I can think of three ways. I'll try them all.
This is the construction for the geometric mean of [tex]a[/tex] and [tex]b[/tex], so [tex]EG=\sqrt{ab}[/tex]. That's no fun, but I suppose it's nice to know the answer going in.
We have three similar triangles, because the two little ones have a common angle and a right angle, so same angles as the big one.
[tex]\dfrac{a}{DE} = \dfrac{EG}{EF} = \dfrac{DE}{a+b}[/tex]
[tex]\dfrac{EG}{DE} = \dfrac{b}{EF} = \dfrac{EF}{a+b}[/tex]
So
[tex]EF^2 = b(a+b)[/tex]
[tex]DE^2 = a(a+b)[/tex]
[tex]EG = \dfrac{b \ DE}{EF}[/tex]
[tex]EG^2 = \dfrac{b^2 DE^2}{EF^2} = \dfrac{ab^2(a+b)}{b(a+b)}=ab[/tex]
There it is again.
I guess the way they really want you to do it is to write the Pythagorean Theorem a few times. Let's abbreviate [tex]d=ED, f=EF, g=EG[/tex]
[tex]g^2 + a^2 = d^2[/tex]
[tex]g^2 + b^2 = f^2[/tex]
[tex]d^2 + f^2 = (a+b)^2[/tex]
[tex]g[/tex] is our unknown. Adding all three equations,
[tex]g^2 + a^2 + g^2 + b^2+ d^2 + f^2= d^2 + f^2 + a^2 + b^2 + 2ab[/tex]
[tex]2g^2=2ab[/tex]
[tex]g^2=ab[/tex]
That's three different ways. Plugging in the numbers,
[tex]\sqrt{2(16)}=4\sqrt{2} \approx 5.657 \quad[/tex] fourth choice.
I hate it when they ruin a nice exact answer with an approximation.
