# SAT Math Practice: How to Find the Perimeter of a Polygon

A polygon is a many-sides figure whose sides are straight lines. 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. Let’s try an easy-level question from Grockit’s SAT database:

**Question 1:** A triangle in the xy-plane has vertices at the coordinates (2,1), (6,1), and (6,6). What is the triangle’s perimeter?

Since two coordinates share an x-value and two coordinates share a y-value, we can tell this triangle is a right triangle, so we can quickly find the lengths of the horizontal and vertical legs by subtracting the x- and y-coordinates. The leg with endpoints (2, 1) and (6, 1) is 6 – 2 = 4 units long, and the leg with endpoints (6, 6) and (6, 1) is 6 – 1 = 5 units long.

Check out this article for advice on how to find an area of a polygon.

To find the length of the hypotenuse, we need to apply the Pythagorean theorem: c^{2} = a^{2} + b^{2}

c^{2} = (4)^{2} + (5)^{2} c^{2} = 41

c = √41

Since the sides of the triangle are 4, 5, and √41, the triangle’s perimeter is equal to 4 + 5 + √41 =

9 + √41 units. Sometimes perimeter questions will involve figures that have already been drawn for you:

**Question 2:**

If x = 60 in triangle ABC provided, how much smaller is the perimeter of triangle ABC than the perimeter of triangle DEF?

If x = 60 then ABC is an equilateral triangle. Since one side is 7, the perimeter is 3*7 = 21. Triangle DEF is an isosceles triangle since equal angles are always across from equal sides. Since DE is 12, then DF is also 12. The perimeter is 12 + 12 + 7 = 31. The difference between the perimeters is 31 – 21 = 10.

Let’s look at how perimeter question can be made more challenging by combining more than one polygon:

**Question 3: **In the figure to the left, *ABDE* is a square with a side length of 6. If side *CB=CD* and the total perimeter of the figure ABCDE must be greater than 25 but less than 30, what is one possible whole number value of *CB*?

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 2*x *+ 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 < 2*x* < 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 whole numbers between 3.5 and 6 are 4 and 5, so either 4 or 5 would be correct here.