Two triangles are similar when their corresponding angles are equal, which is guaranteed as soon as two pairs of angles match, since the third pair must then match too. Similar triangles have corresponding sides in a constant ratio, the scale factor, and equal ratios between any two sides within each triangle. The SAT builds similar triangles in a few recurring ways: a segment drawn parallel to one side of a triangle cuts off a smaller similar triangle; two right triangles that share an acute angle are similar; a person and a tree casting shadows at the same moment form similar right triangles; and the diagonals of a trapezoid or two crossing lines between parallels form a pair of similar triangles. In every case, identify the matching angles, write a proportion between corresponding sides, and cross-multiply. A person 6 feet tall casting a 4 foot shadow at the moment a tree casts a 30 foot shadow gives 6/4 = h/30, so the tree is 45 feet tall.
The two special right triangles are fixed shapes whose side ratios are on the reference sheet. In a 45-45-90 triangle the legs are equal and the hypotenuse is a leg times √2; it is half a square, so a square of side s has diagonal s √2. In a 30-60-90 triangle the side opposite the 30 degree angle is the shortest, the hypotenuse is exactly twice it, and the side opposite the 60 degree angle is the short leg times √3; it is half an equilateral triangle, so an equilateral triangle of side s has height (s/2) √3. The method is always to find the short leg first: given a hypotenuse of 10, the short leg is 5 and the long leg is 5 √3; given a long leg of 6, the short leg is 6/√3 = 2 √3 and the hypotenuse is 4 √3.
Trigonometric ratios at this level go in both directions. Given a triangle, you write the ratio; given a ratio, you rebuild the triangle. If sin A = 5/13, the opposite side is 5 and the hypotenuse 13, so the adjacent side is √(169 - 25) = 12, and cos A = 12/13, tan A = 5/12. Since the two acute angles of a right triangle sum to 90 degrees, and the side opposite one is adjacent to the other, the sine of an angle equals the cosine of its complement: sin x = cos(90° - x). This lets the test ask for cos B when it gives sin A in the same triangle (they are equal), or to solve sin(x) = cos(3x + 10) by setting x + 3x + 10 = 90. Tangent is also the slope of the hypotenuse when a leg lies along the x-axis, a link the coordinate questions exploit.
A circle's circumference is 2 π r and its area π r², both on the reference sheet. A central angle, with its vertex at the center, cuts off an arc and a sector. The arc's length is the fraction theta/360 of the circumference, and the sector's area is the same fraction of the circle's area: for a radius of 6 and a central angle of 60 degrees, the arc is (60/360)(12 π) = 2 π and the sector is (60/360)(36 π) = 6 π. The chord joining the arc's endpoints, the radius to each endpoint, and the central angle form an isosceles triangle, which is where the special triangles reappear inside circle problems. A full circle is 360 degrees; the Advanced lesson introduces radians, where a full circle is 2 π.
In the coordinate plane a circle with center (h, k) and radius r has equation (x - h)² + (y - k)² = r². Reading the center reverses the signs inside the parentheses, and the radius is the square root of the constant on the right. To test whether a point is on the circle, substitute it: (2, 1) is on (x - 2)² + (y + 3)² = 16 because 0 + 16 = 16. To write the equation from a center and a point on the circle, compute the radius as the distance between them first. When the equation is given expanded, as x² + y² - 4x + 6y - 3 = 0, completing the square in x and in y recovers the standard form; that conversion is the Advanced lesson's territory, but recognizing that an expanded equation is still a circle belongs here.
Coordinate geometry supplies the distances. The distance between (x₁, y₁) and (x₂, y₂) is √((x₂ - x₁)² + (y₂ - y₁)²), the Pythagorean theorem with the horizontal and vertical changes as legs. The midpoint is the average of the coordinates, ((x₁ + x₂)/2, (y₁ + y₂)/2). Together they let you find a circle's radius from its center and a point, the length of a diagonal, whether a triangle is isosceles, or the perimeter of a polygon drawn on a grid. The slope of a segment, rise over run, tells you whether two sides are parallel or perpendicular and therefore whether a quadrilateral is a rectangle.
Volume questions at this level combine solids or run backwards. A composite solid, such as a cylinder with a cone on top, is the sum of the pieces, each from the reference sheet. A backwards question gives the volume and asks for a dimension: a cone of volume 24 π and height 8 has (1/3) π r² (8) = 24 π, so r² = 9 and r = 3. Density questions multiply volume by density to get mass or divide mass by density to get volume, and the units carry the setup.