Question

in exercises 11–14, n is the incenter of △abc. use the given information to find the indicated measure. (see example 3.) 11. nd = 6x - 2 ne = 3x + 7 find nf. 12. ng = x + 3 nh = 2x - 3 find nj. 13. nk = 2x - 2 nl = -x + 10 find nm. 14. nq = 2x nr = 3x - 2 find ns.

Explanation:

Problem 11

Step1: Recall incenter property

The incenter \( N \) of a triangle is equidistant from all sides, so \( ND = NE = NF \). Set \( ND = NE \):
\( 6x - 2 = 3x + 7 \)

Step2: Solve for \( x \)

Subtract \( 3x \) from both sides:
\( 3x - 2 = 7 \)
Add 2 to both sides:
\( 3x = 9 \)
Divide by 3:
\( x = 3 \)

Step3: Find \( NF \) (equal to \( ND \) or \( NE \))

Substitute \( x = 3 \) into \( ND \):
\( ND = 6(3) - 2 = 18 - 2 = 16 \)
Thus, \( NF = 16 \).

Step1: Recall incenter property

The incenter \( N \) is equidistant from all sides, so \( NG = NH = NJ \). Set \( NG = NH \):
\( x + 3 = 2x - 3 \)

Step2: Solve for \( x \)

Subtract \( x \) from both sides:
\( 3 = x - 3 \)
Add 3 to both sides:
\( x = 6 \)

Step3: Find \( NJ \) (equal to \( NG \) or \( NH \))

Substitute \( x = 6 \) into \( NG \):
\( NG = 6 + 3 = 9 \)
Thus, \( NJ = 9 \).

Step1: Recall incenter property

The incenter \( N \) is equidistant from all sides, so \( NK = NL = NM \). Set \( NK = NL \):
\( 2x - 2 = -x + 10 \)

Step2: Solve for \( x \)

Add \( x \) to both sides:
\( 3x - 2 = 10 \)
Add 2 to both sides:
\( 3x = 12 \)
Divide by 3:
\( x = 4 \)

Step3: Find \( NM \) (equal to \( NK \) or \( NL \))

Substitute \( x = 4 \) into \( NK \):
\( NK = 2(4) - 2 = 8 - 2 = 6 \)
Thus, \( NM = 6 \).

Answer:

\( 16 \)

Problem 12