Math Problem Statement

Consider ΔTUV in the figure below. The perpendicular bisectors of its sides are AD, BD, and CD. They meet at a single point D. (In other words, D is the circumcenter of ΔTUV.) Suppose CD = 36, UV = 102, and UD = 60. Find BV, VD, and TC.

Solution

The triangle TUV\triangle TUV has perpendicular bisectors meeting at the circumcenter DD. Given:

  • CD=36CD = 36
  • UV=102UV = 102
  • UD=60UD = 60

We need to find:

  • BVBV
  • VDVD
  • TCTC

Key Idea:

Since DD is the circumcenter, it is equidistant from all vertices of the triangle. This means:

UD=VD=CD=RUD = VD = CD = R

where RR is the circumradius.

Therefore:

  • CD=36CD = 36R=36R = 36
  • VD=36VD = 36

BVBV:

The entire length of side UV=102UV = 102 is split symmetrically since BDBD is a perpendicular bisector.

BV=UV2=1022=51BV = \frac{UV}{2} = \frac{102}{2} = 51

TCTC:

The circumradius also applies for TCTC. Since it is symmetric and similar logic applies:

TC=R=36TC = R = 36

Final Answers:

  • BV=51BV = 51
  • VD=36VD = 36
  • TC=36TC = 36

Would you like me to explain the properties of the circumcenter further or explore a different geometric concept?

Related Questions:

  1. What is the circumradius formula for a triangle given its sides?
  2. How is the circumcenter related to the triangle's angles?
  3. Why does the perpendicular bisector pass through the circumcenter?
  4. What is the difference between the circumcenter and the centroid?
  5. How can you construct the circumcenter using a compass and straightedge?

Tip: The circumcenter lies inside the triangle for acute triangles, on the hypotenuse for right triangles, and outside for obtuse triangles.

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Math Problem Analysis

Mathematical Concepts

Geometry
Properties of the Circumcenter
Triangle Perpendicular Bisectors

Formulas

The circumcenter is equidistant from all vertices of the triangle
Perpendicular bisector divides a side of a triangle into two equal parts

Theorems

Circumcenter Theorem

Suitable Grade Level

Grades 8-10