Respuesta :

Answer:

We use Pythagoras' theorem:

[tex] \bf {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \\ \bf {AC}^{2} = ( \frac{12}{13})^{2} + ( \frac{5}{13} )^{2} \\ \\ \bf {AC}^{2} = \frac{ {12}^{2} }{ {13}^{2} } + \frac{ {5}^{2} }{ {13}^{2} } \\ \\ \bf {AC}^{2} = \frac{144}{169} + \frac{25}{169} \\ \\ \bf {AC}^{2} = \frac{169}{169} \\ \\ \bf {AC}^{2} = 1 \\ \\ \bf AC = \sqrt{1} \Rightarrow \red{ \boxed{ \bf AC = 1} } [/tex]

GabrieII

Answer:

AC = 1

Step-by-step explanation:

This is Pythagoras' theorem, which states that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse.

⇒ Th.P ⇒ AC² = AB² + BC²

<=> AC² = (12/13)² + (5/13)²

<=> AC² = 144+25/169

<=> AC² = 169/169

<=> AC² = 1

⇒ AC = √1 = 1.

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