• Jun 24th 2007, 11:01 PM
sanee66
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?

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• Jun 24th 2007, 11:11 PM
CaptainBlack
Quote:

Originally Posted by sanee66
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?

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In a right triangle the "legs" are the sides other than the hypotenuse, so in
$\triangle ABC$ (by Pythagoras' theorem) $AC=\sqrt{5^2+12^2} = 13$.

In $\triangle D E F$ we have the hypotenuse $DE$ is $10$, and leg $EF=8$, so:

$
DE^2 = EF^2+DF^2,
$

so:

$
DF=\sqrt{10^2-8^2}=6.
$

RonL