Question

In triangle $ABC,$ $AB = BC = 17$ and $AC = 16.$ Find the circumradius of triangle $ABC.$

Answer 1

Answer:

$R = 9.63\ units$

Step-by-step explanation:

Given

$AB = BC = 17$

$AC = 16$

Required

Determine the circumradius, R

The circumradius is calculated as follows;

$R = \frac{AB * BC * AC}{\sqrt{(AB + BC + AC)(AB + BC - AC)(AB + AC - BC)(AC + BC - AB)}}$

Substitute the values of AB, BC and AC

$R = \frac{17 * 17 * 16}{\sqrt{(17 + 17 + 16)(17 + 17 - 16)(17 + 16 - 17)(16 + 17 - 17)}}$

Evaluate the denominator

$R = \frac{17 * 17 * 16}{\sqrt{(50)(18)(16)(16)}}$

$R = \frac{4624}{\sqrt{230400}}$

Take square root of 230400

$R = \frac{4624}{480}$

$R = 9.63\ units$ (Approximated)

Hence, the circumradius is 9.63

Answer 2

Answer:

289/30

Step-by-step explanation:

Let the circumcenter be point O. We start by drawing line median BM. Since AB = BC, median BM is perpendicular to side AC.

Therefore, BM is part of the perpendicular bisector of AC and thus, must pass through point O.

We have AM = 8, so the Pythagorean Theorem applied to triangle ABM gives us BM = 15.

Let OA = x, our circumradius. Since O is equidistant from A and B, we have OB = x as well.

Therefore,

OM = BM - BO = 15 - x.

From right triangle OAM, we have (OA)^2 = (OM)^2 + (AM)^2.

Solving for x, we have 30x = 225 + 64.

So, x = 289/30.