Question

find hi.
hi =
submit
work it out
not feeling ready yet? these can help:
side lengths and angle measures in similar figures (81)
midsegments of triangles

Explanation:

Step1: Identify Similar Triangles

The lines \( IJ \) and \( IK \) have lengths \( IJ = 39 - 13 = 26 \)? Wait, no, actually, the segments \( KJ = 13 \) and \( KI = 39 \)? Wait, looking at the diagram, the two triangles \( \triangle KHG \) and \( \triangle KIG \)? Wait, no, the midline or the basic proportionality theorem (Thales' theorem) applies here. Since \( JH \) is parallel to \( KG \)? Wait, the markings on the sides (the red ticks) indicate that \( JH \) is parallel to \( KG \)? Wait, actually, the segments \( KJ \) and \( KI \): \( KJ = 13 \), \( KI = 39 \), so the ratio of \( KJ \) to \( KI \) is \( \frac{13}{39}=\frac{1}{3} \). Wait, no, \( KI \) is the entire side, \( KJ = 13 \), \( JI = 39 - 13 = 26 \)? Wait, maybe the triangles are similar by the Basic Proportionality Theorem (Thales' theorem), which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So, \( JH \parallel KG \), so \( \frac{KJ}{KI}=\frac{HG}{HI} \)? Wait, no, let's correct. Let's denote the triangle as \( \triangle KIG \), with \( J \) on \( KI \) and \( H \) on \( IG \), and \( JH \parallel KG \). Then by Thales' theorem, \( \frac{KJ}{KI}=\frac{HG}{HI} \)? Wait, no, \( KI \) is the side from \( K \) to \( I \), length \( 39 \), \( KJ = 13 \), so \( \frac{KJ}{KI}=\frac{13}{39}=\frac{1}{3} \). Then \( HG = 11 \), so \( \frac{HG}{HI + HG}=\frac{1}{3} \)? Wait, no, maybe the ratio is \( \frac{KJ}{JI}=\frac{HG}{HI} \). Wait, \( KJ = 13 \), \( JI = 39 - 13 = 26 \), so \( \frac{KJ}{JI}=\frac{13}{26}=\frac{1}{2} \)? No, that doesn't make sense. Wait, the correct approach: the two triangles \( \triangle KJH \) and \( \triangle KIG \) are similar? Wait, no, \( JH \parallel KG \), so \( \triangle KJH \sim \triangle KIG \) by AA similarity (corresponding angles equal). So the ratio of corresponding sides is equal. So \( \frac{KJ}{KI}=\frac{KH}{KG} \)? Wait, no, \( KI = 39 \), \( KJ = 13 \), so the ratio of similarity is \( \frac{13}{39}=\frac{1}{3} \). Then \( HG = 11 \), and \( HI \) is the side we need to find. Wait, actually, \( IG = HI + HG = HI + 11 \). Since \( \triangle KJH \sim \triangle KIG \), the ratio of \( KJ \) to \( KI \) is \( \frac{13}{39}=\frac{1}{3} \), so the ratio of \( HG \) to \( IG \) should be \( \frac{1}{3} \)? Wait, no, \( HG \) is corresponding to \( IG \)? No, \( HG \) is part of \( IG \). Wait, maybe I got the sides wrong. Let's re-express:

Let’s consider the triangle \( \triangle KIG \), with \( J \) on \( KI \) and \( H \) on \( IG \), and \( JH \parallel KG \). Then by the Basic Proportionality Theorem (Thales' theorem), \( \frac{KJ}{KI}=\frac{HG}{IG} \). Wait, \( KJ = 13 \), \( KI = 39 \), so \( \frac{13}{39}=\frac{11}{IG} \). Solving for \( IG \): \( IG = \frac{11 \times 39}{13} = 11 \times 3 = 33 \). Then \( IG = HI + HG \), so \( HI = IG - HG = 33 - 11 = 22 \). Wait, that makes sense. Let's check: \( \frac{KJ}{KI}=\frac{13}{39}=\frac{1}{3} \), and \( \frac{HG}{IG}=\frac{11}{33}=\frac{1}{3} \), which matches. So \( IG = 33 \), so \( HI = IG - HG = 33 - 11 = 22 \).

Step1: Apply Thales' Theorem

The line \( JH \) is parallel to \( KG \) (by the markings and the diagram), so by Thales' theorem (Basic Proportionality Theorem) in \( \triangle KIG \), \( \frac{KJ}{KI} = \frac{HG}{IG} \).

Given \( KJ = 13 \), \( KI = 39 \), \( HG = 11 \). Substitute these values:
\( \frac{13}{39} = \frac{11}{IG} \)

Step2: Solve for \( IG \)

Simplify \( \frac{13}{39} = \frac{1}{3} \), so:
\( \frac{1}{3} = \frac{11}{IG} \)
Cross-multiplying gives \( IG = 11 \times 3 = 33 \).

Step3: Find \( HI \)

Since \( IG = HI + HG \), substitute \( IG = 33 \) and \( HG = 11 \):
\( 33 = HI + 11 \)
Subtract 11 from both sides:
\( HI = 33 - 11 = 22 \).

Answer:

\( 22 \)