Square based pyramid - right-angled Geometry - GCSE question

[mrarthur]

VABCD is a square-based pyramid.
ABCD is a square with side 20cm.
The angle between any sloping edge and the plane ABCD is 55 degrees.
Calculate the surface area of the pyramid.

Total area = base area (400) + 4 x triangular sides.
The issue I am having is in calculating the perpendicular height of the triangular sides. More specifically I can see two ways to do that and these two ways are giving me different answers. Working attached: either find the diagonal of the base, find the distance from one base corner to V at the top of pyramid, then use that to find the perp height of the triangles; or find the perp height directly.
Where am I going wrong?

 

[mrarthur]

Wow, it makes my attached working completely illegible. It will also let me reupload. I'll come back to this when technical issues overcome, but if you are happy to work without diagrams:
X is the centre of the base, M is the midpoint of AB - so VM is the length we are looking for.

Working 1:
Triangle VXM has a right-angle at X, angle VMX=55, MX=10 (because half the length of side of base)
Therefore trig: VM=10/cos55=
VM=17.43

Working 2:
Triangle AMX has a right-angle at M, AM=10, MX=10
Therefore pythag: AX=squrt(200)
Triangle VAX has a right-angle at X, angle VAX=55 and AX=squrt(200)
Therefore trig: VA=squrt(200)/cos55=24.66
Triangle VAM has a right-angle at M, AM=10 (because half the length of side of base) and VA=24.66
Therefore pythag: VM=squrt(squ24.66-squ10)
VM=22.54

 

[mrarthur]

the pyramid itself.

It let me upload a bit of diagram. All right-angled triangles referenced in my working above should be relatively easily found in this.

 

[mmm4444bot]

Wow, it makes my attached working completely illegible.

You're trying to put too much information into a single image. The site cannot display an 8.5x11inch image, so it is reduced.

Break your work into smaller images, and crop off any large areas of blank space. (If you use Windows, the Paint app is handy for cropping, rotating, and saving individual image files.) Then upload the files; each image will display much larger. :cool:

 

[Milton24]

Probably too late to be useful to you, but for others finding this:
Consider the two RA triangles VMX and VAX, and note they share side VX.
The angle VMX is 55° (given), but MX ≠ AX so the ratio VX/MX ≠ VX/AX and angle VAX cannot be 55° (as used in the third line of your second solution). This is the likely cause of the difference in the two solutions.