is:
Area = π * (16 cm)^2
= π * 256 cm^2
≈ 804.25 cm^2
So, the area of the intersection between the plane and the sphere is approximately 804.25 square centimeters.