A surveyor maps a triangular plot of land, LMN. A boundary line segment is drawn from poin… — SAT

A surveyor maps a triangular plot of land, $LMN$ . A boundary line segment is drawn from point $P$ on side $LN$ to point $Q$ on side $LM$ , creating a smaller triangular section $LPQ$ . The measure of angle $LMN$ is equal to the measure of angle $LPQ$ . The surveyed lengths are $LM = 18$ meters, $LP = 8$ meters, and $LQ = 12$ meters. What is the length, in meters, of segment $PN$ ?

Answer: 19 meters

Answer

By identifying that triangle

LMN

and triangle

LPQ

share the angle at vertex

L

and have another pair of congruent angles (

\angle LMN = \angle LPQ

), we establish similarity between the two triangles:

\triangle LMN \sim \triangle LPQ

. Using the proportional relationship of their corresponding sides, we write

\frac{LM}{LP} = \frac{LN}{LQ}

. Substituting the given values

LM = 18

LP = 8

, and

LQ = 12

yields

\frac{18}{8} = \frac{LN}{12}

, which solves to

LN = 27

. Finally, subtracting

LP = 8

from the total length of segment

LN

gives the length of segment

PN

19

meters.

Step-by-Step Solution

Identify similar triangles.

Triangle

LMN

is similar to triangle

LPQ

(

\triangle LMN \sim \triangle LPQ

They share angle

L

(

\angle MLN = \angle PLQ

) and we are given that

\angle LMN = \angle LPQ

. By the Angle-Angle (AA) similarity criterion, the two triangles are similar.

Set up the ratio of corresponding sides.

\frac{LM}{LP} = \frac{LN}{LQ}

Corresponding sides of similar triangles are proportional.

Solve for the length of side

LN

LN = 27

meters

Substituting

LM = 18

LP = 8

, and

LQ = 12

into the proportion yields

\frac{18}{8} = \frac{LN}{12}

. Solving for

LN

gives

LN = 12 \times \frac{18}{8} = 27

Calculate the length of segment

PN

PN = 19

meters

Point

P

lies on segment

LN

, so the length of segment

PN

is the difference between

LN

and

LP

, which is

27 - 8 = 19

Key Concept

Triangle similarity using the Angle-Angle (AA) criterion and proportional side ratios.

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