Question

determining a segment length
segment tq is 26 units long. what is the length of qv?
8 units
26 units
31 units
32 units

Explanation:

Step1: Find the value of \(x\)

Since \(SR\) is the perpendicular bisector of \(TV\), \(ST = SV\).
So, \(3x + 2=4x - 1\).
Subtract \(3x\) from both sides: \(2=x - 1\).
Add \(1\) to both sides: \(x = 3\).

Step2: Find the length of \(TV\)

First, find \(ST\): substitute \(x = 3\) into \(3x+2\), \(ST=3\times3 + 2=9 + 2=11\).
Since \(TQ = 26\) and by the property of perpendicular bisector (in a kite - like figure formed here, \(TQ=VQ\) as \(SR\) is the perpendicular bisector), and also \(TV\) is bisected by \(SR\).
We know that in the figure (by properties of perpendicular bisector and congruent segments \(ST = SV\), \(TQ=VQ\)), \(TV\) can be found. But another way:
Since \(x = 3\), \(SV=4x-1=4\times3 - 1=11\).
In right - triangle \(TRS\) and \(VRS\) (right - angled at \(R\)), and \(ST = SV\), \(SR = SR\) (common side), so \(TR = VR\).
We know that \(TQ = 26\), and in the figure (using the property of the perpendicular bisector and congruent triangles formed), \(VQ=TQ\) (by SSS or SAS congruence of triangles \(STQ\) and \(SVQ\)). But if we consider the relation between the segments:
Since \(x = 3\), \(ST=11\), and in right - triangle \(STR\) and \(VQR\) (noting the symmetry), \(TV\) is related to \(TQ\) and \(SV\). But a simpler way:
Since \(ST = SV\) (proven from \(3x + 2=4x - 1\) with \(x = 3\)), and \(TQ = 26\).
We use the fact that \(VQ=TQ\) (by congruence of \(\triangle STQ\) and \(\triangle SVQ\) ( \(ST = SV\), \(SQ=SQ\), \(\angle TSQ=\angle VSQ\) (because \(SR\) is the perpendicular bisector and \(ST = SV\)))).

Answer:

A. 8 units