Question

in the figure below, points w, s, and e are collinear.
find the value of x.
wn = 13 cm
ne = 13 cm
ws = (9x - 181) cm
es = 8 cm

Explanation:

Step1: Identify triangle congruence or right triangle

Since \(WN = NE = 13\) cm and \(\angle WSN=\angle ESN = 90^\circ\) (right angle at \(S\)), triangles \(WSN\) and \(ESN\) are congruent by HL (Hypotenuse - Leg) theorem. So \(WS = ES\)? Wait, no, wait. Wait, \(WN = NE = 13\), \(SN\) is common, and right angles. So \(WS = ES\)? Wait, no, \(ES = 8\), but \(WS=(9x - 181)\). Wait, maybe I made a mistake. Wait, actually, since \(N\) is equidistant from \(W\) and \(E\) (\(WN = NE\)) and \(SN\) is perpendicular to \(WE\) (since \(\angle WSN\) and \(\angle ESN\) are right angles), then \(S\) is the midpoint? Wait, no, if \(SN\) is the perpendicular bisector, then \(WS = ES\). But \(ES = 8\), so \(WS = 8\)? Wait, no, that can't be. Wait, maybe the figure is such that \(WN = 13\), \(WS=(9x - 181)\), \(SN\) is perpendicular, so by Pythagoras in triangle \(WSN\): \(WN^2=WS^2 + SN^2\), and in triangle \(ESN\): \(NE^2=ES^2 + SN^2\). Since \(WN = NE = 13\), then \(WS^2 + SN^2=ES^2 + SN^2\), so \(WS^2=ES^2\), so \(WS = ES\) (since lengths are positive). Wait, \(WS=(9x - 181)\), \(ES = 8\), so \(9x - 181=8\)? Wait, no, that would give \(9x=189\), \(x = 21\). Wait, let's check.

Wait, let's do Pythagoras. In right triangle \(WSN\): \(WN^2=WS^2 + SN^2\) => \(13^2=(9x - 181)^2 + SN^2\). In right triangle \(ESN\): \(NE^2=ES^2 + SN^2\) => \(13^2=8^2 + SN^2\). So set the two equations equal: \((9x - 181)^2 + SN^2=8^2 + SN^2\). Subtract \(SN^2\) from both sides: \((9x - 181)^2=8^2\). Take square roots (since length is positive, \(9x - 181 = 8\) or \(9x - 181=-8\)). But \(9x - 181\) is a length, so it must be positive. So \(9x - 181 = 8\) => \(9x=189\) => \(x = 21\). Or \(9x - 181=-8\) => \(9x=173\) => \(x=\frac{173}{9}\approx19.22\), but then \(WS\) would be negative, which is impossible. So \(9x - 181 = 8\).

Step1: Set up equation from Pythagoras

Since \(WN = NE = 13\), \(\angle WSN=\angle ESN = 90^\circ\), so by Pythagoras:
\(WN^2 = WS^2 + SN^2\) and \(NE^2 = ES^2 + SN^2\).
Subtracting the two equations (or equating them) gives \(WS^2 = ES^2\).

Step2: Solve for \(x\)

Since \(WS=(9x - 181)\) and \(ES = 8\), and \(WS>0\) (length), we have:
\(9x - 181 = 8\)
Add 181 to both sides:
\(9x=8 + 181=189\)
Divide by 9:
\(x=\frac{189}{9}=21\)

Answer:

\(x = 21\)