Well, XY is 10 you are right about that.
So long as the method is correct I am not too concerned.
I am sure you can fix the careless errors I thought there was likely to be some.
I take it you were happy with the method ?
There is probably a much easier way, I have a reputation for taking the scenic route. :))
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
The method looks good Melody, but, almost at the top, shouldn't XY = 10 rather than 8 ?
Well, XY is 10 you are right about that.
So long as the method is correct I am not too concerned.
I am sure you can fix the careless errors I thought there was likely to be some.
I take it you were happy with the method ?
There is probably a much easier way, I have a reputation for taking the scenic route. :))