Understanding the Pythagorean Theorem and the Distance Formula

Area of the Inner Square	Area of the Outer Square	Area of the Four Triangles
$c^{2}$	$(a + b)^{2}$	$4 \cdot \frac{a b}{2} = 2 a b$

Area of the Inner Square

Area of the Outer Square

Area of the Four Triangles

c^{2}

(a + b)^{2}

4 \cdot \frac{a b}{2} = 2 a b

Triple	Substitute in $a^{2} + b^{2} = c^{2}$	Simplify
$(3, 4, 5)$	$3^{2} + 4^{2} = ? 5^{2}$	$9 + 16 = 25 ✓$
$(5, 12, 13)$	$5^{2} + 12^{2} = ? 13^{2}$	$25 + 144 = 169 ✓$
$(8, 15, 17)$	$8^{2} + 15^{2} = ? 17^{2}$	$64 + 225 = 289 ✓$
$(7, 24, 25)$	$7^{2} + 24^{2} = ? 25^{2}$	$49 + 576 = 625 ✓$

Triple

Substitute in

a^{2} + b^{2} = c^{2}

Simplify

(3, 4, 5)

3^{2} + 4^{2} = ? 5^{2}

9 + 16 = 25 ✓

(5, 12, 13)

5^{2} + 12^{2} = ? 13^{2}

25 + 144 = 169 ✓

(8, 15, 17)

8^{2} + 15^{2} = ? 17^{2}

64 + 225 = 289 ✓

(7, 24, 25)

7^{2} + 24^{2} = ? 25^{2}

49 + 576 = 625 ✓

$(a, b, c)$	$k$	$(k a, k b, k c)$
$(3, 4, 5)$	$3$	$(3 (3), 3 (4), 3 (5))$ $⇕$ $(9, 12, 15)$
$(5, 12, 13)$	$2$	$(2 (5), 2 (12), 2 (13))$ $⇕$ $(10, 24, 26)$

(a, b, c)

k

(k a, k b, k c)

(3, 4, 5)

3

(3 (3), 3 (4), 3 (5))

⇕

(9, 12, 15)

(5, 12, 13)

2

(2 (5), 2 (12), 2 (13))

⇕

(10, 24, 26)

How long is the diagonal of the following rectangular box?

We can see that the diagonal of the box is the hypotenuse of a right triangle. Let's begin our math by labeling its vertices.

The length of AB can be found by using the Pythagorean Theorem. a^2+b^2 &= c^2 &⇓ AC^2+BC^2 &= AB^2 We already know that BC=40, since the box is 40 centimeters high. However, we do not know the length of AC. That being said, AC is the hypotenuse of the right triangle formed at the base of the box. Let's label the vertex where the right angle is formed.

Remember, the box is 40 centimeters long and 20 centimeters wide. This means that AD=40 and DC=20. Knowing these lengths means that we can find the length of AC. Let's apply the Pythagorean Theorem. We will keep the answer in radical form.

Now that know the length of AC, nothing stops us from finding the length of the diagonal of the box. Let's substitute 40 for BC and sqrt(2000) for AC.

The diagonal of the box is 60 centimeters long.

Kevin accidentally hit the bottom of his building's gutter with a ball. This caused the upper part of the gutter to become unattached. The building is $13$ meters high. If the bottom part of the gutter moved $5$ meters away from the wall, how many meters did the top part move down?

Run Kevin, run! We are interested in finding how many meters the top part of the gutter moved down. Let's begin by making a drawing to illustrate what we have and what we want.

The gutter and the building form a right triangle whose shorter leg is 5 meters long and whose hypotenuse is the gutter. Our mission is to find CD but it seems we do not have enough information. However, since the gutter was next to the building before, its length must be equal to the building's height. This means that BC=13.

Now that we know two side lengths of △ ABC, we can find the length of AC by applying the Pythagorean Theorem. Let's do it.

To find the length of CD, subtract AC from AD.

The top part of the gutter moved down 1 meter.

The Pythagorean Theorem and the Distance Formula

Catch-Up and Review

Designing a Putting Green with Right Triangles

Extra

Relation Between the Side Lengths of a Right Triangle

Pythagorean Theorem

Proof

Mini Golf Tournament

Hint

Solution

Finding the Missing Length

Determining Whether a Triangle is a Right Triangle

Converse Pythagorean Theorem

Flags

Hint

Solution

Is It a Right Triangle?

Extra

Pythagorean Triple

Studying Triples

Distance Between Two Points

Distance Formula

A Long Stroke

Hint

Solution

Finding the Distance Between Two Points

Contributions by Pythagoras

The Pythagorean Theorem and the Distance Formula

Recommended exercises

	14 Theory slides
	13 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions