Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

You design a bean bag toss board using a coordinate plate. You plot vertices of the board at C (3,2) D (3,6) E (5,2) F (5,6). What is the perimeter of the bean bag toss board?

Sagot :

MrRoyal

The perimeter of the bean bag toss board is 12 units

What is the perimeter of the bean bag toss board?

The points are given as:

C (3,2) D (3,6) E (5,2) F (5,6)

Calculate the distance between adjacent points using:

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

So, we have:

[tex]CD = \sqrt{(3 - 3)^2 + (2 - 6)^2} = 4[/tex]

[tex]DF = \sqrt{(3 - 5)^2 + (6 - 6)^2} = 2[/tex]

[tex]FE = \sqrt{(5 - 5)^2 + (2 - 6)^2} = 4[/tex]

[tex]CE = \sqrt{(3 - 5)^2 + (2 - 2)^2} = 2[/tex]

The perimeter is then calculated as:

P = CD + DF + FE + CE

This gives

P = 4 + 2 + 4 + 2

Evaluate

P =12

Hence, the perimeter of the bean bag toss board is 12 units

Read more about perimeters at:

https://brainly.com/question/24571594

#SPJ1

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.