SAT Math — similar triangles with unknowns and overlapping figures

The College Board describes the SAT's lines-angles-triangles skill as using concepts and theorems about congruence and similarity of triangles to solve problems, determining which statements may be required to prove a relationship, and knowing that a scale factor changes lengths but leaves angle measures unchanged. In practice the similar triangles on the test are rarely drawn side by side. They overlap, sharing a vertex, or cross at a point, and the student has to see the two triangles before writing a proportion. This material trains that step, with the unknowns written as expressions rather than single numbers.

The twelve quiz questions cover the standard configurations. A segment parallel to one side of a triangle creates a smaller similar triangle inside it; students use it to find x from the pieces of two sides, to solve for x when the parallel segments are x + 1 and 2x + 4, and to find the area of the region between the two parallel segments. Two segments crossing between parallel lines form an hourglass, where the vertical angles and alternate interior angles give similarity and the matching vertices have to be read carefully. The altitude to the hypotenuse of a right triangle creates two smaller similar triangles and a geometric-mean relationship. A shadow problem, with the person standing where the tree's shadow ends, applies similarity to measurement. The midsegment appears with algebraic lengths, and a dilation question checks that angle measures do not scale.

Two questions deal with proof rather than calculation, following the framework's line about which statements are required. One asks which additional fact makes two triangles congruent when two pairs of sides are equal, and shows why side-side-angle and equal areas fall short. The other shows that two angles and a non-included side are enough.

Seven questions include a drawing with the given lengths or expressions marked and the unknown left blank; each drawing uses the stated measurements, so the picture matches the numbers. Every question also states its information in words. The explanations name the mistake behind each wrong option: a flipped ratio, a piece compared with a whole side, the wrong vertices matched, an area ratio used as a length ratio, or the value of x reported when a length was asked for.

The flashcards review the similarity criteria (AA, SAS, SSS), the congruence criteria, the side-splitter theorem, the midsegment theorem, the geometric mean in a right triangle, and how perimeters and areas scale.

The material offers a quiz and flashcards. It is practice written by Zestly and is not affiliated with or endorsed by the College Board.

  • Recognize similar triangles in overlapping and hourglass figures and match corresponding vertices
  • Set up and solve proportions with algebraic side lengths
  • Use the altitude to the hypotenuse and the midsegment theorem
  • Relate area ratios to length ratios for similar triangles
  • Decide which facts are enough to prove two triangles congruent (SAS and AAS versus SSA)

Practice material written by Zestly, based on the College Board Assessment Framework for the Digital SAT Suite (version 3.01, August 2024), Geometry and Trigonometry domain, skill Lines, angles, and triangles (congruence and similarity).

Sample question

Triangle ABC with point D on side AB and point E on side AC; segment DE is drawn parallel to BC. AD is labeled x, DB is labeled 6, AE is labeled 4 and EC is labeled 8.

In triangle $ABC$, point $D$ lies on $\overline{AB}$ and point $E$ lies on $\overline{AC}$, and $\overline{DE}$ is parallel to $\overline{BC}$. If $AD = x$, $DB = 6$, $AE = 4$ and $EC = 8$, what is the value of $x$?

See the answer

$3$

A segment parallel to one side of a triangle divides the other two sides proportionally: $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{x}{6} = \frac{4}{8}$ and $x = 3$. (Equivalently, triangle $ADE$ is similar to triangle $ABC$ with $\frac{AD}{AB} = \frac{AE}{AC} = \frac{4}{12}$.) The value $12$ flips one ratio ($\frac{x}{6} = \frac{8}{4}$); $6$ compares $AD$ with the whole side $AB$ on one side but with the piece $EC$ on the other; $2$ compares a piece with a whole side ($\frac{x}{6} = \frac{4}{12}$).

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