Question

geometry-alonso (lms)
5-1: mathxl (a) (lms graded)
use the figure to the right to find jm.
jm = \square units (type an integer or a decimal.)
(figure shows a triangle jmk with a perpendicular segment nb from n to jk, with jb = bk (marked with tick marks), and sides nj = 4y - 10, nk = 2y)

Explanation:

Step1: Identify the property

Since \( NB \) is perpendicular to \( JK \) and \( B \) is the midpoint of \( JK \) (marked by the congruent segments), triangle \( JNK \) is isosceles with \( JM = KM \)? Wait, no, actually, \( JM \) and \( KM \)? Wait, the sides \( JM \) and \( KM \)? Wait, the lengths given are \( JM = 4y - 10 \) and \( KM = 2y \)? Wait, no, looking at the figure, \( JM \) and \( KM \)? Wait, no, \( JN = 4y - 10 \) and \( KN = 2y \)? Wait, no, the figure shows \( JN = 4y - 10 \) and \( KN = 2y \), and \( NB \) is the perpendicular bisector, so \( JN = KN \)? Wait, no, in an isosceles triangle, if a perpendicular is drawn from the vertex to the base, it bisects the base and the two equal sides. Wait, actually, since \( B \) is the midpoint (the marks show \( JB = BK \)) and \( NB \perp JK \), then triangle \( JNK \) is isosceles with \( JN = KN \). So \( 4y - 10 = 2y \).

Step2: Solve for \( y \)

\( 4y - 10 = 2y \)
Subtract \( 2y \) from both sides: \( 4y - 2y - 10 = 0 \)
\( 2y - 10 = 0 \)
Add 10 to both sides: \( 2y = 10 \)
Divide by 2: \( y = 5 \)

Step3: Find \( JM \)? Wait, no, wait, maybe \( JM \) is equal to \( KM \)? Wait, no, maybe \( JM \) is \( JN \)? Wait, the problem says "find \( JM \)". Wait, maybe the sides \( JN \) and \( KN \) are equal? Wait, no, maybe \( JM \) is \( JN \)? Wait, let's re-examine. The figure has \( J \), \( K \), \( N \), with \( NB \perp JK \), \( JB = BK \), so triangle \( JNK \) is isosceles with \( JN = KN \). So \( JN = 4y - 10 \), \( KN = 2y \). So set them equal: \( 4y - 10 = 2y \), solve for \( y = 5 \). Then, maybe \( JM \) is \( JN \)? Wait, no, maybe \( JM \) is \( JN \), so substitute \( y = 5 \) into \( 4y - 10 \): \( 4(5) - 10 = 20 - 10 = 10 \)? Wait, no, \( 2y = 10 \), so \( KN = 10 \), and \( JN = 10 \). Wait, but maybe \( JM \) is \( JN \), so \( JM = 10 \)? Wait, no, maybe I misread. Wait, the problem says "find \( JM \)". Wait, maybe the sides \( JM \) and \( KM \) are \( JN \) and \( KN \). So solving \( 4y - 10 = 2y \) gives \( y = 5 \), then \( JM = 4y - 10 = 4*5 - 10 = 10 \), or \( KM = 2y = 10 \). Wait, but maybe \( JM \) is \( JN \), so \( JM = 10 \). Wait, no, maybe the figure is a triangle where \( JM \) and \( KM \) are the two equal sides? Wait, no, the labels: \( J \), \( K \), \( N \), with \( N \) at the top, \( JK \) as the base, \( NB \) perpendicular to \( JK \), \( JB = BK \). So \( JN = KN \), so \( 4y - 10 = 2y \), solve for \( y = 5 \), then \( JM \) (which is \( JN \)) is \( 4*5 - 10 = 10 \), or \( KM = 2*5 = 10 \). Wait, but maybe \( JM \) is \( JN \), so \( JM = 10 \). Wait, no, maybe I made a mistake. Wait, let's check again.

Wait, the problem says "Use the figure to the right to find \( JM \)". The figure shows \( JN = 4y - 10 \), \( KN = 2y \), and \( NB \) is the perpendicular bisector, so \( JN = KN \). So:

\( 4y - 10 = 2y \)

Subtract \( 2y \): \( 2y - 10 = 0 \)

Add 10: \( 2y = 10 \)

Divide by 2: \( y = 5 \)

Then, \( JM \) is \( JN \), so \( JM = 4y - 10 = 4(5) - 10 = 20 - 10 = 10 \). Wait, but \( KM = 2y = 10 \), so both are 10. So \( JM = 10 \).

Answer:

\( \boxed{10} \)