Question

consider \\(\triangle tuv\\) in the figure below.
the perpendicular bisectors of its sides are \\(\overline{ad}\\), \\(\overline{bd}\\), and \\(\overline{cd}\\). they meet at a single point \\(d\\).
(in other words, \\(d\\) is the circumcenter of \\(\triangle tuv\\).)

suppose \\(cd = 36\\), \\(bv = 58\\), and \\(ud = 60\\).
find \\(uv\\), \\(vd\\), and \\(tc\\).
note: the figure is not drawn to scale.

\\(uv = \\)
\\(vd = \\)
\\(tc = \\)

Explanation:

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<reasoning>

Find UV using perpendicular bisector properties

\[

$$\begin{aligned} &BD \text{ is the perpendicular bisector of } UV \implies BV = UB = 58\\ &UV = UB + BV = 58 + 58 = 116 \end{aligned}$$

\]

Find VD using circumcenter properties

\[

$$\begin{aligned} &D \text{ is the circumcenter of } \triangle TUV \implies TD = UD = VD\\ &UD = 60 \implies VD = 60 \end{aligned}$$

\]

Find TC using the Pythagorean theorem

\[

$$\begin{aligned} &CD \text{ is the perpendicular bisector of } TU \implies \angle TCD = 90^\circ\\ &\text{In right } \triangle TCD: TC^2 + CD^2 = TD^2\\ &TC^2 + 36^2 = 60^2\\ &TC^2 = 3600 - 1296 = 2304\\ &TC = \sqrt{2304} = 48 \end{aligned}$$

\]
</reasoning>

<answer>
Consider \(\triangle TUV\) in the figure below.
The perpendicular bisectors of its sides are \(\overline{AD}\), \(\overline{BD}\), and \(\overline{CD}\). They meet at a single point \(D\).
(In other words, \(D\) is the circumcenter of \(\triangle TUV\).)

Suppose \(CD = 36\), \(BV = 58\), and \(UD = 60\).
Find \(UV\), \(VD\), and \(TC\).

\(UV =\) <blank>116</blank>
\(VD =\) <blank>60</blank>
\(TC =\) <blank>48</blank>
</answer>

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Answer:

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<reasoning>

Find UV using perpendicular bisector properties

\[

$$\begin{aligned} &BD \text{ is the perpendicular bisector of } UV \implies BV = UB = 58\\ &UV = UB + BV = 58 + 58 = 116 \end{aligned}$$

\]

Find VD using circumcenter properties

\[

$$\begin{aligned} &D \text{ is the circumcenter of } \triangle TUV \implies TD = UD = VD\\ &UD = 60 \implies VD = 60 \end{aligned}$$

\]

Find TC using the Pythagorean theorem

\[

$$\begin{aligned} &CD \text{ is the perpendicular bisector of } TU \implies \angle TCD = 90^\circ\\ &\text{In right } \triangle TCD: TC^2 + CD^2 = TD^2\\ &TC^2 + 36^2 = 60^2\\ &TC^2 = 3600 - 1296 = 2304\\ &TC = \sqrt{2304} = 48 \end{aligned}$$

\]
</reasoning>

<answer>
Consider \(\triangle TUV\) in the figure below.
The perpendicular bisectors of its sides are \(\overline{AD}\), \(\overline{BD}\), and \(\overline{CD}\). They meet at a single point \(D\).
(In other words, \(D\) is the circumcenter of \(\triangle TUV\).)

Suppose \(CD = 36\), \(BV = 58\), and \(UD = 60\).
Find \(UV\), \(VD\), and \(TC\).

\(UV =\) <blank>116</blank>
\(VD =\) <blank>60</blank>
\(TC =\) <blank>48</blank>
</answer>

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