In a right triangle ABC, the legs are AB = 5 and BC = 12, and in the right triangle DEF, the hypotenuse is DE = 10, and EF = 8. Which is DF + AC?19
15
13
12
In a right triangle the "legs" are the sides other than the hypotenuse, so in
$\displaystyle \triangle ABC$ (by Pythagoras' theorem) $\displaystyle AC=\sqrt{5^2+12^2} = 13$.
In $\displaystyle \triangle D E F$ we have the hypotenuse $\displaystyle DE$ is $\displaystyle 10$, and leg $\displaystyle EF=8$, so:
$\displaystyle
DE^2 = EF^2+DF^2,
$
so:
$\displaystyle
DF=\sqrt{10^2-8^2}=6.
$
RonL