Question

in the figure below, points m, q, and p are collinear.
find the value of x.
mn = 14.08 cm
np = 14.08
mq = x
pq = 6.4 cm

Explanation:

Step1: Identify Right Triangles

Triangles \( \triangle MNQ \) and \( \triangle PNQ \) are right triangles (since \( NQ \perp MP \)) with \( MN = NP = 14.08 \) cm, \( PQ = 6.4 \) cm, and \( MQ = x \). By the Hypotenuse - Leg (HL) congruence criterion, \( \triangle MNQ \cong \triangle PNQ \), so \( MQ = PQ \)? Wait, no, wait. Wait, actually, in right triangles, if hypotenuse \( MN = NP \) and leg \( NQ \) is common, then the other legs \( MQ \) and \( PQ \)? Wait, no, that's not right. Wait, let's use the Pythagorean theorem. For \( \triangle MNQ \): \( MN^{2}=MQ^{2}+NQ^{2} \), and for \( \triangle PNQ \): \( NP^{2}=PQ^{2}+NQ^{2} \). Since \( MN = NP \), we can set the two equations equal: \( MQ^{2}+NQ^{2}=PQ^{2}+NQ^{2} \), so \( MQ^{2}=PQ^{2} \)? No, that would imply \( MQ = PQ \), but that can't be. Wait, no, I must have misread. Wait, the figure: points \( M, Q, P \) are collinear, \( NQ \) is perpendicular to \( MP \), so \( \triangle MNQ \) and \( \triangle PNQ \) are right - angled at \( Q \). Given \( MN = NP = 14.08 \) cm, so the hypotenuses are equal, and \( NQ \) is a common leg. Therefore, by HL, \( \triangle MNQ\cong\triangle PNQ \), which would mean \( MQ = PQ \)? But \( PQ = 6.4 \) cm? Wait, no, that seems contradictory. Wait, no, maybe I made a mistake. Wait, let's check again. Wait, the lengths: \( MN = 14.08 \), \( NP = 14.08 \), \( PQ = 6.4 \), \( MQ = x \). Since \( \triangle MNQ \) and \( \triangle PNQ \) are right - angled at \( Q \), and \( MN = NP \), \( NQ \) is common, so \( \triangle MNQ\cong\triangle PNQ \) (HL). Therefore, \( MQ = PQ \), so \( x = 6.4 \)? But that seems too simple. Wait, or maybe the other way: Wait, no, maybe the problem is that \( N \) is equidistant from \( M \) and \( P \), so \( N \) lies on the perpendicular bisector of \( MP \). Since \( NQ \perp MP \), \( Q \) is the midpoint of \( MP \)? Wait, no, the perpendicular bisector theorem: if a point is equidistant from the endpoints of a segment, it lies on the perpendicular bisector. So since \( MN = NP \), \( N \) is on the perpendicular bisector of \( MP \), and since \( NQ \perp MP \), \( Q \) is the midpoint of \( MP \). Wait, but then \( MQ = PQ \), so \( x = 6.4 \)? But that seems odd. Wait, no, maybe I misread the problem. Wait, the problem says \( MN = 14.08 \) cm, \( NP = 14.08 \) (I assume cm), \( MQ = x \), \( PQ = 6.4 \) cm. So using Pythagoras in \( \triangle MNQ \): \( MN^{2}=MQ^{2}+NQ^{2}\) and in \( \triangle PNQ \): \( NP^{2}=PQ^{2}+NQ^{2}\). Since \( MN = NP \), we have \( MQ^{2}+NQ^{2}=PQ^{2}+NQ^{2}\), so \( MQ^{2}=PQ^{2}\), which implies \( MQ = PQ \) (since lengths are positive). Therefore, \( x = 6.4 \) cm? Wait, but that seems too straightforward. Wait, maybe the figure is such that \( Q \) is between \( M \) and \( P \), and \( NQ \) is the altitude. Since \( MN = NP \), the triangle \( MNP \) is isosceles with \( MN = NP \), so the altitude from \( N \) to \( MP \) (which is \( NQ \)) bisects \( MP \), so \( MQ = PQ \). Therefore, \( x = PQ = 6.4 \) cm.

Step2: Conclude the Value of \( x \)

Since \( \triangle MNQ\cong\triangle PNQ \) (HL congruence, right - angled at \( Q \), hypotenuse \( MN = NP \), leg \( NQ \) common), the corresponding legs \( MQ \) and \( PQ \) are equal. So \( x = PQ = 6.4 \) cm.

Answer:

\( x = 6.4 \) cm