# Additional topics in math on the SAT test

## Included topics: complex numbers; volume word problems; congruence and similarity; right triangle trigonometry and word problems; circle theorems; angles, arc lengths and trig functions; circle equations.

Additional topics in math include 6 geometry and trigonometry topics (volume word problems; congruence and similarity; right triangle trigonometry and word problems; circle theorems; angles, arc lengths and trig functions; circle equations) and complex numbers.

10% of the questions on the SAT fall into the additional topics classification. The other questions are classified into the three subscores (Heart of algebra, Passport to advanced mathematics and Problem solving and data analysis).

### Complex numbers SAT topic

**A complex number** has a real component and an imaginary component, it is written in a form of a + bi, where a and b are real numbers, and i is an imaginary number satisfying i^{2} = −1.

**An imaginary number **is a number that, when squared, has a negative result (i^{2} = −1). Since an imaginary number i is equal to a square root of a negative number -1 it does not have a tangible value (negative numbers don’t have real square roots since a square is either positive or zero).

**The unit imaginary number, i**, equals the square root of minus 1, so that ** i=√-1**. As said above, when squared imaginary number has a negative result, so that

**.**

__i__^{2}=(√-1)^{2}=-1__ ____For example:__ 3i is an imaginary number, and its square is (3√-1)^{2}=9*-1=-9.

**To add or subtract complex numbers,** combine like terms (real terms with real terms and imaginary terms with imaginary terms).

**To multiply complex numbers,** multiply the numbers with foil formula, replace i^{2 }with -1 and combine like terms.

**To divide complex numbers, **you need to cancel the denominator by turning the imaginary component in the denominator to a real number. This is done by multiplying the numerator and the denominator by the conjugate of the denominator. The next steps are multiplying the numbers with foil formula, replacing i^{2 }with -1 and combining like terms.

Click here to learn the required skills and solve questions about complex numbers on the SAT test.

### Volume word problems SAT topic

**Volume word problems** require making calculations of volumes of three-dimensional shapes using volume formulas. To calculate the volume, plug the given dimension into the relevant volume formula.

**Volume formulas of five basic shapes:**

The volume formulas of five basic shapes are given at the beginning of the math sections of the SAT exam: right rectangular prism, right circular cylinder, sphere, right circular cone and rectangular pyramid.**Right rectangular prism** **volume** formula is V= lwh.**Right circular cylinder** **volume** formula is V=πr^{2}h.**A sphere volume** formula is V= ^{4}/_{3} πr^{3}.**A right circular cone volume** formula is V= ^{1}/_{3} πr^{2}h.**A rectangular pyramid volume** formula is V=^{1}/_{3} lwh.

**Calculating the effect of changes in dimensions on volume: **The power of the dimension determines the size of the change in the volume value.

- If a dimension in the volume formula is raised to a first power, the volume changes by the same factor as the shape.
- If a dimension in the volume formula is raised to a second power, when the shape changes by a factor the volume changes by a square of the factor.
- If a dimension in the volume formula is raised to a third power, when the shape changes by a factor the volume changes by a third degree of the factor.

**Comparing volumes of two shapes:** In these questions we are given ratios between the dimensions of two shapes and we are required to compare their volumes. We need to calculate the total effect on the volume of all the ratios between the shapes.

### Volume formulas worksheet

Volume formulas worksheet is given below, it includes 5 formulas:

Right rectangular prism volume formula is **V= lwh.**

Right circular cylinder volume formula is **V=πr ^{2}h**.

A sphere volume formula is **V= ^{4}/_{3} πr^{3}**.

A right circular cone volume formula is **V= ^{1}/_{3} πr^{2}h**.

A rectangular pyramid volume formula is **V= ^{1}/_{3} lwh**.

**In the formulas below: **

l=length, w=width, h=height, V=volume, A=area

Circle measures: π=3.14159, diameter=2*radius

### Congruence and similarity SAT topic

**Congruence and similarity questions **include congruent angles and similar triangles topics.

**Similar triangles** have the same angle measures and their corresponding side lengths are related by a constant ratio.

**Sum of angles in a triangle: **The sum of the measures in degrees of the angles of a triangle is 180.

**Angle measures in triangles:** In an isosceles triangle the angles opposite the two equal sides are equal; the angles of an equilateral triangle are equal to 60°.

**Congruent angles** are angles that have the same angle measure.

**Vertical Angles** are the angles opposite each other when two lines cross, vertical angles are equal.

**Supplementary angles** are those angles that measure up to 180 degrees. Angles that lie on the same side of a straight line, are always supplementary.

**Alternate angles** are angles located in opposite positions relative to a transversal intersecting two parallel lines. Alternate angles are equal.

**Corresponding angles** are angles located in the same position relative to parallel lines intersected by a transversal. Corresponding angles are equal.

Two intersecting lines create vertical angles and supplementary angles.

Two parallel lines combined with two intersecting lines form 2 similar triangles and equal alternate angles.

A parallel line inside a triangle forms 2 similar triangles and equal corresponding angles.

### Right triangle trigonometry and word problems SAT topic

**Right triangle** is a triangle with a right angle (equal to 90°). The side opposite the right angle (the longest side of the right triangle) is called a hypotenuse.

**Right triangle trigonometry and right triangle word problems** require calculating side lengths and angle measures in right triangles.

**Pythagorean theorem:**

**Pythagorean theorem** states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides: **a ^{2} + b^{2} = c^{2}**.

**Pythagorean triples** are combinations of side lengths a, b and c that satisfy the Pythagorean theorem. If you remember the triples values, you know the size of the third side without the need to calculate it. The most common Pythagorean triples are : 3, 4 and 5; 5, 12 and 13; 7, 24 and 25.

**Trigonometric ratios (sine, cosine and tangent):**

**Trigonometric ratios (functions)** represent connections between angle degrees and side lengths in a right triangle:

- The sine of an angle (sin) in a right triangle is defined as the ratio of the length of the side that is
__opposite__to the angle, to the length of the hypotenuse. - The cosine of an angle (cos) in a right triangle is defined as the ratio of the length of the side that is
__adjacent__to the angle, to the length of the hypotenuse. - The tangent of an angle (tan) in a right triangle is defined as the ratio of the length of the side that is
__opposite__to the angle, side that is__adjacent__to the angle.

**Complementary angles** are two angles with the sum of 90 degrees. Sine of an angle (α) in a right triangle is equal to cosine of its complementary angle (90-α).

**Similar triangles** have the same angle measures and their corresponding side lengths are related by a constant ratio therefore they also have similar sine, cosine and tangent.

**Special right triangles:**

**Special right triangles** are right triangles whose sides are in a particular ratio.

**In a 30°, 60°, 90° right triangle**the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.

In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.**In a 45°, 45°, 90° right triangle**the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.

In 45°, 45°, 90° triangle the sides are s, s and s√2.

### Circle theorems SAT topic

**Circle theorems topic includes two parts: **

Calculating arc length and sector areas using central angles measured in degrees.

Calculating angle measures in degrees inside a circle.

**The following formulas are provided at the beginning of each SAT math section:**

Circumference of a circle formula is C=2πr.

Area of a circle formula is A=πr^{2}.

Number of degrees of arc in a circle is 360.

**Pi (π)** is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal π (approximately 3.14).

**A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle** (one of its angles is a central angle and the other 2 angles are equal).

**The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii. **The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

**The relationship between central angle, arc length and sector area is given by the following ratios:**

central angle = arc length = sector area

______________ _____________________ ____________

360° circle circumference circle area

Click here to learn the required skills and solve questions about circle theorems on the SAT test.

### Angles, arc lengths and trig functions SAT topic

**Angles, arc length and trig functions topic includes 3 parts: **

Calculations of angles in radians.

Calculation of arc lengths and sector areas in radians.

Calculation of sine, cosine and tangent in radians.

**A radian** is defined as the angle subtended from the center of a circle which intercepts an arc equal in length to the radius of the circle. To find the number of radians in a circle, we need to divide the length of the circumference by the radius: 2πr/r=2π, meaning that __the number of radians of arc in a circle is 2π__.

**The relationship between radian and degree measures:**

2π radians is equal to 360 degrees, meaning that one radian is equal to 360/2π=180/π=180/3.14≈57 degrees.

Radian measure = degree measure

_________________ _________________

π 180°

**The relationship between central angle in radians, arc length and sector area:**

central angle = arc length = sector area

_____________ ____________________ ____________

2π circle circumference circle area

**Special right triangles in circles:**

In these questions we are given a circle which center is located at the axis intersection point (0,0).

Special right triangles are right triangles whose sides are in a particular ratio.

Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

The radian measures of angles of special right triangles are:

30° angle radian measure π/6; 45° angle radian measure π/4, 60° angle radian measure π/3; 90° angle radian measure: π/2.

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles.

**Calculating trigonometric functions (sin, cos and tan) with radian angle measures in right triangles in circles:**

We calculate trigonometric functions under the assumption of unit circle, meaning that the radius is equal to 1. Since the hypotenuse is equal to the radius, we know that the hypotenuse is equal to 1.

The trigonometric functions values are:

sin(A)=opposite/hypotenuse= opposite/1=opposite.

cos(A)=adjacent/hypotenuse= adjacent/1=adjacent.

tan(A)=opposite/adjacent.

### Circle equations SAT topic

**A standard form of a circle** contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)^{2}+(y-k)^{2}=r^{2}.

**To find the location of the circle in the xy plane**, we need to draw it according to its standard equation.

If we open brackets of the standard form equation, we get** the expended form**: x^{2}-2hx+y^{2}-2ky+(h^{2}+k^{2}-r^{2})=0, in which there are 3 constants h^{2}, k^{2} and -r^{2}.

**Rewriting the expended equation to a standard form:** We can identify the first 4 components of the equation since each one of them has a different type of variable (x^{2},x,y^{2} and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

Click here to learn the required skills and solve questions about circle equations on the SAT test.