Pythagorean Theorem worksheet with nine right triangles for calculating missing side lengths.

To solve the missing lengths in each right triangle using the Pythagorean theorem, we use the formula:

\[
a^2 + b^2 = c^2
\]

where \( c \) is the hypotenuse (the longest side), and \( a \) and \( b \) are the other two sides. We will solve each problem step by step.

---

1)

Given:
- \( AB = 12 \, \text{cm} \)
- \( AC = 15 \, \text{cm} \)
- \( BC = ? \)

Here, \( AC \) is the hypotenuse. Using the Pythagorean theorem:

\[
AB^2 + BC^2 = AC^2
\]

Substitute the known values:

\[
12^2 + BC^2 = 15^2
\]

\[
144 + BC^2 = 225
\]

Solve for \( BC^2 \):

\[
BC^2 = 225 - 144
\]

\[
BC^2 = 81
\]

Take the square root of both sides:

\[
BC = \sqrt{81} = 9
\]

So, \( BC = 9 \, \text{cm} \).

Answer: \( BC = 9.0 \, \text{cm} \)

---

2)

Given:
- \( QR = 2 \, \text{in} \)
- \( PR = 5 \, \text{in} \)
- \( PQ = ? \)

Here, \( PR \) is the hypotenuse. Using the Pythagorean theorem:

\[
PQ^2 + QR^2 = PR^2
\]

Substitute the known values:

\[
PQ^2 + 2^2 = 5^2
\]

\[
PQ^2 + 4 = 25
\]

Solve for \( PQ^2 \):

\[
PQ^2 = 25 - 4
\]

\[
PQ^2 = 21
\]

Take the square root of both sides:

\[
PQ = \sqrt{21} \approx 4.6
\]

So, \( PQ \approx 4.6 \, \text{in} \).

Answer: \( PQ = 4.6 \, \text{in} \)

---

3)

Given:
- \( UV = 6 \, \text{mm} \)
- \( WV = 9 \, \text{mm} \)
- \( UW = ? \)

Here, \( WV \) is the hypotenuse. Using the Pythagorean theorem:

\[
UW^2 + UV^2 = WV^2
\]

Substitute the known values:

\[
UW^2 + 6^2 = 9^2
\]

\[
UW^2 + 36 = 81
\]

Solve for \( UW^2 \):

\[
UW^2 = 81 - 36
\]

\[
UW^2 = 45
\]

Take the square root of both sides:

\[
UW = \sqrt{45} \approx 6.7
\]

So, \( UW \approx 6.7 \, \text{mm} \).

Answer: \( UW = 6.7 \, \text{mm} \)

---

4)

Given:
- \( ZY = 14 \, \text{yd} \)
- \( XY = 3 \, \text{yd} \)
- \( XZ = ? \)

Here, \( XZ \) is the hypotenuse. Using the Pythagorean theorem:

\[
ZY^2 + XY^2 = XZ^2
\]

Substitute the known values:

\[
14^2 + 3^2 = XZ^2
\]

\[
196 + 9 = XZ^2
\]

\[
XZ^2 = 205
\]

Take the square root of both sides:

\[
XZ = \sqrt{205} \approx 14.3
\]

So, \( XZ \approx 14.3 \, \text{yd} \).

Answer: \( XZ = 14.3 \, \text{yd} \)

---

5)

Given:
- \( FG = 10 \, \text{ft} \)
- \( EG = 8 \, \text{ft} \)
- \( EF = ? \)

Here, \( FG \) is the hypotenuse. Using the Pythagorean theorem:

\[
EF^2 + EG^2 = FG^2
\]

Substitute the known values:

\[
EF^2 + 8^2 = 10^2
\]

\[
EF^2 + 64 = 100
\]

Solve for \( EF^2 \):

\[
EF^2 = 100 - 64
\]

\[
EF^2 = 36
\]

Take the square root of both sides:

\[
EF = \sqrt{36} = 6
\]

So, \( EF = 6 \, \text{ft} \).

Answer: \( EF = 6.0 \, \text{ft} \)

---

6)

Given:
- \( JK = 13 \, \text{m} \)
- \( KL = 17 \, \text{m} \)
- \( JL = ? \)

Here, \( JL \) is the hypotenuse. Using the Pythagorean theorem:

\[
JK^2 + KL^2 = JL^2
\]

Substitute the known values:

\[
13^2 + 17^2 = JL^2
\]

\[
169 + 289 = JL^2
\]

\[
JL^2 = 458
\]

Take the square root of both sides:

\[
JL = \sqrt{458} \approx 21.4
\]

So, \( JL \approx 21.4 \, \text{m} \).

Answer: \( JL = 21.4 \, \text{m} \)

---

7)

Given:
- \( LM = 8 \, \text{in} \)
- \( LN = 16 \, \text{in} \)
- \( MN = ? \)

Here, \( LN \) is the hypotenuse. Using the Pythagorean theorem:

\[
LM^2 + MN^2 = LN^2
\]

Substitute the known values:

\[
8^2 + MN^2 = 16^2
\]

\[
64 + MN^2 = 256
\]

Solve for \( MN^2 \):

\[
MN^2 = 256 - 64
\]

\[
MN^2 = 192
\]

Take the square root of both sides:

\[
MN = \sqrt{192} \approx 13.9
\]

So, \( MN \approx 13.9 \, \text{in} \).

Answer: \( MN = 13.9 \, \text{in} \)

---

8)

Given:
- \( TU = 1 \, \text{cm} \)
- \( SU = 4 \, \text{cm} \)
- \( ST = ? \)

Here, \( SU \) is the hypotenuse. Using the Pythagorean theorem:

\[
ST^2 + TU^2 = SU^2
\]

Substitute the known values:

\[
ST^2 + 1^2 = 4^2
\]

\[
ST^2 + 1 = 16
\]

Solve for \( ST^2 \):

\[
ST^2 = 16 - 1
\]

\[
ST^2 = 15
\]

Take the square root of both sides:

\[
ST = \sqrt{15} \approx 3.9
\]

So, \( ST \approx 3.9 \, \text{cm} \).

Answer: \( ST = 3.9 \, \text{cm} \)

---

9)

Given:
- \( CD = 7 \, \text{yd} \)
- \( DE = 11 \, \text{yd} \)
- \( CE = ? \)

Here, \( CE \) is the hypotenuse. Using the Pythagorean theorem:

\[
CD^2 + DE^2 = CE^2
\]

Substitute the known values:

\[
7^2 + 11^2 = CE^2
\]

\[
49 + 121 = CE^2
\]

\[
CE^2 = 170
\]

Take the square root of both sides:

\[
CE = \sqrt{170} \approx 13.0
\]

So, \( CE \approx 13.0 \, \text{yd} \).

Answer: \( CE = 13.0 \, \text{yd} \)

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1) & BC = 9.0 \, \text{cm} \\
2) & PQ = 4.6 \, \text{in} \\
3) & UW = 6.7 \, \text{mm} \\
4) & XZ = 14.3 \, \text{yd} \\
5) & EF = 6.0 \, \text{ft} \\
6) & JL = 21.4 \, \text{m} \\
7) & MN = 13.9 \, \text{in} \\
8) & ST = 3.9 \, \text{cm} \\
9) & CE = 13.0 \, \text{yd} \\
\end{array}
}
\]