Question

consider $\triangle xyz$ in the figure below.
the perpendicular bisectors of its sides are $\overline{ts}$, $\overline{us}$, and $\overline{vs}$. they meet at a single point $s$.
(in other words, $s$ is the circumcenter of $\triangle xyz$.)

suppose $vs = 48$, $yu = 66$, and $zs = 80$.
find $xs$, $yz$, and $vy$.
note that the figure is not drawn to scale.

your answer

figure of triangle xyz with circumcenter s, perpendicular bisectors ts, us, vs, right angles at t, u, v

$xs = \square$
$yz = \square$
$vy = \square$
buttons: reset, start over

Explanation:

Step1: Recall circumcenter properties

The circumcenter \( S \) is equidistant from all vertices of \( \triangle XYZ \), so \( XS = YS = ZS \). Given \( ZS = 80 \), we have \( XS = ZS = 80 \).

Step2: Find \( YZ \) using perpendicular bisector

\( US \) is the perpendicular bisector of \( YZ \), so \( YU = UZ = 66 \). Then \( YZ = YU + UZ = 66 + 66 = 132 \).

Step3: Find \( VY \) using Pythagorean theorem

\( VS \) is the perpendicular bisector of \( XY \), so \( \triangle VYS \) is a right triangle with \( VS = 48 \) and \( YS = ZS = 80 \). Let \( VY = x \). By Pythagoras:
\( VY^2 + VS^2 = YS^2 \)
\( x^2 + 48^2 = 80^2 \)
\( x^2 = 80^2 - 48^2 = 6400 - 2304 = 4096 \)
\( x = \sqrt{4096} = 64 \).

Answer:

\( XS = 80 \)
\( YZ = 132 \)
\( VY = 64 \)