Solving geometry problems on the SAT Math section requires your knowledge and skills working with lines, angles, triangles, the Pythagorean Theorem, co-ordinate geometric functions, areas and perimeters, volumes and 3-D geometry, similar figures and circles.

Here are some key formulas and tips: You can make flash cards of these formulas to help you highlight the most important information and details.

**Areas:**Memorize the formulas or use the ones given on the first page of Math section for calculating areas:**Rectangle**: base x height or length x width =*b x h.***Parallelogram**: base x height =(The height or altitude is different from the side. It is always shorter than the second side of the parallelogram, as a perpendicular is the shortest distance from a corner point to the base line.*b x h***Rhombus**: ½ x product of diagonals =*½ x d*_{1}x d_{2}*.*- Square: side x side =
**s**^{2}. **Triangle:**½ x base x height**=***½ x b x h.***Equilateral triangle: ¼ x side squared x****√3****Trapezoid = ½ x height x sum of bases = ½ h (b**_{1}+b_{2}).**Circle:****π x radius squared =****πr**^{2}**Π**is the ratio of the circumference of a circle and its diameter,**π = c/d**. The approximate value of**π**is 3.14 or 22/7. Remember that π is an irrational number and cannot be expressed as a fraction or terminating decimal. All answers involving**π**should be left in terms of**π**unless if you are given a specific number for**π**.

**Perimeter:**Add up all of the distance or length around the outside of the figure.**Any polygon**= sum of all sides.**Circle**= π x diameter =**π***d*

**Right triangles and Pythagorean theorem:**- Pythagorean theorem: (leg)
^{2}+ (leg)^{2}= hypotenuse^{2 }=*a*^{2}+ b^{2}= c^{2} - Pythagorean triples: Any multiples of the set of the Pythagorean triple of 3, 4, 5 also form a Pythagorean triple, i.e. 6, 8, 10 or 30, 40, 50. Memorizing the sets of
**Pythagorean triples**that follow 3. 4, 5 will help immensely in solving a problem quickly:- 5, 12, 13;
- 8, 15, 17;
- 7, 24, 25.

- Pythagorean theorem: (leg)
**30**^{o}-60^{o}-90^{o}triangle:- The leg opposite the 30
^{o }angle is ½ of the hypotenuse. - The leg opposite the 60
^{o}angle is ½ of the hypotenuse x √3. - The height in an equilateral triangle forms a 30
^{o}-60^{o}-90^{o}triangle and is therefore =**½ of the side x****√3.**

- The leg opposite the 30
**45**^{o}-45^{o}-90^{o}triangle (isosceles right triangle):- Each leg =
**½ of the hypotenuse x****√2**. - The hypotenuse =
**leg x****√2.**This can be applied to calculating the diagonal of a square.

- Each leg =
**Co-ordinate geometry**:- To find the midpoint, you simply average the
*x*coordinates and average the*y*coordinates.

- To find the midpoint, you simply average the
**Polygons:**- The sum of the measures of the angles of a polygon of
sides =*n***(**. For example, the sum of a pentagon (5 sides) = (5 – 2) x 180*n*– 2)180^{o}^{o}= 3 x 180^{o}= 540 - Corresponding angles of similar polygons are congruent.
- Corresponding sides of similar polygons are in proportion so you can use ratios to determine the unknown side or angles.

- The sum of the measures of the angles of a polygon of
**Circles:**- The central angle = in degrees to its intercepted arc.
- An inscribed angle = in degrees to ½ of its intercepted arc.
- An angle formed by two chords intersecting in a circle is equal in degrees to ½ sum of its intercepted arcs.

Having a good grasp of the above well-known formulas, rules and properties will help you approach geometry questions more effectively and solve the problems faster. Try some SAT practice test questions with these key points to see how you can improve your SAT Math score.