48 Pythagorean Theorem Worksheet with Answers [Word + PDF] - Free Printable

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 side opposite the right angle), and \( a \) and \( b \) are the other two sides.

Let's solve each problem step by step.

---

Problem 1:

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

Here, \( AC \) is the hypotenuse (\( c \)), and \( AB \) and \( BC \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
AB^2 + BC^2 = AC^2
\]
\[
12^2 + BC^2 = 15^2
\]
\[
144 + BC^2 = 225
\]
\[
BC^2 = 225 - 144
\]
\[
BC^2 = 81
\]
\[
BC = \sqrt{81}
\]
\[
BC = 9 \, \text{cm}
\]

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

---

Problem 2:

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

Here, \( PR \) is the hypotenuse (\( c \)), and \( QR \) and \( PQ \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
QR^2 + PQ^2 = PR^2
\]
\[
2^2 + PQ^2 = 5^2
\]
\[
4 + PQ^2 = 25
\]
\[
PQ^2 = 25 - 4
\]
\[
PQ^2 = 21
\]
\[
PQ = \sqrt{21}
\]
\[
PQ \approx 4.6 \, \text{in}
\]

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

---

Problem 3:

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

Here, \( WV \) is the hypotenuse (\( c \)), and \( UW \) and \( UV \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
UW^2 + UV^2 = WV^2
\]
\[
UW^2 + 6^2 = 9^2
\]
\[
UW^2 + 36 = 81
\]
\[
UW^2 = 81 - 36
\]
\[
UW^2 = 45
\]
\[
UW = \sqrt{45}
\]
\[
UW \approx 6.7 \, \text{mm}
\]

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

---

Problem 4:

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

Here, \( XZ \) is the hypotenuse (\( c \)), and \( ZY \) and \( XY \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
ZY^2 + XY^2 = XZ^2
\]
\[
14^2 + 3^2 = XZ^2
\]
\[
196 + 9 = XZ^2
\]
\[
XZ^2 = 205
\]
\[
XZ = \sqrt{205}
\]
\[
XZ \approx 14.3 \, \text{yd}
\]

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

---

Problem 5:

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

Here, \( EG \) is the hypotenuse (\( c \)), and \( FG \) and \( EF \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
FG^2 + EF^2 = EG^2
\]
\[
8^2 + EF^2 = 10^2
\]
\[
64 + EF^2 = 100
\]
\[
EF^2 = 100 - 64
\]
\[
EF^2 = 36
\]
\[
EF = \sqrt{36}
\]
\[
EF = 6 \, \text{ft}
\]

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

---

Problem 6:

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

Here, \( JL \) is the hypotenuse (\( c \)), and \( KL \) and \( JK \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
KL^2 + JK^2 = JL^2
\]
\[
17^2 + 13^2 = JL^2
\]
\[
289 + 169 = JL^2
\]
\[
JL^2 = 458
\]
\[
JL = \sqrt{458}
\]
\[
JL \approx 21.4 \, \text{m}
\]

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

---

Problem 7:

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

Here, \( LN \) is the hypotenuse (\( c \)), and \( LM \) and \( MN \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
LM^2 + MN^2 = LN^2
\]
\[
8^2 + MN^2 = 16^2
\]
\[
64 + MN^2 = 256
\]
\[
MN^2 = 256 - 64
\]
\[
MN^2 = 192
\]
\[
MN = \sqrt{192}
\]
\[
MN \approx 13.9 \, \text{in}
\]

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

---

Problem 8:

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

Here, \( SU \) is the hypotenuse (\( c \)), and \( TU \) and \( ST \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
TU^2 + ST^2 = SU^2
\]
\[
1^2 + ST^2 = 4^2
\]
\[
1 + ST^2 = 16
\]
\[
ST^2 = 16 - 1
\]
\[
ST^2 = 15
\]
\[
ST = \sqrt{15}
\]
\[
ST \approx 3.9 \, \text{cm}
\]

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

---

Problem 9:

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

Here, \( CE \) is the hypotenuse (\( c \)), and \( CD \) and \( DE \) are the legs (\( a \) and \( b \)).

Using the Pythagorean theorem:
\[
CD^2 + DE^2 = CE^2
\]
\[
7^2 + 11^2 = CE^2
\]
\[
49 + 121 = CE^2
\]
\[
CE^2 = 170
\]
\[
CE = \sqrt{170}
\]
\[
CE \approx 13.0 \, \text{yd}
\]

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

---

Final Answers:

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