Hi Dina. I believe the diagram is showing you the side of the cylinder, instead of the top (:

Yes it is the curved portion of the cylinder. From that view it will look like a square/rectangle

If you draw a line from the centre of the cylinder/sphere to any of the 4 corners of the cylinder, the length of that line is also a radius of the sphere, r.

Let this corner be C. Let the centre of the sphere be O. Let the centre of one of the circular flat surface of the cylinder be B.

Now you should be able to see a right angled triangle OBC.

Then,

OB² + BC² = OC² (Pythagoras' Theorem)
OB² + r² = 5²

OB² = 25 - r²

OB = √(25 - r²)

Now realise that OA is actually half the height of the cylinder.


Height of the cylinder = 2√(25 - r²)


Curved surface area , A

= 2πrh

= 2πr(2√(25 - r²))

= 4πr√(25 - r²)

This is what we call a cross sectional area. The A large cylinder, when viewed from very far at the same height level, will appear to look like a rectangle, because the apparent curvature of the cylinder cannot be distinguished from large distances.

Need help on part b?

Actually the cross sectional area of the cylinder is from the top view.

Actually from the top view, the upright cylinder would be circular in shape (since the top and bottom of the cylinder are plain circles rather than curved surfaces)

EDIT: Failed to realise you wrote "cross-section". You are correct.

I was referring to uniform cross section actually. Front view is considered cross section area, just that for the purposes of this question the cut is delivered right across the centre of the circle from the top.