Chapter 7: Right Triangles and the Pythagorean Theorem
Learning Objectives
- State and apply the Pythagorean Theorem and its converse
- Identify and use Pythagorean triples to recognize right triangles quickly
- Apply the 45-45-90 and 30-60-90 special right triangle ratios
- Use the geometric mean to find segment lengths in right triangles with an altitude to the hypotenuse
- Derive and apply the distance formula from the Pythagorean Theorem
- Solve real-world problems involving right triangles and distances
7.1 The Pythagorean Theorem
A right triangle has one angle equal to 90°. The two sides that form the right angle are called the legs, labeled $a$ and $b$. The side opposite the right angle — always the longest side — is the hypotenuse, labeled $c$.
Pythagorean Theorem
In a right triangle with legs $a$ and $b$ and hypotenuse $c$:
$$a^2 + b^2 = c^2$$
Proof idea (area argument): Arrange 4 identical right triangles with legs $a$, $b$ inside a large square of side $(a+b)$. The 4 triangles leave a tilted square of side $c$ in the center. Comparing areas: $(a+b)^2 = 4 \cdot \tfrac{1}{2}ab + c^2$, which simplifies to $a^2 + 2ab + b^2 = 2ab + c^2$, giving $a^2 + b^2 = c^2$.
Converse of the Pythagorean Theorem
If the sides of a triangle satisfy $a^2 + b^2 = c^2$ (where $c$ is the longest side), then the triangle is a right triangle with the right angle opposite side $c$.
Pythagorean Triples
A Pythagorean triple is a set of three positive integers $(a, b, c)$ that satisfies $a^2 + b^2 = c^2$. Recognizing them saves calculation time. Any multiple of a triple is also a triple.
| Triple | Check | Common Multiples |
|---|---|---|
| 3, 4, 5 | $9 + 16 = 25$ ✓ | 6-8-10, 9-12-15, 15-20-25 |
| 5, 12, 13 | $25 + 144 = 169$ ✓ | 10-24-26 |
| 8, 15, 17 | $64 + 225 = 289$ ✓ | 16-30-34 |
| 7, 24, 25 | $49 + 576 = 625$ ✓ | 14-48-50 |
Example 7.1 — Finding the Hypotenuse
A right triangle has legs $a = 6$ and $b = 8$. Find the hypotenuse.
$$c^2 = a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100$$
$$c = \sqrt{100} = \mathbf{10}$$
Note: this is the 3-4-5 triple scaled by 2 (i.e., 6-8-10).
Example 7.2 — Using the Converse
Is a triangle with sides 9, 40, and 41 a right triangle?
Check whether $9^2 + 40^2 = 41^2$:
$$81 + 1600 = 1681 \quad \text{and} \quad 41^2 = 1681 \checkmark$$
Yes, it is a right triangle. (9-40-41 is itself a Pythagorean triple.)
A 15-ft ladder leans against a wall. The base of the ladder is 9 ft from the wall. How high up the wall does the ladder reach?
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Right triangle with labeled legs and hypotenuse. A right angle mark appears at vertex A; a dashed altitude is drawn from C to the base.
Figure 7.1 — Right Triangle: legs $a$, $b$ and hypotenuse $c$
7.2 Special Right Triangles
Two right triangle shapes appear so frequently in geometry and trigonometry that their side ratios are worth memorizing.
The 45-45-90 Triangle
Cut a square diagonally. Each half is a right isosceles triangle with angles 45°, 45°, and 90°. If each leg has length $a$, the Pythagorean Theorem gives the hypotenuse:
$$c = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$$
Ratio: legs to hypotenuse = $1 : 1 : \sqrt{2}$
The 30-60-90 Triangle
Cut an equilateral triangle in half with an altitude. Each half is a 30-60-90 right triangle. If the equilateral triangle has side $2a$, then the short leg = $a$, the hypotenuse = $2a$, and the long leg (altitude):
$$b = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \sqrt{3a^2} = a\sqrt{3}$$
Ratio: short leg : long leg : hypotenuse = $1 : \sqrt{3} : 2$
Special Right Triangle Ratios
| Triangle Type | Short Leg | Long Leg | Hypotenuse |
|---|---|---|---|
| 45-45-90 | $a$ | $a$ | $a\sqrt{2}$ |
| 30-60-90 | $a$ | $a\sqrt{3}$ | $2a$ |
Example 7.3 — 45-45-90 Triangle
The hypotenuse of a 45-45-90 triangle is 10. Find the legs.
Using the ratio: hypotenuse $= a\sqrt{2}$, so $a\sqrt{2} = 10$.
$$a = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07$$
Both legs have length $\mathbf{5\sqrt{2} \approx 7.07}$.
Example 7.4 — 30-60-90 Triangle
The hypotenuse of a 30-60-90 triangle is 12. Find both legs.
Hypotenuse $= 2a = 12$, so $a = 6$.
- Short leg (opposite 30°): $a = \mathbf{6}$
- Long leg (opposite 60°): $a\sqrt{3} = 6\sqrt{3} \approx \mathbf{10.39}$
In a 45-45-90 triangle, one leg $= 8$. Find the hypotenuse and the area of the triangle.
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Area: Both legs $= 8$, so Area $= \tfrac{1}{2}(8)(8) = \mathbf{32}$ square units.
AP Tip: The 45-45-90 and 30-60-90 side ratios are not provided on most standardized tests (SAT, ACT, geometry finals). You must memorize $1:1:\sqrt{2}$ and $1:\sqrt{3}:2$. A fast check: in a 30-60-90, the hypotenuse is always twice the short leg.
Side-by-side comparison of a 45-45-90 triangle (purple) and a 30-60-90 triangle (blue), with right angle marks at the respective vertices.
Figure 7.2 — Special Right Triangles: 45-45-90 (left) and 30-60-90 (right)
7.3 Geometric Mean and Altitude-on-Hypotenuse
When an altitude is drawn from the right angle of a right triangle to the hypotenuse, it creates three similar triangles: the original triangle and the two smaller triangles on each side of the altitude. This similarity produces useful proportional relationships involving the geometric mean.
Geometric Mean
The geometric mean of two positive numbers $a$ and $b$ is the positive number $x$ such that:
$$\frac{a}{x} = \frac{x}{b} \implies x = \sqrt{ab}$$
In other words, $x$ is the geometric mean of $a$ and $b$ when $x^2 = ab$.
Right Triangle Altitude Theorem (Geometric Mean)
In a right triangle, let the altitude from the right angle meet the hypotenuse at point $D$, dividing it into segments $p$ and $q$ (so $p + q = c$). Let $h$ be the length of the altitude, and let $a$, $b$ be the legs opposite to $q$ and $p$ respectively. Then:
- Altitude: $h^2 = p \cdot q$ — the altitude is the geometric mean of the two hypotenuse segments
- Each leg: $a^2 = c \cdot q$ and $b^2 = c \cdot p$ — each leg is the geometric mean of the hypotenuse and the adjacent segment
Example 7.5 — Finding Hypotenuse Segments
A right triangle has hypotenuse $c = 25$. The altitude to the hypotenuse has length $h = 12$. Find the two segments of the hypotenuse.
Let the segments be $p$ and $q$. Then $h^2 = pq$ and $p + q = 25$:
$$144 = pq \quad \text{and} \quad p + q = 25$$
So $p$ and $q$ are roots of $x^2 - 25x + 144 = 0$:
$$x = \frac{25 \pm \sqrt{625 - 576}}{2} = \frac{25 \pm \sqrt{49}}{2} = \frac{25 \pm 7}{2}$$
$$\mathbf{p = 16, \quad q = 9}$$
Check: $16 + 9 = 25$ ✓ and $16 \times 9 = 144 = 12^2$ ✓
Example 7.6 — Using the Altitude Theorem
In right $\triangle ABC$, altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$ with $CD = 6$ and $AD = 4$. Find $DB$, $AB$, $AC$, and $BC$.
- $CD^2 = AD \cdot DB \Rightarrow 36 = 4 \cdot DB \Rightarrow DB = \mathbf{9}$
- $AB = AD + DB = 4 + 9 = \mathbf{13}$
- $AC^2 = AB \cdot AD = 13 \cdot 4 = 52 \Rightarrow AC = \sqrt{52} = 2\sqrt{13} \approx \mathbf{7.21}$
- $BC^2 = AB \cdot DB = 13 \cdot 9 = 117 \Rightarrow BC = \sqrt{117} = 3\sqrt{13} \approx \mathbf{10.82}$
Check: $AC^2 + BC^2 = 52 + 117 = 169 = 13^2 = AB^2$ ✓
The altitude to the hypotenuse of a right triangle is 8. One hypotenuse segment is 4. Find the other segment and the hypotenuse.
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7.4 Distance Formula and Applications
The Pythagorean Theorem gives us a direct method for finding the distance between any two points on the coordinate plane. Given points $(x_1, y_1)$ and $(x_2, y_2)$, the horizontal and vertical differences form the legs of a right triangle with the segment connecting the points as the hypotenuse.
Distance and Midpoint Formulas
Distance Formula: The distance $d$ between $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Midpoint Formula: The midpoint $M$ of the segment from $(x_1, y_1)$ to $(x_2, y_2)$ is:
$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$
Example 7.7 — Perimeter of a Triangle on the Coordinate Plane
Find the perimeter of $\triangle PQR$ with vertices $P = (0, 0)$, $Q = (4, 0)$, $R = (4, 3)$.
- $PQ = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$
- $QR = \sqrt{(4-4)^2 + (3-0)^2} = \sqrt{9} = 3$
- $PR = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
Perimeter $= 4 + 3 + 5 = \mathbf{12}$. (This is a 3-4-5 right triangle with the right angle at $Q$.)
Example 7.8 — Diagonal of a Rectangle (TV Screen)
A television screen is 48 inches wide and 27 inches tall. What is the diagonal measurement?
$$d = \sqrt{48^2 + 27^2} = \sqrt{2304 + 729} = \sqrt{3033} \approx \mathbf{55.1 \text{ inches}}$$
This is how television sizes are typically advertised — by the diagonal of the screen.
Find the distance between $(-3, 4)$ and $(5, -2)$.
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Distance formula derivation: the horizontal and vertical legs (dashed auxiliary lines) of the right triangle formed by two coordinate points, with the hypotenuse AC representing the distance.
Figure 7.3 — Distance Formula as Pythagorean Theorem on the Coordinate Plane
Practice Problems
Find the missing side of a right triangle with legs 5 and 12. Then verify it is a Pythagorean triple.
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A 26-ft rope is stretched from the top of a 24-ft pole to a point on the ground. How far from the base of the pole does the rope end?
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Find all missing sides of a 45-45-90 triangle with hypotenuse $= 14$.
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Find all missing sides of a 30-60-90 triangle with long leg $= 9\sqrt{3}$.
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Altitude to the hypotenuse of a right triangle $= 6$. The shorter hypotenuse segment $= 3$. Find the longer segment and the hypotenuse.
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The diagonals of a rhombus are 16 and 12. Find the side length of the rhombus. (Hint: the diagonals of a rhombus are perpendicular bisectors of each other.)
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Find the perimeter and area of the triangle with vertices $A = (-2, 1)$, $B = (4, 1)$, $C = (4, 5)$.
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AP Problem: Two boats leave the same dock. Boat A travels 12 mi north; Boat B travels 16 mi east. What is the straight-line distance between them? If Boat A then travels 5 mi east and Boat B travels 9 mi north, are they still the same distance apart? Show all work.
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Part 2: Boat A is now at $(5, 12)$; Boat B is at $(16, 9)$.
New distance $= \sqrt{(16-5)^2 + (9-12)^2} = \sqrt{11^2 + (-3)^2} = \sqrt{121 + 9} = \sqrt{130} \approx 11.4 \text{ mi}$.
$\sqrt{130} \neq 20$, so no, they are no longer the same distance apart.
📋 Chapter Summary
Pythagorean Theorem
In a right triangle: $a^2 + b^2 = c^2$ where $c$ is the hypotenuse. Used to find any side given the other two.
If $a^2 + b^2 = c^2$, the triangle is right. If $a^2 + b^2 > c^2$, acute. If $a^2 + b^2 < c^2$, obtuse.
Sides in ratio $1 : 1 : \sqrt{2}$. Hypotenuse $= \text{leg} \times \sqrt{2}$. Half of a square.
Sides in ratio $1 : \sqrt{3} : 2$. Hypotenuse $= 2 \times \text{short leg}$. Long leg $= \text{short leg} \times \sqrt{3}$.
Trigonometric Ratios
$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$ $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$
$\theta = \sin^{-1}\!\left(\dfrac{\text{opp}}{\text{hyp}}\right)$ — use when finding an angle given two sides.