**SAT Geometry: What You Need to Know**

**Bonus Material: PrepMaven’s SAT Geometry Guide**

The SAT has two math sections: a no-calculator section (20 questions) and a calculator section (38 questions).

Test-takers are likely to encounter geometry questions on both of these.

This can be intimidating to a lot of students!

For many SAT test-takers, geometry can be a somewhat “dusty” concept, especially for juniors and seniors who haven’t studied triangles and circles for years. For others, geometry might simply be that one area of math that simply never made sense!

Fortunately, SAT geometry is very different than geometry students learn in traditional classroom settings. There aren’t any proofs on the SAT, for one thing.

Plus, SAT geometry accounts for only a *very small portion* of the test. While these questions do cover a fairly broad scope, the topics are finite and should feel familiar after review.

You can apply everything you learn in this post to the practice problems available in our SAT Geometry Guide. Grab it below.

Here’s what we cover in this post:

- SAT Geometry: The Basics
- General Approach to SAT Geometry Questions
- SAT Geometry: The Content
**Bonus**: PrepMaven’s SAT Geometry Guide

**SAT Geometry: The Basics**

The SAT contains two math sections:

- No-Calculator: 20 questions, 25 minutes
- Calculator: 38 questions, 55 minutes

SAT geometry is likely to appear in *both *of these sections. Yet there’s some good news to this:** these questions are only likely to make up about 10% of SAT math questions.**

Here’s what we tend to see:

- 2-4 Geometry questions on the No-Calculator section
- 3-6 Geometry questions on the Calculator section

Plus, these questions test a *finite *amount of geometry content. SAT geometry questions frequently concern the following topics:

- Angles & Polygons
- Volume & Surface Area
- Triangles
- Circles

What does this mean for SAT test-takers?

Two things: **know the content, and know how it is tested**. We’ll discuss this more in the next section of this post!

**General Approach to SAT Geometry**** **

There are a few core strategies students should keep in mind when it comes to SAT Geometry.

### 1) Understand what is expected of you

If you have a solid understanding of which concepts will be tested, then you’ll know which tools to pull from your arsenal. You’ll also be able to more efficiently attack the problems themselves.

### 2) Know the formulas

Second, you should take the time to make sure that you know all of the required formulas inside-and-out. This includes the formulas that are given in the reference box at the beginning of each math section:

You will save yourself valuable time and mental energy if you’re not scrambling to find the right equation for a problem!

### 3) Draw pictures when possible

If a geometry question does not include an image, **make sure to draw pictures**. Sometimes something that sounds difficult in words becomes immediately apparent when you see it sketched out in front of you.

If you are given a picture with certain side lengths and angles marked and others left as variables, make sure to physically write in new measurements as you solve for them. You don’t want to try to keep everything straight in your head!

Keep in mind that figures are not often drawn to scale. Don’t assume an angle measure or side length based off of *how *a picture looks. You must prove a value based off of what you know to be true.

### 4) Take these questions out of order

Geometry problems tend to be some of the more time-consuming problems on the test, so it might make sense to save these for last.

Remember that all questions on the SAT are worth the same number of points, and so it doesn’t make sense to waste minutes on difficult problems. If you are short on time and/or having trouble with the earlier sections, focus on those first before moving on to this section.

**SAT Geometry: The Content**

The main geometry topics that students can expect to see covered include:

**Angles & Polygons****Volume & Surface Area****Triangles****Circles**

We take a deep dive into each of these content areas below. You can also download all of these tips *and *apply them to practice problems now, with our **free SAT Geometry Guide**.

## Topic 1: Angles & Polygons

This might sound like a large topic. That’s because it is! However, as we’ve said a few times in this post, the way the SAT tests this topic is predictable.

In general, these SAT geometry questions cover:

- Points in the XY-Coordinate plane
- Parallel lines
- Polygons

*Points in the XY-Coordinate Plane*

Some SAT geometry questions might ask you to find the distance between two points, or the halfway point between two sets of coordinates.

In order to solve these questions, students should be familiar with the following equations:

**Midpoint formula:**

** **

**Distance formula:**

*Parallel Lines*

*Parallel Lines*

Other questions might show a set of parallel lines intersected by another line called a **transversal line.**

These questions often ask students to solve for one or more of the angles created by the intersection. In order to solve these questions, students should be aware of the following angle relationships:

- Vertical angles are equal
- Corresponding angles are equal
- Alternate interior angles are equal
- Same side interior angles are supplementary (sum to 180°)
- The angles that make up a straight line are supplementary (sum to 180°)

**Shortcut: Remember that when a set of parallel lines are cut by a third line, all small angles are equal to one another and all large angles are equal to one another. Any big angle + a small angle will equal 180°.**

In this graphic, angles 1, 4, 5, and 8 are equal, and angles 2, 3, 6, and 7 are equal. Any of these first angles (i.e. 1, 4, 5, and 8) plus any of these second angles (i.e. 2, 3, 6, and 7) will sum to 180°.

Now let’s look at an example of an SAT geometry question involving parallel lines:

**How to solve:**

This is an easy one!You know that any large angle will be supplementary to any small angle. Since angle 1 is 35°, angle 2 is simply180° – 35°, which equals 145°, or choiceD.

*Polygons*

*Polygons*

Students might also see questions involving polygons. A **regular polygon **is any shape in which all side lengths and angles are equal to one another.

Students should be familiar with the following rules about polygons:

**The sum of all the interior angles in a polygon with***n*sides = 180°(*n-*2).- Accordingly, each interior angle in a regular polygon with
*n*sides = 180°(*n-*2)/n.

- Accordingly, each interior angle in a regular polygon with
**Exterior Angle Theorem**- An exterior angle is formed when any side of a polygon is extended. The exterior angle will always be equal to the supplement of the adjacent angle (i.e. the exterior angle + the adjacent angle will equal 180°).
- If the polygon is a triangle, the exterior angle equals the sum of the non-adjacent angles in the triangle.

Let’s look at an example of a problem involving polygons:

**How to solve:**

This polygon has 4 sides, and so the sum of the interior angles will be equal to 180° x ([4]-2), which comes out to 360°. That means that 45° + x° + x° + x° = 360°. Solving for x, we get 105°, or choiceD.

**Topic 2: Volume and Surface Area **

These SAT geometry questions are likely to test any (or all) of the following:

- The volume of regular solids
- The surface area of regular solids

In general, there’s not too much to memorize with volume and surface area for the SAT.

The reference information at the beginning of each section of SAT math will provide most of the necessary formulas, and any uncommon formulas will most likely be given in the problem.

But remember: you can save valuable time by memorizing the formulas provided in the reference information!

*Volume*

*Volume*

It’s helpful to remember that the volume of all regular solids can be found using the following formula:

**Volume = Area of base x Height**

Most volume questions on the SAT involve right cylinders. Since the base of a cylinder is a circle, these questions will also incorporate concepts involving circles (see the final section of this post for more detail).

Below are the volume formulas that you should know for the test:

- Volume of a Cylinder

- Volume of a Rectangular Prism

- Volume of a Cube

- Volume of a Cone

- Volume of a Sphere

Let’s look at an example of a question involving volume:

**How to solve:**

If the volume of the cylinder is equal to 72 π and the height is 8 yards, then plugging into the formula for the volume of a right cylinder, we get 72π=8πr^2. Solving for r, that gets us 3 yards.

However, the question is asking about diameter, not radius. Since diameter=2r,the answer is 6 yards.

Some volume problems might be more involved, combining multiple shapes into a single question. Let’s look at one of those:

**How to solve:**

While this question might look overwhelming at first glance, it’s really no more difficult than the previous problem. All we need to do is find the volume of the central cylinder and the volume of each of the cones and add those values together.

We know the volume of a cone is (1/3)πr^2(h). Here, the radius for each cone is 5 feet, and the height is likewise 5 feet. That means the volume of each cone is (25/3) π, or ~130.90 cubic feet. Similarly, the volume of a cylinder = πr^2(h). Here, the radius of the cylinder is 5 feet, and the height is 10 feet. That means the volume of the cylinder is 250π, or ~785.40 cubic feet. The total volume of the silo, then, equals 130.90 cubic feet + 130.90 cubic feet + 785.40 cubic feet, or 1,047.2 cubic feet, choiceD.

*Surface Area*

Surface area is just the sum of the area of each of the faces of a polygon.

For most prisms, this is pretty straightforward.

For a cylinder, it’s a bit less intuitive: a cylinder is basically a rectangle wrapped around a circular base (giving that rectangle a length equal to the circumference of that circle).

That means that the equation for the surface area of a cylinder is as follows:

- Surface Area of a Cylinder

**Topic 3: Triangles**

The SAT loves to test triangles and incorporate them into other geometry questions. The major types of triangles that the SAT tests are:

- Isosceles Triangles
- Two sides are equal, and the corresponding angles across from those sides are also congruent.

- Equilateral Triangles
- All sides and all interior angles are equal. Each interior angle is 60°.

- Right Triangles
- One angle is 90°.

Students should also be familiar with a few other rules of triangles:

- All of the interior angles add to 180°
- For any triangle, the sum of any two sides must be greater than the third side. This is called the
**Triangle Inequality Theorem** - Area of a Triangle = (1/2)base(height)
- Side lengths are proportionate to the angles they’re across from. So, the longer the length of a side, the larger the angle across from it

Let’s look at a basic triangle problem:

**How to solve:**

We know that all of the interior angles in a triangle add up to 180°. If a=34, then 34° + b° + c° = 180°. That means that b + c = 180° – 34°, orb + c = 146°.

*Right Triangles*

Right Triangles are made up of two legs and a hypotenuse (the side opposite the right angle). Every right triangle obeys the **Pythagorean theorem, **which states:

Here, a and b are the legs of the triangle and c is the hypotenuse.

You will see certain right triangles come up repeatedly on the SAT.

These are **Pythagorean triples **or sets of three whole numbers that satisfy the Pythagorean theorem and are therefore used frequently to represent the side lengths of right triangles on the SAT.

**Recognizing Pythagorean Triples can save you a lot of time because if you know two sides, you can easily identify the third without having to use the Pythagorean theorem.**

Common Pythagorean triples include:

- 3, 4, 5 (this is the most common triple)
- Any multiple – i.e. [6, 8, 10], [9, 12, 15], [12, 16, 20]

- 5, 12, 13
- Any multiple – 10, 24, 25

- 7, 24, 25

*Special Right Triangles*

You will have to memorize two special right triangle relationships.

**1) 30° – 60° – 90° Triangles**

- The ratio of sides is: x, x√3, 2x
- This is the most common type of special right triangle on the SAT
- The shortest side, x, is opposite the smallest angle, and the largest side, 2x, is opposite the largest angle
- If you cut an equilateral triangle down the middle from its vertex, you will get two 30°-60°-90° triangles

**2) 45° – 45° – 90° Triangles**

- The ratio of sides is: x, x, x√2
- A 45°-45°-90° triangle is also an isosceles triangle, which might help you remember that both legs must be equal

Any time that you see an angle marked as 45°, 30°, or 60°, you should be looking to utilize the rules of special right triangles, even if it’s not immediately obvious!

Let’s look at an example of a question involving special right triangles:

**How to solve:**

Since Angle ABD is equal to 30° and angle ADB is equal to 90°, angle BAD must equal 60°. That means that triangle ABC is an equilateral triangle, and triangles ABD and DBC are congruent 30° – 60° – 90° triangles.

The hypotenuse of triangle DBC is 12. We know from the rules of special right triangles that the hypotenuse of a 30° – 60° – 90° triangle is equal to 2x, where x is the length of the side opposite the 30° angle (in this case, line DC). That means that DC is 6.

Since triangles ABD and DBC are congruent, as proven above, DC=AD. Therefore, line AD is also 6, and the answer is choiceB.

** Similar Triangles**When two triangles have the same angle measures, their sides are proportional.

- If you can prove 2 angles in 2 separate triangles are identical, then the 3rd angle will also be identical
- To solve similar triangle problems, match up the corresponding sides of the triangle and create a proportion to solve for the missing side

Let’s look at an example of an SAT geometry problem that tests students’ knowledge of similar triangles:

**How to solve:**

Because the three shelves are parallel, the three triangles in the figure are similar. Since the shelves break up the largest triangle in the ratio 2:3:1, the ratio of the middle shelf to the largest triangle is 3:6 (the largest value is found by adding all of the partial values together, i.e. 2 + 3 + 1).

Since the height of the largest triangle is 18, the height of the middle shelf can be found by creating a proportion that relates the side lengths of the middle and largest triangles to their respective heights. In other words, (side length of middle shelf)/(side length of largest triangle) = (height of middle shelf)/(height of largest triangle). Subbing in the above values, that gives us 3/6 = x/18. Solving for x, we get9 as our answer.

**Topic 4: Circles **

Circle properties do not appear as frequently as triangle properties on the SAT Math sections. However, students can expect to encounter 1-3 of these questions, so it’s wise to know this content when preparing for SAT Geometry problems.

In general, these geometry questions cover:

- Basic properties of a circle, including area and circumference
- Advanced circle vocabulary, including
*sector*,*chord*,*arc*, and*tangent* - Arc measure/length
- Sector area
- Central angles

*Basic Properties of Circles*

Students should be familiar with the following key formulas and properties of circles:

- Diameter of a Circle =
- Area of a Circle:
- Circumference of a Circle =

- A
*chord*is a line segment that connects two points on a circle - A
*tangent*is a line that touches a circle at exactly one point. A tangent is always perpendicular to the radius it intersects.

*Arc Length and Sector Area*

Sometimes, instead of being asked to calculate the entire circumference or area of a circle, students will be asked to calculate the length of just a piece of the circumference – known as the *arc length – *or the area of one slice of the pie – known as the *sector. *

Sectors and arcs will always be bound by two radii. The angle formed by the two radii is known as the *central angle. *

In the figure to the left, the length along the edge from A to B would be the *arc length*, the wedge-shaped area bound by angle AOB would be the *sector*, and angle AOB would be the *central angle* (i.e. 45°).

The ratio between the central angle and the total number of degrees in the circle (i.e. 360°) will always be the same as the ratio between the area of the sector and the total area of the circle.

Similarly, the ratio between the central angle and the total number of degrees in the circle (i.e. 360°) will always be the same as the ratio between the arc length and the total circumference of the circle.

For this reason, the formulas for arc length and sector area are actually quite simple to remember.

You just take the formula for circumference and area, respectively, and multiply them by the proportion taken up by the central angle. Here’s what that looks like:

- Arc Length = (2πr)(central angle/360°)
- Sector Area = (πr^2 )(central angle/360°)

Let’s look at an example of a question involving arc length:

**How to solve:**

Because angle AOB is marked as a right angle, we know that the central angle is 90°. The question also tells us that the total circumference is 36. Plugging into the equation for arc length, we get Arc Length = (36)( 90°/360°), which simplifies to 9, or choiceA.

*Arc Measure*

Many students confuse arc length and arc measure.

*Arc length* is the actual distance between points A and B on the circle. *Arc measure* is the number of degrees that one must turn to get from A to B.

You can think of it as a partial rotation along the circumference of the circle – a full rotation is 360°.

Central angles have the same measure as the arcs they “carve out.” Inscribed angles are half the measure of the arcs they “carve out.”

In the figure to the left, angle AOB would be the central angle, angle ACB would be the inscribed angle, and the arc measure of minor arc AB would be 70° (which is equivalent to the central angle and twice that of the inscribed angle).

Let’s look at an example question involving arc measure:

**How to solve:**

The measure of an angle inscribed in a circle is half the measure of the central angle that intercepts the same arc. That means that angle A is equal to (x°/2). We also know that angle P is equal to (360° – x°).

The sum of the interior angles of any quadrilateral equals 360°. That means the interior angles of ABPC must sum to 360°, or (x°/2) + (360° – x°) + 20° + 20° = 360°. Solving for x, that gets us80° as our answer.

*The Equation of a Circle*

Students should also be familiar with the standard form for the equation of a circle in the XY-coordinate plane:

- Where (h, k) are the coordinates for the center of the circle
- Where r is the radius of the circle

How is this equation usually tested? Given the equation, you must be able to identify the center and the radius of the circle.

Let’s look at an example of a question involving the standard equation of a circle:

**How to solve:**

Using what we know from the standard form for the equation of a circle, we can conclude that this circle has a center at (6, -5) and a radius of 4. If P is located at (10, -5), then the end of the diameter lies 4 units directly to the right of the center. That means the other end of the diameter will lie 4 units directly to the left of the center, which would put Q at (2, -5), or choiceA.

## Download PrepMaven’s SAT Geometry Guide

There you have it–all of the geometry principles you need to succeed on the SAT Math sections! With our free SAT Geometry Guide, you’ll get all of these principles in one place.

With this worksheet, you’ll get:

- A recap of the content areas, skills, and strategies discussed in this post
- FREE practice questions
- Detailed explanations of SAT geometry questions
- Information about where to find additional practice questions

Annie is a graduate of Harvard University (B.A. in English). Originally from Connecticut, Annie now lives in Los Angeles and continues to mentor children across the country via online tutoring and college counseling. Over the last eight years, Annie has worked with hundreds of students to prepare them for all-things college, including SAT prep, ACT prep, application essays, subject tutoring, and general counseling.

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