[fs-toc-h2]The Foundations
In order to succeed and excel in this topic, there is a foundational concept that rules over all of geometry and trigonometry: Angles.
You’ve most likely learned what they are in school by now, but if not, watch the video below for an introduction on what they are and how they’re used. We’ll go over more advanced topics using angles later on.
Watch this video for more explanation.
[fs-toc-h2]Finding Angles
[fs-toc-h3]Complementary Angles
As shown in the video, any angles combined to make a 90 degree/right angle are known as complementary.
In the figure above, complementary angles \(a\) and \(b\) must add up to 90 degrees, meaning:
\[a + b = 90\]
This equation means that if we know the value of one complementary angle, we can find the value of the other.
In this case, we want to find the measure of angle \(b\). Since we know the measure of angle \(a\) and that both angles create a right angle, we can set up an equation to solve for \(b\).
\[a + b = 90\]
\[38 + b = 90\]
\[b = 52 \text{ degrees}\]
Questions testing this topic may be asked in a couple of different ways. Refer to the example below.
In the figure above, what is the value of \(x\)?
It's given that all three of these angles create a 90 degree angle (denoted by the right angle symbol).
\[x + x + 3x = 90\]
Combine like terms and solve:
\[5x = 90\]
\[x = 18\]
[fs-toc-h3]Supplementary Angles
Any angles combined to make a 180 degree angle/straight line are known as supplementary angles. If you understand everything about complementary angles, you’ll understand everything here, as all the actual concepts are identical. Take a look at the example below.
In the figure abovall points lie on a straight line. What is the value of \(x\)?
It's given that all the angles make up a straight line, meaning their sum will be 180. Add all angles together in an equation set equal to 180 first.
\[2x + 2x + (3x + 10) + (x + 10) = 180\]
Combine like terms and solve for \(x\):
\[8x + 20 = 180\]
\[8x = 160\]
\[x = 20\]
[fs-toc-h3]Vertical Angles
When any two lines intersect, they create two sets of vertical angles, which are the two angles opposite of each other.
If we know the value of one of these angles, we can find the value of the other.
Notice that angle \(x\) and angle \(y\) actually lie together on a straight line, meaning we can take advantage of supplementary angles. Take a look at the figure below and try to solve for angle \(x\).
Since \(x\) and the 49 degree angle lie on a straight line, then:
\[x + 49 = 180\]
\[x = 135 \text{ degrees}\]
Knowing that these two angles add up to 180 means that all the angles created by two intersecting lines add up to 360 degrees, the sum of interior angles within a circle.
[fs-toc-h3]Parallel Lines and Transversals
A transversal is any line that intersects two other lines, usually a set of parallel lines on the SAT.
These three lines create eight separate angles, and knowing which ones are congruent to each other is important. In the figure below, the four angles labeled \(x\) and four labeled \(y\) are congruent due to vertical angles.
Due to supplementary angles, \(x\) and \(y\) both add up to 180 degrees.
Take a look at the example below.
In the figure above, line \(A\) and line \(B\) are parallel, angle \(z\) has a measure of 42 degrees. What is the measure of angle \(w\)?
Recall from before, the two different angles created by a line intersecting two parallel lines are supplementary, meaning they add up to 180.
Since angles \(w\) and \(z\) are not congruent, then that means that they must add up to 180 degrees:
\[w + z = 180\]
\[w + 42 = 180\]
\[w = 138\]
If you weren't completely sure of this, you could always fill out the full figure and identify that both angle \(w\) and \(z\) lie on a straight line.
We highly recommend watching the video below for a bit more depth on parallel lines and transversals!
[fs-toc-h2]Polygons
A polygon is any closed two-dimensional shape made up of straight lines, it's basically what you think of when someone says ‘shape’.
Regular polygons are polygons in which all sides and interior angles are congruent.
The sum of interior angles of a polygon can be found using the formula \(180(n-2)\), where \(n\) is the number of sides the polygon has.
Since here we have a six-sided polygon (hexagon), the sum of all interior angles is:
\[180(6-2) = 180 \times 4 = 720\]
In order to find the measure of each individual angle, divide by the number of sides/vertices:
\[\frac{720}{6} = 120 \text{ degrees}\]
The perimeter of a polygon is simply the sum of all sides. In the figure below, the rectangle has four side lengths of 12, 12, 7, and 7, meaning the perimeter is \(12 + 12 + 7 + 7\) which is equal to 38.
[fs-toc-h3]Finding the Area of a Polygon
Finding the area of a polygon is different from shape to shape. The area is basically the size of the total space within a shape. Let's go over some different polygons, their characteristics, and how to calculate their area.
[fs-toc-h3]Squares and Rectangles
Squares and rectangles are similar in almost every single way. All four interior angles created by these two polygons must be 90 degrees (meaning that the sum of interior angles is 360), and their opposing sides must be parallel.
Where these polygons differ are their sides. The opposing sides on a rectangle are all equal. While this is technically true for a square, for something to actually be classified as a square, all four sides must be equal.
The area of these two shapes is calculated with the formula:
\[ \text{Area} = \text{Length} \times \text{Width} \]
The rectangle above has side lengths of 6 cm and 10 cm, so area would be calculated as:
\[ \text{Area} = 6 \times 10 = 60 \text{ cm}^2 \]
[fs-toc-h3]Parallelograms
You can basically think of a parallelogram as a slightly tilted rectangle. The opposing sides must be parallel and have the same exact measure.
Area is also calculated the exact same way: \(\text{Length} \times \text{Width}\).
The only difference between a rectangle and parallelogram is the interior angles. While a rectangle must have four 90 degree angles, parallelograms don't.
Something worth noting though is that the angles diagonal from each other are congruent, meaning if you're given one angle within a parallelogram, you can find the others. Try to find the other three angles in the figure below.
Since diagonal angles are congruent, we can already fill that angle with 103 degrees.
And as you may recall, the sum of all angles in a parallelogram is 360. So,
\[ x + x + 103 + 103 = 360 \]
Now combine like terms and solve for \(x\):
\[ 2x + 206 = 360 \]
\[ 2x = 154 \]
\[ x = 77 \]
[fs-toc-h2]Triangles
The reason we haven't walked through any official College Board questions is because of this one topic. Within 90% of all geometry questions, triangles are involved in one way shape or form. Some people might hate this, but this means if you truly understand all the different characteristics of triangles, you can solve almost all geometry questions easily.
Before getting into different types of triangles, it's important to know that, no matter the shape and size of a triangle, all interior angles add up to 180 degrees.
[fs-toc-h3]Equilateral Triangle
An equilateral triangle is a triangle in which all three sides and all three interior angles are equal to one another.
Since all three angles are congruent, and the sum of interior angles in a triangle is 180, then all three angles will always have a measure of 60 degrees.
[fs-toc-h3]Isosceles Triangle
An isosceles triangle has two sides congruent in length. The two angles at the base are also equal.
NOTE: Only the two angles that are ‘in between’ the base and a congruent side are equal. The angle in between both congruent sides is not equal to the others.
With this information, if you're given one angle within an isosceles triangle, then you can find the rest. Try finding the missing angles in the example below.
If you don't know what to initially do, always take it step by step. We know that all angles in a triangle add up to 180. We also know that the angles at the base are congruent to each other. We can label these angles as \(x\). So, the sum of both of these angles plus 32 should equal 180.
\[ x + x + 32 = 180 \]
Now combine like terms and solve:
\[ 2x + 32 = 180 \]
\[ 2x = 148 \]
\[ x = 74 \]
[fs-toc-h4]Example 1
Take a look at the official example below. Remember, you probably won't be able to find what you’re solving for in one step. If you aren’t sure where to start, just begin finding as much information as possible until you have enough to actually find what you want.
The first step here would be to find \(z\). It's really the only useful piece of information we can get with what we know. Since we’re given an equation with \(z\), and a value for \(y\), plug it in and solve for the unknown:
\[ 180 - z = 2(75) \]
\[ -z = -30 \]
\[ z = 30 \]
Now that we have that angle, let's fill in the missing information and label the unnamed angles at the bases. I'll use the variable (\(w\)) in this case.
Notice that angle \(w\) and angle \(x\) both lie on a straight line, meaning that both angle measures have a sum of 180 (supplementary angles). This tells us that if we can find the measure of angle \(w\), we can set up an equation and find angle \(x\).
Since we’ve already found one of the angles in the same isosceles triangle (\(z\)), we can find the other two:
\[ w + w + z = 180 \]
\[ 2w + 30 = 180 \]
\[ 2w = 150 \]
\[ w = 75 \text{ degrees} \]
\(w\) and \(x\) are supplementary angles, and now that we know \(w\), we can solve for \(x\):
\[ w + x = 180 \]
\[ 75 + x = 180 \]
\[ x = 105 \text{ degrees} \]
[fs-toc-h3]Right Triangles
The most important triangle within the study of trigonometry, also making it the most tested.
All you need to know for now is that a right triangle has one right angle, the other angles can have any measure (as long as all angles still add up to 180).
We’ll go much further in-depth with right triangles later.
[fs-toc-h3]Area of a Triangle
The area of a triangle is given using the formula \( \frac{1}{2}bh \), where \(b\) is the base and \(h\) is height. The height in any given triangle is perpendicular to the base.
The equation still holds up for any other type of triangle, but a height always has to be drawn in.
[fs-toc-h3]Triangle Similarity
Similar triangles are triangles that share all three equal angles, but not the same side lengths. They’re simply the same exact shape, but aren't the same size.
The ratio between two side lengths on one of the similar triangles is equal to the ratio between the two corresponding side lengths on the other triangle.
For example, in the two triangles above, line \(AB\) and line \(DE\) are corresponding, and line \(BC\) and \(EF\) correspond. This means that:
\[ \frac{AB}{BC} = \frac{DE}{EF} \]
Since we have a valid ratio, if we’re given three out of the four sides, we can use cross-multiplication to solve for the fourth.
In the figure above, both triangles are similar. What is the value of \(x\)?
First step is to find which sides correspond/"match up". Make sure to use \(x\) in your ratios as that's what we are solving for.
21 corresponds to 7, and \(x\) corresponds to 3, meaning we can set these ratios equal to each other:
\[ \frac{21}{x} = \frac{7}{3} \]
NOTE: Make sure each ratio only includes the side lengths of one triangle, and that the corresponding side lengths are directly across from each other: 21 and 7 are on the top, \(x\) and 3 are on the bottom.
Now cross-multiply to solve for \(x\):
\[ \frac{21}{x} = \frac{7}{3} \]
\[ 7x = 63 \]
\[ x = 9 \]
Watch this video for a more in-depth explanation of similarity.
Similar triangles can also show up in a different way, and that's within another triangle. Take a look at the figure below.
When a line splits a triangle and is parallel to another side length, then the triangle that is created is similar to the larger, overall triangle, as they share all three angles.
Knowing this, you can split the triangles into two and solve for whatever the test wants you to (usually a side length).
Let's go over some questions to review everything we’ve learned.
The only way to find \(x\) in this case would be finding the angle that is supplementary to it (the angle on the other side of line \(s\)). To find this angle we have to find the other angles within the triangle that lines \(r\), \(s\), and \(t\) create.
Firstly, let's label all the angles we have to solve for before moving on, this will make it easier to visualize and understand what we need.
With any geometry problem, start solving for what you can if you can't immediately get what you want. The only angle we can solve for here is angle \(y\), so we’ll start there. Angle \(y\) and 106 degrees are supplementary, so setting up an equation will get us the value.
\[ y + 106 = 180 \]
\[ y = 74 \text{ degrees} \]
We can now fill in the triangle with this information and see if there's anything else we can solve for.
While there is still no apparent way to solve for \(x\) yet, we can solve for the angle we labeled as \(z\). When you have any two angles in a triangle, you can always solve for the third, as all angles add up to 180:
\[ z + 74 + 23 = 180 \]
\[ z = 83 \]
We now have enough information to solve for \(x\). The angle of measure 83 degrees and angle \(x\) both lie on a straight line, making them supplementary, meaning we can set up the equation and solve for \(x\):
\[ x + 83 = 180 \]
\[ x = 97 \]
With any of these geometry questions (and almost all SAT questions), don't try to immediately look for ‘how can I get the final answer’. Take the question step by step and look for ‘what can I find with what I have’, and you’ll begin solving a lot of questions correctly.
First step here doesn’t involve any real math, but involves identifying that these two triangles are similar. Angle \(D\) and angle \(E\) are given to be congruent, but angle \(B\) (on both triangles) is also congruent to itself, due to the law of vertical angles.
When two triangles share two angles, that also means that they share the third, and when all three angles are congruent, both triangles are similar.
You’ll also want to label each side length to get a better visualization of everything, there’s a lot of information here to keep track of.
Now find which sides are corresponding and set up a ratio. Sides \(BD\) and \(EB\) are corresponding; sides \(CD\) and \(X\) are corresponding, so:
\[ \frac{BD}{CD} = \frac{EB}{X} \]
Substitute in the given values and cross multiply:
\[ \frac{700}{800} = \frac{1400}{x} \]
\[ 700x = 1,120,000 \]
\[ x = 1600 \]
[fs-toc-h2]Right Triangle Trigonometry
Right triangle trigonometry may seem a bit difficult to some students due to the number of ways they can test it. Sometimes you might not even know they are testing you on some of these topics.
Though, in reality there's only a few subtopics within here you truly have to master. Once you get them down you will be fine.
Do not be discouraged if you miss a trigonometry problem on your next practice test though (even if you've studied it). While all topics benefit from practice, geometry and trigonometry especially require it. Since there are so many different questions and ways they can ask them, it can take a bit of time to get used to, but with enough hard work these topics are no different than the rest of the algebra ones.
[fs-toc-h3]Finding Missing Side Lengths
In a right triangle, there are two ways you can find missing right triangles: Trigonometric ratios and the Pythagorean theorem.
If you need a lesson on trig ratios, we highly recommend the quick Khan Academy lesson over it, just click here.
In order to find a missing angle or side using trig ratios, you have to either know the value of two sides (to find an angle) or one side and one angle (to find a side).
If you have two sides though, while it is possible to use trig ratios to find the third one, using the Pythagorean theorem is almost always much quicker and ‘safer’ (as there is less room for error).
Pythagorean theorem is the equation:
\[a^2 + b^2 = c^2\]
Where \(a\) and \(b\) are the legs of the right triangle (the two shortest sides) and \(c\) is the hypotenuse (the longest side).
When given the values of two of the sides, simply plug them into their respective variables and solve for the missing one. Try solving for the missing leg below.
In this case we’re given one leg and the length of the hypotenuse, meaning we can plug in 3 for \(a\) and 5 for \(c\).
NOTE: The value 3 can be plugged in for either \(a\) or \(b\). Both \(a\) and \(b\) represent the leg lengths which can be switched around.
\[3^2 + b^2 = 5^2\]
Now simplify everything and solve for \(b\):
\[9 + b^2 = 25\]
\[b^2 = 16\]
\[b = 4\]
For more on the Pythagorean theorem, watch this.
[fs-toc-h3]Special Right Triangles
Special right triangles are triangles in which their sides have a set ratio between them. These triangles do in fact show up on the formula sheet, but understanding them is quite important.
[fs-toc-h4]30-60-90 Triangle
For a triangle with angles 30, 60, and 90 degrees, the ratio between all three of their sides is \(1 : \sqrt{3} : 2\) (where 1 is the shortest leg and 2 is the hypotenuse). You can think of the shortest leg as the “starting point”, meaning that the second longest is \(\sqrt{3}\) times larger than the shortest leg, and the hypotenuse is 2 times larger than the shortest leg.
For example, if the length of the shortest leg is 2 cm, then the length of the second leg is \(2\sqrt{3}\) cm, and the hypotenuse is \(2 \times 2\) units long, or 4 cm.
30-60-90 triangles are especially useful when you’re given an equilateral triangle. Since each angle in an equilateral triangle is 60 degrees, dropping down a perpendicular line from the highest point splits the triangle into two 30-60-90 triangles. This strategy can be used to find the height of the equilateral triangle which is needed in order to calculate area.
Try to find the area of the triangle below.
Again, we need to find the height here, so the first step would be dropping down a line that splits the triangle into two equal right triangles, giving us a way to find the height. Since we cut the triangle right in half, the bases of the two right triangles are \( \frac{8}{2} \), or 4 units long.
Since 4 is our shortest side, the second longest side (in this case the height) is \(\sqrt{3}\) times larger than that, giving us a height of \(4\sqrt{3}\).
Now that we have the height, all that's left is to plug everything into the area formula:
\[ \frac{1}{2}bh \]
\[ \frac{1}{2}(8)(4\sqrt{3}) \]
\[ 4(4\sqrt{3}) \]
\[ 16\sqrt{3} \]
[fs-toc-h4]45-45-90 Triangle
For a triangle with angles 45, 45, and 90 degrees, the ratio between their sides is \(1 : 1 : \sqrt{2}\), where both legs equal each other (as it's an isosceles triangle), and the hypotenuse is \(\sqrt{2}\) times greater than the legs.
While the 30-60-90 triangle shows up in the equilateral triangle, the 45-45-90 triangle shows up in a square that is diagonally split. While taking advantage of this concept is less common than the 30-60-90 triangle, it is still important to know as it is essential to solve some problems involving inscribed squares, more on which in the next lesson.
Solving for a missing side (especially the legs) can be a bit confusing to some students especially when the hypotenuse doesn't have \(\sqrt{2}\) attached to it. Take a look at the example below and try to solve for the legs.
Since 28 in this triangle corresponds to \(x\sqrt{2}\) in the model triangle, you can set them equal to each other to solve for \(x\):
\[28 = x\sqrt{2}\]
\[ \frac{28}{\sqrt{2}} = x \]
While some math classes might have you stop here, the SAT requires you to rationalize (get the square root out of the denominator). If you're not familiar with the process, watch the video below for a great explanation.
\[ \frac{28}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{28\sqrt{2}}{2} \]
\[ x = 14\sqrt{2} \]
Let's try one more example involving some other concepts.
[fs-toc-h3]Example 1
An isosceles right triangle has a perimeter of \(80 + 80\sqrt{2}\) inches. What is the length, in inches, of one leg of this triangle?
• 40
• 40\(\sqrt{2}\)
• 80
• 80\(\sqrt{2}\)
The first step here to identify what we’re being tested on. An isosceles right triangle means that the two legs and the two angles (other than the right angle) are equal, meaning we’re dealing with a 45-45-90 triangle.
The perimeter of our model triangle would be all sides added up: \(x + x + x\sqrt{2}\) or \(2x + x\sqrt{2}\).
In this case, the easiest and most likely the most efficient way would be applying PITA (plugging in the answer). While you could set \(2x + x\sqrt{2}\) equal to \(80 + 80\sqrt{2}\) and solve for \(x\), an equation like this would simply take too long.
Since \(x\) corresponds to leg length in this case, you can plug in each answer choice for \(x\) until you get an answer corresponding to the given perimeter, \(80 + 80\sqrt{2}\). We can start off by trying answer choice A. Many may quickly choose A without a second thought, since \(2(40) = 80\), which is the first term of our given perimeter. Be careful though, take a look below on why this doesn't work.
\[2x + x\sqrt{2}\]
Plug in 40 for \(x\):
\[2(40) + 40(\sqrt{2})\]
\[80 + 40\sqrt{2}\]
Which does NOT match up with \(80 + 80\sqrt{2}\).
While it initially may look like it’ll work, the \(\sqrt{2}\) term is incorrect. Let's try choice B now.
\[2x + x\sqrt{2}\]
Plug in \(40\sqrt{2}\) for \(x\):
\[2(40\sqrt{2}) + (40\sqrt{2})\sqrt{2}\]
\[80\sqrt{2} + 40(2)\]
\[80\sqrt{2} + 80\]
In this case, our perimeter does in fact match up with \(80 + 80\sqrt{2}\), meaning choice B is correct. Since our leg length is \(40\sqrt{2}\), the triangle would look something like this.
This example is great at showing that just because a term has \(\sqrt{2}\) attached to it doesn't necessarily mean that it corresponds to the hypotenuse.