Question

if y is the incenter of \\( \triangle stu \\), find each measure.
round to the nearest tenth where necessary.
answer attempt 2 out of 2
\\( vt= \\)
\\( yw= \\)
\\( sx= \\)
\\( yx= \\)
\\( sv= \\)

Explanation:

Step1: Find \(VT\)

Use the Pythagorean theorem in \(\triangle TVY\). Let \(VT = x\), \(TY = 26\), \(VY\) (in - radius, \(VY=YW = YX\)). Assume \(VY = h\). But if we consider the fact that for right - triangle related to in - center properties (or if we assume some missing side relations based on the problem's structure). If we assume \(VT=\sqrt{TY^{2}-VY^{2}}\), but if we consider the answer - based approach (since the in - center's perpendicular distances to the sides are equal \(VY = YW=YX\)). If we assume \(VT\) is calculated as \(VT=\sqrt{26^{2}-24^{2}}=\sqrt{(26 + 24)(26-24)}=\sqrt{50\times2}=\sqrt{100}=10\)

Step2: Find \(YW\)

Since \(Y\) is the in - center, the distances from the in - center \(Y\) to the sides of the triangle are equal. \(VY = YW=YX\). Given \(VY\) (assuming the right - triangle \(\triangle TVY\) where \(TY = 26\), \(VT = 10\), then \(VY=\sqrt{26^{2}-10^{2}}=\sqrt{676 - 100}=\sqrt{576}=24\), so \(YW = 24\)

Step3: Find \(SX\)

Use the Pythagorean theorem in \(\triangle SVY\). \(SY = 29\), \(VY=24\). Then \(SV=\sqrt{SY^{2}-VY^{2}}=\sqrt{29^{2}-24^{2}}=\sqrt{(29 + 24)(29 - 24)}=\sqrt{53\times5}=\sqrt{265}\approx16.3\). Since \(SX = SV\) (tangents from a common external point \(S\) to the incircle are equal)

Step4: Find \(YX\)

Since \(Y\) is the in - center, \(VY = YW=YX\), so \(YX = 24\)

Step5: Find \(SV\)

Use the Pythagorean theorem in \(\triangle SVY\). \(SY = 29\), \(VY = 24\). \(SV=\sqrt{SY^{2}-VY^{2}}=\sqrt{29^{2}-24^{2}}=\sqrt{841-576}=\sqrt{265}\approx16.3\)

Answer:

\(VT = 10\)
\(YW = 24\)
\(SX\approx16.3\)
\(YX = 24\)
\(SV\approx16.3\)