Question

in the figure below, points q, r, and s are collinear.
find the length of \\( \overline { q r } \\).
\\( q p = 7 \mathrm { ft } \\)
\\( p s = 7 \mathrm { ft } \\)
\\( q r = ( 8 x - 92 ) \mathrm { ft } \\)
\\( s r = ( 2 x - 20 ) \mathrm { ft } \\)

Explanation:

Step1: Identify triangle congruence

Since \(QP = PS = 7\) ft and \(PR\) is perpendicular to \(QS\) (right angle at \(R\)), triangles \(QPR\) and \(SPR\) are congruent (by HL congruence, as \(PR\) is common hypotenuse? Wait, no, \(QP = PS\), \(PR\) is perpendicular, so \(QR = SR\) (corresponding parts of congruent triangles). Wait, actually, in right triangles \(QPR\) and \(SPR\), \(QP = PS\) and \(PR\) is common, so by HL, \(\triangle QPR \cong \triangle SPR\), so \(QR = SR\).
So set \(8x - 92 = 2x - 20\)? Wait, no, wait. Wait, \(Q\), \(R\), \(S\) are collinear, and \(PR\) is perpendicular to \(QS\), so \(PR\) is the perpendicular bisector? Wait, \(QP = PS\), so \(P\) is equidistant from \(Q\) and \(S\), so \(PR\) is the perpendicular bisector, so \(QR = SR\). Wait, but the lengths are \(QR = 8x - 92\) and \(SR = 2x - 20\)? Wait, that can't be, because if \(QR = SR\), then \(8x - 92 = 2x - 20\), but solving that gives \(6x = 72\), \(x = 12\), then \(QR = 8*12 - 92 = 96 - 92 = 4\), \(SR = 2*12 - 20 = 24 - 20 = 4\), which works. Wait, but let's check again.

Wait, \(QP = 7\), \(PS = 7\), \(PR\) is perpendicular to \(QS\), so triangles \(QPR\) and \(SPR\) are right triangles with \(QP = PS\) and \(PR\) common, so they are congruent (HL). Therefore, \(QR = SR\). So:

\(8x - 92 = 2x - 20\)

Step2: Solve for \(x\)

Subtract \(2x\) from both sides: \(6x - 92 = -20\)

Add 92 to both sides: \(6x = 72\)

Divide by 6: \(x = 12\)

Step3: Find \(QR\)

Substitute \(x = 12\) into \(QR = 8x - 92\):

\(QR = 8*12 - 92 = 96 - 92 = 4\) ft

Wait, but let's check \(SR\): \(2*12 - 20 = 24 - 20 = 4\) ft, which matches, so that's correct.

Answer:

The length of \(\overline{QR}\) is \(\boxed{4}\) feet.