The Pythagorean Theorem

Welcome to one of the most essential pillars of the SAT Math section: The Pythagorean Theorem. If you are aiming for a 700+ score, you cannot simply "know"

Welcome to one of the most essential pillars of the SAT Math section: The Pythagorean Theorem. If you are aiming for a 700+ score, you cannot simply "know" this formula; you must be able to deploy it instantly, recognize it when it’s hidden in complex diagrams, and understand its deep connection to coordinate geometry.

The Pythagorean Theorem is the foundation of the "Geometry and Trigonometry" domain. On the Digital SAT, you will encounter approximately 5–7 questions related to Geometry and Trigonometry per test. While not every one of those will be a direct "find the missing side" problem, the Pythagorean Theorem is the "engine" behind many other topics, including circle equations, trigonometry, and the distance formula.

In this guide, we are going to move beyond the basic a2+b2=c2a^2 + b^2 = c^2 that you learned in middle school. We will explore how the SAT "disguises" right triangles within circles and rectangles, how to use Pythagorean Triples to save precious seconds, and how to bridge the gap between a flat geometric shape and the xyxy-coordinate plane. By the end of this guide, you will view every right angle on the SAT as an invitation to use this theorem, and you’ll have the tactical speed necessary to breeze through foundational questions to save time for the harder Module 2 problems.

Core Concepts

1. The Fundamental Theorem

The Pythagorean Theorem states that in any right triangle (a triangle with one 9090^\circ angle), the square of the length of the hypotenuse (cc) is equal to the sum of the squares of the lengths of the two legs (aa and bb).

The formula is expressed as: a2+b2=c2a^2 + b^2 = c^2

When it appears: This is the go-to tool whenever you are given two sides of a right triangle and need the third. How to recognize it: Look for the "square" symbol in the corner of a triangle. If the SAT doesn't explicitly say it's a right triangle, look for keywords like "perpendicular," "vertical and horizontal," or "tangent to a circle."

SAT Reference Sheet Note: This formula is provided on the SAT reference sheet. However, if you have to look at the reference sheet to remember a2+b2=c2a^2 + b^2 = c^2, you are losing valuable time. You should have this memorized until it is second nature.

2. Identifying the Hypotenuse

The most common error students make is plugging the wrong numbers into the wrong variables.

  • Legs (aa and bb): The two sides that form the 9090^\circ angle.
  • Hypotenuse (cc): The longest side, located directly across from the right angle.

On the Digital SAT, triangles are often rotated or flipped. Always locate the 9090^\circ angle first; the side it "points to" is always cc.

3. Pythagorean Triples (The "Time Savers")

The SAT is a timed test. While you can always use the formula, the test makers frequently use "Triples"—sets of three integers that perfectly satisfy the theorem. If you recognize these, you can solve a question in 3 seconds instead of 30.

The Big Two:

  1. 3-4-5 Triangle: 32+42=529+16=253^2 + 4^2 = 5^2 \rightarrow 9 + 16 = 25.
  2. 5-12-13 Triangle: 52+122=13225+144=1695^2 + 12^2 = 13^2 \rightarrow 25 + 144 = 169.

The Multiples: The SAT loves to use multiples of these triples. For example:

  • 6-8-10 (a 3-4-5 multiplied by 2)
  • 9-12-15 (a 3-4-5 multiplied by 3)
  • 30-40-50 (a 3-4-5 multiplied by 10)
  • 10-24-26 (a 5-12-13 multiplied by 2)

Other Triples to Know:

  • 8-15-17
  • 7-24-25

4. The Distance Formula Connection

One of the most common ways the SAT tests the Pythagorean Theorem is through the Distance Formula in the xyxy-plane. The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

If you look closely, this is just the Pythagorean Theorem rearranged: d2=(Δx)2+(Δy)2d^2 = (\Delta x)^2 + (\Delta y)^2 Where Δx\Delta x is the horizontal leg and Δy\Delta y is the vertical leg. On the Digital SAT, if you are asked for the distance between points, you can draw a right triangle on the screen (or in your head) and use a2+b2=c2a^2 + b^2 = c^2.

5. Hidden Right Triangles

The SAT rarely gives you a simple triangle and asks for xx. They hide right triangles inside other shapes:

  • Isosceles Triangles: Drawing an altitude (a line from the top vertex to the base) creates two congruent right triangles.
  • Rectangles/Squares: The diagonal of a rectangle creates two right triangles.
  • Circles: A radius drawn to a tangent line always forms a 9090^\circ angle. Also, any triangle inscribed in a semicircle where one side is the diameter is a right triangle.

SAT Strategy Tips

1. Spot the Triple First

Before you reach for your calculator, look at the sides. If you see a leg of 6 and a hypotenuse of 10, don't do 1026210^2 - 6^2. Recognize it immediately as a 6-8-10 triangle. The answer is 8. This saves mental energy for the harder questions at the end of the module.

2. Use the Desmos Calculator for Radicals

On the Digital SAT, the built-in Desmos calculator is your best friend. If you calculate c=149c = \sqrt{149} and the answer choices are in decimal form, use Desmos. If the answer choices are in simplified radical form (like 525\sqrt{2}), you can type 525\sqrt{2} into Desmos to see if its decimal value matches your 50\sqrt{50}.

3. Draw the "Hidden" Line

If a problem shows a trapezoid or an irregular quadrilateral, look for a way to drop a vertical line to create a rectangle and a right triangle. This is a classic SAT move to turn a "hard" geometry problem into a "foundational" Pythagorean problem.

4. Don't Over-Calculate

If the question asks for the area of a square built on the hypotenuse, and you know a2+b2=c2a^2 + b^2 = c^2, remember that c2c^2 is the area of that square. You don't need to find cc and then square it again.

Worked Example: Medium

Problem

Question: In right triangle ABCABC, the length of leg ABAB is 9 and the length of the hypotenuse BCBC is 15. What is the length of leg ACAC?

Solution
  1. Identify the sides: We are given one leg (a=9a = 9) and the hypotenuse (c=15c = 15). We need to find the other leg (bb).
  2. Set up the equation: a2+b2=c2a^2 + b^2 = c^2 92+b2=1529^2 + b^2 = 15^2
  3. Square the numbers: 81+b2=22581 + b^2 = 225
  4. Isolate b2b^2: b2=22581b^2 = 225 - 81 b2=144b^2 = 144
  5. Take the square root: b=144=12b = \sqrt{144} = 12
  6. Shortcut check: Notice that 9 and 15 are both multiples of 3. 9=3×39 = 3 \times 3 15=5×315 = 5 \times 3 This is a 3-4-5 triangle scaled by a factor of 3. The missing side must be 4×3=124 \times 3 = 12.

Final Answer: 12


Worked Example: Hard

Problem

Question: A coordinate plane shows a circle with center (2,3)(2, 3). The point (6,6)(6, 6) lies on the circle. What is the area of the circle?

Solution
  1. Understand the goal: To find the area of a circle (A=πr2A = \pi r^2), we need the radius rr.
  2. Recognize the radius: The radius is the distance from the center (2,3)(2, 3) to the point on the circle (6,6)(6, 6).
  3. Apply the Pythagorean Theorem (Distance Formula): The horizontal distance (Δx\Delta x) is 62=46 - 2 = 4. The vertical distance (Δy\Delta y) is 63=36 - 3 = 3.
  4. Calculate rr: r2=(Δx)2+(Δy)2r^2 = (\Delta x)^2 + (\Delta y)^2 r2=42+32r^2 = 4^2 + 3^2 r2=16+9=25r^2 = 16 + 9 = 25 r=5r = 5
  5. Calculate Area: A=πr2A = \pi r^2 A=π(5)2=25πA = \pi (5)^2 = 25\pi

Final Answer: 25π25\pi


Worked Example: SAT-Hard

Problem

Question: In the figure below (not shown), rectangle PQRSPQRS is inscribed in a circle. If the length of PQPQ is 10 and the area of the rectangle is 240, what is the diameter of the circle?

Solution
  1. Analyze the geometry: If a rectangle is inscribed in a circle, its diagonal is the diameter of the circle.
  2. Find the missing side of the rectangle: Area = length ×\times width 240=10×QR240 = 10 \times QR QR=24QR = 24
  3. Identify the right triangle: The diagonal PRPR forms a right triangle with sides PQPQ and QRQR.
  4. Apply the Pythagorean Theorem to find the diagonal (diameter): PQ2+QR2=PR2PQ^2 + QR^2 = PR^2 102+242=PR210^2 + 24^2 = PR^2 100+576=PR2100 + 576 = PR^2 676=PR2676 = PR^2
  5. Solve for PRPR: PR=676=26PR = \sqrt{676} = 26
  6. Shortcut check: Notice that 10 and 24 are multiples of 2. 10=5×210 = 5 \times 2 24=12×224 = 12 \times 2 This is a 5-12-13 triangle scaled by 2. The diameter is 13×2=2613 \times 2 = 26.

Final Answer: 26

Practice Problems

  1. A ladder 13 feet long is leaning against a vertical wall. If the bottom of the ladder is 5 feet away from the base of the wall, how many feet up the wall does the ladder reach? 2.

  2. Triangle XYZXYZ is a right triangle with the right angle at YY. If the coordinates of XX are (2,1)(-2, 1) and the coordinates of YY are (4,1)(4, 1), and the length of side YZYZ is 8, what is the length of the hypotenuse XZXZ? 3.

  3. An isosceles triangle has two sides of length 10 and a base of length 12. What is the area of the triangle? (Hint: Draw an altitude to create two right triangles).

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Common Mistakes

1. Misidentifying the Hypotenuse

Students often assume the "missing side" is always cc.

  • The Mistake: If a triangle has a leg of 5 and a hypotenuse of 13, a student might do 52+132=c25^2 + 13^2 = c^2.
  • The Correction: Always check if the missing side is a leg or the hypotenuse. In this case, 52+b2=1325^2 + b^2 = 13^2, so b=12b = 12.

2. Forgetting to Take the Square Root

It is very common to solve a2+b2a^2 + b^2 and get, for example, 100, and then select "100" as the answer.

  • The Mistake: c2=100c^2 = 100 \rightarrow Answer = 100.
  • The Correction: Remember that the theorem gives you the side length squared. You must take the square root at the end to find the actual side length.

3. Using Pythagoras on Non-Right Triangles

The theorem only works for triangles with a 9090^\circ angle.

  • The Mistake: Applying a2+b2=c2a^2 + b^2 = c^2 to an acute or obtuse triangle.
  • The Correction: Look for the square symbol or the word "right triangle." If it's not there, you may need to use the Law of Sines or Cosines (though these are rare on the SAT) or look for other geometric properties.

Frequently Asked Questions

Do I really need to memorize Pythagorean Triples?

You don't have to, but you should. The Digital SAT is designed to reward students who find the most efficient path. Recognizing a 3-4-5 triangle saves you from doing three lines of algebra, which reduces the chance of a calculation error and gives you more time for the harder questions in Module 2.

How does this connect to Trigonometry?

Trigonometry (SOH CAH TOA) is essentially the study of the ratios of the sides of a right triangle. The Pythagorean Theorem allows you to find the "missing" side so that you can then calculate the Sine, Cosine, or Tangent of an angle. For example, if you know sin(θ)=35\sin(\theta) = \frac{3}{5}, you know the opposite side is 3 and the hypotenuse is 5. Using Pythagoras, you can find the adjacent side is 4, meaning cos(θ)=45\cos(\theta) = \frac{4}{5}.

Can the sides of a right triangle be decimals or radicals?

Absolutely. While the SAT loves integers and triples, you will often see sides like 7\sqrt{7} or 4.54.5. The formula a2+b2=c2a^2 + b^2 = c^2 works exactly the same way. Just be careful when squaring radicals: (7)2=7(\sqrt{7})^2 = 7.

Key Takeaways

  • The Formula: a2+b2=c2a^2 + b^2 = c^2, where cc is always the hypotenuse (the side across from the 9090^\circ angle).

  • Memorize the Triples: 3-4-5 and 5-12-13 (and their multiples) are the most common side lengths on the SAT.

  • Distance is Pythagoras: The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is just the hypotenuse of a right triangle with legs Δx\Delta x and Δy\Delta y.

  • Look for Hidden Triangles: Drop altitudes in isosceles triangles or draw diagonals in rectangles to reveal the right triangles you need.

  • Check the Units: Ensure all side lengths are in the same units before plugging them into the theorem.

  • Square Root at the End: Don't forget that a2+b2a^2 + b^2 gives you the square of the hypotenuse, not the hypotenuse itself.

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