Polygons on the SAT

Polygons can be a confusing concept for SAT students since the word “polygon” means a many-sided figure, and can refer to common shapes such as triangles and squares, or less-often seen shapes such as hexagons or pentagons.

A regular polygon has sides of equal length. Every time you add a side to a polygon, you add 180 degrees to the sum of its interior angles. That is why a triangle has a sum of 180 (3 sides), a square has a sum of 360 (4 sides), etc.

The perimeter of a polygon is the sum of the lengths of its sides. A regular polygon has side lengths that are all equal. The number of sides a polygon has determines its name. A triangle has 3 sides. A quadrilateral has four sides. A pentagon has five sides, and a hexagon has six sides. A square is a special type of quadrilateral whose four sides all have the same measure and whose interior angles are each equal to 90 degrees. These four types of polygons (triangle, quadrilateral, pentagon, and hexagon) are the ones you’ll see the most often on the SAT.

A common polygon you may see on a medium or harder Math problem is a regular hexagon; a regular hexagon is a six-sided figure in which each interior angle measures 120 degrees and each side length is equal. Regular hexagons are unique because they can be divided in six equilateral triangles of equal areas! Let’s look at a sample SAT polygon question:

1. AE = BD = AB = ED, CB = CD, and AB = 6 , and the perimeter of polygon ABCDE is more than 25 but less than 30, what is a possible value of CB?

(A)3

(B)5

(C)8

(D)12

(E)18

In this question, we know that the perimeter equals the sum of the total sides of the figure. The square has side lengths of 6, so the figure has a starting perimeter of (3)(6) = 18.

CB=CD. Let’s assign a variable, x, to sides CB and CD.

The total perimeter is 2x + 18.

The question says that 2x + 18 > 25, and 2x + 18 < 30. Solve both inequalities to find the possible values for x.

25< 2x + 18 < 30. Now subtract 18 from both sides to get:

7 < 2x < 12. Next, we divide by 2. Since 2 is positive, we don’t need to switch the inequality signs.

3.5 < x < 6. The only option between 3.5 and 6 is choice (B).

Remember: To find the area of irregular shapes, look for ways to break up the figure into smaller recognizable pieces!

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Vivian Kerr is a regular contributor to the Veritas Prep blog, providing advice to help students better prepare for the GMAT and the SAT. 

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