7.3 Right Triangles & Pythagorean Theorem

7.3 Right Triangles & Pythagorean Theorem

Right triangles are the cornerstone of SAT geometry and trigonometry. They appear across standard planar problems, coordinate geometry distance calculations, 3D solid diagonal measurements, and trigonometric modeling. Achieving speed and 100% accuracy requires recognizing standard Pythagorean triples on sight and manipulating special right triangle side ratios effortlessly.

---

1. The Pythagorean Theorem & Triangle Classification

In any right triangle with legs $a$ and $b$ and hypotenuse $c$:

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

+-----------------------------------------------------------------------------+
|                        THE PYTHAGOREAN THEOREM                              |
|                                                                             |
|                                   /|                                        |
|                                  / |                                        |
|                                 /  |                                        |
|                   Hypotenuse c /   | Leg b                                  |
|                               /    |                                        |
|                              /     |                                        |
|                             /______|                                        |
|                              Leg a                                          |
|                                                                             |
|   - Hypotenuse c is ALWAYS the longest side opposite the 90 degree angle.   |
|   - Legs a and b form the perpendicular right angle.                        |
+-----------------------------------------------------------------------------+

The Converse: Classifying Triangle Types

Given three side lengths $a \le b < c$:

• Right Triangle: $a^2 + b^2 = c^2$
• Acute Triangle: $a^2 + b^2 > c^2$ (hypotenuse is shorter than in a right triangle)
• Obtuse Triangle: $a^2 + b^2 < c^2$ (hypotenuse is stretched wider)

Coordinate Plane Distance Formula

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is directly derived from the Pythagorean theorem:

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

---

2. Must-Memorize Pythagorean Triples

A Pythagorean triple consists of three positive integers that satisfy $a^2 + b^2 = c^2$. Recognizing these triples eliminates tedious multi-digit squaring and square roots under timed test conditions.

+-----------------------------------------------------------------------------+
|                     CORE SAT PYTHAGOREAN TRIPLES TABLE                      |
|                                                                             |
|   [ PRIMITIVE TRIPLES ]        [ COMMON SCALED MULTIPLES ]                  |
|   --------------------         ---------------------------                  |
|   (3, 4, 5)                    x2: (6, 8, 10)         x5: (15, 20, 25)      |
|                                x3: (9, 12, 15)        x10: (30, 40, 50)     |
|                                x4: (12, 16, 20)                             |
|                                                                             |
|   (5, 12, 13)                  x2: (10, 24, 26)       x3: (15, 36, 39)      |
|                                                                             |
|   (8, 15, 17)                  x2: (16, 30, 34)                             |
|                                                                             |
|   (7, 24, 25)                  x2: (14, 48, 50)                             |
|                                                                             |
|   [ DISGUISED TRIPLES WITH DECIMALS OR ROOTS ]:                             |
|   - Decimals:  (1.5, 2.0, 2.5)  -->  0.5 * (3, 4, 5)                        |
|   - Radicals:  (3√2, 4√2, 5√2)  -->  √2 * (3, 4, 5)                         |
+-----------------------------------------------------------------------------+

Worked Example 1: Disguised Pythagorean Triple

Problem: A right triangle has a leg of length $1.6\text{ meters}$ and a hypotenuse of length $3.4\text{ meters}$. What is the area of the triangle in square meters?

Step-by-Step Solution:

1. Identify the underlying triple: Multiply the given dimensions by 5 to clear decimals: $$1.6 \times 5 = 8, \quad 3.4 \times 5 = 17$$
2. Recognize the $(8, 15, 17)$ primitive triple: The missing leg in the scaled triangle is $15$. Divide by 5 to return to the original scale: $$\text{Leg}_2 = \frac{15}{5} = 3.0\text{ meters}$$
3. Calculate the area: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(1.6)(3.0) = 2.4\text{ m}^2$$

---

3. Special Right Triangles (Given on SAT Reference Sheet)

Although the reference sheet includes special right triangles, looking up formulas during the test wastes valuable seconds. You must internalize their ratios completely.

+-----------------------------------------------------------------------------+
|                        SPECIAL RIGHT TRIANGLE RATIOS                        |
|                                                                             |
|   [ 45° - 45° - 90° TRIANGLE ]         [ 30° - 60° - 90° TRIANGLE ]         |
|   (Isosceles Right Triangle)           (Half of Equilateral Triangle)       |
|                                                                             |
|               45°                                  30°                      |
|               /|                                   /|                       |
|              / |                                  / |                       |
|        x√2  /  | x                          2x   /  | x√3 (Long leg)        |
|            /   |                                /   |                       |
|           /____|                               /____|                       |
|         45°  x                               60°  x (Short leg)             |
|                                                                             |
|   Side Ratio:  x : x : x√2             Side Ratio:  x : x√3 : 2x            |
+-----------------------------------------------------------------------------+

1. $45^\circ - 45^\circ - 90^\circ$ Mechanics

• Leg to Hypotenuse: Multiply by $\sqrt{2}$ ($\text{Hypotenuse} = \text{leg} \cdot \sqrt{2}$).
• Hypotenuse to Leg: Divide by $\sqrt{2}$ (or multiply by $\frac{\sqrt{2}}{2}$): $$\text{Leg} = \frac{\text{Hypotenuse}}{\sqrt{2}} = \frac{\text{Hypotenuse} \cdot \sqrt{2}}{2}$$
• Square Diagonal Connection: The diagonal of any square with side length $s$ is $d = s\sqrt{2}$.

2. $30^\circ - 60^\circ - 90^\circ$ Mechanics

• Short Leg ($x$): Always opposite the $30^\circ$ angle (the master reference side).
• Long Leg ($x\sqrt{3}$): Opposite the $60^\circ$ angle ($= \text{short leg} \cdot \sqrt{3}$).
• Hypotenuse ($2x$): Opposite the $90^\circ$ angle ($= 2 \times \text{short leg}$).

Step-by-Step Transition Rules for $30^\circ - 60^\circ - 90^\circ$

Worked Example 2: Reversing from the Long Leg

Problem: In a $30^\circ - 60^\circ - 90^\circ$ triangle, the side opposite the $60^\circ$ angle has a length of $18\text{ inches}$. What is the exact length of the hypotenuse?

Step-by-Step Solution:

1. Identify the given side: $\text{Long Leg} = x\sqrt{3} = 18$.
2. Solve for the short leg $x$: $$x = \frac{18}{\sqrt{3}} = \frac{18\sqrt{3}}{3} = 6\sqrt{3}\text{ inches}$$
3. Calculate the hypotenuse ($2x$): $$\text{Hypotenuse} = 2(6\sqrt{3}) = 12\sqrt{3}\text{ inches}$$

---

4. 3D Space Diagonals in Rectangular Prisms

The distance between opposite corners of a three-dimensional rectangular prism (box) requires extending the Pythagorean theorem into three dimensions:

+-----------------------------------------------------------------------------+
|                        3D SPACE DIAGONAL DERIVATION                         |
|                                                                             |
|                             +-----------------+                             |
|                            /|                /|                             |
|                           / |     Space     / |                             |
|                          +--+--------------+  |                             |
|                          |  |   Diagonal   |  | h                           |
|                          |  +--------------+--+                             |
|                          | /    d          | / w                            |
|                          |/                |/                               |
|                          +-----------------+                                |
|                                   l                                         |
|                                                                             |
|   1. Base diagonal:   d_base² = l² + w²                                     |
|   2. Space triangle:  d² = d_base² + h² = (l² + w²) + h²                    |
|   ===> Space Diagonal:  d = sqrt(l² + w² + h²)                              |
+-----------------------------------------------------------------------------+

Worked Example 3: Longest Object Inside a Container

Problem: A shipping company needs to pack a rigid metal rod into a rectangular box measuring $8\text{ cm}$ by $6\text{ cm}$ by $24\text{ cm}$. What is the maximum length of a rod that can fit entirely inside the box?

Step-by-Step Solution:

1. Apply the 3D space diagonal formula: $$d = \sqrt{l^2 + w^2 + h^2} = \sqrt{8^2 + 6^2 + 24^2}$$
2. Compute the base diagonal first: $$\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm (notice the } 6-8-10\text{ triple!)}$$
3. Combine with height: $$d = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26\text{ cm (notice the } 10-24-26\text{ triple!)}$$