Question

Find the permeter of a quadrilateral with vertices at C (2,4), D(1,1), E(5,0), and F(3,4). Round your answer to the nearest hundredth when necessary.

Answer 1

12.76

Explanation:

This is much easier to solve if you draw it out on a graph. The perimeter is the sum of all the sides. I used the Pythagorean theorem to find the hypotenuse.

`{a}^(2) + {b}^(2) = {c}^(2) \\ c = \sqrt{ {a}^(2) + {b}^(2) }`

`dc = \sqrt{ {3}^(2) + {1}^(2) } = √(10) \\ cf = 1 \\ fe = \sqrt{ {4}^(2) + {2}^(2) } = √(20) \\ de = \sqrt{ {1}^(2) + {4}^(2) } = √(17) \\ √(10) + 1 + √(20) + √(17) = 12.76`