In the figure below, the segments BC and BD are tangent to the circle centered at O. Given that =OC2.8 and =OB5.3, find BD

In the figure below the segments BC and BD are tangent to the circle centered at O Given that OC28 and OB53 find BD class=

Respuesta :

Answer:

4.5

Step-by-step explanation:

First things first - the triangle OCB is similar to ODB - because they are symmetrically reflected over OB. They are also both right triangles, because the tangent is perpendicular to the corresponding radius.

BD = BC

We also know that (because it's a right triangle):

[tex]OB = \sqrt{OC^2 + BC^2}\\\\5.3 = \sqrt{2.8^2 + BC^2}\\\\5.3^2 = 2.8^2 + BC^2\\\\BC = \sqrt{5.3^2 - 2.8^2} = \sqrt{28.09 - 7.84}= \sqrt{20.25} = 4.5\\\\BD = BC = 4.5[/tex]