Find the perimeter of the triangle.

[BTGmoderatorLU]

In the figure above, triangle ABC is inscribed in the circle with center O, such that CD is perpendicular to AB. If the length of side AC is 5 and the radius equals r=25/8, find the perimeter of the triangle.

(A) 14
(B) 15
(C) 16
(D) 20
(E) none of the above

The OA is C.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

[EconomistGMATTutor]

Hello LUANDATO.

Let's take a look at your question.

As CD is perpendicular to AB then AD=DB. Let's name this lenght as "y".

So, the right triangles ADC and BDC has two sides of the same lenght, so BC=AC=5.

By the other hand, CO=r=25/8. Let's name OD as "x".

We need to find "y" to get the perimeter.

Using the Pythagoras theorem we have that: $$5^2=y^2+\left(\frac{25}{8}+x\right)^2\ \leftrightarrow\ 25=y^2+\frac{625}{64}+\frac{25}{4}x+x^2$$ and $$\left(\frac{25}{8}\right)^2=y^2+x^2\ \leftrightarrow\ y^2+x^2=\frac{625}{64}.$$ Replacing this last equation in the first one we have: $$25=\frac{625}{64}+\frac{625}{64}+\frac{25}{4}x\ \leftrightarrow\ 25=\frac{625}{32}+\frac{25}{4}x\ \leftrightarrow\ x=\frac{7}{8}.$$ Now, we can find the value of "y". $$y^2+\left(\frac{7}{8}\right)^2=\frac{625}{64}\ \leftrightarrow\ y^2=\frac{576}{64}\ \leftrightarrow\ y=3.$$ Now, the perimeter of the triangle is $$P=AC+AB+BC=5+2\cdot3+5=16.$$ So, the correct answer is C.

I hope this explanation can help you.

I'm available if you'd like a follow up.

Regards.