Perimeter Is the sum of the lengths of the sides. When solving a perimeter problem, it is helpful to draw and label a figure to model the region.

Example 1: A garden is in the shape of an isosceles triangle, a triangle that has two sides of equal measure. The length of the third side of the triangle is 2 feet greater than the length of each of the equal sides. If the perimeter of the garden is 86 feet, find the length of each side of the garden. Let x = the length of each of the two equal sides x + 2 = the length of the third side 86 = x + x + x = 3x = 3x = x Check: Perimeter = = = 86 √ xx x + 2 = 28 = 30

Example 2: The perimeter of a rectangle is 40 feet. The length is 2 feet more than 5 times the width. Find the dimensions of the rectangle. Let w = the width, in feet, of the rectangle 5w + 2 = the length, in feet, of the rectangle Note: Perimeter = 40 feet (sum of all the sides) w + (5w + 2) + w + (5w + 2) = 40 w + 5w w + 5w + 2 = 40 12w + 4 = w = w = 3 5w + 2 ww = 3 = 17 Check: Perimeter = = = 40 √

Perimeter and Expressions 2x -1 Perimeter = 4s = 4(2x – 1) Example 3: If the length of each side of a square is represented by 2x – 1, represent the perimeter of the square as an expression in terms of x. Example 4: If the length of each side of a regular pentagon is represented by 2x + 3, represent the perimeter of a pentagon as an expression in terms of x. 2x + 3 Perimeter = 5s = 5(2x + 3)

Example 5: The width of a rectangle is 3 yards less than its length. The perimeter is 130 yards. Find the length and the width of the rectangle. Let x = the length, in yards, of the rectangle x – 3 = the width, in yards, of the rectangle **Use the formula for perimeter of a rectangle, P = 2l + 2w, to solve this problem. P = 2l + 2w 130 = 2(x) + 2(x – 3) 130 = 2x + 2x – = 4x – = 4x = x = 34 = 31 Check: Perimeter = = = 130 √ x x x - 3