So what formulas do you need to have memorized for the SAT math section before the day of the test? In this complete guide, I’ll cover every critical formula you MUST know before you sit down for the test. I’ll also explain them in case you need to jog your memory about how a formula works. If you understand every formula in this list, you’ll save yourself valuable time on the test and probably get a few extra questions correct.

Formulas Given on the SAT, Explained

This is exactly what you’ll see at the beginning of both math sections (the calculator and no calculator section). It can be easy to look right past it, so familiarize yourself with the formulas now to avoid wasting time on test day.

You are given 12 formulas on the test itself and three geometry laws. It can be helpful and save you time and effort to memorize the given formulas, but it is ultimately unnecessary, as they are given on every SAT math section.

You are only given geometry formulas, so prioritize memorizing your algebra and trigonometry formulas before the test day (we’ll cover these in the next section). You should focus most of your study effort on algebra anyways because geometry has been de-emphasized on the new SAT and now makes up just 10% (or less) of the questions on each test.

Nonetheless, you do need to know what the given geometry formulas mean. The explanations of those formulas are as follows:

Area of a Circle

A=πr²

> π is a constant that can, for the purposes of the SAT, be written as 3.14 (or 3.14159)

> r is the radius of the circle (any line drawn from the center point straight to the edge of the circle)

Circumference of a Circle

C=2πr (or C=πd)

> d is the diameter of the circle. It is a line that bisects the circle through the midpoint and touches two ends of the circle on opposite sides. It is twice the radius.

Area of a Rectangle

A=lw

> l is the length of the rectangle

> w is the width of the rectangle

Area of a Triangle

> b is the length of the base of the triangle (the edge of one side)

> h is the height of the triangle

i.  In a right triangle, the height is the same as a side of the 90-degree angle. For non-right triangles, the height will drop down through the interior of the triangle, as shown above.

The Pythagorean Theorem

a²+b²+c²

> In a right triangle, the two smaller sides (a and b) are each square. Their sum is equal to the square of the hypotenuse (c, the longest side of the triangle).

Properties of Special Right Triangle: Isosceles Triangle

> An isosceles triangle has two sides that are equal in length and two equal angles opposite those sides.

> An isosceles right triangle always has a 90-degree angle and two 45 degree angles.

> The side lengths are determined by the formula: x, x, x√2, with the hypotenuse (the side opposite 90 degrees) having a length of one of the smaller sides *√2.

i. E.g., An isosceles right triangle may have side lengths of 12, 12, and 12√2.

Properties of Special Right Triangle: 30, 60, 90 Degree Triangle

> A 30, 60, 90 triangle describes the degree measures of the triangle’s three angles.

> The side lengths are determined by the formula: x, x√3, and 2x

i. The side opposite 30 degrees is the smallest, with a measurement of x.

ii. The side opposite 60 degrees is the middle length, with a measurement of x√3.

iii. The side opposite 90 degree is the hypotenuse (longest side), with a length of 2x.

iv. For example, a 30-60-90 triangle may have side lengths of 5, 5√3, and 10.

Volume of a Rectangular Solid

V=lwh

> l is the length of one of the sides.

> h is the height of the figure.

> w is the width of one of the sides.

Volume of a Cylinder

V=πr²h

> r is the radius of the circular side of the cylinder.

> h is the height of the cylinder.

Volume of a Sphere

> r is the radius of the sphere.

Volume of a Cone

> r is the radius of the circular side of the cone.

> h is the height of the pointed part of the cone (as measured from the center of the circular part of the cone).

Volume of a Pyramid

> l is the length of one of the edges of the rectangular part of the pyramid.

> h is the height of the figure at its peak (as measured from the center of the rectangular part of the pyramid).

> w is the width of one of the edges of the rectangular part of the pyramid.

Law: the number of degrees in a circle is 360

Law: the number of radians in a circle is 2π

Law: the number of degrees in a triangle is 180

Gear up that brain because here come the formulas you have to memorize.

Formulas Not Given on the Test

For most of the formulas on this list, you’ll simply need to buckle down and memorize them (sorry). Some of them, however, can be useful to know but are ultimately unnecessary to memorize, as their results can be calculated via other means. (It’s still useful to know these, though, so treat them seriously).

We’ve broken the list into “Need to Know” and “Good to Know,” depending on if you are a formula-loving test taker or a fewer-formulas-the-better kind of test taker.

Slopes and Graphs

How to write the equation of a line

> The equation of a line is written as:y=mx+b

i. If you get an equation that is NOT in this form (ex. mx−y=b), then re-write it into this format! It is very common for the SAT to give you an equation in a different form and then ask you about whether the slope and intercept are positive or negative. If you don’t re-write the equation into y=mx+b, and incorrectly interpret what the slope or intercept is, you will get this question wrong.

> m is the slope of the line.

> b is the y-intercept (the point where the line hits the y-axis).

> If the line passes through the origin (0,0), the line is written as y=mx.

Good to Know

> Midpoint formula

i. Given two points, A(x₁,y₁), B(x₂,y₂), find the midpoint of the line that connects them:

> Distance formula

i. Given two points, A(x1,y1),B(x2,y2), find the distance between them:

You don’t need this formula, as you can simply graph your points and then create a right triangle from them. The distance will be the hypotenuse, which you can find via the Pythagorean Theorem.

Circles

Good to Know

> Length of an arc

i. Given a radius and a degree measure of an arc from the center, find the length of the arc

ii. Use the formula for the circumference multiplied by the angle of the arc divided by the total angle measure of the circle (360)

Area of an arc sector

> Given a radius and a degree measure of an arc from the center, find the area of the arc sector

i. Use the formula for the area multiplied by the angle of the arc divided by the total angle measure of the circle

Averages

Need to Know

> The average is the same thing as the mean

> Find the average/mean of a set of numbers/terms

Probabilities

Need to Know

> Probability is a representation of the odds of something happening.

Good to Know

> A probability of 1 is guaranteed to happen. A probability of 0 will never happen.

Percentages

Need to Know

Trigonometry

Trigonometry is a new addition to the new 2016 SAT math section. Though it makes up less than 5% of math questions, you won’t be able to answer the trigonometry questions without knowing the following formulas.

Difference of Squares

What you see below is a double bar chart. Take a close look at it and study it, so you can become familiar with its features

After you have examined the double bar chart above, here are a few good observations you can make from the graph:

The scale is on the left of the graph and it is 10 units.

The title is “Scores on a Fractions Test with or without Preparation”

Preparation helped students to score higher

The lowest score without preparation is 45 and the highest score is 85

The lowest score with preparation is 55 and the highest score is 100.

The students who made the most improvement are Darline, Carla, and John. Their score improved by 15.

The student who made the least improvement is Peter. Although Jetser’s score is still slow, he improved by 10 points while Peter’s improved only by 5 points

How to construct a double bar graph:

1. Decide what title you will give the graph
2. Decide if you want horizontal or vertical bars
3. Choose a scale
4. Put the label on the axes
5. Draw the bars

Example
Use the table below to construct a double bar chart. We will follow all steps outlined above to construct this graph

1. The title can be clearly seen from the table.
2. We will choose vertical bars
3. Since the scores differ from one another mostly by 5, 10, 15, or 20, it makes sense to chose a scale of 10

If the variation between scores were like 1, 2, 3, 4, or 5, it would have been better to choose a scale of 1 or 2

4. We put names on the x-axis and scores on the y-axis. If we had decided to make horizontal bars, we would have put names on the y-axis and scores on the x-axis

5. Finally, we drawbars. The double bar graph is shown below:

Also Read   Making Line Graphs

The title is ” Scores on a basic math test”

The scale is at the left of the graph. The distance between each square is 10, so the scale is 10

It shows scores for a student over a period of time or after taking 10 basic math test.

A wealth of information can be deducted from the graph

The lowest score is 40 and the highest score is 100

The greatest increase or improvement happened between test #2 and test #3. As you can see the student went from a score of 40 to a score of 70.

The lowest decrease or happened between test #6 and test #7. The student went from a score of 90 to a score of 60

Looking at the graphs, it looks like although the score dropped a couple of times, over time the student made consistent progress.

Example:

Use the following data to make a line graph

We get:

The graph has an upward trend and shows that over time, gas prices kept increasing.

Looking at the graph, you can see that the line is steeper between 2007 and 2008.

This means that the highest increase was experienced between 2007 and 2008. Gas prices went up by a dollar

Also Read  Double Bar Graphs

Example #1:

Add 2x² + 3x + 4 and 3x² + x + 1

Step #1:

Model both polynomials with tiles

Step #2:

Combine all tiles that are alike or the same and count them

You got a total of 5 light blue square tiles, so 5x²

You got a total of 4 green rectangle tiles, so 4x

You got a total of 5 blue small square tiles, so 5

Putting it all together, we get 5x² + 4x + 5

I hope from the above modeling, it is clear that we can only combine tiles of the same type

For example, you could not add light blue square tiles to green rectangle tiles just like it would not make sense to add 5 potatoes to 5 apples

Try adding 5 potatoes to 5 apples and tell me if you got 10 apples or 10 potatoes. It just does not make sense

Keep this important fact in mind when adding polynomials

We call tiles that are alike or are the same type “like terms”, so this means again that you can only add like terms

Basically, likes terms are terms with the same variable and the same exponent

For example, 2x² and 5x² are like terms because they have the same variable which is x and the same exponent which is 2.

To add like the term, just add the coefficients, or the numbers attached to the term, or the number on the left side of the term

2x² + 5x² = (2 + 5)x² = 7x²

Example #2:

Add 6x² + 2x + 4 to 10x² + 5x + 6

Combine all like terms. You could use parentheses to keep things organized

(6x² + 10x²) + ( 2x + 5x) + (4 + 6)

Add the coefficient

(6 + 10)x² + (2 + 5)x + 4 + 6

We get 16x² + 7x + 10

Example #1: Done by combining like terms and adding the coefficients

Add 2x² + 3x + 4 to 3x² + x + 1

Combine all like terms. You could use parentheses to keep things organized

(2x² + 3x²) + ( 3x + x) + (4 + 1)

Add the coefficient

(2 + 3)x² + (3 + 1)x + 4 + 1

We get 5x² + 4x + 5

Notice that if the term is x, you can rewrite it as 1x, so your coefficient is 1

• Isolate the absolute value on one side of the equation.
• Is the number on the other side of the equation negative? If you answered yes, then the equation has no solution. If you answered no, then go on to step 3.
• Write two equations without absolute values. The first equation will set the quantity inside the bars equal to the number on the other side of the equal sign; the second equation will set the quantity inside the bars equal to the opposite of the number on the other side.
• Solve the two equations.

Follow these steps to solve an absolute value equality which contains two absolute values (one on each side of the equation):

• Write two equations without absolute values.  The first equation will set the quantity inside the bars on the left side equal to the quantity inside the bars on the right side.  The second equation will set the quantity inside the bars on the left side equal to the opposite of the quantity inside the bars on the right side.
• Solve the two equations.

Let’s look at some examples.

Example 1: Solve |2x – 1| + 3 = 6

Step 1: Isolate the absolute value

|2x – 1| + 3 = 6

|2x – 1| = 3

Step 2: Is the number on the other side of the equation negative?

No, it’s a positive number, 3, so continue on to step 3

Step 3: Write two equations without absolute value bars

2x – 1 = 3

2x – 1 = -3

Step 4: Solve both equations

2x – 1 = 3         :     2x – 1 = -3

2x = 4               :        2x = -2

x = 2                 :        x = -1

Example 2: Solve |3x – 6| – 9 = -3

Step 1: Isolate the absolute value

|3x – 6| – 9 = -3

|3x – 6| = 6

Step 2: Is the number on the other side of the equation negative?

No, it’s a positive number, 6, so continue on to step 3

Step 3: Write two equations without absolute value bars

3x – 6 = 6

3x – 6 = -6

Step 4: Solve both equations

3x – 6 = 6        :     3x – 6 = -6

3x = 12            :         3x = 0

x = 4                :           x = 0

Example 3: Solve |5x + 4| + 10 = 2

Step 1: Isolate the absolute value

|5x + 4| + 10 = 2

|5x + 4| = -8

Step 2: Is the number on the other side of the equation negative?

Yes, it’s a negative number, -8. There is no solution to this problem.

Example 4:  Solve |x – 7| = |2x – 2|

Step 1: Write two equations without absolute value bars

x – 7 = 2x – 2

x – 7 = -(2x – 2)

Step 4: Solve both equations

x – 7 = 2x – 2         :            x – 7 = -2x + 2

-x – 7 = -2              :              3x – 7= 2

-x = 5                     :                   3x = 9

x = -5                     :                      x = 3

Example 5:  Solve |x – 3| = |x + 2|

Step 1: Write two equations without absolute value bars

x – 3 = x + 2

x – 3 = -(x + 2)

Step 4: Solve both equations

x – 3 = x + 2                                            :           x – 3 = -x – 2

– 3 = -2                                                    :           2x – 3= -2

false statement                                        :            2x = 1

No solution from this equation                 :             x = 1/2

So the only solution to this problem is x = 1/2

Example 6:  Solve |x – 3| = |3 – x|

Step 1: Write two equations without absolute value bars

x – 3 = 3 – x

x – 3 = -(3 – x)

Step 4: Solve both equations

x – 3 = 3 – x                :       x – 3 = -(3 – x)

2x – 3 = 3                   :      x – 3= -3 + x

2x = 6                         :      -3 = -3

x = 3                           :      All real numbers are solutions to this equation

Since 3 is included in the set of real numbers, we will just say that the solution to this equation is All Real Numbers

It provides an easier way to write numbers and make multiplication and division of very large or very small numbers a lot easier.

A number is in this format if we can write it as:

a × 10ⁿ

with 1 ≤ a < 10 and n is an integer.

1 ≤ a < 10 means that a is a number between 1 and 10

Thus, a can be 1,2,3,4,5,6,7,8, and 9

Let’s start with something simple.Write 500 in this useful notation:

500 = 5 × 100 = 5 × 10²

You can also claim as we saw before that there is a decimal point after 0 and write 500.0

Then, move the decimal point 2 places to the left between 5 and 0 to get 5.000, which is the same as 5.

Since you moved it two places to the left, you know that your exponent is 2.

Your base is always 10

Thus, 500 = 5 × 10²

More examples of scientific notation:

1) 75000

75000 = 75000.0

Move the decimal point 4 places to the left between 7 and 5.

We get 7.5000, which is the same as 7.5

Since we moved it 4 places to the left, your exponent is 4 and your base is still 10.

Thus, 75000 = 7.5 × 10⁴

Sometimes, instead of moving your decimal point to the left, you have to move it to the right as the following example demonstrates:

When you move your decimal point to the right, your exponent is negative.

2) 0.002

Move your decimal point 3 places to the right after the 2 to get 0002. and 0002. is the same as 2. or 2

Since you had to move it 3 places to the right, your exponent is -3 and the base is still 10

Thus, 0.002 = 2 × 10^-3

3) 0.000065

Move the decimal point 5 places to the right

The answer is 6.5 × 10^-5

4) 650000

Move the decimal point 5 places to the left

The answer is 6.5 × 10⁵

In order to make a cube like the one shown above, you basically use the following cube template:

Looking at the cube template, it is easy to see that the cube has six sides and each side is a square

The area of one square is a × a = a²

Since there are six sides, the total surface area, call it SA is:

SA = a² + a² + a² + a² + a² + a²

SA = 6 × a²

Example #1:

Find the surface area if the length of one side is 3 cm

Surface area = 6 × a²

Surface area = 6 × 3²

Surface area = 6 × 3 × 3

Surface area = 54 cm²

Example #2:

Find the surface area if the length of one side is 5 cm

Surface area = 6 × a²

Surface area = 6 × 5²

Surface area = 6 × 5 × 5

Surface area = 150 cm²

Example #3:

Find the surface area if the length of one side is 1/2 cm

Surface area = 6 × a²

Surface area = 6 × (1/2)²

Surface area = 6 × 1/2 × 1/2

Surface area = 6 × 1/4

Surface area = 6/4 cm²

Surface area = 3/2 cm²

Surface area = 1.5 cm²

Formula for Area under the Curve

SOLVED EXAMPLES

Question 1: Calculate the area under the curve of a function, f(x) = 7- x², the limit is given as  x = -1 to 2 ?
Solution:

Given function is, f(x) = 7- x² and limit is x = -1 to 2