Respuesta :

gmany
Look at the picture.

ΔABD and ΔBCD are similar, therefore:

[tex]\dfrac{y}{5}=\dfrac{11}{y}\ \ \ |\cross\ multiply\\\\y^2=55\to y=\sqrt{55}[/tex]

Use Pythagorean theorem:

[tex]x^2=5^2+(\sqrt{55})^2\\\\x^2=25+55\\\\x^2=80\to x=\sqrt{80}\\\\x=\sqrt{16\cdot5}\\\\x=\sqrt{16}\cdot\sqrt5\\\\\boxed{x=4\sqrt5}[/tex]
Ver imagen gmany

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