Question

6.4. angle and perpendicular bisectors
1.
find sv if s is the circumcenter of \\(\triangle qpr\\), \\(qr = 7.79\\), and \\(ps = 4.25\\). if necessary, round your answer to the nearest tenth.
\\(sv = \underline{\quad}\\)

Explanation:

Step1: Recall circumcenter properties

The circumcenter \( S \) of a triangle is equidistant from all vertices, so \( SQ = SR = SP \). Also, \( S \) lies on the perpendicular bisector of \( QR \), so \( V \) is the midpoint of \( QR \), meaning \( QV = VR=\frac{QR}{2} \).

Step2: Calculate \( VR \)

Given \( QR = 7.79 \), then \( VR=\frac{7.79}{2}=3.895 \).

Step3: Recognize right triangle \( \triangle SVR \)

We know \( SR = PS = 4.25 \) (since \( S \) is circumcenter, \( SR = SP \)). In right triangle \( \triangle SVR \), we can use the Pythagorean theorem \( SV=\sqrt{SR^{2}-VR^{2}} \).

Step4: Substitute values and calculate

Substitute \( SR = 4.25 \) and \( VR = 3.895 \) into the formula:
\( SV=\sqrt{4.25^{2}-3.895^{2}}=\sqrt{18.0625 - 15.171025}=\sqrt{2.891475}\approx1.7 \) (rounded to nearest tenth)

Answer:

\( 1.7 \)