Similar Triangles Worksheets - Math Monks - Free Printable

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.

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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 \( XW = x \) and \( WZ = 8 - x \).
- \( \triangle XYZ \sim \triangle WYZ \).

We need to find \( x \).

Since \( \triangle XYZ \sim \triangle WYZ \):
\[
\frac{XY}{WY} = \frac{XZ}{YZ}
\]
Substitute the known values:
\[
\frac{9}{x} = \frac{8}{12}
\]
Simplify:
\[
\frac{9}{x} = \frac{2}{3}
\]
Cross-multiply:
\[
9 \cdot 3 = 2 \cdot x \implies 27 = 2x \implies x = \frac{27}{2} = 13.5
\]

Answer:
\[
\boxed{13.5}
\]

---

Problem 4

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

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

Since \( \triangle PQS \sim \triangle RQS \):
\[
\frac{PQ}{RS} = \frac{PS}{QS}
\]
Substitute the known values:
\[
\frac{3}{5} = \frac{d}{62}
\]
Cross-multiply:
\[
3 \cdot 62 = 5 \cdot d \implies 186 = 5d \implies d = \frac{186}{5} = 37.2
\]

Answer:
\[
\boxed{37.2}
\]

---

Problem 5

Given:
- Triangle \( \triangle XYT \) with \( XY = 16 \) m and \( YT = 8 \) m.
- Point \( Z \) divides \( XT \) into segments \( XZ = d \) and \( ZT = 6 \) m.
- \( \triangle XYT \sim \triangle XZW \).

We need to find \( d \).

Since \( \triangle XYT \sim \triangle XZW \):
\[
\frac{XY}{XZ} = \frac{YT}{ZW}
\]
Substitute the known values:
\[
\frac{16}{d} = \frac{8}{6}
\]
Simplify:
\[
\frac{16}{d} = \frac{4}{3}
\]
Cross-multiply:
\[
16 \cdot 3 = 4 \cdot d \implies 48 = 4d \implies d = \frac{48}{4} = 12
\]

Answer:
\[
\boxed{12}
\]

---

Problem 6

Given:
- Triangle \( \triangle ABC \) with \( BC = 12 \) mi and \( AC = 18 \) mi.
- Triangle \( \triangle QCR \) with \( QR = 24 \) mi.
- \( \triangle ABC \sim \triangle QCR \).

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

Since \( \triangle ABC \sim \triangle QCR \):
\[
\frac{AB}{QR} = \frac{AC}{CR}
\]
Substitute the known values:
\[
\frac{h}{24} = \frac{18}{36}
\]
Simplify:
\[
\frac{h}{24} = \frac{1}{2}
\]
Cross-multiply:
\[
h = 24 \cdot \frac{1}{2} = 12
\]

Answer:
\[
\boxed{12}
\]

---

Problem 7

Given:
- Triangle \( \triangle ABE \) with \( BE = 8 \) mi (since \( ED = 4 \) mi and \( DB = 4 \) mi).
- Point \( C \) divides \( AB \) into segments \( AC = d \) and \( CB = 6 - d \).
- \( \triangle ABE \sim \triangle CDE \).

We need to find \( d \).

Since \( \triangle ABE \sim \triangle CDE \):
\[
\frac{AB}{CD} = \frac{BE}{DE}
\]
Substitute the known values:
\[
\frac{d}{6} = \frac{8}{4}
\]
Simplify:
\[
\frac{d}{6} = 2
\]
Cross-multiply:
\[
d = 6 \cdot 2 = 12
\]

Answer:
\[
\boxed{12}
\]

---

Problem 8

Given:
- Right triangle \( \triangle PQS \) with \( PQ = 320 \) mi and \( QS = 60 \) mi.
- Right triangle \( \triangle PQS \sim \triangle RQS \).
- \( RS = 162 \) mi.

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

Since \( \triangle PQS \sim \triangle RQS \):
\[
\frac{PQ}{RS} = \frac{PS}{QS}
\]
Substitute the known values:
\[
\frac{320}{162} = \frac{d}{60}
\]
Cross-multiply:
\[
320 \cdot 60 = 162 \cdot d \implies 19200 = 162d \implies d = \frac{19200}{162} = \frac{3200}{27} \approx 118.52
\]

Answer:
\[
\boxed{\frac{3200}{27}}
\]

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Final Answers:

\[
\boxed{4.8, 40, 13.5, 37.2, 12, 12, 12, \frac{3200}{27}}
\]