Pythagorean Theorem

8.G.1Apply Pythagorean01/04

Apply the Pythagorean theorem.

The relationship

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

Step01of 04

In a right triangle, $a$ and $b$ are the legs (the two shorter sides forming the right angle), and $c$ is the hypotenuse (the longest side, OPPOSITE the right angle).

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

Memorable triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25.

8.G.1 — Apply the Pythagorean theorem.

The relationship

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

Step 01 of 04

In a right triangle, $a$ and $b$ are the legs (the two shorter sides forming the right angle), and $c$ is the hypotenuse (the longest side, OPPOSITE the right angle).

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

Memorable triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25.

Step 02 of 04

Find the hypotenuse. Legs 6 and 8.

c^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 \Rightarrow c = 10

The 6-8-10 triangle is just a scaled-up 3-4-5 (multiply each side by 2).

Step 03 of 04

Find a leg. Hypotenuse 13, one leg 5.

5^{2} + b^{2} = 1 3^{2}

b^{2} = 169 - 25 = 144 \Rightarrow b = 12

Recognize: 5-12-13.

Step 04 of 04

Real-world example. A ladder 13 ft long leans against a wall; its base is 5 ft from the wall. How high up the wall does it reach?

The wall is one leg, the ground distance is the other, the ladder is the hypotenuse. So $5^{2} + h^{2} = 1 3^{2}$ , height $h = 12$ ft.

Key insight

Pythagorean theorem:

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

only when one angle is right and

c

is opposite that angle. Memorize the common triples (and recognize their multiples). Always identify which side is the hypotenuse first.

8.G.2 — Distance between two points.

Step 01 of 04

The distance formula is the Pythagorean theorem applied to the changes in x and y.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The horizontal difference is one leg, the vertical difference is the other, the straight-line distance is the hypotenuse.

Step 02 of 04

Worked example. Distance between $(2, 1)$ and $(8, 9)$ .

d = (8 - 2)^{2} + (9 - 1)^{2} = 36 + 64 = 100 = 10

That's a 6-8-10 triangle (scaled 3-4-5).

Step 03 of 04

Negative coordinates. The formula handles them automatically — squaring removes any negatives in the differences.

Distance between $(- 3, 2)$ and $(1, - 1)$ :

d = (1 - (- 3))^{2} + (- 1 - 2)^{2} = 16 + 9 = 5

Step 04 of 04

Order doesn't matter. You can pick either point as the "first." $(x_{2} - x_{1})^{2}$ equals $(x_{1} - x_{2})^{2}$ after squaring. Just be consistent within each parenthesis.

Visualize: a right triangle with the two points as the endpoints of the hypotenuse; the legs are along the grid lines.

Key insight

Distance formula = Pythagorean theorem with horizontal and vertical changes as the legs. Squaring handles negatives automatically. Order doesn't matter.

8.G.3 — Converse and 3D applications.

Step 01 of 04

Converse of the Pythagorean theorem. If three sides satisfy $a^{2} + b^{2} = c^{2}$ (with $c$ the longest), the triangle IS a right triangle, with the right angle opposite $c$ .

This lets you VERIFY whether a triangle has a right angle without measuring angles directly.

Step 02 of 04

Worked example. Is a triangle with sides 9, 12, 15 a right triangle?

9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2} ✓

Yes — it's a 9-12-15 right triangle (scaled 3-4-5).

Counter-example. Sides 5, 6, 7? $5^{2} + 6^{2} = 25 + 36 = 61 \neq = 49 = 7^{2}$ . Not a right triangle (it's actually acute since $a^{2} + b^{2} > c^{2}$ ).

Step 03 of 04

3D application — diagonal of a box. A box has length $l$ , width $w$ , height $h$ . Find the longest diagonal (corner to opposite corner through the inside).

Apply Pythagorean TWICE. First, the diagonal of the BASE: $d_{1}^{2} = l^{2} + w^{2}$ . Then, the space diagonal uses $d_{1}$ and $h$ as legs:

d^{2} = d_{1}^{2} + h^{2} = l^{2} + w^{2} + h^{2}

d = l^{2} + w^{2} + h^{2}

Step 04 of 04

Worked example. Box 4 × 3 × 12. Space diagonal:

d = 4^{2} + 3^{2} + 1 2^{2} = 16 + 9 + 144 = 169 = 13

The 3D version generalizes: diagonal of an n-dimensional box is $x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2}$ .

Key insight

Converse: if

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

(with c largest), the triangle IS right. 3D space diagonal of a box is

l^{2} + w^{2} + h^{2}

— Pythagorean applied twice.

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