Pythagorean Theorem practice worksheet with six right triangles and problems to solve for the unknown side x.

Problem: Solve for \( x \) using the Pythagorean Theorem

The Pythagorean Theorem states:
\[
a^2 + b^2 = c^2
\]
where \( c \) is the hypotenuse (the longest side of the right triangle), and \( a \) and \( b \) are the other two sides.

Let's solve each problem step by step.

---

#### Problem #1
Given:
- One leg = 14
- Other leg = 48
- Hypotenuse = \( x \)

Using the Pythagorean Theorem:
\[
14^2 + 48^2 = x^2
\]

Calculate:
\[
14^2 = 196
\]
\[
48^2 = 2304
\]
\[
196 + 2304 = 2500
\]

So:
\[
x^2 = 2500
\]

Take the square root of both sides:
\[
x = \sqrt{2500} = 50
\]

Answer:
\[
\boxed{50}
\]

---

#### Problem #2
Given:
- One leg = 6
- Other leg = 10
- Hypotenuse = \( x \)

Using the Pythagorean Theorem:
\[
6^2 + 10^2 = x^2
\]

Calculate:
\[
6^2 = 36
\]
\[
10^2 = 100
\]
\[
36 + 100 = 136
\]

So:
\[
x^2 = 136
\]

Take the square root of both sides:
\[
x = \sqrt{136} = 2\sqrt{34}
\]

Answer:
\[
\boxed{2\sqrt{34}}
\]

---

#### Problem #3
Given:
- One leg = 12
- Other leg = 16
- Hypotenuse = \( x \)

Using the Pythagorean Theorem:
\[
12^2 + 16^2 = x^2
\]

Calculate:
\[
12^2 = 144
\]
\[
16^2 = 256
\]
\[
144 + 256 = 400
\]

So:
\[
x^2 = 400
\]

Take the square root of both sides:
\[
x = \sqrt{400} = 20
\]

Answer:
\[
\boxed{20}
\]

---

#### Problem #4
Given:
- One leg = 8
- Other leg = 10
- Hypotenuse = \( x \)

Using the Pythagorean Theorem:
\[
8^2 + 10^2 = x^2
\]

Calculate:
\[
8^2 = 64
\]
\[
10^2 = 100
\]
\[
64 + 100 = 164
\]

So:
\[
x^2 = 164
\]

Take the square root of both sides:
\[
x = \sqrt{164} = 2\sqrt{41}
\]

Answer:
\[
\boxed{2\sqrt{41}}
\]

---

#### Problem #5
Given:
- One leg = 5
- Hypotenuse = 13
- Other leg = \( x \)

Using the Pythagorean Theorem:
\[
5^2 + x^2 = 13^2
\]

Calculate:
\[
5^2 = 25
\]
\[
13^2 = 169
\]

So:
\[
25 + x^2 = 169
\]

Solve for \( x^2 \):
\[
x^2 = 169 - 25 = 144
\]

Take the square root of both sides:
\[
x = \sqrt{144} = 12
\]

Answer:
\[
\boxed{12}
\]

---

#### Problem #6
Given:
- One leg = 24
- Hypotenuse = 32
- Other leg = \( x \)

Using the Pythagorean Theorem:
\[
24^2 + x^2 = 32^2
\]

Calculate:
\[
24^2 = 576
\]
\[
32^2 = 1024
\]

So:
\[
576 + x^2 = 1024
\]

Solve for \( x^2 \):
\[
x^2 = 1024 - 576 = 448
\]

Take the square root of both sides:
\[
x = \sqrt{448} = 8\sqrt{7}
\]

Answer:
\[
\boxed{8\sqrt{7}}
\]

---

#### Problem #7
Given:
- \( a = 35 \)
- \( b = 35 \)
- \( c = 37 \)

Verify if this satisfies the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]

Calculate:
\[
35^2 = 1225
\]
\[
35^2 = 1225
\]
\[
1225 + 1225 = 2450
\]

But:
\[
c^2 = 37^2 = 1369
\]

Since \( 2450 \neq 1369 \), this does not satisfy the Pythagorean Theorem. There seems to be an error in the problem statement. Assuming the problem is correct as stated, we can conclude that it does not form a right triangle.

Answer:
\[
\text{Not a right triangle}
\]

---

#### Problem #8
Given:
- \( a = 7 \)
- \( b = ? \)
- \( c = 25 \)

Using the Pythagorean Theorem:
\[
7^2 + b^2 = 25^2
\]

Calculate:
\[
7^2 = 49
\]
\[
25^2 = 625
\]

So:
\[
49 + b^2 = 625
\]

Solve for \( b^2 \):
\[
b^2 = 625 - 49 = 576
\]

Take the square root of both sides:
\[
b = \sqrt{576} = 24
\]

Answer:
\[
\boxed{24}
\]

---

#### Problem #9
Given:
- \( a = 12 \)
- \( b = 16 \)
- \( c = ? \)

Using the Pythagorean Theorem:
\[
12^2 + 16^2 = c^2
\]

Calculate:
\[
12^2 = 144
\]
\[
16^2 = 256
\]
\[
144 + 256 = 400
\]

So:
\[
c^2 = 400
\]

Take the square root of both sides:
\[
c = \sqrt{400} = 20
\]

Answer:
\[
\boxed{20}
\]

---

#### Problem #10
Given:
- \( a = ? \)
- \( b = 24 \)
- \( c = 26 \)

Using the Pythagorean Theorem:
\[
a^2 + 24^2 = 26^2
\]

Calculate:
\[
24^2 = 576
\]
\[
26^2 = 676
\]

So:
\[
a^2 + 576 = 676
\]

Solve for \( a^2 \):
\[
a^2 = 676 - 576 = 100
\]

Take the square root of both sides:
\[
a = \sqrt{100} = 10
\]

Answer:
\[
\boxed{10}
\]

---

Final Answers:

1. \(\boxed{50}\)
2. \(\boxed{2\sqrt{34}}\)
3. \(\boxed{20}\)
4. \(\boxed{2\sqrt{41}}\)
5. \(\boxed{12}\)
6. \(\boxed{8\sqrt{7}}\)
7. \(\text{Not a right triangle}\)
8. \(\boxed{24}\)
9. \(\boxed{20}\)
10. \(\boxed{10}\)