Question

consider △pqr in the figure below.
the perpendicular bisectors of its sides are \\(\overline{sv}\\), \\(\overline{tv}\\), and \\(\overline{uv}\\). they meet at a single point \\(v\\).
(in other words, \\(v\\) is the circumcenter of \\(\triangle pqr\\).)

suppose \\(uv = 48\\), \\(ps = 62\\), and \\(rv = 80\\).
find \\(pu\\), \\(pr\\), and \\(qv\\).
note that the figure is not drawn to scale.

figure of triangle pqr with perpendicular bisectors sv, tv, uv meeting at v, and right angles at u, t, s. there is a box to fill in pu, pr, qv values.

Explanation:

Step1: Find \( PU \)

Since \( \overline{UV} \) is the perpendicular bisector of \( \overline{PQ} \), triangle \( UVP \) is a right triangle with \( \angle U = 90^\circ \), \( RV = 80 \) (circumradius, so \( PV = RV = 80 \)) and \( UV = 48 \). Using the Pythagorean theorem: \( PU = \sqrt{PV^2 - UV^2} = \sqrt{80^2 - 48^2} \). Calculate \( 80^2 = 6400 \), \( 48^2 = 2304 \), so \( 6400 - 2304 = 4096 \), and \( \sqrt{4096} = 64 \). So \( PU = 64 \).

Step2: Find \( PR \)

Since \( \overline{SV} \) is the perpendicular bisector of \( \overline{PR} \), \( PS = SR = 62 \). Thus, \( PR = PS + SR = 62 + 62 = 124 \).

Step3: Find \( QV \)

The circumcenter \( V \) is equidistant from all vertices, so \( QV = RV = 80 \).

Answer:

\( PU = 64 \)
\( PR = 124 \)
\( QV = 80 \)