I think the answer is 15*sqrt3+15
Now let me try and explain how I got this.
< KJA = 60° Ext < of a triangle is equal sum of opposite interior angles
triangle JKF is isosceles (2 equal angles) therefore KJ=JF=10
consider triangle AJK
Sin30=1/2=AJ/10 so AJ=5
Using Pythagoras' Theorum
$$AK^2=10^2-5^2
AK^2=100-25
AK^2=75
AK=5\sqrt3$$
AF=5+10=15
$$\\KF^2=15^2+(5\sqrt3)^2\\\\
KF^2=225+75\\\\
KF^2=300\\\\
KF=10\sqrt3\\\\$$
$$\\Perimeter = 5\sqrt3+10\sqrt3+15\\\\
Perimeter = 15\sqrt3+15\;\; units\\\\$$
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