Question

geometry
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assignment
name_____________________
date_____________________ pe
find the missing length indicated.

1. find eg
2. find ge
3. find yw
4. find zx
5. find uv
6. find tu

Explanation:

Step1: Identify the theorem

This is a midline theorem (also called midsegment theorem) problem. The midline of a triangle is parallel to the third side and half its length. In the first diagram, \( KJ \) is the midline? Wait, no, let's check the markings. The segments \( FK = KG \) (since there are two marks), \( FJ = JE \) (two marks), so \( KJ \) is parallel to \( GE \) and \( KJ=\frac{1}{2}GE \)? Wait, no, maybe \( KJ \) is the midline. Wait, the length of \( KJ \) is 6, \( FJ = 7 \), so \( FE = 14 \) (since \( FJ = JE = 7 \)). Wait, maybe I got it wrong. Let's take problem 1: Find \( EG \).

Looking at triangle \( FGE \), with \( K \) on \( FG \) (since \( FK = KG \), so \( K \) is the midpoint) and \( J \) on \( FE \) ( \( FJ = JE \), so \( J \) is the midpoint). Then by midline theorem, \( KJ \parallel EG \) and \( KJ=\frac{1}{2}EG \). Wait, \( KJ = 6 \), so \( EG = 2\times KJ = 12 \)? Wait, no, maybe the other way. Wait, \( FG \) has length \( FK + KG = 5 + 5 = 10 \)? Wait, the diagram shows \( FK \) with length 5 (marked with two ticks), so \( FK = KG = 5 \), so \( FG = 10 \). \( FJ = 7 \), so \( JE = 7 \), so \( FE = 14 \). Then \( KJ \) is the midline, so \( KJ \parallel EG \) and \( KJ=\frac{1}{2}EG \). Wait, \( KJ = 6 \), so \( EG = 12 \)? Wait, maybe I mixed up the sides. Alternatively, maybe \( KJ \) is parallel to \( EG \), and \( FJ/F E= KJ/EG \). Since \( FJ = 7 \), \( FE = 14 \) (because \( J \) is the midpoint), so \( 7/14 = 6/EG \), so \( 1/2 = 6/EG \), so \( EG = 12 \). Wait, that makes sense. So \( EG = 12 \).

Step2: Apply midline theorem

For problem 1: \( K \) is midpoint of \( FG \) ( \( FK = KG \) ), \( J \) is midpoint of \( FE \) ( \( FJ = JE \) ). By midline theorem, \( KJ \parallel EG \) and \( KJ = \frac{1}{2}EG \). Given \( KJ = 6 \), so \( EG = 2 \times KJ = 12 \).

Answer:

For problem 1, \( EG = 12 \) (assuming the above steps are correct; the exact answer depends on the diagram's markings, but using midline theorem, if \( KJ = 6 \) and \( KJ \) is midline, then \( EG = 12 \)).