Question

find gh.
gh =
submit

Explanation:

Step1: Identify Similar Triangles

Since \( IG \parallel FJ \) (implied by the midline or proportional segments), triangles \( \triangle IGH \) and \( \triangle FJH \) are similar by the Basic Proportionality Theorem (Thales' theorem).

Step2: Set Up Proportion

Let \( GH = x \). The ratio of corresponding sides in similar triangles is equal. So, \(\frac{IH}{JH}=\frac{GH}{FH}\). First, find \( IH = JH - JI = 49 - 28 = 21 \). Then, \(\frac{21}{49}=\frac{x}{35}\).

Step3: Solve for \( x \)

Cross - multiply: \( 49x = 21\times35 \). Calculate \( 21\times35 = 735 \). Then, \( x=\frac{735}{49}=15 \)? Wait, no, wait. Wait, maybe I mixed up the segments. Wait, actually, the segment \( FH = 35 \), \( JH = 49 \), \( JI = 28 \), so \( IH=49 - 28 = 21 \). Wait, maybe the correct proportion is \(\frac{JI}{JH}=\frac{FJ - FG}{FH}\)? No, better to use the midline or the proportionality. Wait, another approach: The line \( IG \) divides \( JH \) and \( FH \) proportionally. So \(\frac{JI}{JH}=\frac{FG}{FH}\)? No, wait, \( JI = 28 \), \( JH = 49 \), so the ratio of \( JI \) to \( JH \) is \(\frac{28}{49}=\frac{4}{7}\). Then, the ratio of \( GH \) to \( FH \) should be the same as the ratio of \( IH \) to \( JH \). Wait, \( IH=49 - 28 = 21 \), so \(\frac{IH}{JH}=\frac{21}{49}=\frac{3}{7}\)? No, I think I made a mistake. Wait, let's start over.

Wait, the two triangles: \( \triangle IGH \) and \( \triangle FJH \). The sides \( JI = 28 \), \( JH = 49 \), so \( \frac{JI}{JH}=\frac{28}{49}=\frac{4}{7} \), so the remaining part \( IH=\frac{3}{7}\) of \( JH \)? No, \( 49-28 = 21 \), and \( \frac{21}{49}=\frac{3}{7} \), \( \frac{28}{49}=\frac{4}{7} \). Now, since the triangles are similar, the ratio of \( GH \) to \( FH \) is equal to the ratio of \( IH \) to \( JH \). So \( \frac{GH}{35}=\frac{21}{49} \). Simplify \( \frac{21}{49}=\frac{3}{7} \). Then \( GH = 35\times\frac{3}{7}=15 \)? Wait, that can't be. Wait, maybe the correct proportion is \(\frac{JI}{JH}=\frac{FH - GH}{FH}\). So \(\frac{28}{49}=\frac{35 - GH}{35}\). Cross - multiply: \( 28\times35=49\times(35 - GH) \). \( 980 = 1715-49GH \). \( 49GH=1715 - 980=735 \). \( GH=\frac{735}{49}=15 \)? No, that's not right. Wait, maybe I have the similar triangles wrong. Let's look at the diagram again. The triangle \( FJH \) with a line \( IG \) parallel to \( FJ \) (since the arrows indicate parallel or midline). So by the Basic Proportionality Theorem, \( \frac{JI}{JH}=\frac{FG}{FH} \). Wait, \( JI = 28 \), \( JH = 49 \), \( FH = 35 \). So \(\frac{28}{49}=\frac{FG}{35}\). Then \( FG=\frac{28\times35}{49}=20 \). Wait, no, \( FH = 35 \), so \( GH=FH - FG \)? No, maybe \( FG \) is not the segment. Wait, maybe the segment \( FH = 35 \), and the ratio of \( JI \) to \( JH \) is \( \frac{28}{49}=\frac{4}{7} \), so the length of \( GH \) is \( 35\times\frac{4}{7}=20 \). Ah, that makes sense. So \( \frac{JI}{JH}=\frac{28}{49}=\frac{4}{7} \), so \( GH = 35\times\frac{4}{7}=20 \). Yes, that's correct. So the proportion is \( \frac{JI}{JH}=\frac{GH}{FH} \), so \( \frac{28}{49}=\frac{GH}{35} \), cross - multiply: \( 49GH = 28\times35 \), \( 49GH = 980 \), \( GH=\frac{980}{49}=20 \).

Answer:

20