Triangles | SAT Geometry | SAT Math

Introduction

Triangles questions on the SAT have some overlap with angles questions in that they often introduce similar triangles and ask the student to assess the angles found in those triangles. Beyond that, SAT questions in this category explore the proportional sides involved in similar triangles and, otherwise, delve primarily into questions surrounding right triangles and the Pythagorean theorem.

Although similar triangles and right triangles make up the vast majority of SAT questions about this topic, we will touch on a few other formulas and concepts in an effort to cover, as much as possible, all the kinds of questions about triangles that a student might encounter.

Approach Question

Triangle $MNO$ is similar to triangle $PQR$ , where angles $M$ and $P$ correspond to each other, as do angles $N$ and $Q$ and angles $O$ and $R$ . The length of $\overline{PQ}$ is $95$ and the length of $\overline{QR}$ is $38$ . If the length of $\overline{NO}$ is $8$ , what is the length of $\overline{MN}$ ?

A. 20
B. 19
C. 18
D. 16

Explanation

Although we normally recommend drawing any geometry figure if the SAT does not provide the figure, you do not necessarily need to do so with an SAT similar triangles question. (If you have any doubts, though, do draw the figure!) This is because the SAT will always make the correspondences clear; corresponding angles will typically follow alphabetical order, as they do here. And since the angles (and therefore the vertices of the triangle) are in alphabetical order, the sides will correspond in a similar way. In this case, side $\overline{MN}$ will correspond to side $\overline{PQ}$ ; side $\overline{MO}$ will correspond to side $\overline{PR}$ ; and side $\overline{NO}$ will correspond to side $\overline{QR}$ .

As explained under Definitions below, similar triangles have a more complicated relationship among their sides than among their angles. The corresponding sides of similar triangles have proportional relationships; in other words, they exist in the same ratio to each other. If a side of a triangle is three times as large as its corresponding side in a smaller, similar triangle, then all of the sides in the larger triangle are three times as large as their corresponding sides in the smaller triangle.

This means that we need more information to solve for a side in a pair of similar triangles than we do to solve for an angle. First, we need enough information to establish the ratio between the side lengths of the two triangles; second, we need the exact length of one of the corresponding sides in order to find the length of its corresponding side in the other triangle.

Let’s start with the first need: the ratio of corresponding sides. We need to look carefully at all the sides whose lengths are mentioned in order to find two corresponding sides. It turns out that the second side mentioned (of length 38) and the third side mentioned (of length 8) are the corresponding pair (because $\overline{QR}$ must correspond to $\overline{NO}$ ); we’ll use the first side mentioned later. For now, we should establish and simplify the ratio between the two sides mentioned: the ratio is $38$ to $8$ , which can be reduced to $19$ to $4$ . (Remember that ratios, like fractions, can be reduced by dividing both parts by their greatest common factor).

Now we’re ready to follow standard procedure for any proportional relationship: set up a proportion! For a proportion, we need two fractions, so let’s render our first ratio, $19 : 4$ , as the fraction $\frac{19}{4}$ . The other fraction in our proportion must include one numeric value and the unknown for which we hope to solve. In addition, the quantities in the numerator of each fraction must be of the same kind or category, and the denominators must contain the same kind or category as well. In this case, if each fraction relates corresponding sides, then both numerators must represent the smaller values and both numerators must represent the larger values (or vice versa). An example would look like something like this:

(smaller side of triangle A)/(smaller side of triangle B) = (larger side of triangle A)/(larger side of triangle B)

Of these four quantities, exactly one must be the unknown for which we are solving. So the fraction might look like this:

(known side of triangle A)/known side of triangle B) = (unknown side of triangle A)/(known side of triangle B)

The placements could vary, but the quantities cannot: in a solvable proportion, there must always be three knowns and one unknown.

So, back to the problem at hand, our three known lengths are $38$ , $8$ , and $95$ . We have already related the first two because they are corresponding and reduced the relation to $19 : 4$ , or $\frac{19}{4}$ . Now we are ready to write the proportion, making sure that the $95$ goes in the numerator because, like the $19$ , it represents the larger triangle. The unknown $x$ fills in the last empty spot, giving us:

$\frac{19}{4} = \frac{95}{x}$

Once we have a proportion set up, cross-multiplying is the way to proceed. In cross-multiplication, we multiply the first numerator by the second denominator and set the result equal to the product of the first denominator times the second numerator. In this case, we get:

$(19) (x) 19 x x = (4) (95) = 380 = 20$

The answer is 20. One last note: if you were to divide $95$ by $38$ in the original problem to check the ratio, you would get $2.5$ . It turns out, not coincidentally, that $8 \times 2.5 = 20$ ( $8$ is also an original number from the problem.) Few students will notice this ratio at the outset, but it confirms the logic of the right answer to see the consistent ratio of $2.5 : 1$ .

Definitions

Hypotenuse: The longest side of a right triangle; the side across from the right angle. Represented by c in the Pythagorean theorem.
Leg: One of the shorter two sides of a right triangle. Represented by a and b in the Pythagorean theorem.
Isosceles: A triangle with two congruent sides and two congruent angles.
Equilateral: A triangle in which all three sides are congruent and all three angles are congruent. Each angle measures 60°.
Scalene: A triangle in which no side is equal to any other, and therefore no angle is equal to any other.
Similar: A description of two or more triangles whose corresponding angle pairs are congruent, but whose corresponding sides (typically) are not. Two similar triangles, if drawn to scale, will look like copies of each other, with similar dimensions except that one is larger than the other. In similar triangles, corresponding sides are proportional, that is, these pairs of sides all share the same ratio of their lengths. This means that proportions are helpful algebra tools for working with similar triangles.
Congruent: Two shapes (in this case triangles) that are identical to each other in every way. The SAT doesn’t often test for triangle congruence, but for the rare cases, consider: congruence can be proved by various theorems comparing their sides and angles; some notable examples are abbreviated SSS, SAS, AAS, and ASA.

Vertex

The points at which any two sides of a triangle meet; the “corners” of a triangle. A letter denoting the vertex of a triangle can also be used to denote the interior angle at that vertex.

Topics for Cross-Reference

Variations

There are not many variations on the SAT’s most common triangle themes, but on occasion you might see questions involving triangles being drawn inside, or making up some part of, another shape. For example, a triangle might be inscribed in a circle–drawn so that all three of its vertices touch the circle. In this module, you’ll see an example wherein a hexagon is divided up into triangles to help us find its area. Triangles and quadrilaterals overlap quite a bit in standardized test questions; we’ll explore this confluence in the lesson on quadrilaterals.

Strategy Insights

Here’s a visual exercise. Place your hands together so that only the bases of your hands are touching, with your fingertips about an inch apart from each other (index finger an inch away from the other index finger, etc.). Notice how the angle created next to the base of your hands is rather small. Now, keeping the base of your hands together, spread out your hands so your right-hand fingers are several inches apart from your left-hand fingers. Do you see how the angle at the base of your hands grew much larger? This visual tells us an important truth: The size of an angle is related to the length of its opposite side, and vice versa. A small angle will correspond to a small side across from it, which is why the side across from the $30°$ angle must be the smallest in a $30 - 60 - 90$ triangle. A large angle corresponds to a large side across from it, which is why the hypotenuse must be the largest side in a right triangle: it is, by definition, across from the largest angle.
Much triangle work begins with knowing the right formulas. There are several triangle-related formulas provided in the SAT reference list; even though they are given to you on the test, we will reproduce them in the “Flashcard Fodder” section because we think you should memorize them anyway. That way, you don’t have to take time consulting the reference box unless absolutely necessary.

Flashcard Fodder

Triangle formulas provided by the SAT on test day

Area of a triangle = $(1/2) (ba se) (h e i g h t)$

The sum of the measures of the interior angles of a triangle is $180°$ .

Pythagorean theorem for a right triangle: If $a$ , $b$ , and $c$ are the side lengths of the triangle, and $c$ is the hypotenuse, then $a^{2} + b^{2} = c^{2}$ .

$30° - 60° - 90°$ triangles have side lengths in a ratio of $1 : 3 : 2$ , corresponding to their opposite angle.

$45° - 45° - 90°$ triangles have side lengths in a ratio of $1 : 1 : 2$ , corresponding to their opposite angle.

**Triangle formulas NOT provided by the SAT on test day

Pythagorean triples:

$3 : 4 : 5$

$5 : 12 : 13$

$7 : 24 : 25$

$8 : 15 : 17$

Area of an equilateral triangle: $\frac{s ^{2} 3}{4}$ , where s is the length of one side

Sample Questions

Difficulty 1

A right triangle has one leg of length $12$ cm and a hypotenuse of $13$ cm. What is the length of its other leg, in cm? (Note: this is a free-response question.)

(spoiler)

The answer is 5. Among other formulas, the SAT provides you with the Pythagorean theorem ( $a^{2} + b^{2} = c^{2}$ ) for right triangles. (We recommend memorizing the reference formulas so you don’t have to spend time consulting them on test day!) In this formula, $a$ and $b$ are interchangeable, representing the two legs of the triangle. The most common mistake is to confuse one of the legs with the hypotenuse, plugging in the wrong value for $c$ . For this reason, it’s helpful to think of the formula in the other direction ( $c^{2} = a^{2} + b^{2}$ ) and make sure to identify $c$ first. In this case, we know the value of $c$ : it’s $13$ . So we can write the formula either as $1 3^{2} = 1 2^{2} + b^{2}$ or $1 3^{2} = a^{2} + 1 2^{2}$ . Either way, solving for our variable will yield $5$ ( $169 - 144 = 25$ , which is $5^{2}$ ).

Note: $5 : 12 : 13$ is a Pythagorean triple, so if you have that memorized, you don’t need to use the Pythagorean theorem at all!

Difficulty 2

Triangle $A BC$ is similar to triangle $FG H$ , where angle $A$ corresponds to angle $F$ . If the measure of angle $B$ is $48°$ and the measure of angle $H$ is $72°$ , what is the measure of angle $F$ ?

A. 48°
B. 60°
C. 72°
D. 90°

(spoiler)

The answer is 60°. Similar triangles have congruent corresponding angles, so when the question tells us that angle $A$ corresponds to angle $F$ , we know that those two angles are congruent. In addition, we may assume that the alphabetical order of the angles shows us the other corresponding pairs: $B$ and $G$ are congruent, as are $C$ and $H$ . That means that to discover the measure of angle $F$ , we could also calculate the measure of angle $A$ . The information given pertains to two different triangles, so we need to replace one of the angles with its corresponding partner from the other triangle. For example, if angle $B$ has a measure of $48°$ , then so does angle $G$ . Now that we have the measure of angles $G$ and $H$ , we can subtract their measure from $180°$ : $180 - 72 - 48 = 60$ .

Difficulty 3

A certain triangle has a $45°$ angle, a $90°$ angle, and two sides of length of $52$ . What is the perimeter of the triangle?

A. $152$
B. $102 + 10$
C. $152 + 15$
D. $25$

(spoiler)

The answer is $102 + 10$ . This problem provides everything we need to know to complete the triangle. First, we know that the angle not mentioned must be $45°$ , because $180 - 90 - 45 = 45$ . So we have an isosceles right triangle, also known as a $45 - 45 - 90$ right triangle. Consulting the SAT reference formula list, we see that the ratio of sides in a $45 - 45 - 90$ is $x : x : x 2$ .

What does this mean for the third side, whose length is not mentioned here? The clue is that we already know the length of two equal sides, so those must be the legs; the hypotenuse is missing. The tricky part is to apply the ratio to the known length of the legs in order to find the length of the hypotenuse. If we see that the two lengths of $52$ must correspond to the $x$ ’s in the ratio, then the hypotenuse must correspond to the $x 2$ . That means we need to multiply the leg length by $2$ to find the hypotenuse. What is $2$ times itself? Simply $2$ , because we are essentially squaring a square root, thereby canceling it. This means that $52 \times 2 = 5 x 2 = 10$ . The hypotenuse is $10$ . We can now add that hypotenuse to the two legs and discover our perimeter of $102 + 10$ .

In case it helps you to see that ratio calculation done a different way, consider setting up a proportion as follows:

$\frac{1}{5 2} = \frac{2}{x}$

In this proportion, the ratio parts ( $1$ and $2$ ) are in the numerator and the side lengths ( $52$ and $x$ for the unknown hypotenuse) are in the denominator. Once we set up a proportion we should always cross-multiply; doing so gives us $x = (52) (2) = 10$ .

Final note: the only answer choice without a square root, $25$ , is a trap answer: this is the area of the triangle, not the perimeter.

Difficulty 4

Triangle $D EF$ is similar to triangle $G H I$ such that $D$ , $E$ , and $F$ correspond to $G$ , $H$ , and $I$ , respectively. The measure of angle $G$ is $35°$ and $D E = (2/3) (G H)$ . What is the measure of angle $D$ ?

A. 35°
B. 45°
C. 55°
D. 90°

(spoiler)

The answer is 35°. The only reason this question is in difficulty $4$ is the distractor material regarding the relative length of sides $D E$ and $G H$ . Why is this a distractor? The relationship between two corresponding sides is meant to get you thinking about the proportional relationship of corresponding sides in similar triangles. But, if you use the UnCLES method well and keep a laser focus on what the question is asking for, you’ll notice that it asks for the measure of an angle, not the length of a side.

Since we know that all corresponding angles in similar triangles are congruent, we simply need to find the measure of the angle corresponding to angle $D$ . That would be angle $G$ , whose measure we already have! The question is deceptively simple and the answer is $35°$ .

Difficulty 5

A honeybee begins building a honeycomb with circular cells, but the surface tension soon pulls the cells into a hexagonal shape. If each hexagon in the honeycomb has a side length of $3$ mm, what is the area of each honeycomb? (Note: ignore the three-dimensional depth of the honeycomb.)

A. $63 m m^{2}$
B. $123 m m^{2}$
C. $13.5 3 m m^{2}$
D. $273 m m^{2}$

(spoiler)

The answer is $13.5 3 m m^{2}$ . Why is a hexagon question in this lesson? Because a regular hexagon divides nicely into six equilateral triangles, as shown in the figure below. Admittedly, you are less likely to see this type of question on the SAT than you are the other practice questions in this lesson; most triangle questions of the hardest difficulty deal with right triangle trigonometry, which we will cover in another lesson. But the SAT does occasionally include a polygon with more sides than a quadrilateral, and in some cases those polygons can be broken down (at least in part) into triangles.

Each equilateral triangle has an area that can be discovered by dividing the triangle into two congruent $30 : 60 : 90$ right triangles. Pausing your reading for the moment, can you draw an equilateral triangle with sides of length and figure out its area using this method? It’s great practice in using a special right triangle like a $30 : 60 : 90$ .

However, there is a faster way. The formula for the area of an equilateral triangle is $\frac{s ^{2} 3}{4}$ , where $s$ is the length of one side. Since these six triangles all have sides of length $3$ because they each share a side with the hexagon, we can plug three into the formula. That yields $\frac{9}{4} 3$ . But don’t forget that there are six of these triangles in the hexagon! Multiplying by $6$ and converting to a decimal yields the correct answer.

By the way, unless the SAT specifically tells you otherwise, you should always leave the irrational numbers $π$ , $2$ , and $3$ as they are when making calculations. The answer choices will typically leave these values as part of each answer. Since the exact value of an irrational number like $3$ is a decimal that goes on forever, the SAT could not possibly ask you for the exact decimal value!

For Reflection

How will you approach triangles questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
There are a number of formulas and strategies in this lesson. Have you identified which ones you know and which you don’t? Make a plan to master the tools needed for triangles.

Introduction

Approach Question

Triangle $MNO$ is similar to triangle $PQR$ , where angles $M$ and $P$ correspond to each other, as do angles $N$ and $Q$ and angles $O$ and $R$ . The length of $\overline{PQ}$ is $95$ and the length of $\overline{QR}$ is $38$ . If the length of $\overline{NO}$ is $8$ , what is the length of $\overline{MN}$ ?

A. 20
B. 19
C. 18
D. 16

Explanation

(smaller side of triangle A)/(smaller side of triangle B) = (larger side of triangle A)/(larger side of triangle B)

Of these four quantities, exactly one must be the unknown for which we are solving. So the fraction might look like this:

(known side of triangle A)/known side of triangle B) = (unknown side of triangle A)/(known side of triangle B)

The placements could vary, but the quantities cannot: in a solvable proportion, there must always be three knowns and one unknown.

$\frac{19}{4} = \frac{95}{x}$

$(19) (x) 19 x x = (4) (95) = 380 = 20$

Definitions

Hypotenuse: The longest side of a right triangle; the side across from the right angle. Represented by c in the Pythagorean theorem.
Leg: One of the shorter two sides of a right triangle. Represented by a and b in the Pythagorean theorem.
Isosceles: A triangle with two congruent sides and two congruent angles.
Equilateral: A triangle in which all three sides are congruent and all three angles are congruent. Each angle measures 60°.
Scalene: A triangle in which no side is equal to any other, and therefore no angle is equal to any other.
Similar: A description of two or more triangles whose corresponding angle pairs are congruent, but whose corresponding sides (typically) are not. Two similar triangles, if drawn to scale, will look like copies of each other, with similar dimensions except that one is larger than the other. In similar triangles, corresponding sides are proportional, that is, these pairs of sides all share the same ratio of their lengths. This means that proportions are helpful algebra tools for working with similar triangles.
Congruent: Two shapes (in this case triangles) that are identical to each other in every way. The SAT doesn’t often test for triangle congruence, but for the rare cases, consider: congruence can be proved by various theorems comparing their sides and angles; some notable examples are abbreviated SSS, SAS, AAS, and ASA.

Vertex

The points at which any two sides of a triangle meet; the “corners” of a triangle. A letter denoting the vertex of a triangle can also be used to denote the interior angle at that vertex.

Topics for Cross-Reference

Variations

Strategy Insights

Here’s a visual exercise. Place your hands together so that only the bases of your hands are touching, with your fingertips about an inch apart from each other (index finger an inch away from the other index finger, etc.). Notice how the angle created next to the base of your hands is rather small. Now, keeping the base of your hands together, spread out your hands so your right-hand fingers are several inches apart from your left-hand fingers. Do you see how the angle at the base of your hands grew much larger? This visual tells us an important truth: The size of an angle is related to the length of its opposite side, and vice versa. A small angle will correspond to a small side across from it, which is why the side across from the $30°$ angle must be the smallest in a $30 - 60 - 90$ triangle. A large angle corresponds to a large side across from it, which is why the hypotenuse must be the largest side in a right triangle: it is, by definition, across from the largest angle.
Much triangle work begins with knowing the right formulas. There are several triangle-related formulas provided in the SAT reference list; even though they are given to you on the test, we will reproduce them in the “Flashcard Fodder” section because we think you should memorize them anyway. That way, you don’t have to take time consulting the reference box unless absolutely necessary.

Flashcard Fodder

Triangle formulas provided by the SAT on test day

Area of a triangle = $(1/2) (ba se) (h e i g h t)$

The sum of the measures of the interior angles of a triangle is $180°$ .

Pythagorean theorem for a right triangle: If $a$ , $b$ , and $c$ are the side lengths of the triangle, and $c$ is the hypotenuse, then $a^{2} + b^{2} = c^{2}$ .

$30° - 60° - 90°$ triangles have side lengths in a ratio of $1 : 3 : 2$ , corresponding to their opposite angle.

$45° - 45° - 90°$ triangles have side lengths in a ratio of $1 : 1 : 2$ , corresponding to their opposite angle.

**Triangle formulas NOT provided by the SAT on test day

Pythagorean triples:

$3 : 4 : 5$

$5 : 12 : 13$

$7 : 24 : 25$

$8 : 15 : 17$

Area of an equilateral triangle: $\frac{s ^{2} 3}{4}$ , where s is the length of one side

Sample Questions

Difficulty 1

A right triangle has one leg of length $12$ cm and a hypotenuse of $13$ cm. What is the length of its other leg, in cm? (Note: this is a free-response question.)

(spoiler)

Note: $5 : 12 : 13$ is a Pythagorean triple, so if you have that memorized, you don’t need to use the Pythagorean theorem at all!

Difficulty 2

Triangle $A BC$ is similar to triangle $FG H$ , where angle $A$ corresponds to angle $F$ . If the measure of angle $B$ is $48°$ and the measure of angle $H$ is $72°$ , what is the measure of angle $F$ ?

A. 48°
B. 60°
C. 72°
D. 90°

(spoiler)

Difficulty 3

A certain triangle has a $45°$ angle, a $90°$ angle, and two sides of length of $52$ . What is the perimeter of the triangle?

A. $152$
B. $102 + 10$
C. $152 + 15$
D. $25$

(spoiler)

In case it helps you to see that ratio calculation done a different way, consider setting up a proportion as follows:

$\frac{1}{5 2} = \frac{2}{x}$

Final note: the only answer choice without a square root, $25$ , is a trap answer: this is the area of the triangle, not the perimeter.

Difficulty 4

Triangle $D EF$ is similar to triangle $G H I$ such that $D$ , $E$ , and $F$ correspond to $G$ , $H$ , and $I$ , respectively. The measure of angle $G$ is $35°$ and $D E = (2/3) (G H)$ . What is the measure of angle $D$ ?

A. 35°
B. 45°
C. 55°
D. 90°

(spoiler)

Difficulty 5

A honeybee begins building a honeycomb with circular cells, but the surface tension soon pulls the cells into a hexagonal shape. If each hexagon in the honeycomb has a side length of $3$ mm, what is the area of each honeycomb? (Note: ignore the three-dimensional depth of the honeycomb.)

A. $63 m m^{2}$
B. $123 m m^{2}$
C. $13.5 3 m m^{2}$
D. $273 m m^{2}$

(spoiler)

For Reflection

How will you approach triangles questions on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
There are a number of formulas and strategies in this lesson. Have you identified which ones you know and which you don’t? Make a plan to master the tools needed for triangles.