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

Answer:

12.76

Step-by-step 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.

${a}^{2} + {b}^{2} = {c}^{2} \\ c = \sqrt{ {a}^{2} + {b}^{2} }$

$dc = \sqrt{ {3}^{2} + {1}^{2} } = \sqrt{10} \\ cf = 1 \\ fe = \sqrt{ {4}^{2} + {2}^{2} } = \sqrt{20} \\ de = \sqrt{ {1}^{2} + {4}^{2} } = \sqrt{17} \\ \sqrt{10} + 1 + \sqrt{20} + \sqrt{17} = 12.76$