Minimalist abstract design with scattered dots on a white background.

The worksheet provided involves solving for unknown lengths using the Pythagorean Theorem. The Pythagorean Theorem states:

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

where \( a \) and \( b \) are the legs of a right triangle, and \( c \) is the hypotenuse.

Let's solve each problem step by step.

---

Problem 1:


Given:
- One leg = 26
- Hypotenuse = 48
- Find the other leg (\( x \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 26 \), \( b = x \), and \( c = 48 \).

\[
26^2 + x^2 = 48^2
\]

Calculate the squares:
\[
676 + x^2 = 2304
\]

Solve for \( x^2 \):
\[
x^2 = 2304 - 676
\]
\[
x^2 = 1628
\]

Take the square root of both sides:
\[
x = \sqrt{1628}
\]

Simplify \( \sqrt{1628} \):
\[
\sqrt{1628} = \sqrt{4 \times 407} = 2\sqrt{407}
\]

Thus:
\[
x = 2\sqrt{407}
\]

Answer:
\[
\boxed{x = 2\sqrt{407}}
\]

---

Problem 2:


Given:
- One leg = 2
- Hypotenuse = 3
- Find the other leg (\( x \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 2 \), \( b = x \), and \( c = 3 \).

\[
2^2 + x^2 = 3^2
\]

Calculate the squares:
\[
4 + x^2 = 9
\]

Solve for \( x^2 \):
\[
x^2 = 9 - 4
\]
\[
x^2 = 5
\]

Take the square root of both sides:
\[
x = \sqrt{5}
\]

Answer:
\[
\boxed{x = \sqrt{5}}
\]

---

Problem 3:


Given:
- One leg = 4
- Hypotenuse = 16
- Find the other leg (\( y \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 4 \), \( b = y \), and \( c = 16 \).

\[
4^2 + y^2 = 16^2
\]

Calculate the squares:
\[
16 + y^2 = 256
\]

Solve for \( y^2 \):
\[
y^2 = 256 - 16
\]
\[
y^2 = 240
\]

Take the square root of both sides:
\[
y = \sqrt{240}
\]

Simplify \( \sqrt{240} \):
\[
\sqrt{240} = \sqrt{16 \times 15} = 4\sqrt{15}
\]

Thus:
\[
y = 4\sqrt{15}
\]

Answer:
\[
\boxed{y = 4\sqrt{15}}
\]

---

Problem 4:


Given:
- One leg = 3
- Hypotenuse = 5
- Find the other leg (\( x \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 3 \), \( b = x \), and \( c = 5 \).

\[
3^2 + x^2 = 5^2
\]

Calculate the squares:
\[
9 + x^2 = 25
\]

Solve for \( x^2 \):
\[
x^2 = 25 - 9
\]
\[
x^2 = 16
\]

Take the square root of both sides:
\[
x = \sqrt{16}
\]
\[
x = 4
\]

Answer:
\[
\boxed{x = 4}
\]

---

Problem 5:


Given:
- One leg = 5
- Hypotenuse = 13
- Find the other leg (\( h \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 5 \), \( b = h \), and \( c = 13 \).

\[
5^2 + h^2 = 13^2
\]

Calculate the squares:
\[
25 + h^2 = 169
\]

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

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

Answer:
\[
\boxed{h = 12}
\]

---

Problem 6:


Given:
- One leg = 6
- Hypotenuse = 10
- Find the other leg (\( x \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 6 \), \( b = x \), and \( c = 10 \).

\[
6^2 + x^2 = 10^2
\]

Calculate the squares:
\[
36 + x^2 = 100
\]

Solve for \( x^2 \):
\[
x^2 = 100 - 36
\]
\[
x^2 = 64
\]

Take the square root of both sides:
\[
x = \sqrt{64}
\]
\[
x = 8
\]

Answer:
\[
\boxed{x = 8}
\]

---

Problem 7:


Given:
- One leg = 8
- Hypotenuse = 10
- Find the other leg (\( y \)).

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 8 \), \( b = y \), and \( c = 10 \).

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

Calculate the squares:
\[
64 + y^2 = 100
\]

Solve for \( y^2 \):
\[
y^2 = 100 - 64
\]
\[
y^2 = 36
\]

Take the square root of both sides:
\[
y = \sqrt{36}
\]
\[
y = 6
\]

Answer:
\[
\boxed{y = 6}
\]

---

Problem 8:


Given:
- Trapezoid with bases 12 and 5, height 13.
- Find the length of the slant side (\( x \)).

The slant side forms a right triangle with the height and half the difference of the bases as the legs. The difference in the bases is:
\[
12 - 5 = 7
\]

Half of this difference is:
\[
\frac{7}{2} = 3.5
\]

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 13 \) (height), \( b = 3.5 \), and \( c = x \).

\[
13^2 + 3.5^2 = x^2
\]

Calculate the squares:
\[
169 + 12.25 = x^2
\]
\[
x^2 = 181.25
\]

Take the square root of both sides:
\[
x = \sqrt{181.25}
\]

Simplify \( \sqrt{181.25} \):
\[
\sqrt{181.25} = \sqrt{\frac{725}{4}} = \frac{\sqrt{725}}{2}
\]

Thus:
\[
x = \frac{\sqrt{725}}{2}
\]

Answer:
\[
\boxed{x = \frac{\sqrt{725}}{2}}
\]

---

Problem 9:


Given:
- Rhombus with diagonals 12 and 16.
- Find the side length (\( x \)).

The diagonals of a rhombus bisect each other at right angles, forming four right triangles. Each triangle has legs of:
\[
\frac{12}{2} = 6 \quad \text{and} \quad \frac{16}{2} = 8
\]

Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 6 \), \( b = 8 \), and \( c = x \).

\[
6^2 + 8^2 = x^2
\]

Calculate the squares:
\[
36 + 64 = x^2
\]
\[
x^2 = 100
\]

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

Answer:
\[
\boxed{x = 10}
\]

---

Problem 10:


Given:
- Right triangle with one leg = 12, hypotenuse = 13.
- Find the other leg (\( y \)) and the altitude to the hypotenuse (\( z \)).

#### Step 1: Find \( y \)
Using the Pythagorean Theorem:
\[
a^2 + b^2 = c^2
\]
Here, \( a = 12 \), \( b = y \), and \( c = 13 \).

\[
12^2 + y^2 = 13^2
\]

Calculate the squares:
\[
144 + y^2 = 169
\]

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

Take the square root of both sides:
\[
y = \sqrt{25}
\]
\[
y = 5
\]

#### Step 2: Find \( z \) (altitude to the hypotenuse)
The area of the triangle can be calculated in two ways:
1. Using the legs:
\[
\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = \frac{1}{2} \times 12 \times 5 = 30
\]

2. Using the hypotenuse and the altitude:
\[
\text{Area} = \frac{1}{2} \times \text{hypotenuse} \times \text{altitude} = \frac{1}{2} \times 13 \times z
\]

Set the two expressions for the area equal:
\[
30 = \frac{1}{2} \times 13 \times z
\]

Solve for \( z \):
\[
30 = \frac{13z}{2}
\]
\[
60 = 13z
\]
\[
z = \frac{60}{13}
\]

Answers:
\[
\boxed{y = 5, z = \frac{60}{13}}
\]

---

Final Answers:

1. \( x = 2\sqrt{407} \)
2. \( x = \sqrt{5} \)
3. \( y = 4\sqrt{15} \)
4. \( x = 4 \)
5. \( h = 12 \)
6. \( x = 8 \)
7. \( y = 6 \)
8. \( x = \frac{\sqrt{725}}{2} \)
9. \( x = 10 \)
10. \( y = 5, z = \frac{60}{13} \)

\boxed{\text{All problems solved.}}