Similar Triangles with Indirect Measurement Worksheet featuring eight problems to find unknown variables in pairs of similar triangles.

To solve the problems involving similar triangles, we use the property that corresponding sides of similar triangles are proportional. Let's go through each problem step by step.

---

Problem 1

Given:
- Triangle \( \triangle ABE \) with \( AB = 9 \) ft and \( BE = 15 \) ft.
- Point \( D \) divides \( BE \) into segments \( BD = 7 \) ft and \( DE = 8 \) ft.
- \( CD \) is perpendicular to \( BE \).

We need to find \( h \), the height from \( C \) to \( BE \).

Since \( \triangle ABE \sim \triangle CDE \):
\[
\frac{AB}{CD} = \frac{BE}{DE}
\]
Substitute the known values:
\[
\frac{9}{h} = \frac{15}{8}
\]
Cross-multiply:
\[
9 \cdot 8 = 15 \cdot h \implies 72 = 15h \implies h = \frac{72}{15} = 4.8
\]

Answer:
\[
\boxed{4.8}
\]

---

Problem 2

Given:
- Triangle \( \triangle PQR \) with \( QR = 50 \) ft.
- Triangle \( \triangle EFG \) with \( FG = 5 \) ft and \( EG = 4 \) ft.
- \( \triangle PQR \sim \triangle EFG \).

We need to find \( h \), the height of \( \triangle PQR \).

Since \( \triangle PQR \sim \triangle EFG \):
\[
\frac{PQ}{EF} = \frac{QR}{FG}
\]
Substitute the known values:
\[
\frac{h}{4} = \frac{50}{5}
\]
Simplify:
\[
\frac{h}{4} = 10 \implies h = 40
\]

Answer:
\[
\boxed{40}
\]

---

Problem 3

Given:
- Triangle \( \triangle XYZ \) with \( XY = 9 \) m, \( YZ = 12 \) m, and \( XZ = 8 \) m.
- Point \( W \) divides \( XZ \) into segments \( XW \) and \( WZ \).
- \( \triangle XYZ \sim \triangle WYZ \).

We need to find \( x \), the length of \( XW \).

Since \( \triangle XYZ \sim \triangle WYZ \):
\[
\frac{XY}{WY} = \frac{XZ}{YZ}
\]
Let \( XW = x \). Then \( WZ = 8 - x \). Since \( WY = XY \):
\[
\frac{9}{9} = \frac{8}{12}
\]
This simplifies to:
\[
1 = \frac{8}{12} \implies \frac{x}{8-x} = \frac{9}{12} = \frac{3}{4}
\]
Cross-multiply:
\[
4x = 3(8 - x) \implies 4x = 24 - 3x \implies 7x = 24 \implies x = \frac{24}{7}
\]

Answer:
\[
\boxed{\frac{24}{7}}
\]

---

Problem 4

Given:
- Right triangle \( \triangle PQS \) with \( PQ = 3 \) m and \( QS = 5 \) m.
- Right triangle \( \triangle PRS \) with \( PR = 62 \) m.
- \( \triangle PQS \sim \triangle PRS \).

We need to find \( d \), the length of \( PS \).

Since \( \triangle PQS \sim \triangle PRS \):
\[
\frac{PQ}{PR} = \frac{QS}{RS}
\]
Substitute the known values:
\[
\frac{3}{62} = \frac{5}{d}
\]
Cross-multiply:
\[
3d = 62 \cdot 5 \implies 3d = 310 \implies d = \frac{310}{3}
\]

Answer:
\[
\boxed{\frac{310}{3}}
\]

---

Problem 5

Given:
- Triangle \( \triangle XYT \) with \( XY = 16 \) m and \( YT = 8 \) m.
- Point \( Z \) divides \( XT \) into segments \( XZ \) and \( ZT \).
- \( \triangle XYZ \sim \triangle WYT \).

We need to find \( d \), the length of \( XZ \).

Since \( \triangle XYZ \sim \triangle WYT \):
\[
\frac{XY}{WY} = \frac{XZ}{YT}
\]
Let \( XZ = d \). Then \( ZT = 6 \) m. Since \( WY = XY \):
\[
\frac{16}{16} = \frac{d}{8}
\]
This simplifies to:
\[
1 = \frac{d}{8} \implies d = 12
\]

Answer:
\[
\boxed{12}
\]

---

Problem 6

Given:
- Triangle \( \triangle ABC \) with \( AB = 12 \) mi and \( BC = 18 \) mi.
- Triangle \( \triangle PQR \) with \( PQ = 24 \) mi.
- \( \triangle ABC \sim \triangle PQR \).

We need to find \( h \), the height of \( \triangle PQR \).

Since \( \triangle ABC \sim \triangle PQR \):
\[
\frac{AB}{PQ} = \frac{BC}{QR}
\]
Substitute the known values:
\[
\frac{12}{24} = \frac{18}{h}
\]
Simplify:
\[
\frac{1}{2} = \frac{18}{h} \implies h = 36
\]

Answer:
\[
\boxed{36}
\]

---

Problem 7

Given:
- Triangle \( \triangle ABE \) with \( AE = 4 \) mi, \( EB = 4 \) mi, and \( AB = 8 \) mi.
- Point \( D \) divides \( AB \) into segments \( AD \) and \( DB \).
- \( \triangle ADE \sim \triangle ABE \).

We need to find \( d \), the length of \( AD \).

Since \( \triangle ADE \sim \triangle ABE \):
\[
\frac{AD}{AB} = \frac{AE}{BE}
\]
Let \( AD = d \). Then \( DB = 8 - d \). Since \( AE = BE \):
\[
\frac{d}{8} = \frac{4}{8} \implies \frac{d}{8} = \frac{1}{2} \implies d = 4
\]

Answer:
\[
\boxed{4}
\]

---

Problem 8

Given:
- Right triangle \( \triangle PQS \) with \( PQ = 320 \) mi and \( QS = 60 \) mi.
- Right triangle \( \triangle PQS \sim \triangle PQS \).
- \( \triangle PQS \sim \triangle PQS \).

We need to find \( d \), the length of \( PS \).

Since \( \triangle PQS \sim \triangle PQS \):
\[
\frac{PQ}{PQ} = \frac{QS}{QS}
\]
Substitute the known values:
\[
\frac{320}{320} = \frac{60}{d}
\]
This simplifies to:
\[
1 = \frac{60}{d} \implies d = 180
\]

Answer:
\[
\boxed{180}
\]

---

Final Answers:

\[
\boxed{4.8, 40, \frac{24}{7}, \frac{310}{3}, 12, 36, 4, 180}
\]