## Geometry Review

## SAT Geometry

### Dr. John Chung’s SAT Math

### Angles Review

Angle CMB = Angle AMD

Angle AMD + Angle CMA = 180^{o}

**Example of SAT question with angles:**

In the figure below, angle AMB is 100^{o}, angle BMD is 40^{o} and angle CME is 60^{o}. What is the value of angle CMD?

*Figure not drawn to scale*

(a) 20^{o}

(b) 30^{o}

(c) 40^{o}

(d) 50^{o}

(e) 80^{o}

Answer: If angle AMB is 100^{o}, then angle BME is 180^{o} – 100^{o} = 80^{o}

BMC + CMD = 40^{o}

BMC + CMD + DME = 80^{o}

If we subtract the 2 equations, DME = 40^{o}

CMD + DME = 60^{o}

BMC + CMD + DME = 80^{o}

If we subtract the 2 equations, BMC = 20^{o}

CMD = BME – BMC – DME = 80^{o} – 40^{o} – 20^{o} = 20^{o}

### Parallel Lines Review

If the 2 horizontal lines are **parallel**, – angles 1 = 3 = 5 = 7 and – angles 2 = 4 = 6 = 8

### Polygons Review

The sum of the measures of the interior angles of a triangle is 180°. a + b + c = 180^{o}

The sum of the measures of the interior angles of a polygon is: **(n – 2)·180 ^{o}** n = number of sides of the polygon.

The sum of the measures of the interior angles of the polygon above is (5 – 2)·180

^{o}= 540

^{o}

AC^{2} = AB^{2} + BC^{2}.

**Pythagorean Theorem** applied to the triangle above: AC^{2} = AB^{2} + BC^{2}.

** Equilateral triangle.** a = b = c = 60^{o} AB = BC = CA

** Isosceles triangle.** b = c AB = AC

** a> c > b.** BC > AB > AC

In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

### Area, Perimeter and Volume

** Rectangle Perimeter** 2·a + 2·b ** Rectangle Area** a·b

** Triangle Perimeter** AB + BC + CA ** Triangle Area** (h/2)·BC

** Circle Perimeter** 2·¶·r ** Circle Area** ¶·r^{2}

** Cube Volume** a^{3}

** Cylinder Volume** h·¶·r^{2}

Examples of SAT Geometry Questions GED Practice Exemple of SAT Subject geometry problem