Will all slices created by a slicing plane that passes through a sphere be the same size

The suggestion proposed considering special cases, so that’s what we’ll do.

One special case we could consider is when x=0. If h=r, then we have a hemisphere. You may have learnt that the surface area of a sphere is 4πr2, so the answer in this case is 2πr2. Likewise, we can consider the most extreme cases: when h=2r, we get an area of 4πr2 (the entire sphere), and when h=0, we get zero area.

So if we do manage to get a general formula, we will be able to check it in these special cases.

Another possible idea when x=0 is to consider what happens when h is very small. In this case, the surface area looks a lot like a circle, but the radius of the circle does not seem particularly easy to work out, so maybe we’ll leave this one for a moment. (There are things we can do to approximate the radius using more advanced techniques, but they are currently beyond us.)

Another thing we could consider is the situation where h is very small. Then the part of the sphere between the planes looks very much like the frustum of a cone, and we know how to find the (surface) area of such a shape – see Cones.

If we recall our answer from that problem, we find the formulae for the surface area

π(R+r)s=π(R+r)hcosθ=π(R2−r2)sinθ.

Here, R is the radius of the base of the frustum, r is the radius of the top, h is the perpendicular height, s is the slant length and θ is the angle the slant makes with the vertical.

Step-by-step explanation:

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Step-by-step explanation:

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step-by-step explanation: divide by 2,