Finding Missing Lengths in Similar Right Triangles: Worksheet ... - Free Printable

To solve the problems involving right triangles, we will use the Pythagorean Theorem and trigonometric relationships (sine, cosine, tangent). Let's go through each problem step by step.

---

Problem 1:

#### Given:
- A right triangle with one leg = 3 units, hypotenuse = 5 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a \) and \( b \) are the legs, and \( c \) is the hypotenuse. Here, \( a = 3 \), \( c = 5 \), and we need to find \( b \).

\[
3^2 + b^2 = 5^2
\]
\[
9 + b^2 = 25
\]
\[
b^2 = 25 - 9
\]
\[
b^2 = 16
\]
\[
b = \sqrt{16} = 4
\]

#### Answer:
\[
\boxed{4}
\]

---

Problem 2:

#### Given:
- A right triangle with one leg = 8 units, hypotenuse = 10 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 8 \), \( c = 10 \), and we need to find \( b \).

\[
8^2 + b^2 = 10^2
\]
\[
64 + b^2 = 100
\]
\[
b^2 = 100 - 64
\]
\[
b^2 = 36
\]
\[
b = \sqrt{36} = 6
\]

#### Answer:
\[
\boxed{6}
\]

---

Problem 3:

#### Given:
- A right triangle with one leg = 7 units, hypotenuse = 25 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 7 \), \( c = 25 \), and we need to find \( b \).

\[
7^2 + b^2 = 25^2
\]
\[
49 + b^2 = 625
\]
\[
b^2 = 625 - 49
\]
\[
b^2 = 576
\]
\[
b = \sqrt{576} = 24
\]

#### Answer:
\[
\boxed{24}
\]

---

Problem 4:

#### Given:
- A right triangle with one leg = 12 units, hypotenuse = 13 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 12 \), \( c = 13 \), and we need to find \( b \).

\[
12^2 + b^2 = 13^2
\]
\[
144 + b^2 = 169
\]
\[
b^2 = 169 - 144
\]
\[
b^2 = 25
\]
\[
b = \sqrt{25} = 5
\]

#### Answer:
\[
\boxed{5}
\]

---

Problem 5:

#### Given:
- A right triangle with one leg = 6 units, hypotenuse = 10 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 6 \), \( c = 10 \), and we need to find \( b \).

\[
6^2 + b^2 = 10^2
\]
\[
36 + b^2 = 100
\]
\[
b^2 = 100 - 36
\]
\[
b^2 = 64
\]
\[
b = \sqrt{64} = 8
\]

#### Answer:
\[
\boxed{8}
\]

---

Problem 6:

#### Given:
- A right triangle with one leg = 5 units, hypotenuse = 13 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 5 \), \( c = 13 \), and we need to find \( b \).

\[
5^2 + b^2 = 13^2
\]
\[
25 + b^2 = 169
\]
\[
b^2 = 169 - 25
\]
\[
b^2 = 144
\]
\[
b = \sqrt{144} = 12
\]

#### Answer:
\[
\boxed{12}
\]

---

Problem 7:

#### Given:
- A right triangle with one leg = 9 units, hypotenuse = 15 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 9 \), \( c = 15 \), and we need to find \( b \).

\[
9^2 + b^2 = 15^2
\]
\[
81 + b^2 = 225
\]
\[
b^2 = 225 - 81
\]
\[
b^2 = 144
\]
\[
b = \sqrt{144} = 12
\]

#### Answer:
\[
\boxed{12}
\]

---

Problem 8:

#### Given:
- A right triangle with one leg = 1 unit, hypotenuse = √2 units.
- Find the missing side.

#### Solution:
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 1 \), \( c = \sqrt{2} \), and we need to find \( b \).

\[
1^2 + b^2 = (\sqrt{2})^2
\]
\[
1 + b^2 = 2
\]
\[
b^2 = 2 - 1
\]
\[
b^2 = 1
\]
\[
b = \sqrt{1} = 1
\]

#### Answer:
\[
\boxed{1}
\]

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Final Answers:

\[
\boxed{4, 6, 24, 5, 8, 12, 12, 1}
\]