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

Here are the step-by-step solutions for each problem on the worksheet.

1.
This is a right triangle. The legs are 5 and 12, and $x$ is the hypotenuse (the longest side).
Formula: $a^2 + b^2 = c^2$
$$5^2 + 12^2 = x^2$$
$$25 + 144 = x^2$$
$$169 = x^2$$
Take the square root of 169.
$$x = 13$$

2.
This is an isosceles right triangle because the two legs are equal ($x$ and 6) and the angles are 45-45-90. However, looking at the diagram, the legs are labeled $x$ and $6$, but usually in these specific diagrams, if it looks like the legs are equal, they might be. Let's look closer. Actually, the legs are $x$ and $6$. Wait, looking at the hypotenuse $6\sqrt{2}$, this is a special 45-45-90 triangle where the hypotenuse is $\text{leg} \times \sqrt{2}$.
So, $\text{leg} \times \sqrt{2} = 6\sqrt{2}$.
Dividing both sides by $\sqrt{2}$, we get $\text{leg} = 6$.
Therefore, $x = 6$.
*Check with Pythagorean theorem:*
$$x^2 + 6^2 = (6\sqrt{2})^2$$
$$x^2 + 36 = 36 \cdot 2$$
$$x^2 + 36 = 72$$
$$x^2 = 36$$
$$x = 6$$

3.
This is a rectangle. The diagonal $x$ forms a right triangle with the length (24) and width (7).
$$7^2 + 24^2 = x^2$$
$$49 + 576 = x^2$$
$$625 = x^2$$
Take the square root of 625.
$$x = 25$$

4.
This is a square with side length 8. The diagonal is $x$.
$$8^2 + 8^2 = x^2$$
$$64 + 64 = x^2$$
$$128 = x^2$$
$$x = \sqrt{128}$$
Simplify the radical: $\sqrt{128} = \sqrt{64 \cdot 2} = 8\sqrt{2}$.
$$x = 8\sqrt{2}$$

5.
This is a square with side length $x$ and diagonal $16\sqrt{2}$.
In a square, $\text{diagonal} = \text{side} \cdot \sqrt{2}$.
$$x \cdot \sqrt{2} = 16\sqrt{2}$$
Divide both sides by $\sqrt{2}$.
$$x = 16$$
*Check with Pythagorean theorem:*
$$x^2 + x^2 = (16\sqrt{2})^2$$
$$2x^2 = 256 \cdot 2$$
$$2x^2 = 512$$
$$x^2 = 256$$
$$x = 16$$

6.
This is a right triangle. One leg is 6, the hypotenuse is 18, and the other leg is $x$.
$$x^2 + 6^2 = 18^2$$
$$x^2 + 36 = 324$$
Subtract 36 from both sides.
$$x^2 = 288$$
$$x = \sqrt{288}$$
Simplify the radical: $\sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2}$.
$$x = 12\sqrt{2}$$

7.
Find the diagonal of a rectangle with length 3 and width 4.
$$3^2 + 4^2 = d^2$$
$$9 + 16 = d^2$$
$$25 = d^2$$
$$d = 5 \text{ feet}$$

8.
Find the diagonal of a square with perimeter 16 meters.
First, find the side length. A square has 4 equal sides.
$$\text{Side} = \frac{\text{Perimeter}}{4} = \frac{16}{4} = 4 \text{ meters}.$$
Now find the diagonal ($d$) using sides of 4.
$$4^2 + 4^2 = d^2$$
$$16 + 16 = d^2$$
$$32 = d^2$$
$$d = \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \text{ meters}.$$

9.
A ladder acts as the hypotenuse.
Hypotenuse (ladder) = 28 ft.
Leg (height) = 20 ft.
Leg (distance from base) = $x$.
$$x^2 + 20^2 = 28^2$$
$$x^2 + 400 = 784$$
$$x^2 = 384$$
$$x = \sqrt{384}$$
Simplify: $\sqrt{384} = \sqrt{64 \cdot 6} = 8\sqrt{6}$.
Approximate value: $8 \cdot 2.45 \approx 19.6$ feet.
Exact answer: $8\sqrt{6}$ feet.

10.
The pole and ground form the legs, the wire is the hypotenuse.
Leg 1 (pole) = 30 ft.
Leg 2 (ground distance) = 40 ft.
Hypotenuse (wire) = $x$.
$$30^2 + 40^2 = x^2$$
$$900 + 1600 = x^2$$
$$2500 = x^2$$
$$x = 50 \text{ feet}$$

Final Answer:
1. 13
2. 6
3. 25
4. $8\sqrt{2}$
5. 16
6. $12\sqrt{2}$
7. 5 feet
8. $4\sqrt{2}$ meters
9. $8\sqrt{6}$ feet (or approx. 19.6 feet)
10. 50 feet