SAT — mathMedium10 min read

Special Right Triangles (30-60-90 and 45-45-90)

In the world of the Digital SAT, efficiency is just as important as accuracy. While the Pythagorean Theorem ($a^2 + b^2 = c^2$) is a reliable tool for solv

Tutor AI Team
Tutor AI Team

In the world of the Digital SAT, efficiency is just as important as accuracy. While the Pythagorean Theorem (a2+b2=c2a^2 + b^2 = c^2) is a reliable tool for solving right triangles, it can be slow and cumbersome when you are dealing with specific, recurring angle measures. This is where Special Right Triangles come in. These triangles—the 45∘−45∘−90∘45^\circ-45^\circ-90^\circ and the 30∘−60∘−90∘30^\circ-60^\circ-90^\circ—are the "shortcuts" of SAT Geometry.

On a typical SAT Math section, you can expect to encounter 1 to 3 questions that directly or indirectly require the use of special right triangle ratios. These questions appear in the Geometry and Trigonometry domain, often appearing as "Medium" to "Hard" difficulty because the SAT likes to hide these triangles inside other shapes, such as squares, hexagons, or circles.

Why do these triangles matter so much? Because the SAT is a timed test. If you use the Pythagorean Theorem to find a missing side, you might spend 45 seconds calculating squares and square roots. If you recognize a special right triangle, you can find that same side in 5 seconds just by looking at the ratios.

In this guide, we will move beyond simple memorization. Even though these formulas are provided on the SAT Reference Sheet at the start of every math section, the most successful students (those scoring 700+) don't waste time scrolling back to the reference page. They internalize these ratios so they can spot them instantly, even when the SAT tries to "disguise" them. By the end of this guide, you will know how to identify these triangles, apply their ratios in reverse, and solve complex multi-step problems involving circles and polygons.

Core Concepts

Special right triangles are defined by their interior angles. Because the angles are fixed, the ratios of their side lengths are also fixed, regardless of how large or small the triangle is.

1. The 45∘−45∘−90∘45^\circ-45^\circ-90^\circ Triangle (The Isosceles Right Triangle)

The 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle is created by cutting a square in half diagonally. Because two of the angles are equal (45∘45^\circ), the sides opposite those angles must also be equal. This makes it an isosceles right triangle.

The Ratio: In a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle, the sides follow the ratio: x:x:x2x : x : x\sqrt{2}

  • Leg 1: xx
  • Leg 2: xx
  • Hypotenuse: x2x\sqrt{2}

When it appears:

  • Squares: If an SAT question mentions a square and its diagonal, you are looking at two 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangles. The diagonal is the hypotenuse.
  • Isosceles Right Triangles: If the problem states a right triangle has two equal sides or two 45∘45^\circ angles.

How to solve:

  • If you have a leg and need the hypotenuse, multiply by 2\sqrt{2}.
  • If you have the hypotenuse and need a leg, divide by 2\sqrt{2}.

2. The 30∘−60∘−90∘30^\circ-60^\circ-90^\circ Triangle

The 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle is created by cutting an equilateral triangle in half with an altitude. This triangle is a favorite of SAT question writers because it has three distinct side lengths.

The Ratio: In a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle, the sides follow the ratio: x:x3:2xx : x\sqrt{3} : 2x

  • Short Leg (opposite 30∘30^\circ): xx
  • Long Leg (opposite 60∘60^\circ): x3x\sqrt{3}
  • Hypotenuse (opposite 90∘90^\circ): 2x2x

When it appears:

  • Equilateral Triangles: When an altitude is drawn in an equilateral triangle, it creates two 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangles.
  • Regular Hexagons: Hexagons can be divided into six equilateral triangles, which in turn can be divided into 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangles.
  • Trigonometry Problems: Questions involving sin⁡(30∘)\sin(30^\circ), cos⁡(60∘)\cos(60^\circ), or tan⁡(30∘)\tan(30^\circ) are secretly special right triangle questions.

How to solve: The "Short Leg" (xx) is your anchor. Always try to find the short leg first.

  • If you have the short leg, double it to get the hypotenuse.
  • If you have the short leg, multiply by 3\sqrt{3} to get the long leg.
  • If you have the hypotenuse, divide by 22 to get the short leg.
  • If you have the long leg, divide by 3\sqrt{3} to get the short leg.

3. The Reference Sheet vs. Mastery

On the Digital SAT, you can click the "Reference" button to see these triangles:

  1. A 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle with sides s,s,s2s, s, s\sqrt{2}.
  2. A 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle with sides x,x3,2xx, x\sqrt{3}, 2x.

Tutor Tip: While it's great the formulas are there, relying on them is a sign of a "Medium" level student. A "Hard" level student knows that if the hypotenuse of a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle is 1010, the leg is 102\frac{10}{\sqrt{2}}, which simplifies to 525\sqrt{2}. You must be able to rationalize denominators quickly: 102⋅22=1022=52\frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}

SAT Strategy Tips

1. Spot the "Hidden" Triangles

The SAT rarely gives you a standalone triangle and asks for a side. Instead, they hide them.

  • The Square Diagonal: If you see a square with side ss, the diagonal is always s2s\sqrt{2}.
  • The Equilateral Altitude: If you see an equilateral triangle with side ss, the altitude (height) is always s32\frac{s\sqrt{3}}{2}.

2. Use Desmos Wisely

On the Digital SAT, the built-in Desmos calculator is your best friend. If you are stuck on a radical expression like 123\frac{12}{\sqrt{3}}, type it into Desmos. Then, type in the answer choices (like 434\sqrt{3}). If the decimals match, you’ve found your answer. This is a great way to avoid algebraic errors when rationalizing denominators.

3. Look for "Key Numbers"

If you see the numbers 2,3,30∘,45∘,\sqrt{2}, \sqrt{3}, 30^\circ, 45^\circ, or 60∘60^\circ in a problem, your brain should immediately scream "Special Right Triangles!" Even if the triangle isn't drawn, these numbers are your signal to apply these ratios.

4. Work from the "Known" to the "Unknown"

In multi-step problems, identify which side you actually know.

  • In a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ, if you know the long leg is 99, don't try to jump straight to the hypotenuse. Go to the short leg first (9/3=339/\sqrt{3} = 3\sqrt{3}), then double it to get the hypotenuse (636\sqrt{3}).

Worked Example: Medium

Problem

A square has a perimeter of 3232. What is the length of the diagonal of the square?

Solution
  1. Find the side length: A square has four equal sides. If the perimeter is 3232, then each side s=324=8s = \frac{32}{4} = 8.
  2. Identify the triangle: Drawing a diagonal in a square creates two 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangles.
  3. Apply the ratio: The legs of the triangle are the sides of the square (x=8x = 8). The diagonal is the hypotenuse.
  4. Calculate: According to the x:x:x2x : x : x\sqrt{2} ratio, the hypotenuse is 828\sqrt{2}.

Final Answer: 828\sqrt{2}

Worked Example: Hard

Problem

An equilateral triangle has an altitude of 636\sqrt{3}. What is the area of the triangle?

Solution
  1. Identify the triangle: An altitude in an equilateral triangle splits it into two 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangles.
  2. Map the sides: The altitude is the long leg (opposite the 60∘60^\circ angle). So, x3=63x\sqrt{3} = 6\sqrt{3}.
  3. Find the short leg (xx): Dividing both sides by 3\sqrt{3}, we get x=6x = 6.
  4. Find the base of the equilateral triangle: The short leg (xx) is half of the total base. Therefore, the full base of the equilateral triangle is 2x=2(6)=122x = 2(6) = 12.
  5. Calculate Area: The formula for the area of a triangle is A=12bhA = \frac{1}{2}bh. A=12(12)(63)A = \frac{1}{2}(12)(6\sqrt{3}) A=6(63)=363A = 6(6\sqrt{3}) = 36\sqrt{3}

Final Answer: 36336\sqrt{3}

Worked Example: SAT-Hard

Problem

In the xyxy-plane, a circle with center (0,0)(0,0) contains the point PP. If the measure of the angle formed by the positive xx-axis and the line segment OPOP (where OO is the origin) is 30∘30^\circ, and the radius of the circle is 1010, what are the coordinates of point PP?

Solution
  1. Visualize: Point PP lies on the circle. Segment OPOP is the radius, so its length is 1010.
  2. Create a triangle: Drop a vertical line from point PP to the xx-axis. This creates a right triangle where the hypotenuse is 1010 and the angle at the origin is 30∘30^\circ.
  3. Identify the triangle: This is a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle.
  4. Find the side lengths:
    • The hypotenuse is 2x=102x = 10, so the short leg x=5x = 5.
    • The short leg is opposite the 30∘30^\circ angle, which corresponds to the vertical distance (yy-coordinate). So, y=5y = 5.
    • The long leg is x3=53x\sqrt{3} = 5\sqrt{3}. This corresponds to the horizontal distance (xx-coordinate). So, x=53x = 5\sqrt{3}.
  5. Determine coordinates: Since the angle is 30∘30^\circ from the positive xx-axis, the point is in Quadrant I.

Final Answer: (53,5)(5\sqrt{3}, 5)

Practice Problems

  1. In △ABC\triangle ABC, ∠B\angle B is a right angle and m∠A=45∘m\angle A = 45^\circ. If the length of ACAC is 1212, what is the length of ABAB? A) 66 B) 626\sqrt{2} C) 636\sqrt{3} D) 12212\sqrt{2}

  2. A regular hexagon is inscribed in a circle with a radius of 44. What is the area of the hexagon? A) 12312\sqrt{3} B) 24324\sqrt{3} C) 32332\sqrt{3} D) 48348\sqrt{3}

  3. In the figure below (not shown), △XYZ\triangle XYZ is a right triangle with m∠X=60∘m\angle X = 60^\circ and m∠Y=90∘m\angle Y = 90^\circ. If a point WW lies on XZXZ such that YWYW is an altitude of △XYZ\triangle XYZ, and XY=8XY = 8, what is the length of YWYW? (Grid-in your answer)

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Common Mistakes

1. Swapping 2\sqrt{2} and 3\sqrt{3}

Students often remember that there is a square root involved but mix them up.

  • The Fix: Remember that a 45−45−9045-45-90 comes from a 2-dimensional square (use 2\sqrt{2}). A 30−60−9030-60-90 has 3 different angles (use 3\sqrt{3}).

2. Misplacing the 3\sqrt{3} on the Hypotenuse

In a 30−60−9030-60-90 triangle, students sometimes think the hypotenuse is x3x\sqrt{3}.

  • The Fix: The hypotenuse must be the longest side. Since 22 is greater than 3\sqrt{3} (which is approx 1.731.73), the hypotenuse must be 2x2x.

3. Forgetting to Divide

When given the hypotenuse of a 45−45−9045-45-90 triangle, many students multiply by 2\sqrt{2} to find the leg.

  • The Fix: The hypotenuse is the longest side. If you are moving from a long side to a shorter side, you must divide. If you are moving from a short side to a longer side, you multiply.

4. Not Simplifying Radicals

The SAT might give an answer as 626\sqrt{2}, but your calculation resulted in 122\frac{12}{\sqrt{2}}.

  • The Fix: Always rationalize your denominators or use Desmos to check for decimal equivalence.

Frequently Asked Questions

Do I really need to memorize these if they are on the reference sheet?

A: Yes. The Digital SAT is fast-paced. If you have to click away from your question to look at the reference sheet, you lose your "flow" and precious seconds. Memorizing these allows you to "see" the answer as soon as you read the problem.

How do these triangles connect to Trigonometry?

A: They are the foundation of the Unit Circle. For example, sin⁡(45∘)=oppositehypotenuse=xx2=12=22\sin(45^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}. If you know your special right triangles, you don't need to memorize the Unit Circle values for 30∘,45∘,30^\circ, 45^\circ, and 60∘60^\circ.

What if the triangle isn't a special right triangle?

A: If the angles aren't 30,45,6030, 45, 60 or the sides don't fit the ratios, you must use the Pythagorean Theorem (a2+b2=c2a^2 + b^2 = c^2) or standard Trigonometry (SOH CAH TOA) using your calculator.

Key Takeaways

  • ✓

    45∘−45∘−90∘45^\circ-45^\circ-90^\circ Ratio: x:x:x2x : x : x\sqrt{2} (Leg : Leg : Hypotenuse).

  • ✓

    30∘−60∘−90∘30^\circ-60^\circ-90^\circ Ratio: x:x3:2xx : x\sqrt{3} : 2x (Short Leg : Long Leg : Hypotenuse).

  • ✓

    The Anchor: In 30−60−9030-60-90 problems, always find the short leg (xx) first; it makes finding the other two sides much easier.

  • ✓

    The Square Shortcut: The diagonal of a square is always side×2side \times \sqrt{2}.

  • ✓

    The Equilateral Shortcut: The height of an equilateral triangle is always side×32\frac{side \times \sqrt{3}}{2}.

  • ✓

    Direction Matters: Multiply by the radical to get a larger side; divide by the radical to get a smaller side.

  • ✓

    Check with Desmos: If radicals confuse you, use the calculator to compare decimal values of your answer and the given choices.

Tags:
#sat#math#geometry-and-trigonometry

Related Topics

Ready to Ace Your SAT math?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store