+118723 I am not going to do this very elegently but perhaps i can do it.
First I am going to cut this square pyramid in half so I need a few more points.
X can be the midpoint of AB
Y is the midpoint of CD and
Z is the midpoint of EF
Consider triangle VXB <VXB=90, XB=5 VB=8 find VX
vx= sqrt( 64-25) = sqrt(39)
Now consider triangle VXY VX=VY=sqrt(39), XY=8 Find <VXY (I'm going to call it $$\theta$$ )
$$\\39=39+64-(2*\sqrt{39}*8cos\theta)\\\\
64=16\sqrt{39}*cos\theta\\\\
4/\sqrt{39}=cos\theta\\\\$$
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{4}}}{{\sqrt{{\mathtt{39}}}}}}\right)} = {\mathtt{50.169\: \!945\: \!446\: \!964^{\circ}}}$$
so <VXY= 50.17°
NOW lets consider triangle XZY <ZYX=30°, <ZXY=50.17° XY=8 Find XZ
$$\\\frac{XZ}{sin30}=\frac{8}{sin(180-30-50.17)}\\\\
\frac{XZ}{sin30}=\frac{8}{sin(98.3)}\\\\
XZ=\frac{8}{sin(98.3)}\times sin30\\\\$$
$${\frac{{\mathtt{8}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{98.3}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)} = {\mathtt{4.042\: \!340\: \!325\: \!221\: \!254\: \!6}}$$
so $$XZ\approx 4.04 cm$$
NOW I am going to consider triangle AVB
VX=sqrt(39), XZ=4.04 so VZ=sqrt(39)-4.04 = 2.205 cm approx
NOW triangle EFV is similar to triangle ABV so
$$\\\frac{EF}{AB}=\frac{VZ}{VX}\\\\
\frac{EF}{10}=\frac{2.205}{\sqrt{39}}\\\\
EF=\frac{2.205}{\sqrt{39}}\times 10\\\\$$
$${\frac{{\mathtt{2.205}}}{{\sqrt{{\mathtt{39}}}}}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{3.530\: \!825\: \!791\: \!402\: \!171\: \!3}}$$
So if I have not done anything incorrectly the answer is EF= 3.53cm (approximately)
BLAST, I FOUND THE WRONG ONE. BUMMER!!! (ﾉ °益°)ﾉ 彡
nevermind:
$$\\\frac{VE}{VA}=\frac{VZ}{VX}\\\\
\frac{VE}{8}=\frac{2.205}{\sqrt{39}}\\\\
VE=\frac{2.205}{\sqrt{39}}\times 8\\\\$$
$${\frac{{\mathtt{2.205}}}{{\sqrt{{\mathtt{39}}}}}}{\mathtt{\,\times\,}}{\mathtt{8}} = {\mathtt{2.824\: \!660\: \!633\: \!121\: \!737}}$$
AE=AV-EV
AE=8-2.825
$${\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2.825}} = {\frac{{\mathtt{207}}}{{\mathtt{40}}}} = {\mathtt{5.175}}$$
AE is approx 5.18 cm
+118723 @@ End of Day Wrap Thurs 5/3/15 Sydney, Australia Time 11:10 pm ♪ ♫
G'day all,
This day our magnificant answerers were Alan, Rodala, Sasini, SmartCookie, TheJonyMyster, Labby, Geno3141, cmoney, Mellophonist, Heureka, Krisivier and Bertie. Thanks you, you are all wonderful.
Interest posts:
1) Changing the subject of a formula Middle level students should try it. Thanks Alan. 
2) Simultaneous equations Thanks Alan
3) A slightly unusual quadratic Thanks Alan
4) Population growth and decay Thanks Geno.
5) Probability Thanks Alan.
6) Permutations - I have no idea how to do this one. Not answered yet. 
7) Find SA when given V and r of a cylinder Thanks CPhill
8) Polar co-ordinates Thanks Alan and Heureka, and Bertie
9) This geometry one looks interesting No answer yet.
10) Finding the 4th vertex of a parallelogram. Thanks CPhill and Heureka
11) Normal distribution Melody
♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫
+118723 Fri 6/3/15
1) 3D Geometry. Melody
2) What is the co for in cosine? Melody
3) Permutation, a bit tricky. Melody
4) Finding average speed Thanks Alan
5) Selections. Everyone can try this question. It is very cool 
Melody
6) Counting. This is another counting one that everyone can try. It is just a matter of being methodical. Thanks Heureka.
7) Negative indices What do you do with them? Thanks CPhill
8) Logarithm equation Melody
9) Scaling. We have has quite a few of these lately but the concept is a really good one for everyone to understand. Thanks CPhill.
10) Average and instantaneous velocity Melody
11) Ratio of angles to sides of triangle Thanks Chris
12) What is a year ? Thanks Heureka
13) Finding the exponential function Thanks Geno
14) Ratios with simultaneous equations Melody
15) Simultaneous Equations by elimination. Heureka has laid this out beautifully. Thank you 
♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫
