Exam prep · SAT · Unit guide

SAT: Math: Geometry and Trigonometry

Area, volume, angle rules, circle formulas, and right-triangle trig ratios explained

On the exam: Area, volume, angles, circles, right-triangle trigonometry. About 15%.

Think you can read a geometry diagram and skip the formula sheet? On the SAT, geometry and trigonometry make up about fifteen percent of the math questions, and this video covers area, volume, angles, circles, and right-triangle trigonometry the way the test actually asks about them.

Area and Volume Formulas

The SAT gives you a formula sheet on every math section, but you still need to know which formula fits which shape. Area measures a flat region, like length times width for a rectangle or half of base times height for a triangle. Volume measures three-dimensional space, like length times width times height for a rectangular box, or four-thirds pi r cubed for a sphere.

Key points.

• Rectangle area = length × width
• Triangle area = half × base × height
• Circle area = π × radius squared
• Volume = length × width × height (box)

Angle Relationships

When two parallel lines are cut by a third line, called a transversal, the angles formed follow set rules: corresponding angles are equal, and angles on a straight line add up to one hundred eighty degrees. Inside any triangle, the three interior angles always sum to one hundred eighty degrees as well, a fact that shows up constantly once you start working with multi-shape figures.

Key points.

• Angles on a line sum to 180 degrees
• Vertical angles are equal
• Triangle's three angles sum to 180 degrees
• Parallel lines cut by a transversal create equal angles

Circles: Equation, Arcs, and Sectors

A circle centered at the point h, k with radius r satisfies the equation x minus h squared plus y minus k squared equals r squared. For a slice of that circle, arc length is the fraction of the way around, theta over three sixty, times the full circumference, two pi r. Sector area works the same way, theta over three sixty times pi r squared.

(x−h)² + (y−k)² = r²
Arc length = (θ/360) × 2πr
Sector area = (θ/360) × πr²

Key points.

• Center (h,k), radius r defines a circle
• Arc length is a fraction of circumference
• Sector area is a fraction of total area
• θ is the central angle in degrees

Right-Triangle Trig Ratios

In a right triangle, every acute angle has three basic ratios built from its sides. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent, often remembered as SOH-CAH-TOA. These ratios depend only on the angle, so a thirty-degree angle always gives the same sine, no matter the triangle's size.

Sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent. SOH-CAH-TOA

Key points.

• Sine = opposite ÷ hypotenuse
• Cosine = adjacent ÷ hypotenuse
• Tangent = opposite ÷ adjacent
• Ratios depend only on the angle, not triangle size

Special Right Triangles

Two triangle shapes come up again and again. In a forty-five, forty-five, ninety triangle, the legs are equal length x, and the hypotenuse is x times the square root of two. In a thirty, sixty, ninety triangle, the sides are x, x root three, and two x. Both follow from the Pythagorean theorem, a squared plus b squared equals c squared.

45-45-90: legs x, x; hyp = x√2
30-60-90: sides x, x√3, 2x
a² + b² = c²

Key points.

• 45-45-90 triangle: legs x, x, hypotenuse x√2
• 30-60-90 triangle: sides x, x√3, 2x
• Pythagorean theorem: a² + b² = c²
• Works only in right triangles

Worked Example: Finding Sine from a Triangle

Here's a typical question. A right triangle has legs of nine and twelve. First, apply the Pythagorean theorem: nine squared plus twelve squared equals c squared. That's eighty-one plus one forty-four, which is two twenty-five, so c, the hypotenuse, is fifteen. Now find the sine of the angle opposite the twelve-length leg: that's opposite over hypotenuse, twelve over fifteen, which simplifies to four over five.

9² + 12² = c²
81 + 144 = 225
c = √225 = 15
sin(θ) = 12 ÷ 15
sin(θ) = 4/5

Key points.

• Legs 9 and 12, find hypotenuse
• Apply a² + b² = c²
• c = 15 after taking square root
• sin(θ) = opposite ÷ hypotenuse = 4/5

Radius vs. Diameter

A common slip is mixing up radius and diameter. The radius runs from the center to the edge and plugs directly into area and sector formulas. The diameter runs all the way across through the center and equals twice the radius; it's the value used in circumference, pi times diameter. Confusing the two is one of the easiest ways to miss an otherwise simple circle question.

Remember this

The one thing to carry forward: match the right formula or ratio to the shape in front of you, whether that's area, a circle's arc, or a trig ratio in a right triangle. That's the core of the SAT's geometry and trigonometry questions. Now try quizzing yourself on these formulas and a few practice problems to make sure they stick.

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