Question

there is a diagram with triangles. triangle fgh has sides fg (with an arrow), fh = 6, gh = 4. triangle jih has sides ji (with an arrow), jh = 16, ih = x. the two triangles share the intersection point h, and the arrows on fg and ji indicate they are parallel.

Explanation:

Step1: Identify Similar Triangles

Since \( FG \parallel JI \) (indicated by the arrows), triangles \( FHG \) and \( JHI \) are similar by the Basic Proportionality Theorem (AA similarity, as vertical angles at \( H \) are equal and alternate interior angles are equal due to parallel lines).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \(\frac{FH}{JH}=\frac{GH}{IH}\). Substituting the known values \( FH = 6 \), \( JH = 16 \), \( GH = 4 \), we get \(\frac{6}{16}=\frac{4}{x}\).

Step3: Solve for \( x \)

Cross - multiply: \( 6x=16\times4 \). Then \( 6x = 64 \), and \( x=\frac{64}{6}=\frac{32}{3}\approx10.67 \)? Wait, no, wait. Wait, maybe I mixed up the sides. Wait, actually, \( FH \) corresponds to \( JH \), and \( GH \) corresponds to \( IH \), but maybe the correspondence is \( \triangle FHG \sim \triangle JHI \), so \( \frac{FH}{JH}=\frac{GH}{IH} \), but let's check again. Wait, \( FH = 6 \), \( JH = 16 \), \( GH = 4 \), \( IH=x \). So cross - multiplying: \( 6x=16\times4 \)? No, that's wrong. Wait, no, similar triangles: \( \frac{FH}{IH}=\frac{GH}{JH} \)? Wait, no, let's label the triangles. \( F \) corresponds to \( I \), \( G \) corresponds to \( J \), because \( FG \parallel JI \), so the order is \( F - G - H \) and \( J - I - H \). So \( \triangle FHG \sim \triangle IHJ \)? Wait, maybe the correct proportion is \(\frac{FH}{IH}=\frac{GH}{JH}\). So \( FH = 6 \), \( IH=x \), \( GH = 4 \), \( JH = 16 \). Then \(\frac{6}{x}=\frac{4}{16}\). Ah, that's the mistake! I had the proportion reversed. So \(\frac{6}{x}=\frac{4}{16}\). Then cross - multiply: \( 4x=6\times16 \), \( 4x = 96 \), so \( x = 24 \). Wait, that makes more sense. Because if \( FG \parallel JI \), the smaller triangle has sides 6 and 4, the larger has 16 and \( x \), so the ratio of the smaller to larger should be consistent. So \( \frac{6}{16}=\frac{4}{x} \) was wrong. The correct correspondence is \( FH \) (smaller triangle) to \( JH \) (larger triangle), and \( GH \) (smaller) to \( IH \) (larger). Wait, no, \( FH = 6 \), \( JH = 16 \) (so \( JH \) is longer than \( FH \)), \( GH = 4 \), \( IH=x \). So the ratio of the sides of the smaller triangle (\( \triangle FHG \)) to the larger triangle (\( \triangle JHI \)) is \( \frac{FH}{JH}=\frac{GH}{IH} \), so \( \frac{6}{16}=\frac{4}{x} \), but that gives \( x=\frac{64}{6}\approx10.67 \), which seems odd. Wait, maybe the triangles are \( \triangle FHG \) and \( \triangle JHI \), with \( FH = 6 \), \( JH = 16 \), \( GH = 4 \), \( IH=x \), and since \( FG \parallel JI \), the alternate interior angles: \( \angle F=\angle I \), \( \angle G=\angle J \), so \( \triangle FHG \sim \triangle IHJ \)? No, let's use the correct method. The vertical angle at \( H \) is equal, and \( \angle FHG=\angle JHI \) (vertical angles), and \( \angle F=\angle I \) (alternate interior angles, \( FG \parallel JI \), transversal \( FI \)), so \( \triangle FHG \sim \triangle IHJ \) by AA similarity. Therefore, the ratio of corresponding sides: \( \frac{FH}{IH}=\frac{GH}{JH} \). So \( FH = 6 \), \( IH=x \), \( GH = 4 \), \( JH = 16 \). So \(\frac{6}{x}=\frac{4}{16}\). Then \( 4x=6\times16 \), \( 4x = 96 \), \( x = 24 \). Yes, that makes sense. Because if \( FH = 6 \), \( JH = 16 \), the ratio of similarity is \( \frac{6}{x}=\frac{4}{16}\), and \( \frac{4}{16}=\frac{1}{4}\), so \( \frac{6}{x}=\frac{1}{4}\), so \( x = 24 \).

Answer:

\( x = 24 \)