The Pythagorean Theorem

Proofs of the Pythagorean Theorem

The Pythagorean Theorem is one of the most important ideas in geometry. It tells us the relationship between the side lengths of a right triangle.

If a right triangle has legs of lengths \(a\) and \(b\), and hypotenuse of length \(c\), then:

$$a^2 + b^2 = c^2$$

In this lesson, we will not only use this formula, but also understand why it is true. A proof explains why a math statement must always work.

Important vocabulary:

• Right triangle: a triangle with one \(90^\circ\) angle.
• Legs: the two sides that make the right angle.
• Hypotenuse: the side opposite the right angle, and the longest side.
• Square of a number: the number times itself, such as \(5^2 = 25\).

When we see \(a^2\), \(b^2\), and \(c^2\) in the theorem, we can think of them as the areas of squares built on each side of the triangle.

The big idea: In any right triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two legs.

$$\text{area on side } a + \text{area on side } b = \text{area on side } c$$

That is exactly what \(a^2 + b^2 = c^2\) means.

Why do we study proofs?

• Proofs help us understand, not just memorize.
• Proofs show that the theorem works for every right triangle.
• Proofs connect geometry and algebra.

Visual Proof Idea 1: Squares on the Sides

Imagine a right triangle with legs \(a\) and \(b\), and hypotenuse \(c\). Now build a square on each side:

• a square with side length \(a\), so its area is \(a^2\)
• a square with side length \(b\), so its area is \(b^2\)
• a square with side length \(c\), so its area is \(c^2\)

The Pythagorean Theorem says that the two smaller square areas together exactly match the largest square area.

So if the triangle has side lengths \(3\), \(4\), and \(5\), then:

$$3^2 + 4^2 = 5^2$$ $$9 + 16 = 25$$

The two smaller squares have areas \(9\) and \(16\), and together they make \(25\), the area of the largest square.

This is a visual way to think about the theorem: it is really an area relationship.

Geometric Proof Idea 2: Rearranging Triangles Inside a Square

One famous proof uses a large square and four copies of the same right triangle.

Suppose each triangle has legs \(a\) and \(b\), and hypotenuse \(c\).

First, make a large square with side length \(a+b\). Its area is:

$$ (a+b)^2 $$

Now place four identical right triangles inside this large square. There are two useful ways to arrange them.

Arrangement 1: The four triangles leave a small square in the center with side length \(c\).

The area of the large square is equal to:

• the area of 4 triangles, plus
• the area of the center square

Each triangle has area:

$$\frac{1}{2}ab$$

So four triangles have total area:

$$4\left(\frac{1}{2}ab\right) = 2ab$$

The center square has area:

$$c^2$$

So:

$$ (a+b)^2 = 2ab + c^2 $$

Now expand the left side:

$$ a^2 + 2ab + b^2 = 2ab + c^2 $$

Subtract \(2ab\) from both sides:

$$ a^2 + b^2 = c^2 $$

That proves the Pythagorean Theorem.

This proof is powerful because it uses only area and rearranging shapes.

Geometric Proof Idea 3: Comparing Two Ways to Find the Same Area

Another way to understand the proof is to compare two expressions for the same large square.

The large square has side length \(a+b\), so its area is:

$$ (a+b)^2 $$

But if it is made from four right triangles and a center square of side \(c\), then its area is also:

$$4\left(\frac{1}{2}ab\right) + c^2$$

Simplify:

$$2ab + c^2$$

Since both expressions describe the same area, they must be equal:

$$ (a+b)^2 = 2ab + c^2 $$

Then:

$$ a^2 + 2ab + b^2 = 2ab + c^2 $$ $$ a^2 + b^2 = c^2 $$

This kind of proof is called a proof by area comparison.

What the proof shows

The theorem is not just a pattern from a few examples like \(3\)-\(4\)-\(5\). The proof shows that whenever a triangle is a right triangle, the side lengths must satisfy:

$$a^2 + b^2 = c^2$$

And if side lengths satisfy this equation, that is a strong sign that the triangle is a right triangle.

Worked Example 1: Understanding the area model

A right triangle has legs \(6\) and \(8\), and hypotenuse \(10\). Show how the theorem works using areas.

Step 1: Find the area of the square on each leg.

$$6^2 = 36$$ $$8^2 = 64$$

Step 2: Add the two areas.

$$36 + 64 = 100$$

Step 3: Find the area of the square on the hypotenuse.

$$10^2 = 100$$

Conclusion: The two smaller square areas add to the largest square area, so this matches the Pythagorean Theorem.

Worked Example 2: Using the rearrangement proof with numbers

Use the rearrangement proof for a right triangle with legs \(3\) and \(4\), and hypotenuse \(5\).

Step 1: Find the area of the large square with side length \(3+4=7\).

$$7^2 = 49$$

Step 2: Find the total area of the four triangles.

One triangle has area:

$$\frac{1}{2}(3)(4) = 6$$

Four triangles have area:

$$4 \cdot 6 = 24$$

Step 3: Find the center square area.

$$49 - 24 = 25$$

Step 4: Compare with \(c^2\).

$$5^2 = 25$$

Conclusion: The center square really has area \(c^2\), and the proof works for this triangle.

Worked Example 3: Proving the theorem with algebra

Suppose a large square has side length \(a+b\). Inside it are four right triangles with area \(\frac{1}{2}ab\) each, and a center square with area \(c^2\). Prove the theorem.

Step 1: Write the area of the large square.

$$ (a+b)^2 $$

Step 2: Write the area of the pieces inside.

$$4\left(\frac{1}{2}ab\right) + c^2$$

Step 3: Set them equal.

$$ (a+b)^2 = 4\left(\frac{1}{2}ab\right) + c^2 $$

Step 4: Simplify.

$$ a^2 + 2ab + b^2 = 2ab + c^2 $$

Step 5: Subtract \(2ab\) from both sides.

$$ a^2 + b^2 = c^2 $$

Conclusion: This proves the Pythagorean Theorem.

Worked Example 4: Explaining why a triangle is right

A triangle has side lengths \(5\), \(12\), and \(13\). Use the theorem to explain why it is a right triangle.

Step 1: Check whether the shorter sides satisfy the equation.

$$5^2 + 12^2 = 25 + 144 = 169$$

Step 2: Compare with the longest side squared.

$$13^2 = 169$$

Since:

$$5^2 + 12^2 = 13^2$$

the side lengths fit the Pythagorean Theorem.

Conclusion: This triangle is a right triangle.

Common mistakes to avoid

• Using the theorem on a triangle that is not a right triangle.
• Forgetting that \(c\) is always the hypotenuse, the longest side.
• Adding side lengths instead of adding squares of side lengths.
• Mixing up area and side length. For example, \(a^2\) is the area of a square with side \(a\).

How proofs connect to later math

Understanding this proof helps with many future ideas:

• finding missing sides in right triangles
• finding distances on a coordinate plane
• understanding diagonal lengths in rectangles, boxes, and other shapes
• seeing how algebra can prove geometry facts

Quick check for understanding

1. What do \(a\), \(b\), and \(c\) stand for in a right triangle?
2. Why do we build squares on the sides of the triangle in the proof?
3. In the rearrangement proof, what is the area of one triangle?
4. Why does comparing two ways to find the area of the same large square prove the theorem?

Brief Summary

The Pythagorean Theorem says that in a right triangle, \(a^2 + b^2 = c^2\). Proofs help us understand that this is always true, not just true for a few examples.

A visual proof shows that the areas of the two smaller squares add to the area of the square on the hypotenuse. A geometric proof uses four identical right triangles inside a large square and compares areas in two different ways.

When you understand the proof, you understand the theorem more deeply. You are not just using a formula—you know why it works.

Finding the Hypotenuse

When you look at a right triangle, one side is special. It is called the hypotenuse.

The hypotenuse is the side across from the right angle, and it is always the longest side of the triangle.

To find the hypotenuse, we use the Pythagorean Theorem. This theorem only works for right triangles.

The Pythagorean Theorem says:

$$a^2+b^2=c^2$$

In this formula:

• \(a\) and \(b\) are the lengths of the two shorter sides, called the legs.

• \(c\) is the length of the hypotenuse.

So if you know the lengths of both legs, you can find the hypotenuse by following these steps.

1. Square each leg length.

2. Add the squares.

3. Take the square root of the sum.

That means the formula for the hypotenuse can also be written as:

$$c=\sqrt{a^2+b^2}$$

Let’s look at this carefully. If a triangle has legs of length 3 and 4, then:

$$c=\sqrt{3^2+4^2}$$

$$c=\sqrt{9+16}$$

$$c=\sqrt{25}=5$$

So the hypotenuse is 5 units long.

Important reminder: Do not add the side lengths first and then square. You must square each leg first, then add.

For example, this is not correct:

$$c=(3+4)^2$$

The correct method is:

$$c=\sqrt{3^2+4^2}$$

How to recognize the hypotenuse

• Find the right angle first.

• The side directly across from that right angle is the hypotenuse.

• It will always be the longest side.

This matters because in the formula, the hypotenuse must be the side labeled \(c\).

Worked Example 1: Basic whole-number sides

A right triangle has legs of lengths 6 units and 8 units. Find the hypotenuse.

Step 1: Write the formula.

$$c=\sqrt{a^2+b^2}$$

Step 2: Substitute the known values.

$$c=\sqrt{6^2+8^2}$$

Step 3: Square each number.

$$c=\sqrt{36+64}$$

Step 4: Add.

$$c=\sqrt{100}$$

Step 5: Take the square root.

$$c=10$$

Answer: The hypotenuse is 10 units.

Worked Example 2: When the answer is not a whole number

A right triangle has legs of lengths 5 cm and 12 cm. Find the hypotenuse.

Use the formula:

$$c=\sqrt{5^2+12^2}$$

$$c=\sqrt{25+144}$$

$$c=\sqrt{169}$$

$$c=13$$

Answer: The hypotenuse is 13 cm.

Now let’s try one where the square root does not make a perfect whole number.

A right triangle has legs of lengths 7 m and 9 m. Find the hypotenuse.

$$c=\sqrt{7^2+9^2}$$

$$c=\sqrt{49+81}$$

$$c=\sqrt{130}$$

Since \(130\) is not a perfect square, the exact answer is \(\sqrt{130}\).

As a decimal,

$$\sqrt{130}\approx 11.4$$

Answer: The hypotenuse is about 11.4 m.

Worked Example 3: Real-world problem

A ladder is leaning against a wall. The bottom of the ladder is 9 feet from the wall, and the top touches the wall 12 feet above the ground. How long is the ladder?

This forms a right triangle.

• One leg is 9 ft.

• The other leg is 12 ft.

• The ladder is the hypotenuse because it is across from the right angle.

Use the Pythagorean Theorem:

$$c=\sqrt{9^2+12^2}$$

$$c=\sqrt{81+144}$$

$$c=\sqrt{225}$$

$$c=15$$

Answer: The ladder is 15 feet long.

Worked Example 4: Using decimals

A right triangle has legs of lengths 2.5 inches and 6 inches. Find the hypotenuse.

$$c=\sqrt{2.5^2+6^2}$$

$$c=\sqrt{6.25+36}$$

$$c=\sqrt{42.25}$$

$$c=6.5$$

Answer: The hypotenuse is 6.5 inches.

Tips for success

• Make sure the triangle is a right triangle.

• Label the legs as \(a\) and \(b\), and the hypotenuse as \(c\).

• Square the leg lengths before adding.

• Take the square root at the end.

• Include the correct units in your final answer.

• If needed, round your decimal answer as directed.

Common mistakes to avoid

• Using the theorem on a triangle that is not a right triangle.

• Forgetting which side is the hypotenuse.

• Adding first instead of squaring first.

• Forgetting to take the square root.

• Writing an answer that is shorter than one of the legs. The hypotenuse must be the longest side.

Quick check

If the legs are 8 and 15, then:

$$c=\sqrt{8^2+15^2}$$

$$c=\sqrt{64+225}$$

$$c=\sqrt{289}=17$$

So the hypotenuse is 17.

Summary

To find the hypotenuse of a right triangle, use the Pythagorean Theorem: $$a^2+b^2=c^2$$

When solving for the hypotenuse, use:

$$c=\sqrt{a^2+b^2}$$

Square the two leg lengths, add them, and then take the square root. Always remember that the hypotenuse is across from the right angle and is the longest side in the triangle.

Finding a Missing Leg is an important skill when using the Pythagorean Theorem. In a right triangle, if you know the length of the hypotenuse and one leg, you can find the other leg.

This lesson will show you how to set up the equation, solve it step by step, and check that your answer makes sense.

First, remember the Pythagorean Theorem:

For any right triangle, if the legs are \(a\) and \(b\), and the hypotenuse is \(c\), then

$$a^2+b^2=c^2$$

The legs are the two shorter sides that form the right angle. The hypotenuse is the longest side, and it is always across from the right angle.

When you are finding a missing leg, you already know:

• the hypotenuse
• one leg

So your job is to solve for the other leg.

The basic idea is to substitute the known side lengths into the Pythagorean Theorem and then isolate the missing leg.

If \(a\) is missing, and you know \(b\) and \(c\), then:

$$a^2+b^2=c^2$$

Subtract \(b^2\) from both sides:

$$a^2=c^2-b^2$$

Then take the square root:

$$a=\sqrt{c^2-b^2}$$

You can use the same process if \(b\) is the missing leg.

Important: Do not just subtract the side lengths. You must subtract the squares of the side lengths first.

For example, if the hypotenuse is 13 and one leg is 5, you do not do \(13-5=8\). Instead, you do:

$$a^2=13^2-5^2$$

That means:

$$a^2=169-25=144$$

Then:

$$a=\sqrt{144}=12$$

Steps for finding a missing leg

1. Identify the hypotenuse. It is the longest side.
2. Write the Pythagorean Theorem: \(a^2+b^2=c^2\).
3. Substitute the known values.
4. Subtract the square of the known leg from the square of the hypotenuse.
5. Take the square root to find the missing leg.
6. Check that your answer is shorter than the hypotenuse.

Worked Example 1

A right triangle has a hypotenuse of 10 units and one leg of 6 units. Find the other leg.

Step 1: Set up the equation.

$$a^2+6^2=10^2$$

Step 2: Square the known numbers.

$$a^2+36=100$$

Step 3: Subtract 36 from both sides.

$$a^2=64$$

Step 4: Take the square root.

$$a=8$$

Answer: The missing leg is 8 units.

Check:

$$6^2+8^2=36+64=100=10^2$$

The answer works.

Worked Example 2

A right triangle has a hypotenuse of 15 units and one leg of 9 units. Find the missing leg.

Step 1: Write the equation.

$$a^2+9^2=15^2$$

Step 2: Square the numbers.

$$a^2+81=225$$

Step 3: Subtract 81.

$$a^2=144$$

Step 4: Take the square root.

$$a=12$$

Answer: The missing leg is 12 units.

Worked Example 3

A right triangle has a hypotenuse of 25 units and one leg of 7 units. Find the other leg.

Step 1: Set up the equation.

$$a^2+7^2=25^2$$

Step 2: Square the known values.

$$a^2+49=625$$

Step 3: Subtract 49 from both sides.

$$a^2=576$$

Step 4: Take the square root.

$$a=24$$

Answer: The missing leg is 24 units.

Worked Example 4

A right triangle has a hypotenuse of 13.4 units and one leg of 8 units. Find the missing leg.

Step 1: Write the equation.

$$a^2+8^2=13.4^2$$

Step 2: Square the numbers.

$$a^2+64=179.56$$

Step 3: Subtract 64.

$$a^2=115.56$$

Step 4: Take the square root.

$$a=\sqrt{115.56}\approx 10.75$$

Answer: The missing leg is about 10.75 units.

Sometimes your answer will be a whole number, and sometimes it will be a decimal. Both are possible.

Common mistakes to avoid

• Using the wrong side as the hypotenuse: The hypotenuse is always the longest side.
• Subtracting before squaring: You must square first, then subtract.
• Forgetting the square root: If you find \(a^2=49\), the leg is \(a=7\), not 49.
• Getting a leg longer than the hypotenuse: That cannot happen in a right triangle.

Quick check strategy

After you solve, ask yourself:

• Is my missing side shorter than the hypotenuse?
• Did I square the side lengths first?
• Did I take the square root at the end?

Practice thinking

If the hypotenuse is \(20\) and one leg is \(16\), then the missing leg is found by:

$$a^2=20^2-16^2$$$$a^2=400-256=144$$$$a=12$$

So the missing leg is 12.

Summary

To find a missing leg in a right triangle, use the Pythagorean Theorem:

$$a^2+b^2=c^2$$

When the hypotenuse and one leg are known, subtract the square of the known leg from the square of the hypotenuse. Then take the square root to find the missing leg.

This can be written as:

$$\text{missing leg}=\sqrt{(\text{hypotenuse})^2-(\text{known leg})^2}$$

If you follow the steps carefully, you can solve missing leg problems with confidence.

Converse of the Pythagorean Theorem

In earlier work, you may have used the Pythagorean Theorem to find a missing side length in a right triangle. It says that if a triangle is a right triangle, then the side lengths follow this rule:

\(a^2 + b^2 = c^2\)

Here, \(a\) and \(b\) are the lengths of the two shorter sides, and \(c\) is the longest side, called the hypotenuse.

The converse of the Pythagorean Theorem works in the opposite direction. Instead of starting with a right triangle, we start with the side lengths. Then we test whether those side lengths make a right triangle.

Converse of the Pythagorean Theorem: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

In symbols, if the side lengths are \(a\), \(b\), and \(c\), with \(c\) as the longest side, then:

$$a^2 + b^2 = c^2$$

If this equation is true, the triangle is a right triangle. If it is not true, then the triangle is not a right triangle.

Important: Before testing, always make sure you know which side is the longest. That side must be used as \(c\).

How to use the converse

1. Write the side lengths in order from smallest to largest.
2. Let the longest side be \(c\).
3. Square the two shorter sides and add them.
4. Square the longest side.
5. Compare the results.

• If \(a^2 + b^2 = c^2\), the triangle is a right triangle.
• If \(a^2 + b^2 \ne c^2\), the triangle is not a right triangle.

This idea is helpful when you are given three side lengths and need to decide what kind of triangle they make.

Worked Example 1

Determine whether a triangle with side lengths 3, 4, and 5 is a right triangle.

Step 1: Identify the longest side. The longest side is 5, so let \(c = 5\).

Step 2: Substitute into the equation.

$$3^2 + 4^2 = 5^2$$

Step 3: Calculate both sides.

$$9 + 16 = 25$$ $$25 = 25$$

The equation is true, so the triangle is a right triangle.

Worked Example 2

Determine whether a triangle with side lengths 6, 8, and 10 is a right triangle.

Step 1: The longest side is 10, so \(c = 10\).

Step 2: Test the side lengths.

$$6^2 + 8^2 = 10^2$$

Step 3: Compute the squares.

$$36 + 64 = 100$$ $$100 = 100$$

The equation is true, so this triangle is a right triangle.

Notice that 6, 8, and 10 are a larger version of 3, 4, and 5. Many right triangles follow patterns like this.

Worked Example 3

Determine whether a triangle with side lengths 5, 6, and 7 is a right triangle.

Step 1: The longest side is 7, so \(c = 7\).

Step 2: Test the side lengths.

$$5^2 + 6^2 = 7^2$$

Step 3: Compute the squares.

$$25 + 36 = 49$$ $$61 = 49$$

The equation is not true, so the triangle is not a right triangle.

Worked Example 4

A triangle has side lengths 9, 12, and 15. Is it a right triangle?

Step 1: The longest side is 15, so \(c = 15\).

Step 2: Check the equation.

$$9^2 + 12^2 = 15^2$$

Step 3: Square each number.

$$81 + 144 = 225$$ $$225 = 225$$

Since the equation is true, the triangle is a right triangle.

Why the longest side matters

The converse only works correctly if you use the longest side as \(c\). That is because in a right triangle, the hypotenuse is always the longest side.

For example, if the sides are 8, 15, and 17, then 17 must be used as \(c\), not 8 or 15.

$$8^2 + 15^2 = 17^2$$ $$64 + 225 = 289$$ $$289 = 289$$

So the triangle is a right triangle.

A common mistake

Some students square the wrong side for \(c\). For example, with side lengths 7, 24, and 25, you should not test \(25^2 + 7^2 = 24^2\). That does not make sense because 24 is not the longest side.

The correct test is:

$$7^2 + 24^2 = 25^2$$ $$49 + 576 = 625$$ $$625 = 625$$

So the triangle is a right triangle.

What if the equation does not work?

If the side lengths do not satisfy \(a^2 + b^2 = c^2\), then the triangle is not a right triangle.

For example, with side lengths 4, 5, and 6:

$$4^2 + 5^2 = 6^2$$ $$16 + 25 = 36$$ $$41 \ne 36$$

So this is not a right triangle.

Quick check strategy

• Find the longest side.
• Square it.
• Square the other two sides and add them.
• See if the numbers match exactly.

Practice thinking

When you see three side lengths, ask yourself:

• Which side is longest?
• What is the sum of the squares of the two shorter sides?
• Does that equal the square of the longest side?

If yes, it is a right triangle. If no, it is not.

Summary

The converse of the Pythagorean Theorem helps you decide whether a triangle is a right triangle when you know all three side lengths. Put the longest side in the \(c\) position, then test whether \(a^2 + b^2 = c^2\). If the equation is true, the triangle is right; if not, it is not a right triangle.

Distance on the Coordinate Plane

Sometimes on a coordinate grid, two points are not directly above, below, or beside each other. When that happens, we cannot find the distance by simple counting alone. Instead, we can use the Pythagorean Theorem to find the shortest distance between the points.

This lesson will show how distance on the coordinate plane connects to right triangles, how the distance formula is built, and how to use it step by step.

Review: The Coordinate Plane

A point on the coordinate plane is written as \\((x, y)\\).

• The x-coordinate tells how far left or right the point is.
• The y-coordinate tells how far up or down the point is.

For example, the point \\((3, 5)\\) is 3 units to the right and 5 units up from the origin.

When Finding Distance Is Easy

If two points have the same y-coordinate, they lie on a horizontal line. Then the distance is just the difference between their x-coordinates.

For example, the distance between \\((2, 4)\\) and \\((7, 4)\\) is

$$|7 - 2| = 5$$

If two points have the same x-coordinate, they lie on a vertical line. Then the distance is the difference between their y-coordinates.

For example, the distance between \\((3, 1)\\) and \\((3, 9)\\) is

$$|9 - 1| = 8$$

But what if the points are diagonal from each other? That is when we use a right triangle.

Using a Right Triangle to Find Distance

Suppose we want the distance between the points \\((x_1, y_1)\\) and \\((x_2, y_2)\\).

Imagine drawing a horizontal segment and a vertical segment between the points. This forms a right triangle.

• The horizontal leg has length \\(|x_2 - x_1|\\).
• The vertical leg has length \\(|y_2 - y_1|\\).
• The distance between the two points is the hypotenuse.

Now use the Pythagorean Theorem:

$$a^2 + b^2 = c^2$$

So the distance \\d\\ between the points is

$$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Taking the square root of both sides gives the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

This formula gives the shortest distance between two points on the coordinate plane.

Important Idea

It does not matter which point you call \\((x_1, y_1)\\) or \\((x_2, y_2)\\). The answer will be the same, because squaring removes any negative signs.

For example, \\((7 - 2)^2 = 5^2\\) and \\((2 - 7)^2 = (-5)^2\\), and both equal 25.

Steps for Using the Distance Formula

1. Write the coordinates of both points.
2. Find the difference in the x-coordinates.
3. Find the difference in the y-coordinates.
4. Square both differences.
5. Add the squares.
6. Take the square root.

Worked Example 1: A Simple Distance

Find the distance between \\((1, 2)\\) and \\((4, 6)\\).

Step 1: Use the formula

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Step 2: Substitute the values

$$d = \sqrt{(4 - 1)^2 + (6 - 2)^2}$$

Step 3: Simplify

$$d = \sqrt{3^2 + 4^2}$$ $$d = \sqrt{9 + 16}$$ $$d = \sqrt{25}$$ $$d = 5$$

Answer: The distance is 5 units.

This example makes a 3-4-5 right triangle.

Worked Example 2: Distance with Negative Coordinates

Find the distance between \\((-2, 3)\\) and \\((4, -1)\\).

Step 1: Substitute into the formula

$$d = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}$$

Step 2: Simplify inside the parentheses

$$d = \sqrt{6^2 + (-4)^2}$$

Step 3: Square and add

$$d = \sqrt{36 + 16}$$ $$d = \sqrt{52}$$

Step 4: Simplify the square root if possible

$$d = \sqrt{4 \cdot 13} = 2\sqrt{13}$$

Answer: The distance is \\(2\sqrt{13}\\) units.

If a decimal is needed, \\((2\sqrt{13}) \\approx 7.2\\) units.

Worked Example 3: Horizontal and Vertical Changes

Find the distance between \\((5, -2)\\) and \\((-1, 7)\\).

Step 1: Find the change in x and y

$$x\text{-change} = -1 - 5 = -6$$ $$y\text{-change} = 7 - (-2) = 9$$

Step 2: Use the distance formula

$$d = \sqrt{(-6)^2 + 9^2}$$

Step 3: Simplify

$$d = \sqrt{36 + 81}$$ $$d = \sqrt{117}$$

Step 4: Simplify the radical

$$d = \sqrt{9 \cdot 13} = 3\sqrt{13}$$

Answer: The distance is \\(3\sqrt{13}\\) units.

Worked Example 4: Deciding Whether a Shape Has Equal Sides

Suppose points \\A(1, 1)\\, \\B(5, 1)\\, and \\C(5, 4)\\ are connected. Is triangle \\ABC\\ a right triangle?

We can check the side lengths.

Side \\AB\\:

$$AB = \sqrt{(5 - 1)^2 + (1 - 1)^2} = \sqrt{4^2 + 0^2} = 4$$

Side \\BC\\:

$$BC = \sqrt{(5 - 5)^2 + (4 - 1)^2} = \sqrt{0^2 + 3^2} = 3$$

Side \\AC\\:

$$AC = \sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = 5$$

The side lengths are 3, 4, and 5. Since

$$3^2 + 4^2 = 5^2$$

triangle \\ABC\\ is a right triangle.

This shows that distance on the coordinate plane can help us understand shapes, not just find lengths.

Common Mistakes to Avoid

• Mixing up x- and y-coordinates: Subtract x-values from x-values and y-values from y-values.
• Forgetting negative signs: Be careful when subtracting a negative number, such as \\((4 - (-2))\\).
• Adding before squaring: You must square each difference first.
• Not using the square root: After adding the squares, take the square root to get the distance.

Quick Check

Try these on your own:

• Find the distance between \\((0, 0)\\) and \\((6, 8)\\).
• Find the distance between \\((-3, -2)\\) and \\((1, 1)\\).
• Are the points \\((2, 2)\\), \\((6, 2)\\), and \\((6, 5)\\) the vertices of a right triangle?

Answers:

• \\(10\\) units
• \\(5\\) units
• Yes, because the side lengths are 4, 3, and 5.

Summary

The shortest distance between two points on the coordinate plane can be found by making a right triangle and using the Pythagorean Theorem.

If the points are \\((x_1, y_1)\\) and \\((x_2, y_2)\\), then the distance is

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

This formula works for any two points. Always subtract carefully, square each difference, add, and then take the square root.

Lesson: Pythagorean Theorem in Three Dimensions

So far, you may have used the Pythagorean Theorem to find the missing side of a right triangle in a flat, 2D shape. In this lesson, you will learn how to use that same idea in three dimensions.

We will focus on finding the internal diagonal of a rectangular prism. A rectangular prism is a box-shaped solid with length, width, and height.

Even though the shape is 3D, the main idea is still the same: we use the Pythagorean Theorem more than once.

1. Review: The Pythagorean Theorem in 2D

In a right triangle, if the legs are \(a\) and \(b\), and the hypotenuse is \(c\), then:

$$a^2 + b^2 = c^2$$

This theorem only works for right triangles.

For example, if a right triangle has side lengths 3 and 4, then the hypotenuse is:

$$3^2 + 4^2 = c^2$$ $$9 + 16 = c^2$$ $$25 = c^2$$ $$c = 5$$

2. Moving from 2D to 3D

Now imagine a rectangular prism with:

• length \(l\)
• width \(w\)
• height \(h\)

There are two important diagonals we can think about:

• The diagonal of the base, which lies flat on the bottom rectangle
• The internal diagonal, which goes from one corner of the prism to the opposite corner through the inside of the solid

To find the internal diagonal, we first find the base diagonal, then use the Pythagorean Theorem again with the height.

3. Step 1: Find the diagonal of the base

The base of the prism is a rectangle. A rectangle can be split into a right triangle by drawing a diagonal.

If the base has length \(l\) and width \(w\), then the base diagonal, which we will call \(d_b\), is:

$$d_b^2 = l^2 + w^2$$ $$d_b = \sqrt{l^2 + w^2}$$

4. Step 2: Use the height to find the internal diagonal

Now picture a right triangle inside the prism:

• one leg is the base diagonal \(d_b\)
• the other leg is the height \(h\)
• the hypotenuse is the internal diagonal, which we will call \(D\)

Using the Pythagorean Theorem again:

$$D^2 = d_b^2 + h^2$$

Since \(d_b^2 = l^2 + w^2\), we can substitute:

$$D^2 = l^2 + w^2 + h^2$$

So the formula for the internal diagonal of a rectangular prism is:

$$D = \sqrt{l^2 + w^2 + h^2}$$

This is the main formula for 3D Pythagorean problems involving rectangular prisms.

5. Why this formula works

The Pythagorean Theorem is still being used with right triangles. In 3D, we just build one right triangle after another.

1. Use length and width to make a right triangle on the base.
2. Use the base diagonal and the height to make another right triangle.

That is why all three squared dimensions get added together:

$$l^2 + w^2 + h^2$$

6. Worked Example 1: Finding an internal diagonal

A rectangular prism has length 3 cm, width 4 cm, and height 12 cm. Find the internal diagonal.

Step 1: Use the 3D formula.

$$D = \sqrt{l^2 + w^2 + h^2}$$

Substitute the values:

$$D = \sqrt{3^2 + 4^2 + 12^2}$$ $$D = \sqrt{9 + 16 + 144}$$ $$D = \sqrt{169}$$ $$D = 13$$

Answer: The internal diagonal is 13 cm.

7. Worked Example 2: Using two steps

A box has length 6 m, width 8 m, and height 5 m. Find the internal diagonal.

Step 1: Find the base diagonal.

$$d_b = \sqrt{6^2 + 8^2}$$ $$d_b = \sqrt{36 + 64}$$ $$d_b = \sqrt{100}$$ $$d_b = 10$$

Step 2: Use the height and base diagonal.

$$D = \sqrt{10^2 + 5^2}$$ $$D = \sqrt{100 + 25}$$ $$D = \sqrt{125}$$

Simplify if possible:

$$\sqrt{125} = \sqrt{25 \cdot 5} = 5\sqrt{5}$$

Answer: The internal diagonal is \(5\sqrt{5}\) m, which is about 11.2 m.

This example shows that answers are not always whole numbers.

8. Worked Example 3: Finding a missing dimension

A rectangular prism has width 4 in, height 12 in, and internal diagonal 13 in. What is the length?

We use:

$$D^2 = l^2 + w^2 + h^2$$

Substitute the known values:

$$13^2 = l^2 + 4^2 + 12^2$$ $$169 = l^2 + 16 + 144$$ $$169 = l^2 + 160$$ $$9 = l^2$$ $$l = 3$$

Answer: The length is 3 inches.

9. Worked Example 4: A spatial reasoning problem

A spider is at one corner inside a rectangular room shaped like a box. The room is 9 ft long, 12 ft wide, and 8 ft high. What is the straight-line distance from the spider to the opposite corner of the room?

This is the internal diagonal of a rectangular prism.

$$D = \sqrt{9^2 + 12^2 + 8^2}$$ $$D = \sqrt{81 + 144 + 64}$$ $$D = \sqrt{289}$$ $$D = 17$$

Answer: The straight-line distance is 17 feet.

10. How to solve 3D Pythagorean problems

When you see a box-shaped solid, follow these steps:

1. Identify the three dimensions: length, width, and height.
2. Decide whether you need the base diagonal or the internal diagonal.
3. If finding the internal diagonal, use:

$$D = \sqrt{l^2 + w^2 + h^2}$$

4. Substitute carefully.
5. Square each value before adding.
6. Take the square root at the end.
7. Include units in your answer.

11. Common mistakes to avoid

• Forgetting to square all three dimensions. Make sure you compute \(l^2\), \(w^2\), and \(h^2\).
• Adding first, then squaring. Do not do \((l+w+h)^2\). That is not correct.
• Using the theorem when there is no right triangle. The Pythagorean Theorem only works because the edges of a rectangular prism meet at right angles.
• Stopping after finding the base diagonal. If the problem asks for the internal diagonal, you need one more step.
• Forgetting units. Always write cm, m, in, ft, or whatever unit is given.

12. Quick check

Try these on your own:

• A prism has dimensions 2, 3, and 6. Find the internal diagonal.
• A prism has dimensions 5, 12, and 4. Find the internal diagonal.
• A prism has internal diagonal 15, width 8, and height 12. Find the missing length.

Answers:

• \(\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7\)
• \(\sqrt{5^2+12^2+4^2}=\sqrt{25+144+16}=\sqrt{185}\)
• \(15^2=l^2+8^2+12^2\Rightarrow 225=l^2+64+144\Rightarrow 225=l^2+208\Rightarrow l^2=17\Rightarrow l=\sqrt{17}\)

13. Summary

The Pythagorean Theorem can be used in 3D when working with rectangular prisms. First, think about right triangles inside the solid.

For a prism with length \(l\), width \(w\), and height \(h\), the internal diagonal is:

$$D = \sqrt{l^2 + w^2 + h^2}$$

This formula helps you find the straight-line distance from one corner of a box to the opposite corner. It can also help you find a missing dimension when the diagonal is known.